# Complete Arithmetic Previous Questions (CAT PYQs 2017-22) [Download PDF]

Arithmetic is a crucial component of the Quantitative Aptitude section in the CAT (Common Admission Test) exam. Prospective CAT candidates recognize the significance of excelling in this section to perform well in the test. If you’re embarking on your CAT preparation journey and aiming to conquer Arithmetic, you’ve landed in the right place. In this blog post, we’re excited to present an extensive collection of Arithmetic questions from the CAT papers of 2017 to 2022. These questions aren’t just practice problems; they are actual CAT questions thoughtfully chosen to provide you with insight into what you may face on the CAT exam.

Arithmetic in CAT may appear challenging, but with proper guidance and practice, you can master it. Our blog will simplify the complexities of CAT Arithmetic by delving into Previous Year Questions (PYQs) from 2017 to 2022. We’ll provide detailed, step-by-step solutions and expert insights, ensuring that you not only solve the problems but also build the confidence to handle any Arithmetic question that the CAT exam presents.

Prepare to delve into the realm of CAT Arithmetic, equip yourself with effective tools and strategies, and embark on a successful journey toward CAT excellence. Let’s get started

Download Complete Arithmetic Previous Questions (CAT PYQs 2017-22)

**Question 1: **Arun’s present age in years is 40% of Barun’s. In another few years, Arun’s age will be half of Barun’s. By what percentage will Barun’s age increase during this period?

**1) Answer: 20**

**Solution:**

Let Arun’s current age be A. Hence, Barun’s current age is 2.5A

Let Arun’s age be half of Barun’s age after X years.

Therefore, 2*(X+A) = 2.5A + X

Or, X = 0.5A

Hence, Barun’s age increased by 0.5A/2.5A = 20%

**Question 2: **A person can complete a job in 120 days. He works alone on Day 1. On Day 2, he is joined by another person who also can complete the job in exactly 120 days. On Day 3, they are joined by another person of equal efficiency. Like this, everyday a new person with the same efficiency joins the work. How many days are required to complete the job?

**2) Answer: 15**

**Solution:**

Let the rate of work of a person be x units/day. Hence, the total work = 120x.

It is given that one first day, one person works, on the second day two people work and so on.

Hence, the work done on day 1, day 2,… will be x, 2x, 3x, … respectively.

The sum should be equal to 120x.

$120x = x* \frac{n(n+1)}{2}$

$n^2 + n – 240 = 0$

n = 15 is the only positive solution.

Hence, it takes 15 days to complete the work.

**Question 3: **An elevator has a weight limit of 630 kg. It is carrying a group of people of whom the heaviest weighs 57 kg and the lightest weighs 53 kg. What is the maximum possible number of people in the group?

**3) Answer: 11**

**Solution:**

It is given that the maximum weight limit is 630. The lightest person’s weight is 53 Kg and the heaviest person’s weight is 57 Kg.

In order to have maximum people in the lift, all the remaining people should be of the lightest weight possible, which is 53 Kg.

Let there be n people.

53 + n(53) + 57 = 630

n is approximately equal to 9.8. Hence, 9 people are possible.

Therefore, a total of 9 + 2 = 11 people can use the elevator.

**Question 4: **A man leaves his home and walks at a speed of 12 km per hour, reaching the railway station 10 minutes after the train had departed. If instead he had walked at a speed of 15 km per hour, he would have reached the station 10 minutes before the train’s departure. The distance (in km) from his home to the railway station is

**4) Answer: 20**

**Solution:**

We see that the man saves 20 minutes by changing his speed from 12 Km/hr to 15 Km/hr.

Let d be the distance

Hence,

$\frac{d}{12} – \frac{d}{15} = \frac{1}{3}$

$\frac{d}{60} = \frac{1}{3}$

d = 20 Km.

**Question 5: **Ravi invests 50% of his monthly savings in fixed deposits. Thirty percent of the rest of his savings is invested in stocks and the rest goes into Ravi’s savings bank account. If the total amount deposited by him in the bank (for savings account and fixed deposits) is Rs 59500, then Ravi’s total monthly savings (in Rs) is

**5) Answer: 70000**

**Solution:**

Let his total savings be 100x.

He invests 50x in fixed deposits. 30% of 50x, which is 15x is invested in stocks and 35x goes to savings bank.

It is given 85x = 59500

x = 700

Hence, 100x = 70000

**Question 6: **If a seller gives a discount of 15% on retail price, she still makes a profit of 2%. Which of the following ensures that she makes a profit of 20%?

a) Give a discount of 5% on retail price.

b) Give a discount of 2% on retail price.

c) Increase the retail price by 2%.

d) Sell at retail price.

**6) Answer (D)**

**Solution:**

Let the retail price be M and cost price be C.

Given,

0.85 M = 1.02 C

M = 1.2 C

If he wants 20% profit he has to sell at 1.2C, which is nothing but the retail price.

**Question 7: **A man travels by a motor boat down a river to his office and back. With the speed of the river unchanged, if he doubles the speed of his motor boat, then his total travel time gets reduced by 75%. The ratio of the original speed of the motor boat to the speed of the river is

a) $\sqrt{6}:\sqrt{2}$

b) $\sqrt{7}:2$

c) $2\sqrt{5}:3$

d) 3:2

**7) Answer (B)**

**Solution:**

Let the speed of the river be $x$ and the speed of the boat be $u$. Let $d$ be the one way distance and $t$ be the initial time taken.

Given,

$t = \frac{d}{u – x} + \frac{d}{u + x}$ … i

Also,

$\frac{t}{4} = \frac{d}{2u – x} + \frac{d}{2u + x}$

$t = \frac{4d}{2u – x} + \frac{4d}{2u + x}$ … ii

Equating both i and ii,

$\dfrac{d}{u – x}$+ $\dfrac{d}{u + x}$ = $\dfrac{4d}{2u – x} + \dfrac{4d}{2u + x}$

$\dfrac{2u}{u^2 – x^2} = \dfrac{16u}{4u^2 – x^2}$

$4u^2 – x^2 = 8u^2 – 8x^2$

$\frac{u^2}{x^2} = \frac{7}{4}$

$\frac{u}{x} = \frac{\sqrt{7}}{2}$

**Question 8: **Suppose, C1, C2, C3, C4, and C5 are five companies. The profits made by Cl, C2, and C3 are in the ratio 9 : 10 : 8 while the profits made by C2, C4, and C5 are in the ratio 18 : 19 : 20. If C5 has made a profit of Rs 19 crore more than C1, then the total profit (in Rs) made by all five companies is

a) 438 crore

b) 435 crore

c) 348 crore

d) 345 crore

**8) Answer (A)**

**Solution:**

Given,

C1 : C2 : C3 = 9 : 10 : 8 … i

C2 : C4 : C5 = 18 : 19 : 20 … ii

Let’s multiply i by 9 and ii by 5

C1 : C2 : C3 = 81 : 90 : 72

C2 : C4 : C5 = 90 : 95 : 100

Therefore, C1 : C2 : C3 : C4 : C5 = 81 : 90 : 72 : 95 : 100

Given,

100x – 81x = 19

x = 1

Hence, total profit = 100 + 95 + 72 + 90 + 81 = 438

**Question 9: **The number of girls appearing for an admission test is twice the number of boys. If 30% of the girls and 45% of the boys get admission, the percentage of candidates who do not get admission is

a) 35

b) 50

c) 60

d) 65

**9) Answer (D)**

**Solution:**

Let the number of girls be 2x and number of boys be x.

Girls getting admission = 0.6x

Boys getting admission = 0.45x

Number of students not getting admission = 3x – 0.6x -0.45x = 1.95x

Percentage = (1.95x/3x) * 100 = 65%

**Question 10: **A stall sells popcorn and chips in packets of three sizes: large, super, and jumbo. The numbers of large, super, and jumbo packets in its stock are in the ratio 7 : 17 : 16 for popcorn and 6 : 15 : 14 for chips. If the total number of popcorn packets in its stock is the same as that of chips packets, then the numbers of jumbo popcorn packets and jumbo chips packets are in the ratio

a) 1 : 1

b) 8 : 7

c) 4 : 3

d) 6 : 5

**10) Answer (A)**

**Solution:**

The ratio of L, S, J for popcorn = 7 : 17 : 16

Let them be 7$x$, 17$x$ and 16$x$

The ratio of L, S, J for chips = 6 : 15 : 14

Let them 6$y$, 15$y$ and 14$y$

Given, 40$x$ = 35$y$, $x = \frac{7y}{8}$

Jumbo popcor = 16$x$ = 16 * $\frac{7y}{8}$= 14$y$

Hence, the ratio of jumbo popcorn and jumbo chips = 1 : 1

**Question 11: **In a market, the price of medium quality mangoes is half that of good mangoes. A shopkeeper buys 80 kg good mangoes and 40 kg medium quality mangoes from the market and then sells all these at a common price which is 10% less than the price at which he bought the good ones. His overall profit is

a) 6%

b) 8%

c) 10%

d) 12%

**11) Answer (B)**

**Solution:**

Let the cost of good mangoes be 2x per kg. The cost of medium mangoes be x per kg.

CP of good mangoes = 160x

CP of medium mangoes = 40x

His selling price = 0.9*2x = 1.80x

Therefore, total revenue generated by selling all the mangoes = 120*1.8x = 216x

Hence, the profit % = $\frac{16x}{200x} * 100 $ = 8%

**Question 12: **If Fatima sells 60 identical toys at a 40% discount on the printed price, then she makes 20% profit. Ten of these toys are destroyed in fire. While selling the rest, how much discount should be given on the printed price so that she can make the same amount of profit?

a) 30%

b) 25%

c) 24%

d) 28%

**12) Answer (D)**

**Solution:**

Let the cost price be C and the marked price be M.

Given,

0.6 M = 1.2 C

M = 2C

CP of 60 toys = 60C

Now only 50 are remaining.

Hence,

M (1 – d) * 50 = 72C

1- d = 0.72

d = .28

Hence 28%

**Question 13: **A class consists of 20 boys and 30 girls. In the mid-semester examination, the average score of the girls was 5 higher than that of the boys. In the final exam, however, the average score of the girls dropped by 3 while the average score of the entire class increased by 2. The increase in the average score of the boys is

a) 9.5

b) 10

c) 4.5

d) 6

**13) Answer (A)**

**Solution:**

Let, the average score of boys in the mid semester exam is A.

Therefore, the average score of girls in the mid semester exam be A+5.

Hence, the total marks scored by the class is $20\times (A) + 30\times (A+5) = 50\times A + 150$

The average score of the entire class is $\dfrac{(50\times A + 150)}{50} = A + 3$

wkt, class average increased by 2, class average in final term $= (A+3) + 2 = A + 5$

Given, that score of girls dropped by 3, i.e $(A+5)-3 = A+2$

Total score of girls in final term $= 30\times(A+2) = 30A + 60$

Total class score in final term $= (A + 5)\times50 = 50A + 250$

the total marks scored by the boys is $(50A + 250) – (30A – 60) = 20A + 190$

Hence, the average of the boys in the final exam is $\dfrac{(20G + 190)}{20} = A + 9.5$

Hence, the increase in the average marks of the boys is $(A+9.5) – A = 9.5$

**Question 14: **In a 10 km race, A, B, and C, each running at uniform speed, get the gold, silver, and bronze medals, respectively. I f A beats B by 1 km and B beats C by 1 km, then by how many metres does A beat C?

**14) Answer: 1900**

**Solution:**

By the time A traveled 10 KM, B traveled 9 KM

Hence $Speed_A : Speed_B = 10:9$

Similarly $Speed_B : Speed_C = 10:9$

Hence $Speed_A : Speed_B : Speed_C = 100:90:81$

Hence by the time A traveled 10 KMs , C should have traveled 8.1 KMs

So A beat C by 1.9 KMs = 1900 Mts

**Question 15: **Bottle 1 contains a mixture of milk and water in 7: 2 ratio and Bottle 2 contains a mixture of milk and water in 9: 4 ratio. In what ratio of volumes should the liquids in Bottle 1 and Bottle 2 be combined to obtain a mixture of milk and water in 3:1 ratio?

a) 27:14

b) 27:13

c) 27:16

d) 27:18

**15) Answer (B)**

**Solution:**

The ratio of milk and water in Bottle 1 is 7:2 and the ratio of milk and water in Bottle 2 is 9:4

Therefore, the proportion of milk in Bottle 1 is $\frac{7}{9}$ and the proportion of milk in Bottle 2 is $\frac{9}{13}$

Let the ratio in which they should be mixed be equal to X:1.

Hence, the total volume of milk is $\frac{7X}{9}+\frac{9}{13}$

The total volume of water is $\frac{2X}{9}+\frac{4}{13}$

They are in the ratio 3:1

Hence, $\frac{7X}{9}+\frac{9}{13} = 3*(\frac{2X}{9}+\frac{4}{13})$

Therefore, $91X+81=78X+108$

Therefore $X = \frac{27}{13}$

**Question 16: **Arun drove from home to his hostel at 60 miles per hour. While returning home he drove half way along the same route at a speed of 25 miles per hour and then took a bypass road which increased his driving distance by 5 miles, but allowed him to drive at 50 miles per hour along this bypass road. If his return journey took 30 minutes more than his onward journey, then the total distance traveled by him is

a) 55 miles

b) 60 miles

c) 65 miles

d) 70 miles

**16) Answer (C)**

**Solution:**

Let the distance between the home and office be $2x$ miles

Time taken for going in the morning = $\frac{2x}{60}$ hrs

Time taken for going back in the evening = $\frac{x}{25} + \frac{x+5}{50}$. hrs

It is given that he took 30 minutes (0.5 hrs) more in the evening

Hence $\frac{2x}{60}$ hrs + 0.5 = $\frac{x}{25} + \frac{x+5}{50}$

Solving for x, we get x = 15 miles.

Total distance traveled = 2x + x + x + 5 = 4x + 5 = 65 Miles

**Question 17: **Out of the shirts produced in a factory, 15% are defective, while 20% of the rest are sold in the domestic market. If the remaining 8840 shirts are left for export, then the number of shirts produced in the factory is

a) 13600

b) 13000

c) 13400

d) 14000

**17) Answer (B)**

**Solution:**

Let the total number of shirts be x.

Hence number of non defective shirts = x – 15% of x = 0.85x

Number

of shirts left for export = No of non defective shirts – number of

shirts sold in domestic market

= No of non defective shirts – 20% of No of non defective shirts

= 80% of No of non defective shirts

Hence 8840 = 0.8 * (0.85x) .

Solving for x we get, x = 13000

**Question 18: **The average height of 22 toddlers increases by 2 inches when two of them leave this group. If the average height of these two toddlers is one-third the average height of the original 22, then the average height, in inches, of the remaining 20 toddlers is

a) 30

b) 28

c) 32

d) 26

**18) Answer (C)**

**Solution:**

Let the average height of 22 toddlers be 3x.

Sum of the height of 22 toddlers = 66x

Hence average height of the two toddlers who left the group = x

Sum of the height of the remaining 20 toddlers = 66x – 2x = 64x

Average height of the remaining 20 toddlers = 64x/20 = 3.2x

Difference = 0.2x = 2 inches => x = 10 inches

Hence average height of the remaining 20 toddlers = 3.2x = 32 inches

**Question 19: **The manufacturer of a table sells it to a wholesale dealer at a profit of 10%. The wholesale dealer sells the table to a retailer at a profit of 30% Finally, the retailer sells it to a customer at a profit of 50%. If the customer pays Rs 4290 for the table, then its manufacturing cost (in Rs) is

a) 1500

b) 2000

c) 2500

d) 3000

**19) Answer (B)**

**Solution:**

Let the manufacturing price of the table = $x$

Hence the price at which the wholesaler bought from the manufacturer = $1.1 \times x$

The price at which the retailer bought from the wholesaler = $1.3 \times 1.1 \times x$

The price at which the customer bought from the retailer = $1.5 \times 1.3 \times 1.1 \times x$

$1.5 \times 1.3 \times 1.1 \times x = 4290$

=> x = 2000

**Question 20: **A tank has an inlet pipe and an outlet pipe. If the outlet pipe is closed then the inlet pipe fills the empty tank in 8 hours. If the outlet pipe is open then the inlet pipe fills the empty tank in 10 hours. If only the outlet pipe is open then in how many hours the full tank becomes half-full?

a) 20

b) 30

c) 40

d) 45

**20) Answer (A)**

**Solution:**

Let the time taken by the outlet pipe to empty = x hours

Then, $\frac{1}{8} – \frac{1}{x} = \frac{1}{10}$

=> $x = 40$

Hence time taken by the outlet pipe to make the tank half-full = 40/2 = 20 hour

**Question 21: **Mayank buys some candies for Rs 15 a dozen and an equal number of different candies for Rs 12 a dozen. He sells all for Rs 16.50 a dozen and makes a profit of Rs 150. How many dozens of candies did he buy altogether?

a) 50

b) 30

c) 25

d) 45

**21) Answer (A)**

**Solution:**

Let the number of dozens of candies he bought of each variety be x

Hence total cost = 12x + 15x = 27x

Total selling price = 16.50*2x = 33x

Profit = 33x – 27x = 6x

Given 6x = 150 => x = 25

Hence he bought 50 dozens of candies in total

**Question 22: **In a village, the production of food grains increased by 40% and the per capita production of food grains increased by 27% during a certain period. The percentage by which the population of the village increased during the same period is nearest to

a) 16

b) 13

c) 10

d) 7

**22) Answer (C)**

**Solution:**

Let initial population and production be x,y and final population be z

Final production = 1.4y, final percapita = 1.27 times initial percapita

=> $\frac{1.4y}{z} $ = $ 1.27 \times \frac{y}{x}$

=> $\frac{z}{x} = \frac{1.4}{1.27} \approx 1.10$

Hence the percentage increase in population = 10%

**Question 23: **If a, b, c are three positive integers such that a and b are in the ratio 3 : 4 while b and c are in the ratio 2:1, then which one of the following is a possible value of (a + b + c)?

a) 201

b) 205

c) 207

d) 210

**23) Answer (C)**

**Solution:**

a : b = 3:4 and b : c = 2:1 => a:b:c = 3:4:2

=> a = 3x, b = 4x , c = 2x

=> a + b + c = 9x

=> a + b + c is a multiple of 9.

From the given options only, option C is a multiple of 9

**Question 24: **A motorbike leaves point A at 1 pm and moves towards point B at a uniform speed. A car leaves point B at 2 pm and moves towards point A at a uniform speed which is double that of the motorbike. They meet at 3:40 pm at a point which is 168 km away from A. What is the distance, in km, between A and B7

a) 364

b) 378

c) 380

d) 388

**24) Answer (B)**

**Solution:**

Let the distance traveled by the car be x KMs

Distance traveled by the bike = 168 KMs

Speed of car is double the speed of bike

=> $\frac{x}{3:40 – 2:00}$ = $2 \times \frac{168}{3:40 – 1:00}$

=> $\frac{x}{100}$ = $2 \times \frac{168}{160}$

=> x = 210

Hence the distance between A and B is x + 168 = 378 KMs

**Question 25: **Amal can complete a job in 10 days and Bimal can complete it in 8 days. Amal, Bimal and Kamal together complete the job in 4 days and are paid a total amount of Rs 1000 as remuneration. If this amount is shared by them in proportion to their work, then Kamal’s share, in rupees, is

a) 100

b) 200

c) 300

d) 400

**25) Answer (A)**

**Solution:**

Let the time take by kamal to complete the task be x days.

Hence we have $\frac{1}{10} + \frac{1}{8} + \frac{1}{x} = \frac{1}{4}$

=> x = 40 days.

Ratio of the work done by them = $\frac{1}{10} : \frac{1}{8} : \frac{1}{40}$ = 4 : 5 : 1

Hence the wage earned by Kamal = 1/10 * 1000 = 100

**Question 26: **Consider three mixtures — the first having water and liquid A in the ratio 1:2, the second having water and liquid B in the ratio 1:3, and the third having water and liquid C in the ratio 1:4. These three mixtures of A, B, and C, respectively, are further mixed in the proportion 4: 3: 2. Then the resulting mixture has

a) The same amount of water and liquid B

b) The same amount of liquids B and C

c) More water than liquid B

d) More water than liquid A

**26) Answer (C)**

**Solution:**

The proportion of water in the first mixture is $\frac{1}{3}$

The proportion of Liquid A in the first mixture is $\frac{2}{3}$

The proportion of water in the second mixture is $\frac{1}{4}$

The proportion of Liquid B in the second mixture is $\frac{3}{4}$

The proportion of water in the third mixture is $\frac{1}{5}$

The proportion of Liquid C in the third mixture is $\frac{4}{5}$

As they are mixed in the ratio 4:3:2, the final amount of water is $4 \times \frac{1}{3} + 3 \times \frac{1}{4} + 2 \times \frac{1}{5} = \frac{149}{60}$

The final amount of Liquid A in the mixture is $4\times\frac{2}{3} = \frac{8}{3}$

The final amount of Liquid B in the mixture is $3\times\frac{3}{4} = \frac{9}{4}$

The final amount of Liquid C in the mixture is $2\times\frac{4}{5} = \frac{8}{5}$

Hence, the ratio of Water : A : B : C in the final mixture is $\frac{149}{60}:\frac{8}{3}:\frac{9}{4}:\frac{8}{5} = 149:160:135:96$

From the given choices, only option C is correct.

**Question 27: **Humans and robots can both perform a job but at different efficiencies. Fifteen humans and five robots working together take thirty days to finish the job, whereas five humans and fifteen robots working together take sixty days to finish it. How many days will fifteen humans working together (without any robot) take to finish it?

a) 45

b) 36

c) 32

d) 40

**27) Answer (C)**

**Solution:**

Let the efficiency of humans be ‘h’ and the efficiency of robots be ‘r’.

In the first case,

Total work = (15h + 5r) * 30……(i)

In the second case,

Total work = (5h + 15r) * 60……(ii)

On equating (i) and (ii), we get

(15h + 5r) * 30 = (5h + 15r) * 60

Or, 15h + 5r = 10h + 30r

Or, 5h = 25r

Or, h = 5r

Total work = (15h + 5r) * 30 = (15h + h) * 30 = 480h

Time taken by 15 humans = $\dfrac{\text{480h}}{\text{15h}}$ days= 32 days.

Hence, option C is the correct answer.

**Question 28: **Point P lies between points A and B such that the length of BP is thrice that of AP. Car 1 starts from A and moves towards B. Simultaneously, car 2 starts from B and moves towards A. Car 2 reaches P one hour after car 1 reaches P. If the speed of car 2 is half that of car 1, then the time, in minutes, taken by car 1 in reaching P from A is

**28) Answer: 12**

**Solution:**

Let the distance between A and B be 4x.

Length of BP is thrice the length of AP.

=> AP = x and BP = 3x

Let the speed of car 1 be s and the speed of car 2 be 0.5s.

Car 2 reaches P one hour (60 minutes) after Car 1 reaches P.

=> x/s + 60 = 3x/0.5s

x/s + 60 = 6x/s

5x/s = 60

x/s = 12

Time taken by car 1 in reaching P from A = x/s = 12 minutes.

Therefore, 12 is the correct answer.

**Question 29: **Train T leaves station X for station Y at 3 pm. Train S, traveling at three quarters of the speed of T, leaves Y for X at 4 pm. The two trains pass each other at a station Z, where the distance between X and Z is three-fifths of that between X and Y. How many hours does train T take for its journey from X to Y?

**29) Answer: 15**

**Solution:**

Train T starts at 3 PM and train S starts at 4 PM.

Let the speed of train T be t.

=> Speed of train S = 0.75t.

When the trains meet, train t would have traveled for one more hour than train S.

Let us assume that the 2 trains meet x hours after 3 PM. Trains S would have traveled for x-1 hours.

Distance traveled by train T = xt

Distance traveled by train S = (x-1)*0.75t = 0.75xt-0.75t

We know that train T has traveled three fifths of the distance. Therefore, train S should have traveled two-fifths the distance between the 2 cities.

=> (xt)/(0.75xt-0.75t) = 3/2

2xt = 2.25xt-2.25t

0.25x = 2.25

x = 9 hours.

Train T takes 9 hours to cover three-fifths the distance. Therefore, to cover the entire distance, train T will take 9*(5/3) = 15 hours.

Therefore, 15 is the correct answer.

**Question 30: **Two types of tea, A and B, are mixed and then sold at Rs. 40 per kg. The profit is 10% if A and B are mixed in the ratio 3 : 2, and 5% if this ratio is 2 : 3. The cost prices, per kg, of A and B are in the ratio

a) 17 : 25

b) 18 : 25

c) 19 : 24

d) 21 : 25

**30) Answer (C)**

**Solution:**

The selling price of the mixture is Rs.40/kg.

Let a be the price of 1 kg of tea A in the mixture and b be the price per kg of tea B.

It has been given that the profit is 10% if the 2 varieties are mixed in the ratio 3:2

Let the cost price of the mixture be x.

It has been given that 1.1x = 40

x = 40/1.1

Price per kg of the mixture in ratio 3:2 = $\frac{3a+2b}{5} $

$\frac{3a+2b}{5} = \frac{40}{1.1}$

$3.3a+2.2b=200$ ——–(1)

The profit is 5% if the 2 varieties are mixed in the ratio 2:3.

Price per kg of the mixture in ratio 2:3 = $\frac{2a+3b}{5}$

$\frac{2a+3b}{5} = \frac{40}{1.05}$

$2.1a+3.15b=200$ ——(2)

Equating (1) and (2), we get,

$3.3a+2.2b = 2.1a+3.15b$

$1.2a=0.95b$

$\frac{a}{b} = \frac{0.95}{1.2}$

$\frac{a}{b} = \frac{19}{24}$

Therefore, option C is the right answer.

**Question 31: **John borrowed Rs. 2,10,000 from a bank at an interest rate of 10% per annum, compounded annually. The loan was repaid in two equal instalments, the first after one year and the second after another year. The first instalment was interest of one year plus part of the principal amount, while the second was the rest of the principal amount plus due interest thereon. Then each instalment, in Rs., is

**31) Answer: 121000**

**Solution:**

We have to equate the installments and the amount due either at the time of borrowing or at the time when the entire loan is repaid. Let us bring all values to the time frame in which all the dues get settled, i.e, by the end of 2 years.

John borrowed Rs. 2,10,000 from the bank at 10% per annum. This loan will amount to 2,10,000*1.1*1.1 = Rs.2,54,100 by the end of 2 years.

Let the amount paid as installment every year be Rs.x.

John would pay the first installment by the end of the first year. Therefore, we have to calculate the interest on this amount from the end of the first year to the end of the second year. The loan will get settled the moment the second installment is paid.

=> 1.1x + x = 2,54,100

2.1x = 2,54,100

=> x = Rs. 1,21,000.

Therefore, 121000 is the correct answer.

**Question 32: **When they work alone, B needs 25% more time to finish a job than A does. They two finish the job in 13 days in the following manner: A works alone till half the job is done, then A and B work together for four days, and finally B works alone to complete the remaining 5% of the job. In how many days can B alone finish the entire job?

a) 20

b) 22

c) 16

d) 18

**32) Answer (A)**

**Solution:**

Let us assume that A can complete ‘a’ units of work in a day and B can complete ‘b’ units of work in a day.

A works alone till half the work is completed.

A and B work together for 4 days and B works alone to complete the last 5% of the work.

=> A and B in 4 days can complete 45% of the work.

Let us assume the total amount of work to be done to be 100 units.

4a + 4b = 45 ———(1)

B needs 25% more time than A to finish a job.

=> 1.25*b = a ———-(2)

Substituting (2) in (1), we get,

5b+4b = 45

9b = 45

b = 5 units/day

B alone can finish the job in 100/5 = 20 days.

Therefore, option A is the right answer.

**Question 33: **Raju and Lalitha originally had marbles in the ratio 4:9. Then Lalitha gave some of her marbles to Raju. As a result, the ratio of the number of marbles with Raju to that with Lalitha became 5:6. What fraction of her original number of marbles was given by Lalitha to Raju?

a) $\frac{1}{5}$

b) $\frac{6}{19}$

c) $\frac{1}{4}$

d) $\frac{7}{33}$

**33) Answer (D)**

**Solution:**

Let the number of marbles with Raju and Lalitha initially be 4x and 9x.

Let the number of marbles that Lalitha gave to Raju be a.

It has been given that (4x+a)/(9x-a) = 5/6

24x + 6a = 45x – 5a

11a = 21x

a/x = 21/11

Fraction of original marbles given to Raju by Lalitha = a/9x (Since Lalitha had 9x marbles initially).

a/9x = 21/99

= 7/33.

Therefore, option D is the right answer.

**Question 34: **A wholesaler bought walnuts and peanuts, the price of walnut per kg being thrice that of peanut per kg. He then sold 8 kg of peanuts at a profit of 10% and 16 kg of walnuts at a profit of 20% to a shopkeeper. However, the shopkeeper lost 5 kg of walnuts and 3 kg of peanuts in transit. He then mixed the remaining nuts and sold the mixture at Rs. 166 per kg, thus making an overall profit of 25%. At what price, in Rs. per kg, did the wholesaler buy the walnuts?

a) 96

b) 98

c) 86

d) 84

**34) Answer (A)**

**Solution:**

Let the price of peanuts be Rs. 100x per kg

Then, the price of walnuts = Rs. 300x per kg

Cost price of peanuts for the shopkeeper = Rs. 110x per kg

Cost price of walnuts for the shopkeeper = Rs. 360x per kg

Total cost incurred to the shopkeeper while buying = Rs.(8 * 110x + 16 * 360x) = Rs. 6640x

Since, 5kg walnut and 3kg peanuts are lost in transit, the shopkeeper will be remained with (16-5)+(8-3)=16kgs of nuts

Total selling price that the shopkeeper got = Rs. (166 * 16) = Rs. 2656

Profit = 25%

So, cost price = Rs. 2124.80

Therefore, 6640x = 2124.80

On solving, we get x = 0.32

Therefore, price of walnuts = Rs. (300 * 0.32) = Rs. 96 per kg.

Hence, option A is the correct answer

**Question 35: **In an apartment complex, the number of people aged 51 years and above is 30 and there are at most 39 people whose ages are below 51 years. The average age of all the people in the apartment complex is 38 years. What is the largest possible average age, in years, of the people whose ages are below 51 years?

a) 27

b) 25

c) 26

d) 28

**35) Answer (D)**

**Solution:**

The possible average age of people whose ages are below 51 years will be maximum if the average age of the number of people aged 51 years and above is minimum. Hence, we can say that that there are 30 people having same age 51 years.

Let ‘x’ be the maximum average age of people whose ages are below 51.

Then we can say that,

$\dfrac{51*30+39*x}{30+39} = 38$

$\Rightarrow$ $1530+39x = 2622$

$\Rightarrow$ $x = 1092/39 = 28$

Hence, we can say that option D is the correct answer.

**Question 36: **A trader sells 10 litres of a mixture of paints A and B, where the amount of B in the mixture does not exceed that of A. The cost of paint A per litre is Rs. 8 more than that of paint B. If the trader sells the entire mixture for Rs. 264 and makes a profit of 10%, then the highest possible cost of paint B, in Rs. per litre, is

a) 16

b) 26

c) 20

d) 22

**36) Answer (C)**

**Solution:**

Let the price of paint B be x.

Price of paint A = x+8

We know that the amount of paint B in the mixture does not exceed the amount of paint A. Therefore, paint B can at the maximum compose 50% of the mixture.

The seller sells 10 litres of paint at Rs.264 earning a profit of 10%.

=> The cost price of 10 litres of the paint mixture = Rs. 240

Therefore, the cost of 1 litre of the mixture = Rs.24

We have to find the highest possible cost of paint B.

When we increase the cost of paint B, the cost of paint A will increase too. If the cost price of the mixture is closer to the cost of paint B, then the amount of paint B present in the mixture should be greater than the amount of paint A present in the mixture.

The highest possible cost of paint B will be obtained when the volumes of paint A and paint B in the mixture are equal.

=> (x+x+8)/2 = 24

2x = 40

x = Rs. 20

Therefore, option C is the right answer.

**Question 37: **The distance from A to B is 60 km. Partha and Narayan start from A at the same time and move towards B. Partha takes four hours more than Narayan to reach B. Moreover, Partha reaches the mid-point of A and B two hours before Narayan reaches B. The speed of Partha, in km per hour, is

a) 6

b) 4

c) 3

d) 5

**37) Answer (D)**

**Solution:**

Let the time taken by Partha to cover 60 km be x hours.

Narayan will cover 60 km in x-4 hours.

Speed of Partha = $\frac{60}{x}$

Speed of Narayan = $\frac{60}{x-4}$

Partha reaches the mid-point of A and B two hours before Narayan reaches B.

=> $\dfrac{30}{\frac{60}{x}} + 2 = \dfrac{60}{\frac{60}{(x-4)}}$

$\frac{x}{2} + 2 = x-4$

$\frac{x+4}{2}=x-4$

$x+4=2x-8$

$x=12$

Partha will take 12 hours to cross 60 km.

=> Speed of Partha = $\frac{60}{12}=5$ Kmph.

Therefore, option D is the right answer.

**Question 38: **A tank is fitted with pipes, some filling it and the rest draining it. All filling pipes fill at the same rate, and all draining pipes drain at the same rate. The empty tank gets completely filled in 6 hours when 6 filling and 5 draining pipes are on, but this time becomes 60 hours when 5 filling and 6 draining pipes are on. In how many hours will the empty tank get completely filled when one draining and two filling pipes are on?

**38) Answer: 10**

**Solution:**

Let the efficiency of filling pipes be ‘x’ and the efficiency of draining pipes be ‘-y’.

In the first case,

Capacity of tank = (6x – 5y) * 6……….(i)

In the second case,

Capacity of tank = (5x – 6y) * 60…..(ii)

On equating (i) and (ii), we get

(6x – 5y) * 6 = (5x – 6y) * 60

or, 6x – 5y = 50x – 60y

or, 44x = 55y

or, 4x = 5y

or, x = 1.25y

Capacity of the tank = (6x – 5y) * 6 = (7.5y – 5y) * 6 = 15y

Net efficiency of 2 filling and 1 draining pipes = (2x – y) = (2.5y – y) = 1.5y

Time required = $\dfrac{\text{15y}}{\text{1.5y}}$hours = 10 hours.

Hence, 10 is the correct answer.

**Question 39: **~~A CAT aspirant appears for a certain number of tests. His average score increases by 1 if the first 10 tests are not considered, and decreases by 1 if the last 10 tests are not considered. If his average scores for the first 10 and the last 10 tests are 20 and 30, respectively, then the total number of tests taken by him is~~

**39) Answer: 60**

**Solution:**

Let the total number of tests be ‘n’ and the average by ‘A’

Total score = n*A

When 1st 10 tests are excluded, decrease in total value of scores = (nA – 20 * 10) = (nA – 200)

Also, (n – 10)(A + 1) = (nA – 200)

On solving, we get 10A – n = 190……….(i)

When last 10 tests are excluded, decrease in total value of scores = (nA – 30 * 10) = (nA – 300)

Also, (n – 10)(A – 1) = (nA – 300)

On solving, we get 10A + n = 310……….(ii)

From (i) and (ii), we get n = 60

Hence, 60 is the correct answer.

**Question 40: **In an examination, the maximum possible score is N while the pass mark is 45% of N. A candidate obtains 36 marks, but falls short of the pass mark by 68%. Which one of the following is then correct?

a) $N \leq 200$.

b) $243 \leq N \leq 252$.

c) $201 \leq N \leq 242$.

d) $N \geq 253$.

**40) Answer (B)**

**Solution:**

Total marks = N

Pass marks = 45% of N = 0.45N

Marks obtained = 36

It is given that, obtained marks is 68% less than that pass marks

=>the obtained marks is 32% of the pass marks.

So, 0.32 * 0.45N = 36

On solving, we get N = 250

Hence, option B is the correct answer.

**Question 41: **A water tank has inlets of two types A and B. All inlets of type A when open, bring in water at the same rate. All inlets of type B, when open, bring in water at the same rate. The empty tank is completely filled in 30 minutes if 10 inlets of type A and 45 inlets of type B are open, and in 1 hour if 8 inlets of type A and 18 inlets of type B are open. In how many minutes will the empty tank get completely filled if 7 inlets of type A and 27 inlets of type B are open?

**41) Answer: 48**

**Solution:**

Let the efficiency of type A pipe be ‘a’ and the efficiency of type B be ‘b’.

In the first case, 10 type A and 45 type B pipes fill the tank in 30 mins.

So, the capacity of the tank = $\dfrac{1}{2}$(10a + 45b)……..(i)

In the second case, 8 type A and 18 type B pipes fill the tank in 1 hour.

So, the capacity of the tank = (8a + 18b)……….(ii)

Equating (i) and (ii), we get

10a + 45b = 16a + 36b

=>6a = 9b

From (ii), capacity of the tank = (8a + 18b) = (8a + 12a) = 20a

In the third case, 7 type A and 27 type B pipes fill the tank.

Net efficiency = (7a + 27b) = (7a + 18a) = 25a

Time taken = $\dfrac{20\text{a}}{25\text{a}}$ hour = 48 minutes.

Hence, 48 is the correct answer.

**Question 42: **Points A, P, Q and B lie on the same line such that P, Q and B are, respectively, 100 km, 200 km and 300 km away from A. Cars 1 and 2 leave A at the same time and move towards B. Simultaneously, car 3 leaves B and moves towards A. Car 3 meets car 1 at Q, and car 2 at P. If each car is moving in uniform speed then the ratio of the speed of car 2 to that of car 1 is

a) 2 : 7

b) 2 : 9

c) 1 : 2

d) 1 : 4

**42) Answer (D)**

**Solution:**

Car 3 meets car 1 at Q, which is 200 km from A.

Therefore, at the time of their meeting car 1 must have travelled 200 km and car 3 must have travelled 100 km.

As the time is same, ratio of speed of car 1 to speed of car 3 = 2 : 1.

Car 3 meets car 2 at P, which is 100 km from A.

Therefore, at the time of their meeting car 2 must have travelled 100 km and car 3 must have travelled 200 km.

As the time is same, ratio of speed of car 2 to speed of car 3 = 1 : 2.

Speed of car 1 : speed of car 3 = 2 : 1

And speed of car 2 : speed of car 3 = 1 : 2

So, speed of car 1 : speed of car 2 : speed of car 3 = 4 : 1 : 2

Hence, option D is the correct answer.

**Question 43: **The scores of Amal and Bimal in an examination are in the ratio 11 : 14. After an appeal, their scores increase by the same amount and their new scores are in the ratio 47 : 56. The ratio of Bimal’s new score to that of his original score is

a) 4 : 3

b) 8 : 5

c) 5 : 4

d) 3 : 2

**43) Answer (A)**

**Solution:**

Let the score of Amal and Bimal be 11k and 14k

Let the scores be increased by x

So, after increment, Amal’s score = 11k + x and Bimal’s score = 14k + x

According to the question,

$\dfrac{\text{11k + x}}{\text{14k + x}}$ = $\dfrac{47}{56}$

On solving, we get x = $\dfrac{42}{9}$k

Ratio of Bimal’s new score to his original score

= $\dfrac{\text{14k + x}}{\text{14k}}$

=$\dfrac{14k +\frac{42k}{9}}{14k}$

=$\dfrac{\text{168k}}{\text{14*9k}}$

=$\dfrac{4}{3}$

Hence, option A is the correct answer.

**Question 44: **A 20% ethanol solution is mixed with another ethanol solution, say, S of unknown concentration in the proportion 1:3 by volume. This mixture is then mixed with an equal volume of 20% ethanol solution. If the resultant mixture is a 31.25% ethanol solution, then the unknown concentration of S is

a) 30%

b) 40%

c) 50%

d) 60%

**44) Answer (C)**

**Solution:**

Let the volume of the first and the second solution be 100 and 300.

When they are mixed, quantity of ethanol in the mixture

= (20 + 300S)

Let this solution be mixed with equal volume i.e. 400 of third solution in which the strength of ethanol is 20%.

So, the quantity of ethanol in the final solution

= (20 + 300S + 80) = (300S + 100)

It is given that, 31.25% of 800 = (300S + 100)

or, 300S + 100 = 250

or S = $\frac{1}{2}$ = 50%

Hence, 50 is the correct answer.

**Question 45: **A tank is emptied everyday at a fixed time point. Immediately thereafter, either pump A or pump B or both start working until the tank is full. On Monday, A alone completed ﬁlling the tank at 8 pm. On Tuesday, B alone completed filling the tank at 6 pm. On Wednesday, A alone worked till 5 pm, and then B worked alone from 5 pm to 7 pm, to fill the tank. At what time was the tank ﬁlled on Thursday if both pumps were used simultaneously all along?

a) 4:48 pm

b) 4:12 pm

c) 4:24 pm

d) 4:36 pm

**45) Answer (C)**

**Solution:**

Let ‘t’ pm be the time when the tank is emptied everyday. Let ‘a’ and ‘b’ be the liters/hr filled by pump A and pump B respectively.

On Monday, A alone completed ﬁlling the tank at 8 pm. Therefore, we can say that pump A worked for (8 – t) hours. Hence, the volume of the tank = a*(8 – t) liters.

Similarly, on Tuesday, B alone completed ﬁlling the tank at 6 pm. Therefore, we can say that pump B worked for (6 – t) hours. Hence, the volume of the tank = b*(6 – t) liters.

On Wednesday, A alone worked till 5 pm, and then B worked alone from 5 pm to 7 pm, to fill the tank. Therefore, we can say that pump A worked for (5 – t) hours and pump B worked for 2 hours. Hence, the volume of the tank = a*(5 – t)+2b liters.

We can say that a*(8 – t) = b*(6 – t) = a*(5 – t) + 2b

a*(8 – t) = a*(5 – t) + 2b

$\Rightarrow$ 3a = 2b … (1)

a*(8 – t) = b*(6 – t)

Using equation (1), we can say that

$a*(8-t)=\dfrac{3a}{2}*(6-t)$

$t = 2$

Therefore, we can say that the tank gets emptied at 2 pm daily. We can see that A takes 6 hours and pump B takes 4 hours alone.

Hence, working together both can fill the tank in = \dfrac{6*4}{6+4} = 2.4 hours or 2 hours and 24 minutes.

The pumps started filling the tank at 2:00 pm. Hence, the tank will be filled by 4:24 pm.

**Question 46: **Points A and B are 150 km apart. Cars 1 and 2 travel from A to B, but car 2 starts from A when car 1 is already 20 km away from A. Each car travels at a speed of 100 kmph for the ﬁrst 50 km, at 50 kmph for the next 50 km, and at 25 kmph for the last 50 km. The distance, in km, between car 2 and B when car 1 reaches B is

**46) Answer: 5**

**Solution:**

Time taken to cover first 50 km at 100 km/hr = $\frac{1}{2}$ hr.

Time taken to cover second 50 km at 50 km/hr = 1 hr.

Time taken to cover last 50 km at 25 km/hr = 2 hr.

When car 2 starts, car 1 has already covered 20 km.

So, time taken by car 1 to reach B after car 1 starts = total time – time required to travel first 20 km

= 3 hr 30 min – 12 min = 3 hr 18 min

Distance travelled by car 2 = (50 + 50 + 45) = 145 km

Distance from B = (150 – 145) km = 5 km

Hence, 5 is the correct answer.

**Question 47: **A jar contains a mixture of 175 ml water and 700 ml alcohol. Gopal takes out 10% of the mixture and substitutes it by water of the same amount. The process is repeated once again. The percentage of water in the mixture is now

a) 30.3

b) 35.2

c) 25.4

d) 20.5

**47) Answer (B)**

**Solution:**

Final quantity of alcohol in the mixture = $\dfrac{700}{700+175}*(\dfrac{90}{100})^2*[700+175]$ = 567 ml

Therefore, final quantity of water in the mixture = 875 – 567 = 308 ml

Hence, we can say that the percentage of water in the mixture = $\dfrac{308}{875}\times 100$ = 35.2 %

**Question 48: **There are two drums, each containing a mixture of paints A and B. In drum 1, A and B are in the ratio 18 : 7. The mixtures from drums 1 and 2 are mixed in the ratio 3 : 4 and in this final mixture, A and B are in the ratio 13 : 7. In drum 2, then A and B were in the ratio

a) 251 : 163

b) 239 : 161

c) 220 : 149

d) 229 : 141

**48) Answer (B)**

**Solution:**

It is given that in drum 1, A and B are in the ratio 18 : 7.

Let us assume that in drum 2, A and B are in the ratio x : 1.

It is given that drums 1 and 2 are mixed in the ratio 3 : 4 and in this final mixture, A and B are in the ratio 13 : 7.

By equating concentration of A

$\Rightarrow$ $\dfrac{3*\dfrac{18}{18+7}+4*\dfrac{x}{x+1}}{3+4} = \dfrac{13}{13+7}$

$\Rightarrow$ $\dfrac{54}{25}+\dfrac{4x}{x+1} = \dfrac{91}{20}$

$\Rightarrow$ $\dfrac{4x}{x+1} = \dfrac{239}{100}$

$\Rightarrow$ $x = \dfrac{239}{161}$

Therefore, we can say that in drum 2, A and B are in the ratio $\dfrac{239}{161}$ : 1 or 239 : 161.

**Question 49: **Ramesh and Ganesh can together complete a work in 16 days. After seven days of working together, Ramesh got sick and his eﬃciency fell by 30%. As a result, they completed the work in 17 days instead of 16 days. If Ganesh had worked alone after Ramesh got sick, in how many days would he have completed the remaining work?

a) 14.5

b) 11

c) 13.5

d) 12

**49) Answer (C)**

**Solution:**

Let ‘R’ and ‘G’ be the amount of work that Ramesh and Ganesh can complete in a day.

It is given that they can together complete a work in 16 days. Hence, total amount of work = 16(R+G) … (1)

For first 7 days both of them worked together. From 8th day, Ramesh worked at 70% of his original efficiency whereas Ganesh worked at his original efficiency. It took them 17 days to finish the same work. i.e. Ramesh worked at 70% of his original efficiency for 10 days.

$\Rightarrow$ 16(R+G) = 7(R+G)+10(0.7R+G)

$\Rightarrow$ 16(R+G) = 14R+17G

$\Rightarrow$ R = 0.5G … (2)

Total amount of work left when Ramesh got sick = 16(R+G) – 7(R+G) = 9(R+G) = 9(0.5+G) = 13.5G

Therefore, time taken by Ganesh to complete the remaining work = $\dfrac{13.5G}{G}$ = 13.5 days.

**Question 50: **Gopal borrows Rs. X from Ankit at 8% annual interest. He then adds Rs. Y of his own money and lends Rs. X+Y to Ishan at 10% annual interest. At the end of the year, after returning Ankit’s dues, the net interest retained by Gopal is the same as that accrued to Ankit. On the other hand, had Gopal lent Rs. X+2Y to Ishan at 10%, then the net interest retained by him would have increased by Rs. 150. If all interests are compounded annually, then find the value of X + Y.

**50) Answer: 4000**

**Solution:**

Amount of interest paid by Ishan to Gopal if the borrowed amount is Rs. (X+Y) = $\dfrac{10}{100}*(X+Y)$ = 0.1(X+Y)

Gopal also borrowed Rs. X from Ankit at 8% per annum. Therefore, he has to return Ankit Rs. 0.08X as the interest amount on borrowed sum.

Hence, the interest retained by gopal = 0.1(X+Y) – 0.08X = 0.02X + 0.1Y … (1)

It is given that the net interest retained by Gopal is the same as that accrued to Ankit.

Therefore, 0.08X = 0.02X + 0.1Y

$\Rightarrow$ X = (5/3)Y … (2)

Amount of interest paid by Ishan to Gopal if the borrowed amount is Rs. (X+2Y) = $\dfrac{10}{100}*(X+2Y)$ = 0.1X+0.2Y

In this case the amount of interest retained by Gopal = 0.1X+0.2Y – 0.08X = 0.02X + 0.2Y … (3)

It is given that the interest retained by Gopal increased by Rs. 150 in the second case.

$\Rightarrow$ (0.02X + 0.2Y) – (0.02X + 0.1Y) = 150

$\Rightarrow$ Y = Rs. 1500

By substituting value of Y in equation (2), we can say that X = Rs. 2500

Therefore, (X+Y) = Rs. 4000.

**Question 51: **The strength of a salt solution is p% if 100 ml of the solution contains p grams of salt. If three salt solutions A, B, C are mixed in the proportion 1 : 2 : 3, then the resulting solution has strength 20%. If instead the proportion is 3 : 2 : 1, then the resulting solution has strength 30%. A fourth solution, D, is produced by mixing B and C in the ratio 2 : 7. The ratio of the strength of D to that of A is

a) 3 : 10

b) 1 : 3

c) 1 : 4

d) 2 : 5

**51) Answer (B)**

**Solution:**

Let ‘a’, ‘b’ and ‘c’ be the concentration of salt in solutions A, B and C respectively.

It is given that three salt solutions A, B, C are mixed in the proportion 1 : 2 : 3, then the resulting solution has strength 20%.

$\Rightarrow$ $\dfrac{a+2b+3c}{1+2+3} = 20$

$\Rightarrow$ $a+2b+3c = 120$ … (1)

If instead the proportion is 3 : 2 : 1, then the resulting solution has strength 30%.

$\Rightarrow$ $\dfrac{3a+2b+c}{1+2+3} = 30$

$\Rightarrow$ $3a+2b+c = 180$ … (2)

From equation (1) and (2), we can say that

$\Rightarrow$ $b+2c = 45$

$\Rightarrow$ $b = 45 – 2c$

Also, on subtracting (1) from (2), we get

$a – c = 30$

$\Rightarrow$ $a = 30 + c$

In solution D, B and C are mixed in the ratio 2 : 7

So, the concentration of salt in D = $\dfrac{2b + 7c}{9}$ = $\dfrac{90 – 4c + 7c}{9}$ = $\dfrac{90 + 3c}{9}$

Required ratio = $\dfrac{90 + 3c}{9a}$ = $\dfrac{90 + 3c}{9 (30 + c)}$ = $1 : 3$

Hence, option B is the correct answer.

**Question 52: **On a long stretch of east-west road, A and B are two points such that B is 350 km west of A. One car starts from A and another from B at the same time. If they move towards each other, then they meet after 1 hour. If they both move towards east, then they meet in 7 hrs. The diﬀerence between their speeds, in km per hour, is

**52) Answer: 50**

**Solution:**

Let ‘a’ and ‘b’ be the speed (in km/hr) of cars starting from both A and B respectively.

If they both move in east direction, then B will catch A if and only if b > a.

Relative speed of both the cars when they move in east direction = (b – a) km/hr

It takes them 7 hours to meet. i.e. they travel 350 km in 7 hours with a relative speed of (b-a) km/hr.

Hence, (b – a) = $\dfrac{350}{7}$ = 50 km/hr.

**Question 53: **The average of 30 integers is 5. Among these 30 integers, there are exactly 20 which do not exceed 5. What is the highest possible value of the average of these 20 integers?

a) 3.5

b) 5

c) 4.5

d) 4

**53) Answer (C)**

**Solution:**

It is given that the average of the 30 integers = 5

Sum of the 30 integers = 30*5=150

There are exactly 20 integers whose value is less than 5.

To maximise the average of the 20 integers, we have to assign minimum value to each of the remaining 10 integers

So the sum of 10 integers = 10*6=60

The sum of the 20 integers = 150-60= 90

Average of 20 integers = $\ \frac{\ 90}{20}$ = 4.5

**Question 54: **Amal invests Rs 12000 at 8% interest, compounded annually, and Rs 10000 at 6% interest, compounded semi-annually, both investments being for one year. Bimal invests his money at 7.5% simple interest for one year. If Amal and Bimal get the same amount of interest, then the amount, in Rupees, invested by Bimal is

**54) Answer: 20920**

**Solution:**

The amount with Amal at the end of 1 year = 12000*1.08+10000*1.03*1.03=23569

Interest received by Amal = 23569-22000=1569

Let the amount invested by Bimal = 100b

Interest received by Bimal = 100b*7.5*1/100=7.5b

It is given that the amount of interest received by both of them is the same

7.5b=1569

b=209.2

Amount invested by Bimal = 100b = 20920

**Question 55: **Two ants A and B start from a point P on a circle at the same time, with A moving clock-wise and B moving anti-clockwise. They meet for the first time at 10:00 am when A has covered 60% of the track. If A returns to P at 10:12 am, then B returns to P at

a) 10:25 am

b) 10:45 am

c) 10:18 am

d) 10:27 am

**55) Answer (D)**

**Solution:**

When A and B met for the first time at 10:00 AM, A covered 60% of the track.

So B must have covered 40% of the track.

It is given that A returns to P at 10:12 AM i.e A covers 40% of the track in 12 minutes

60% of the track in 18 minutes

B covers 40% of track when A covers 60% of the track.

B covers 40% of the track in 18 minutes.

B will cover the rest 60% in 27 minutes, hence it will return to B at 10:27 AM

**Question 56: **The strength of a salt solution is p% if 100 ml of the solution contains p grams of salt. Each of three vessels A, B, C contains 500 ml of salt solution of strengths 10%, 22%, and 32%, respectively. Now, 100 ml of the solution in vessel A is transferred to vessel B. Then, 100 ml of the solution in vessel B is transferred to vessel C. Finally, 100 ml of the solution in vessel C is transferred to vessel A. The strength, in percentage, of the resulting solution in vessel A is

a) 15

b) 13

c) 12

d) 14

**56) Answer (D)**

**Solution:**

Each of three vessels A, B, C contains 500 ml of salt solution of strengths 10%, 22%, and 32%, respectively.

The amount of salt in vessels A, B, C = 50 ml, 110 ml, 160 ml respectively.

The amount of water in vessels A, B, C = 450 ml, 390 ml, 340 ml respectively.

In 100 ml solution in vessel A, there will be 10ml of salt and 90 ml of water

Now, 100 ml of the solution in vessel A is transferred to vessel B. Then, 100 ml of the solution in vessel B is transferred to vessel C. Finally, 100 ml of the solution in vessel C is transferred to vessel A

i.e after the first transfer, the amount of salt in vessels A, B, C = 40, 120, 160 ml respectively.

after the second transfer, the amount of salt in vessels A, B, C =40, 100, 180 ml respectively.

After the third transfer, the amount of salt in vessels A, B, C = 70, 100, 150 respectively.

Each transfer can be captured through the following table.

Percentage of salt in vessel A =$\ \frac{\ 70}{500}\times\ 100$

=14%

**Question 57: **A cyclist leaves A at 10 am and reaches B at 11 am. Starting from 10:01 am, every minute a motorcycle leaves A and moves towards B. Forty-five such motorcycles reach B by 11 am. All motorcycles have the same speed. If the cyclist had doubled his speed, how many motorcycles would have reached B by the time the cyclist reached B?

a) 22

b) 23

c) 15

d) 20

**57) Answer (C)**

**Solution:**

It is given that starting from 10:01 am, every minute a motorcycle leaves A and moves towards B.

Forty-five such motorcycles reach B by 11 am.

It means that the forty-fifth motorcycle starts at 10:45 AM at A and reaches B by 11:00 AM i.e 15 minutes.

Since the speed of all the motorcycles is the same, all the motorcycles will take the same duration i.e 15 minutes.

If the cyclist doubles the speed, then he will reach B by 10:30 AM. (Since if the speed is doubled, time is reduced by half)

Since each motorcycle takes 15 minutes to reach B, 15 motorcycles would have reached B by the time the cyclist reaches B

**Question 58: **John jogs on track A at 6 kmph and Mary jogs on track B at 7.5 kmph. The total length of tracks A and B is 325 metres. While John makes 9 rounds of track A, Mary makes 5 rounds of track B. In how many seconds will Mary make one round of track A?

**58) Answer: 48**

**Solution:**

Speed of John = 6kmph

Speed of Mary = 7.5 kmph

Lengths of tracks A and B = 325 m

Let the length of track A be a, then the length of track B = 325-a

9 rounds of John on track A = 5 rounds of Mary on track B

$\ \frac{\ 9\times\ a}{6\ \times\ \ \frac{\ 5}{18}}\ =\ \ \frac{\ 5\cdot\left(325-a\right)}{7.5\times\ \ \frac{\ 5}{18}}$

On solving we get , 13a=1300

a=100

The length of track A = 100m, track B = 225m

Mary makes one round of track A = $\ \frac{\ 100}{7.5\times\ \ \frac{\ 5}{18}}$

= 48 sec

**Question 59: **John gets Rs 57 per hour of regular work and Rs 114 per hour of overtime work. He works altogether 172 hours and his income from overtime hours is 15% of his income from regular hours. Then, for how many hours did he work overtime?

**59) Answer: 12**

**Solution:**

It is given that John works altogether 172 hours i.e including regular and overtime hours.

Let a be the regular hours, 172-a will be the overtime hours

John’s income from regular hours = 57*a

John’s income for working overtime hours = (172-a)*144

It is given that his income from overtime hours is 15% of his income from regular hours

a*57*0.15 = (172-a)*114

a=160

The number of hours for which he worked overtime = 172-160=12 hrs

**Question 60: **A shopkeeper sells two tables, each procured at cost price p, to Amal and Asim at a profit of 20% and at a loss of 20%, respectively. Amal sells his table to Bimal at a profit of 30%, while Asim sells his table to Barun at a loss of 30%. If the amounts paid by Bimal and Barun are x and y, respectively, then (x − y) / p equals

a) 1

b) 1.2

c) 0.50

d) 0.7

**60) Answer (A)**

**Solution:**

CP of the table at which the shopkeeper procured each table = p

It is given that shopkeeper sold the tables to Amal and Asim at a profit of 20% and at a loss of 20%, respectively

The selling price of the tables = 1.2p and 0.8p to Amal and Asim respectively.

Amal sells his table to Bimal at a profit of 30%

So, CP of the table by Bimal (x)= 1.2p*1.3 = 1.56p

Asim sells his table to Barun at a loss of 30%

So, CP of the table by Barun (y)= 0.7*0.8p = 0.56p

(x-y)/p = (1.56p-0.56p)/p = p/p=1

**Question 61: **Mukesh purchased 10 bicycles in 2017, all at the same price. He sold six of these at a profit of 25% and the remaining four at a loss of 25%. If he made a total profit of Rs. 2000, then his purchase price of a bicycle, in Rupees, was

a) 6000

b) 8000

c) 4000

d) 2000

**61) Answer (C)**

**Solution:**

Let the cost of each bicycle= 100b

CP of 10 bicycles = 1000b

It is given that he sold six of these at a profit of 25% and the remaining four at a loss of 25%

SP of 10 bicycles = 125b*6+75b*4

=1050b

Profit = 1050b-1000b =50b

50b=2000

CP = 100b = 4000

**Question 62: **In an examination, the score of A was 10% less than that of B, the score of B was 25% more than that of C, and the score of C was 20% less than that of D. If A scored 72, then the score of D was

**62) Answer: 80**

**Solution:**

Let the score of D = 100d

The score of C = 20% less than that of D = 80d

The score of B = 25% more than C = 100d

The score of A = 10% less than B =90d

90d=72

100d= 72*100/90

= 80

**Question 63: **The salaries of Ramesh, Ganesh and Rajesh were in the ratio 6:5:7 in 2010, and in the ratio 3:4:3 in 2015. If Ramesh’s salary increased by 25% during 2010-2015, then the percentage increase in Rajesh’s salary during this period is closest to

a) 10

b) 7

c) 9

d) 8

**63) Answer (B)**

**Solution:**

Let the salaries of Ramesh, Ganesh and Rajesh in 2010 be 6x, 5x, 7x respectively

Let the salaries of Ramesh, Ganesh and Rajesh in 2015 be 3y, 4y, 3y respectively

It is given that Ramesh’s salary increased by 25% during 2010-2015,3y = 1.25*6x

y=2.5x

Percentage increase in Rajesh’s salary = 7.5-7/7=0.07

=7%

**Question 64: **Anil alone can do a job in 20 days while Sunil alone can do it in 40 days. Anil starts the job, and after 3 days, Sunil joins him. Again, after a few more days, Bimal joins them and they together finish the job. If Bimal has done 10% of the job, then in how many days was the job done?

a) 12

b) 13

c) 15

d) 14

**64) Answer (B)**

**Solution:**

Let the total work be LCM of 20, 40 = 40 units

Efficiency of Anil and Sunil is 2 units and 1 unit per day respectively.

Anil works alone for 3 days, so Anil must have completed 6 units.

Bimal completes 10% of the work while working along with Anil and Sunil.

Bimal must have completed 4 units.

The remaining 30 units of work is done by Anil and Sunil

Number of days taken by them 30/3=10

The total work is completed in 3+10=13 days

**Question 65: **In an examination, Rama’s score was one-twelfth of the sum of the scores of Mohan and Anjali. After a review, the score of each of them increased by 6. The revised scores of Anjali, Mohan, and Rama were in the ratio 11:10:3. Then Anjali’s score exceeded Rama’s score by

a) 26

b) 32

c) 35

d) 24

**65) Answer (B)**

**Solution:**

Let the scores of Rama, Anjali and Mohan be r, a, m.

It is given that Rama’s score was one-twelfth of the sum of the scores of Mohan and Anjali

r=$\ \frac{\ m+a}{12}$ ———-(1)

The scores of Rama, Anjali and Mohan after review = r+6, a+6, m+6

a+6:m+6:r+6 = 11:10:3

Let a+6 = 11x => a= 11x-6

m+6=10x => m=10x-6

r+ 6 =3x => r = 3x-6

Substituting these values in equation (1), we get

3x-6=$\ \frac{\ 21x-12}{12}$

12(3x-6) = 21x-12

x=4

Anjali’s score exceeds Rama’s score by (a-r)=8x=32

**Question 66: **Two cars travel the same distance starting at 10:00 am and 11:00 am, respectively, on the same day. They reach their common destination at the same point of time. If the first car travelled for at least 6 hours, then the highest possible value of the percentage by which the speed of the second car could exceed that of the first car is

a) 20

b) 30

c) 25

d) 10

**66) Answer (A)**

**Solution:**

Let the speed of cars be a and b and the distance =d

Minimum time taken by 1st car = 6 hours,

For maximum difference in time taken by both of them, car 1 has to start at 10:00 AM and car 2 has to start at 11:00 AM.

Hence, car 2 will take 5 hours.

Hence a= $\ \frac{\ d}{6}$ and b = $\ \frac{\ d}{5}$

Hence the speed of car 2 will exceed the speed of car 1 by $\ \dfrac{\ \ \frac{\ d}{5}-\ \frac{\ d}{6}}{\ \frac{\ d}{6}}\times\ 100$ = $\ \dfrac{\ \ \frac{\ d}{30}}{\ \frac{\ d}{6}}\times\ 100$ = 20

**Question 67: **The income of Amala is 20% more than that of Bimala and 20% less than that of Kamala. If Kamala’s income goes down by 4% and Bimala’s goes up by 10%, then the percentage by which Kamala’s income would exceed Bimala’s is nearest to

a) 31

b) 29

c) 28

d) 32

**67) Answer (A)**

**Solution:**

Assuming the income of Bimla = 100a, then the income of Amala will be 120a.

And the income of Kamala will be 120a*100/80=150a

If Kamala’s income goes down by 4%, then new income of Kamala = 150a-150a(4/100) = 150a-6a=144a

If Bimla’s income goes up by 10 percent, her new income will be 100a+100a(10/100)=110a

=> Hence the Kamala income will exceed Bimla income by (144a-110a)*100/110a=31

**Question 68: **The wheels of bicycles A and B have radii 30 cm and 40 cm, respectively. While traveling a certain distance, each wheel of A required 5000 more revolutions than each wheel of B. If bicycle B traveled this distance in 45 minutes, then its speed, in km per hour, was

a) $18 \pi$

b) $14 \pi$

c) $16 \pi$

d) $12 \pi$

**68) Answer (C)**

**Solution:**

Distance covered by A in 1 revolution = 2$\pi\ $*30 = 60$\pi\ $

Distance covered by B in 1 revolution = 2$\pi\ $*40 = 80$\pi\ $

Now, (5000+n)60$\pi\ $ = 80$\pi\ $n

=> 15000= 4n-3n =>n=15000

Then distance travelled by B = 15000*80$\pi\ $ cm = 12$\pi\ $ km

Hence, the speed = $\ \frac{\ 12\pi\times\ 60\ }{45}$ = 16$\pi\ $

**Question 69: **In a race of three horses, the first beat the second by 11 metres and the third by 90 metres. If the second beat the third by 80 metres, what was the length, in metres, of the racecourse?

**69) Answer: 880**

**Solution:**

Assuming the length of race course = x and the speed of three horses be a,b and c respectively.

Hence, $\ \frac{\ x}{a}=\ \frac{\ x-11}{b}$……(1)

and $\ \frac{\ x}{a}=\ \frac{\ x-90}{c}$……(2)

Also, $\ \frac{\ x}{b}=\ \frac{\ x-80}{c}$……(3)

From 1 and 2, we get, $\ \frac{\ x-11}{b}=\ \frac{\ x-90}{c}$ …..(4)

Dividing (3) by (4), we get, $\ \frac{\ x-11}{x}=\ \frac{\ x-90}{x-80}$

=> (x-11)(x-80)=x(x-90)

=> 91x-90x=880 => x=880

**Question 70: **Amala, Bina, and Gouri invest money in the ratio 3 : 4 : 5 in fixed deposits having respective annual interest rates in the ratio 6 : 5 : 4. What is their total interest income (in Rs) after a year, if Bina’s interest income exceeds Amala’s by Rs 250?

a) 6350

b) 6000

c) 7000

d) 7250

**70) Answer (D)**

**Solution:**

Assuming the investment of Amala, Bina, and Gouri be 300x, 400x and 500x, hence the interest incomes will be 300x*6/100=18x, 400x*5/100=20x and 500x*4/100 = 20x

Given, Bina’s interest income exceeds Amala by 20x-18x=2x=250 => x=125

Now, total interest income = 18x+20x+20x=58x = 58*125 = 7250

**Question 71: **One can use three different transports which move at 10, 20, and 30 kmph, respectively to reach from A to B. Amal took each mode of transport for $\frac{1}{3}^{rd}$ of his total journey time, while Bimal took each mode of transport for $\frac{1}{3}^{rd}$ of the total distance. The percentage by which Bimal’s travel time exceeds Amal’s travel time is nearest to

a) 22

b) 20

c) 19

d) 21

**71) Answer (A)**

**Solution:**

Assume the total distance between A and B as d and time taken by Amal = t

Since Amal travelled $\frac{1}{3}^{rd}$ of his total journey time in different speeds

d = $\ \frac{\ t}{3}\times\ 10+\ \frac{\ t}{3}\times\ 20+\frac{\ t}{3}\times\ 30\ \ =\ 20t$

$\text{Total time taken by Bimal} = \ \frac{d_1}{s_1}+\frac{d_2}{s_2}+\frac{d_3}{s_3}$

$=\ \frac{20t}{3}\times\ \frac{1}{10}+\frac{20t}{3}\times\ \frac{1}{20}+\frac{20t}{3}\times\ \frac{1}{30}\ \ =\frac{20t\left(6+3+2\right)}{3\ \times30}\ =\frac{11}{9}t$

Hence, the ratio of time taken by Bimal to time taken by Amal = $\frac{\frac{11t}{9}}{t}=\frac{11}{9}$

Therefore, Bimal will exceed Amal’s time by $\ \ \ \frac{\ \ \frac{\ 11t}{9}-t}{t}\times\ 100 = 22.22%$

**Question 72: **Meena scores 40% in an examination and after review, even though her score is increased by 50%, she fails by 35 marks. If her post-review score is increased by 20%, she will have 7 marks more than the passing score. The percentage score needed for passing the examination is

a) 60

b) 80

c) 70

d) 75

**72) Answer (C)**

**Solution:**

Assuming the maximum marks =100a, then Meena got 40a

After increasing her score by 50%, she will get 40a(1+50/100)=60a

Passing score = 60a+35

Post review score after 20% increase = 60a*1.2=72a

=>Hence, 60a+35+7=72a

=>12a=42 =>a=3.5

=> maximum marks = 350 and passing marks = 210+35=245

=> Passing percentage = 245*100/350 = 70

**Question 73: **A person invested a total amount of Rs 15 lakh. A part of it was invested in a fixed deposit earning 6% annual interest, and the remaining amount was invested in two other deposits in the ratio 2 : 1, earning annual interest at the rates of 4% and 3%, respectively. If the total annual interest income is Rs 76000 then the amount (in Rs lakh) invested in the fixed deposit was

**73) Answer: 9**

**Solution:**

Assuming the amount invested in the ratio 2:1 was 200x and 100x, then the fixed deposit investment = 1500000-300x

Hence, the interest = 200x*4/100 = 8x and 100x*3/100=3x

Interest from the fixed deposit = (1500000-300x)*6/100 = 90000-18x

Hence the total interest = 90000-18x+8x+3x=90000-7x =76000

=> 7x=14000 => x=2000

Hence, the fixed deposit investment = 1500000-300*2000 = 900000 = 9 lakhs

**Question 74: **At their usual efficiency levels, A and B together finish a task in 12 days. If A had worked half as efficiently as she usually does, and B had worked thrice as efficiently as he usually does, the task would have been completed in 9 days. How many days would A take to finish the task if she works alone at her usual efficiency?

a) 36

b) 24

c) 18

d) 12

**74) Answer (C)**

**Solution:**

Assuming A completes a units of work in a day and B completes B units of work in a day and the total work = 1 unit

Hence, 12(a+b)=1………(1)

Also, 9($\ \frac{\ a}{2}$+3b)=1 ………(2)

Using both equations, we get, 12(a+b)= 9($\ \frac{\ a}{2}$+3b)

=> 4a+4b=$\ \frac{\ 3a}{2}$+9b

=> $\ \frac{\ 5a}{2}$=5b

=> a=2b

Substituting the value of b in equation (1),

12($\ \frac{\ 3a}{2}$)=1

=> a=$\ \frac{\ 1}{18}$

Hence, the number of days required = 1/($\ \frac{\ 1}{18}$)=18

**Question 75: **In a class, 60% of the students are girls and the rest are boys. There are 30 more girls than boys. If 68% of the students, including 30 boys, pass an examination, the percentage of the girls who do not pass is

**75) Answer: 20**

**Solution:**

Assuming the number of students =100x

Hence, the number of girls = 60x and the number of boys = 40x

We have, 60x-40x=30 => x=1.5

The number of girls = 60*1.5=90

Number of girls that pass = 68x-30=68*1.5-30 = 102-30=72

The number of girls who do not pass = 90-72=18

Hence the percentage of girls who do not pass = 1800/90=20

**Question 76: **On selling a pen at 5% loss and a book at 15% gain, Karim gains Rs. 7. If he sells the pen at 5% gain and the book at 10% gain, he gains Rs. 13. What is the cost price of the book in Rupees?

a) 95

b) 85

c) 80

d) 100

**76) Answer (C)**

**Solution:**

Assuming the cost price of pen = 100p and the cost price of book = 100b

So, on selling a pen at 5% loss and a book at 15% gain, net gain = -5p+15b = 7 ….1

On selling the pen at 5% gain and the book at 10% gain, net gain = 5p+10b = 13 …..2

Adding 1 and 2 we get, 25b=20

Hence 100b= 20*4=80,

C is the answer.

**Question 77: **A chemist mixes two liquids 1 and 2. One litre of liquid 1 weighs 1 kg and one litre of liquid 2 weighs 800 gm. If half litre of the mixture weighs 480 gm, then the percentage of liquid 1 in the mixture, in terms of volume, is

a) 80

b) 70

c) 85

d) 75

**77) Answer (A)**

**Solution:**

The weight/volume(g/L) for liquid 1 = 1000

The weight/volume(g/L) for liquid 2 = 800

The weight/volume(g/L) of the mixture = 480/(1/2) = 960

Using alligation the ratio of liquid 1 and liquid 2 in the mixture = (960-800)/(1000-960) = 160/40 = 4:1

Hence the percentage of liquid 1 in the mixture = 4*100/(4+1)=80

**Question 78: **Ramesh and Gautam are among 22 students who write an examination. Ramesh scores 82.5. The average score of the 21 students other than Gautam is 62. The average score of all the 22 students is one more than the average score of the 21 students other than Ramesh. The score of Gautam is

a) 53

b) 51

c) 48

d) 49

**78) Answer (B)**

**Solution:**

Assume the average of 21 students other than Ramesh = a

Sum of the scores of 21 students other than Ramesh = 21a

Hence the average of 22 students = a+1

Sum of the scores of all 22 students = 22(a+1)

The score of Ramesh = Sum of scores of all 22 students – Sum of the scores of 21 students other than Ramesh = 22(a+1)-21a=a+22 = 82.5 (Given)

=> a = 60.5

Hence, sum of the scores of all 22 students = 22(a+1) = 22*61.5 = 1353

Now the sum of the scores of students other than Gautam = 21*62 = 1302

Hence the score of Gautam = 1353-1302=51

**Question 79: **Three men and eight machines can finish a job in half the time taken by three machines and eight men to finish the same job. If two machines can finish the job in 13 days, then how many men can finish the job in 13 days?

**79) Answer: 13**

**Solution:**

Consider the work done by a man in a day = a and that by a machine = b

Since, three men and eight machines can finish a job in half the time taken by three machines and eight men to finish the same job, hence the efficiency will be double.

=> 3a+8b = 2(3b+8a)

=> 13a=2b

Hence work done by 13 men in a day = work done by 2 machines in a day.

=> If two machines can finish the job in 13 days, then same work will be done 13 men in 13 days.

Hence the required number of men = 13

**Question 80: **A straight road connects points A and B. Car 1 travels from A to B and Car 2 travels from B to A, both leaving at the same time. After meeting each other, they take 45 minutes and 20 minutes, respectively, to complete their journeys. If Car 1 travels at the speed of 60 km/hr, then the speed of Car 2, in km/hr, is

a) 100

b) 90

c) 80

d) 70

**80) Answer (B)**

**Solution:**

Let the speed of Car 2 be ‘x’ kmph and the time taken by the two cars to meet be ‘t’ hours.

In ‘t’ hours, Car 1 travels $\left(60\ \times\ t\right)\ km$ while Car 2 travels $\left(x\ \times\ t\right)\ km$

It is given that the time taken by Car 1 to travel $\left(x\ \times\ t\right)\ km$ is 45 minutes or (3/4) hours. $\therefore\ \frac{\left(x\ \times\ t\right)}{60}\ =\ \frac{3}{4}\ $ or $t=\frac{180}{4x}$….(i)

Similarly, the time taken by Car 2 to travel $\left(60\ \times\ t\right)\ km$ is 20 minutes or (1/3) hours. $\therefore\ \frac{\left(60\times\ t\right)}{x}=\frac{1}{3}$ or $\therefore\ t=\frac{x}{180}$….(ii)

Equating the values in (i) and (ii), and solving for x:

$\therefore\ \frac{180}{4x}=\frac{x}{180}\ \ \longrightarrow\ \ \ x\ =90\ kmph$

Hence, Option B is the correct answer.

**Question 81: **A person spent Rs 50000 to purchase a desktop computer and a laptop computer. He sold the desktop at 20% profit and the laptop at 10% loss. If overall he made a 2% profit then the purchase price, in rupees, of the desktop is

**81) Answer: 20000**

**Solution:**

Let the price of desktop and laptop be x,y respectively.

Given,

x+y=50000…(i)

12.x+0.9y=50000(1.02)=51000…(ii)

(ii)-0.9(i) gives

0.3x=6000=> x=20000.

**Question 82: **A solution, of volume 40 litres, has dye and water in the proportion 2 : 3. Water is added to the solution to change this proportion to 2 : 5. If one fourths of this diluted solution is taken out, how many litres of dye must be added to the remaining solution to bring the proportion back to 2 : 3?

**82) Answer: 8**

**Solution:**

Initially the amount of Dye and Water are 16,24 respectively.

To make the ratio of Dye to Water to 2:5 the amount of water should be 40l for 16l of Dye=> 16l of water is added.

Now, the Dye and Water arr 16,40 respectively.

After removing 1/4th of solution the amount of Dye and Water will be 12,30l respectively.

To have Dye and Water in the ratio of 2:3, for 30l of water we need 20l of Dye => 8l of Dye should be added.

Hence , 8 is correct answer.

**Question 83: **An alloy is prepared by mixing three metals A, B and C in the proportion 3 : 4 : 7 by volume. Weights of the same volume of the metals A. B and C are in the ratio 5 : 2 : 6. In 130 kg of the alloy, the weight, in kg. of the metal C is

a) 48

b) 84

c) 70

d) 96

**83) Answer (B)**

**Solution:**

Let the volume of Metals A,B,C we 3x, 4x, 7x

Ratio weights of given volume be 5y,2y,6y

.’. 15xy+8xy+42xy=130 => 65xy=130 => xy=2.

.’.`The weight, in kg. of the metal C is 42xy=84.

**Question 84: **Leaving home at the same time, Amal reaches the office at 10:15 am if he travels at 8 km/hr, and at 9:40 am if he travels at 15 km/hr. Leaving home at 9.10 am, at what speed, in km/hr, must he travel so as to reach office exactly at 10 am?

a) 13

b) 12

c) 14

d) 11

**84) Answer (B)**

**Solution:**

The difference in the time take to traverse the same distance $’d’$ at two different speeds is 35 minutes. Equating this: $\frac{d}{8}-\frac{d}{15}\ =\ \frac{35}{60}$

On solving, we obtain $d = 10 kms$. Let $x kmph$ be the speed at which Amal needs to travel to reach the office in 50 minutes; then

$\frac{10}{x}=\frac{50}{60}\ or\ x\ =\ 12\ kmph$.Hence, Option B is the correct answer.

**Question 85: **A train travelled at one-thirds of its usual speed, and hence reached the destination 30 minutes after the scheduled time. On its return journey, the train initially travelled at its usual speed for 5 minutes but then stopped for 4 minutes for an emergency. The percentage by which the train must now increase its usual speed so as to reach the destination at the scheduled time, is nearest to

a) 50

b) 58

c) 67

d) 61

**85) Answer (C)**

**Solution:**

Let the total distance be ‘D’ km and the speed of the train be ‘s’ kmph. The time taken to cover D at speed d is ‘t’ hours. Based on the information: on equating the distance, we get $s\ \times\ t\ =\ \frac{s}{3}\times\ \left(t+\frac{1}{2}\right)$

On solving we acquire the value of $\ t\ =\frac{1}{4}$ or 15 mins. We understand that during the return journey, the first 5 minutes are spent traveling at speed ‘s’ {distance traveled in terms of s = $\ \frac{s}{12}$}. Remaining distance in terms of ‘s’ = $\ \frac{s}{4}-\frac{s}{12}\ =\frac{s}{6}$

The rest 4 minutes of stoppage added to this initial 5 minutes amounts to a total of 9 minutes. The train has to complete the rest of the journey in $15 – 9 = 6 mins$ or {1/10 hours}. Thus, let ‘x’ kmph be the new value of speed. Based on the above, we get $\frac{s}{\frac{6}{x}}\ =\frac{1}{10}\ or\ x\ =\frac{10s}{6}$

Since the increase in speed needs to be calculated: $\frac{\left(\frac{10s}{6}\ -s\right)}{s}\times\ 100\ =\frac{200}{3}\approx\ 67\%$ increase.

Hence, Option C is the correct answer.

**Question 86: **Two persons are walking beside a railway track at respective speeds of 2 and 4 km per hour in the same direction. A train came from behind them and crossed them in 90 and 100 seconds, respectively. The time, in seconds, taken by the train to cross an electric post is nearest to

a) 87

b) 82

c) 78

d) 75

**86) Answer (B)**

**Solution:**

Let the length of the train be $ l kms$ and speed be $ s kmph$. Base on the two scenarios presented, we obtain:

$\frac{l}{s-2}=\frac{90}{3600}$….(i) and $\frac{l}{s-4}=\frac{100}{3600}$…(ii)

On dividing (ii) by (i) and simplifying we acquire the value of $s$ as $22 kmph$. Substituting this value in (i), we have $l=\frac{90}{3600}\times\ 20\ kms$ {keeping it in km and hours for convenience}

Since we need to find $\frac{l}{s}$, let this be equal to $x$. Then, $x\ =\ 90\times\frac{20}{22}\ =81.81\ \approx\ 82\ \sec onds\ $

Hence, Option B is the correct choice.

**Question 87: **In a group of people, 28% of the members are young while the rest are old. If 65% of the members are literates, and 25% of the literates are young, then the percentage of old people among the illiterates is nearest to

a) 62

b) 55

c) 59

d) 66

**87) Answer (D)**

**Solution:**

Let ‘x’ be the strength of group G. Based on the information, $0.65x$ constitutes of literate people {the rest $0.35x$ = illiterate}

Of this $0.65x$, 75% are old people =(0.75)0.65x old literates. The total number of old people in group G is $0.72x$ {72% of the total}. Thus, the total number of old people who are illiterate = $0.72x-0.4875x\ =\ 0.2325x$. This is $\frac{0.2325x}{0.35x}\times\ 100\ \approx\ \ 66\%$ of the total number of illiterates. Hence, Option D is the correct answer.

**Question 88: **Veeru invested Rs 10000 at 5% simple annual interest, and exactly after two years, Joy invested Rs 8000 at 10% simple annual interest. How many years after Veeru’s investment, will their balances, i.e., principal plus accumulated interest, be equal?

**88) Answer: 12**

**Solution:**

Let their individual Amounts be equal after ‘t’ years. Let their initial investments amount to $A_V$ and $A_J$ ;

$A_V\ =10,000\left(1+\frac{5t}{100}\right)$ and $A_J\ =8,000\left(1+\frac{10\left(t-2\right)}{100}\right)$

Equating both: $10,000\left(1+\frac{5t}{100}\right)\ =8,000\left(1+\frac{10\left(t-2\right)}{100}\right)$

On simplifying both sides, we get: $15t\ =\ 180\ ;\ t\ =\ 12$

**Question 89: **Two alcohol solutions, A and B, are mixed in the proportion 1:3 by volume. The volume of the mixture is then doubled by adding solution A such that the resulting mixture has 72% alcohol. If solution A has 60% alcohol, then the percentage of alcohol in solution B is

a) 90%

b) 94%

c) 92%

d) 89%

**89) Answer (C)**

**Solution:**

Initially let’s consider A and B as one component

The volume of the mixture is doubled by adding A(60% alcohol) i.e they are mixed in 1:1 ratio and the resultant mixture has 72% alcohol.

Let the percentage of alcohol in component 1 be ‘x’.

Using allegations , $\frac{\left(72-60\right)}{x-72}=\frac{1}{1}$ => x= 84

Percentage of alcohol in A = 60% => Let’s percentage of alcohol in B = x%

The resultant mixture has 84% alcohol. ratio = 1:3

Using allegations , $\frac{\left(x-84\right)}{84-60}=\frac{1}{3}$

=> x= 92%

**Question 90: **A batsman played n + 2 innings and got out on all occasions. His average score in these n + 2 innings was 29 runs and he scored 38 and 15 runs in the last two innings. The batsman scored less than 38 runs in each of the first n innings. In these n innings, his average score was 30 runs and lowest score was x runs. The smallest possible value of x is

a) 4

b) 3

c) 2

d) 1

**90) Answer (C)**

**Solution:**

Given, $\frac{\text{sum of scores in n matches+38+15}}{n+2}=29$

Given, $\frac{\text{sum of scores in n matches}}{n}=30$

=> 30n + 53 = 29(n+2) => n=5

Sum of the scores in 5 matches = 29*7 – 38-15 = 150

Since the batsmen scored less than 38, in each of the first 5 innings. The value of x will be minimum when remaining four values are highest

=> 37+37+37+37 + x = 150

=> x = 2

**Question 91: **A contractor agreed to construct a 6 km road in 200 days. He employed 140 persons for the work. After 60 days, he realized that only 1.5 km road has been completed. How many additional people would he need to employ in order to finish the work exactly on time?

**91) Answer: 40**

**Solution:**

Let the desired efficiency of each worker ‘6x’ per day.

140*6x*200= 6 km …(i)

In 60 days 60/200*6=1.8 km of work is to be done but actually 1.5km is only done.

Actual efficiency ‘y’= 1.5/1.8 *6x =5x.

Now, left over work = 4.5km which is to be done in 140 days with ‘n’ workers whose efficiency is ‘y’.

=> n*5x*140=4.5 …(ii)

(i)/(ii) gives,

$\frac{\left(140\cdot6x\cdot200\right)}{\left(n\cdot5x\cdot140\right)}=\frac{6}{4.5}$

=> n=180.

.’. Extra 180-140 =40 workers are needed.

**Question 92: **Vimla starts for office every day at 9 am and reaches exactly on time if she drives at her usual speed of 40 km/hr. She is late by 6 minutes if she drives at 35 km/hr. One day, she covers two-thirds of her distance to office in one-thirds of her usual total time to reach office, and then stops for 8 minutes. The speed, in km/hr, at which she should drive the remaining distance to reach office exactly on time is

a) 29

b) 26

c) 28

d) 27

**92) Answer (C)**

**Solution:**

Let distance = d

Given, $\frac{d}{35}-\frac{d}{40}=\frac{6}{60}$

=> d = 28km

The actual time taken to travel 28km = 28/40 = 7/10 hours = 42 min.

Given time taken to travel 58/3 km = 1/3 *42 = 14 min.

Then a break of 8 min.

To reach on time, he should cover remaining 28/3 km in 20 min => Speed = $\frac{\left(\frac{28}{3}\right)}{\frac{20}{60}}=28\ $ km/hr

**Question 93: **In the final examination, Bishnu scored 52% and Asha scored 64%. The marks obtained by Bishnu is 23 less, and that by Asha is 34 more than the marks obtained by Ramesh. The marks obtained by Geeta, who scored 84%, is

a) 357

b) 417

c) 439

d) 399

**93) Answer (D)**

**Solution:**

Let the total marks be 100x

Marks obtained by Bishnu = 52x

Marks obtained by Asha = 64x

Marks obtained by Ramesh = 52x+23

Marks obtained by Ramesh = 64x-34

=> 52x+23 = 64x-34

=> x = $\frac{19}{4}$

Marks obtained by Geeta =84x = 84*19/4 = 399

**Question 94: **A person invested a certain amount of money at 10% annual interest, compounded half-yearly. After one and a half years, the interest and principal together became Rs.18522. The amount, in rupees, that the person had invested is

**94) Answer: 16000**

**Solution:**

Given,

Rate of interest = 10%

Since it is compounded half-yearly, R=5%

n=3

We know, A = $P\left(1+\frac{R}{100}\right)^{^n}$

18522 = $P\left(1+0.05\right)^3$

=> P = 16000

**Question 95: **A man buys 35 kg of sugar and sets a marked price in order to make a 20% profit. He sells 5 kg at this price, and 15 kg at a 10% discount. Accidentally, 3 kg of sugar is wasted. He sells the remaining sugar by raising the marked price by p percent so as to make an overall profit of 15%. Then p is nearest to

a) 22

b) 35

c) 25

d) 31

**95) Answer (C)**

**Solution:**

Let the cost price of 1kg of sugar = Rs 100

The total cost price of 35 kg = Rs3500

Marked up price per kg = Rs 120

GIven, the final profit is 15% => Final SP of 35 kg = 3500 *1.15 = Rs 4025

First 5 kg’s are sold at 20% marked up price => $SP_1=5\cdot100\cdot1.2$ = Rs 600

Next 15 kgs are sold after giving 10% discount => $SP_2=15\cdot100\cdot1.2\cdot0.9\ =\ 1620$

3kgs of sugar got wasted

=> 23 kg of sugar was sold at Rs (600 +1620) = Rs 2220

Remaining 12kg should be sold at Rs 4025 – 2220 = Rs1805

=> SP of 1kg = 1805/12 $\simeq\ 150$

Hence, the seller should further mark up by $\frac{\left(150-120\right)}{120}\cdot100\ =\ 25\%$

**Question 96: **A and B are two railway stations 90 km apart. A train leaves A at 9:00 am, heading towards B at a speed of 40 km/hr. Another train leaves B at 10:30 am, heading towards A at a speed of 20 km/hr. The trains meet each other at

a) 11 : 45 am

b) 11 : 20 am

c) 11 : 00 am

d) 10 : 45 am

**96) Answer (C)**

**Solution:**

The distance travelled by A between 9:00 Am and 10:30 Am is 3/2*40 =60 km.

Now they are separated by 30 km

Let the time taken to meet =t

Distance travelled by A in time t + Distance travelled by B in time t = 30

40t + 20t =30 => t=1/2 hour

Hence they meet at 11:00 AM

**Question 97: **Anil, Sunil, and Ravi run along a circular path of length 3 km, starting from the same point at the same time, and going in the clockwise direction. If they run at speeds of 15 km/hr, 10 km/hr, and 8 km/hr, respectively, how much distance in km will Ravi have run when Anil and Sunil meet again for the first time at the starting point?

a) 4.8

b) 4.6

c) 5.2

d) 4.2

**97) Answer (A)**

**Solution:**

Anil and Sunil will meet at a first point after LCM ( $\frac{3}{15},\frac{3}{10}$) = 3/5 hr

In the mean time, distance travelled by ravi = 8 * 3/5 = 4.8 km

**Question 98: **The distance from B to C is thrice that from A to B. Two trains travel from A to C via B. The speed of train 2 is double that of train 1 while traveling from A to B and their speeds are interchanged while traveling from B to C. The ratio of the time taken by train 1 to that taken by train 2 in travelling from A to C is

a) 5:7

b) 4:1

c) 1:4

d) 7:5

**98) Answer (A)**

**Solution:**

Let the distance from A to B be “x”, then the distance from B to C will be 3x. Now the speed of Train 2 is double of Train 1. Let the speed of Train 1 be “v”, then the speed of Train 2 will be “2v” while travelling from A to B.

Time taken by Train 1 = (x/v)

Time taken by Train 2 = (x/2v)

Now from B to C distance is “3x” and the speed of Train 2 is (v) and the speed of Train 1 is (2v).

Time taken by Train 1 = 3x/2v

Time taken by Train 2 = 3x/v

Total time taken by Train 1 = x/v(1+(3/2)) = (5/2)(x/v)

Total time taken by Train 2 = x/v(3+(1/2))= (7/2)(x/v)

Ratio of time taken = $\frac{5}{\frac{2}{\frac{7}{2}}}=\frac{5}{7}$

**Question 99: **John takes twice as much time as Jack to finish a job. Jack and Jim together take one-thirds of the time to finish the job than John takes working alone. Moreover, in order to finish the job, John takes three days more than that taken by three of them working together. In how many days will Jim finish the job working alone?

**99) Answer: 4**

**Solution:**

Let Jack take “t” days to complete the work, then John will take “2t” days to complete the work. So work done by Jack in one day is (1/t) and John is (1/2t) .

Now let Jim take “m” days to complete the work. According to question, $\frac{1}{t}+\frac{1}{m}=\frac{3}{2t}\ or\ \frac{1}{m}=\frac{1}{2t\ }or\ m=2t$ Hence Jim takes “2t” time to complete the work.

Now let the three of them complete the work in “p” days. Hence John takes “p+3” days to complete the work.

$\frac{1}{2t}\left(m+3\right)=\left(\frac{4}{2t}\right)m$

$\frac{1}{2t}\left(m+3\right)=\left(\frac{4}{2t}\right)m$

or m=1. Hence JIm will take (1+3)=4 days to complete the work. Similarly John will also take 4 days to complete the work

**Question 100: **For the same principal amount, the compound interest for two years at 5% per annum exceeds the simple interest for three years at 3% per annum by Rs 1125. Then the principal amount in rupees is

**100) Answer: 90000**

**Solution:**

For two years the compound interest is $\frac{PR(1)}{100}+\frac{PR(1)}{100}\left(1+\frac{PR(1)}{100}\right)$

For three years the simple interest is $\frac{9PR}{100}$

Now R(1)= 5% and R=3%

Hence $\frac{5P}{100}+\frac{5P}{100}\left(1.05\right)-\frac{9P}{100}=1125$

$\frac{-4P}{100}+\frac{5.25P}{100}=1125$

$\frac{1.25P}{100}=1125$

Solving we get P= 90000

**Question 101: **In a car race, car A beats car B by 45 km. car B beats car C by 50 km. and car A beats car C by 90 km. The distance (in km) over which the race has been conducted is

a) 475

b) 450

c) 500

d) 550

**101) Answer (B)**

**Solution:**

Now car A beats car B by 45km. Let the speed of car A be v(a) and speed of car B be v(b).

$\frac{v\left(a\right)}{v\left(b\right)}=\frac{m}{m-45}$ …..(1)where ‘”m” is the entire distance of the race track.

Moreover $\frac{v\left(b\right)}{v\left(c\right)}=\frac{m}{m-50}$…….(2)

and finally $\frac{v\left(a\right)}{v\left(c\right)}=\frac{m}{m-90}$……(3)

Multiplying (1) and (2) we get (3). $\frac{m}{m-90}=\frac{m}{m-45}\left(\frac{m}{m-50}\right)$

Solving we get m=450 which is the length of the entire race track

**Question 102: **A sum of money is split among Amal, Sunil and Mita so that the ratio of the shares of Amal and Sunil is 3:2, while the ratio of the shares of Sunil and Mita is 4:5. If the difference between the largest and the smallest of these three shares is Rs.400, then Sunil’s share, in rupees, is

**102) Answer: 800**

**Solution:**

Let the amount of money with Amal and Sunil be 6x and 4x. Now the amount of money with Mita be 5x. Difference between the largest and smallest amount is ₹400 i.e. 6x-4x=400 or 2x=400 or x=200 . Amount of money with Sunil is 200(4)=₹800

**Question 103: **A and B are two points on a straight line. Ram runs from A to B while Rahim runs from B to A. After crossing each other. Ram and Rahim reach their destination in one minute and four minutes, respectively. if they start at the same time, then the ratio of Ram’s speed to Rahim’s speed is

a) $\frac{1}{2}$

b) $\sqrt{2}$

c) $2$

d) $2\sqrt{2}$

**103) Answer (C)**

**Solution:**

Let the speed of Ram be v(r) and the speed of Rahim be v(h) respectively. Let them meet after time “t” from the beginning.

Hence Ram will cover v(r)(t) during that time and Rahim will cover v(h)t respectively.

Now after meeting Ram reaches his destination in 1 min i.e. Ram covered v(h)t in 1 minute or v(r)(1)= v(h)(t)

Similarly Rahim reaches his destination in 4 min i.e. Rahim covered v(r)t in 4 minutes or v(h)(4)= v(r)(t)

Dividing both the equations we get $\frac{v\left(r\right)}{4v\left(h\right)}=\frac{v\left(h\right)}{v\left(r\right)}\ or\ \frac{v\left(r\right)}{v\left(h\right)}=2$ Hence the ratio is 2.

**Question 104: **Two circular tracks T1 and T2 of radii 100 m and 20 m, respectively touch at a point A. Starting from A at the same time, Ram and Rahim are walking on track T1 and track T2 at speeds 15 km/hr and 5 km/hr respectively. The number of full rounds that Ram will make before he meets Rahim again for the first time is

a) 5

b) 3

c) 2

d) 4

**104) Answer (B)**

**Solution:**

To complete one round Ram takes 100m/15kmph and Rahim takes 20m/5kmph

They meet for the first time after L.C.M of (100m/15kmph , 20m/5kmph) = 100m/5kmph=20m/kmph.

Distance traveled by Ram =20m/kmph * 15kmph =300m.

So, he must have ran 300/100=3 rounds.

**Note:**

CAT gave both 2 and 3 as correct answers because of the word ‘**before**‘.

**Question 105: **Anil buys 12 toys and labels each with the same selling price. He sells 8 toys initially at 20% discount on the labeled price. Then he sells the remaining 4 toys at an additional 25% discount on the discounted price. Thus, he gets a total of Rs 2112, and makes a 10% profit. With no discounts, his percentage of profit would have been

a) 50

b) 60

c) 54

d) 55

**105) Answer (A)**

**Solution:**

Let the CP of the each toy be “x”. CP of 12 toys will be “12x”. Now the shopkeeper made a 10% profit on CP. This means that

12x(1.1)= 2112 or x=160 . Hence the CP of each toy is ₹160.

Now let the SP of each toy be “m”. Now he sold 8 toys at 20% discount. This means that 8m(0.8) or 6.4m

He sold 4 toys at an additional 25% discount. 4m(0.8)(0.75)=2.4m Now 6.4m+2.4m=8.8m=2112 or m=240

Hence CP= 160 and SP=240. Hence profit percentage is 50%.

**Question 106: **Two trains cross each other in 14 seconds when running in opposite directions along parallel tracks. The faster train is 160 m long and crosses a lamp post in 12 seconds. If the speed of the other train is 6 km/hr less than the faster one, its length, in m, is

a) 184

b) 192

c) 190

d) 180

**106) Answer (C)**

**Solution:**

Speed of the faster train = $\frac{160}{12}=\frac{40}{3}\ $ m/s

Speed of the slower train = $\frac{40}{3}-\left(6\times\ \frac{5}{18}\right)=\frac{35}{3}$ m/s

Sum of speeds (when the trains travel towards each other) = $\frac{40}{3}+\frac{35}{3}=25$ m/s

Let the slower train be $x$ metres long; then: $\frac{160+x}{25}=14$

On solving, $x=190\ m$

**Question 107: **Anu, Vinu and Manu can complete a work alone in 15 days, 12 days and 20 days, respectively. Vinu works everyday. Anu works only on alternate days starting from the first day while Manu works only on alternate days starting from the second day. Then, the number of days needed to complete the work is

a) 5

b) 8

c) 6

d) 7

**107) Answer (D)**

**Solution:**

Let the total amount of work be 60 units.

Then Anu, Vinu, and Manu do 4, 5, and 3 units of work per day respectively.

On the 1st day, Anu and Vinu work. Work done on the 1st day = 9 units

On the 2nd day, Manu and Vinu work. Work done on the 2nd day = 8 units

This cycle goes on. And in 6 days, the work completed is 9+8+9+8+9+8 = 51 units.

On the 7th day, again Anu and Vinu work and complete the remaining 9 units of work. Thus, the number of days taken is 7 days.

**Question 108: **The amount Neeta and Geeta together earn in a day equals what Sita alone earns in 6 days. The amount Sita and Neeta together earn in a day equals what Geeta alone earns in 2 days. The ratio of the daily earnings of the one who earns the most to that of the one who earns the least is

a) 3:2

b) 11:7

c) 11:3

d) 7:3

**108) Answer (C)**

**Solution:**

Let the amounts Neeta, Geeta, and Sita earn in a day be n, g, and s respectively.

Then, it has been given that:

n+g=6s -i

s+n=2g -ii

ii-i, we get: s-g = 2g-6s

7s = 3g.

Let g be 7a. Then s earns 3a.

Then n earns 6s-g = 18a-7a = 11a.

Thus, the ratio is 11a:3a = 11:3

**Question 109: **Onion is sold for 5 consecutive months at the rate of Rs 10, 20, 25, 25, and 50 per kg, respectively. A family spends a fixed amount of money on onion for each of the first three months, and then spends half that amount on onion for each of the next two months. The average expense for onion, in rupees per kg, for the family over these 5 months is closest to

a) 26

b) 18

c) 16

d) 20

**109) Answer (B)**

**Solution:**

Let us assume the family spends Rs. 100 each month for the first 3 months and then spends Rs. 50 in each of the next two months.

Then amount of onions bought = 10, 5, 4, 2, 1, for months 1-5 respectively.

Total amount bought = 22kg.

Total amount spent = 100+100+100+50+50 = 400.

Average expense = $\frac{400}{22}=Rs.18.18\approx\ 18$

**Question 110: **Amal purchases some pens at ₹ 8 each. To sell these, he hires an employee at a fixed wage. He sells 100 of these pens at ₹ 12 each. If the remaining pens are sold at ₹ 11 each, then he makes a net profit of ₹ 300, while he makes a net loss of ₹ 300 if the remaining pens are sold at ₹ 9 each. The wage of the employee, in INR, is

**110) Answer: 1000**

**Solution:**

Let the number of pens purchased be n. Then the cost price is 8n. The total expenses incurred would be 8n+W, where W refers to the wage.

Then SP in the first case = $12\times\ 100+11\times\ \left(n-100\right)$

Given profit is 300 in this case: 1200+11n-1100-8n-W=300 =>3n-W = 200

In second case: 1200+9n-900-8n-W=-300 (Loss). => W-n = 600.

Adding the two equations: 2n = 800

n = 400.

Thus W = 600 + 400 = 1000

**Question 111: **Anil invests some money at a fixed rate of interest, compounded annually. If the interests accrued during the second and third year are ₹ 806.25 and ₹ 866.72, respectively, the interest accrued, in INR, during the fourth year is nearest to

a) 929.48

b) 934.65

c) 931.72

d) 926.84

**111) Answer (C)**

**Solution:**

Let the principal amount be P and the interest rate be r.

Then $P\left(1+r\right)^2-P\left(1+r\right)=806.25$ -(1)

$P\left(1+r\right)^3-P\left(1+r\right)^2=866.72$ -(2)

Dividing (2) by (1), we get:

$\frac{\left(P\left(1+r\right)^3-P\left(1+r\right)^2\right)}{P\left(1+r\right)^2-P\left(1+r\right)}=\frac{866.72}{806.25}$

$\frac{\left(\left(1+r\right)^2-1-r\right)}{1+r-1}=1.075$

$\frac{r^2+r}{r}=1.075$

r=0.075 or 7.5%

$\frac{\left(Interest\ accrued\ in\ 4th\ yr\right)}{Interest\ accrued\ in\ 3rd\ yr}=\frac{X}{866.72}$

$\frac{\left(P\left(1+r\right)^4-P\left(1+r\right)^3\right)}{P\left(1+r\right)^3-P\left(1+r\right)^2}=\frac{X}{866.72}$

Dividing numerator and denominator by $P\left(1+r\right)^2$

$\frac{r^2+2r+1-1-r}{1+r-1}=\frac{X}{866.72}$

$r+1=\frac{X}{866.72}$

$X=1.075\times\ 866.72=931.72$

**Question 112: **Amar, Akbar and Anthony are working on a project. Working together Amar and Akbar can complete the project in 1 year, Akbar and Anthony can complete in 16 months, Anthony and Amar can complete in 2 years. If the person who is neither the fastest nor the slowest works alone, the time in months he will take to complete the project is

**112) Answer: 32**

**Solution:**

Let the total work be 48 units. Let Amar do ‘m’ work, Akbar do ‘k’ work, and Anthony do ‘n’ units of work in a month.

Amar and Akbar complete the project in 12 months. Hence, in a month they do $\frac{48}{12}$=4 units of work.

m+k = 4.

Similarly, k+n = 3, and m+n = 2.

Solving the three equations, we get $m=\frac{3}{2},\ k=\frac{5}{2},\ n=\frac{1}{2}$.

Here, Amar works neither the fastest not the slowest, and he does 1.5 units of work in a month. Hence, to complete the work, he would take $\frac{48}{1.5}=32$months.

**Question 113: **The strength of an indigo solution in percentage is equal to the amount of indigo in grams per 100 cc of water. Two 800 cc bottles are filled with indigo solutions of strengths 33% and 17%, respectively. A part of the solution from the first bottle is thrown away and replaced by an equal volume of the solution from the second bottle. If the strength of the indigo solution in the first bottle has now changed to 21% then the volume, in cc, of the solution left in the second bottle is

**113) Answer: 200**

**Solution:**

Let Bottle A have an indigo solution of strength 33% while Bottle B have an indigo solution of strength 17%.

The ratio in which we mix these two solutions to obtain a resultant solution of strength 21% : $\frac{A}{B}=\frac{21-17}{33-21}=\frac{4}{12}or\ \frac{1}{3}$

Hence, three parts of the solution from Bottle B is mixed with one part of the solution from Bottle A. For this process to happen, we need to displace 600 cc of solution from Bottle A and replace it with 600 cc of solution from Bottle B {since both bottles have 800 cc, three parts of this volume = 600cc}.As a result, 200 cc of the solution remains in Bottle B.

Hence, the correct answer is 200 cc.

**Question 114: **Identical chocolate pieces are sold in boxes of two sizes, small and large. The large box is sold for twice the price of the small box. If the selling price per gram of chocolate in the large box is 12% less than that in the small box, then the percentage by which the weight of chocolate in the large box exceeds that in the small box is nearest to

a) 144

b) 127

c) 135

d) 124

**114) Answer (B)**

**Solution:**

Let the selling price of the Large and Small boxes of chocolates be Rs.200 and Rs.100 respectively. Let us consider that the Large box has $L$ grams of chocolate while the Small box has $S$ grams of chocolate.

The relation between the selling price per gram of chocolate can be represented as: $\frac{200}{L}=0.88\times\ \frac{100}{S}$

On solving we obtain the ratio of the amount of chocolate in each box as: $\frac{L}{S}=\frac{25}{11}$

The percentage by which the weight of chocolate in the large box exceeds that in the small box = $\left(\frac{25}{11}-1\right)\times\ 100\approx\ 127\%$

**Question 115: **Raj invested ₹ 10000 in a fund. At the end of first year, he incurred a loss but his balance was more than ₹ 5000. This balance, when invested for another year, grew and the percentage of growth in the second year was five times the percentage of loss in the first year. If the gain of Raj from the initial investment over the two year period is 35%, then the percentage of loss in the first year is

a) 5

b) 15

c) 17

d) 10

**115) Answer (D)**

**Solution:**

Raj invested Rs 10000 in the first year. Assuming the loss he faced was x%.

The amount after 1 year is 10,000*(1 – x/100). = 10000 – 100*x.

Given the balance was greater than Rs 5000 and hence x < 50 percent.

When Raj invested this amount in the second year he earned a profit which is five times that of the first-year percentage.

Hence the amount after the second year is : (10000 – 100x)(1+$\frac{\left(5\cdot x\right)}{100}$).

Raj gained a total of 35 percent over the period of two years and hence the 35 percent is Rs 3500.

Hence the final amount is Rs 13,500.

(10000 – 100x)(1+$\frac{\left(5\cdot x\right)}{100}$) = 13,500

$\left(100+5\cdot x\right)\cdot\left(100\ -\ x\right)\ =\ 13500$

10000 – 100*x +500*x – 5*$x^2$ = 13500.

$5x^2-400x+3500\ =\ 0$

Solving the equation the roots are :

x = 10, x = 70.

Since x < 50, x = 10 percent.

**Question 116: **Two trains A and B were moving in opposite directions, their speeds being in the ratio 5 : 3. The front end of A crossed the rear end of B 46 seconds after the front ends of the trains had crossed each other. It took another 69 seconds for the rear ends of the trains to cross each other. The ratio of length of train A to that of train B is

a) 3:2

b) 5:3

c) 2:3

d) 2:1

**116) Answer (A)**

**Solution:**

Considering the length of train A = La, length of train B = Lb.

The speed of train A be 5*x, speed of train B be 3*x.

From the information provided :

The front end of A crossed the rear end of B 46 seconds after the front ends of the trains had crossed each other.

In this case, train A traveled a distance equivalent to the length of train B which is Lb at a speed of 5*x+3*x = 8*x because both the trains are traveling in the opposite direction.

Hence (8*x)*(46) = Lb.

In the information provided :

It took another 69 seconds for the rear ends of the trains to cross each other.

In the next 69 seconds

The train B traveled a distance equivalent to the length of train A in this 69 seconds.

Hence (8*x)*(69) = La.

La/Lb = 69/46 = 3/2 = 3 : 2

**Question 117: **From a container filled with milk, 9 litres of milk are drawn and replaced with water. Next, from the same container, 9 litres are drawn and again replaced with water. If the volumes of milk and water in the container are now in the ratio of 16 : 9, then the capacity of the container, in litres, is

**117) Answer: 45**

**Solution:**

Let initial volume be V, final be F for milk.

The formula is given by : $F\ =\ V\cdot\left(1-\frac{K}{V}\right)^n$ n is the number of times the milk is drawn and replaced.

so we get $F=\ V\left(1-\frac{K}{V}\right)^{^2}$

here K =9

we get

$\frac{16}{25}V\ =\ V\ \left(1-\frac{9}{V}\right)^{^2}$

we get $1-\frac{9}{V}=\ \frac{4}{5}or\ -\frac{4}{5}$

If considering $1-\frac{9}{V}=-\frac{4}{5}$

V =5, but this is not possible because 9 liters is drawn every time.

Hence : $1-\frac{9}{V}=\frac{4}{5},\ V\ =\ 45\ liters$

**Question 118: **Anil can paint a house in 60 days while Bimal can paint it in 84 days. Anil starts painting and after 10 days, Bimal and Charu join him. Together, they complete the painting in 14 more days. If they are paid a total of ₹ 21000 for the job, then the share of Charu, in INR, proportionate to the work done by him, is

a) 9000

b) 9200

c) 9100

d) 9150

**118) Answer (C)**

**Solution:**

Let Entire work be W

Now Anil worked for 24 days

Bimal worked for 14 days and Charu worked for 14 days .

Now Anil Completes W in 60 days

so in 24 days he completed 0.4W

Bimal completes W in 84 Days

So in 14 Days Bimal completes = $\frac{W}{6}$

Therefore work done by charu = $W-\frac{W}{6}-\frac{4W}{10}$= $\frac{26W}{10}$=$\frac{13W}{30}$

Therefore proportion of Charu = $\frac{13}{30}\times\ 21000$=9100

**Question 119: **A box has 450 balls, each either white or black, there being as many metallic white balls as metallic black balls. If 40% of the white balls and 50% of the black balls are metallic, then the number of non-metallic balls in the box is

**119) Answer: 250**

**Solution:**

Let the number of white balls be x and black balls be y

So we get x+y =450 (1)

Now metallic black balls = 0.5y

Metallic white balls = 0.4x

From condition 0.4x=0.5y

we get 4x-5y=0 (2)

Solving (1) and (2) we get

x=250 and y =200

Now number of Non Metallic balls = 0.6x+0.5y = 150+100 = 250

**Question 120: **In a football tournament, a player has played a certain number of matches and 10 more matches are to be played. If he scores a total of one goal over the next 10 matches, his overall average will be 0.15 goals per match. On the other hand, if he scores a total of two goals over the next 10 matches, his overall average will be 0.2 goals per match. The number of matches he has played is

**120) Answer: 10**

**Solution:**

Let Total matches played be n and in initial n-10 matches his goals be x

so we get $\frac{\left(x+1\right)}{n}=0.15$

we get x+1 =0.15n (1)

From condition (2) we get :

$\frac{\left(x+2\right)}{n}=0.2$

we get x+2 = 0.2n (2)

Subtracting (1) and (2)

we get 1 =0.05n

n =20

So initially he played n-10 =10 matches

**Question 121: **A person buys tea of three different qualities at ₹ 800, ₹ 500, and ₹ 300 per kg, respectively, and the amounts bought are in the proportion 2 : 3 : 5. She mixes all the tea and sells one-sixth of the mixture at ₹ 700 per kg. The price, in INR per kg, at which she should sell the remaining tea, to make an overall profit of 50%, is

a) 653

b) 688

c) 692

d) 675

**121) Answer (B)**

**Solution:**

Considering the three kinds of tea are A, B, and C.

The price of kind A = Rs 800 per kg.

The price of kind B = Rs 500 per kg.

The price of kind C = Rs 300 per kg.

They were mixed in the ratio of 2 : 3: 5.

1/6 of the total mixture is sold for Rs 700 per kg.

Assuming the ratio of mixture to A = 12kg, B = 18kg, C =30 kg.

The total cost price is 800*12+500*18+300*30 = Rs 27600.

Selling 1/6 which is 10kg for Rs 700/kg the revenue earned is Rs 7000.

In order to have an overall profit of 50 percent on Rs 27600.

Thes selling price of the 60 kg is Rs 27600*1.5 = Rs 41400.

Hence he must sell the remaining 50 kg mixture for Rs 41400 – Rs 7000 = 34400.

Hence the price per kg is Rs 34400/50 = Rs 688

**Question 122: **Anil, Bobby, and Chintu jointly invest in a business and agree to share the overall profit in proportion to their investments. Anil’s share of investment is 70%. His share of profit decreases by ₹ 420 if the overall profit goes down from 18% to 15%. Chintu’s share of profit increases by ₹ 80 if the overall profit goes up from 15% to 17%. The amount, in INR, invested by Bobby is

a) 2000

b) 2400

c) 2200

d) 1800

**122) Answer (A)**

**Solution:**

Let the amount invested by Anil Bobby and Chintu be x, y, and z.

Considering x+y+z = 100*p.

Given Anil’s share was 70 percent = 70*p.

As per the information provided :

His share of profit decreases by ₹ 420 if the overall profit goes down from 18% to 15%.

Since the profits are distributed in the ratio of their investments :

With a 3% decrease in the profits the value of profit earned by A decreased by Rs 420 which was 70 percent of the total invested.

Hence for all three of them would be combinedly losing $\left(420\right)\cdot\left(\frac{10}{7}\right)\ =\ 600$

Hence 3 percent profit was equivalent to Rs 600.

The initial investment is equivalent to Rs 20000.

This is the total amount invested.

Chintu’s profit share increased by Rs 80 when the profit percentage increased by 2 %. A 2 percent increase in profit is equivalent to Rs 20000*2/100 = Rs 400.

Of which Rs 80 is earned by Chintu which is 20% of the total Rs 400.

Hence he invested 20% of the total amount.

Bobby invested the other 10 percent.

10 percent of Rs 20000 = Rs 2000

**Question 123: **Two pipes A and B are attached to an empty water tank. Pipe A fills the tank while pipe B drains it. If pipe A is opened at 2 pm and pipe B is opened at 3 pm, then the tank becomes full at 10 pm. Instead, if pipe A is opened at 2 pm and pipe B is opened at 4 pm, then the tank becomes full at 6 pm. If pipe B is not opened at all, then the time, in minutes, taken to fill the tank is

a) 144

b) 140

c) 264

d) 120

**123) Answer (A)**

**Solution:**

Let A fill the tank at x liters/hour and B drain it at y liters/hour

Now as per Condition 1 :

We get Volume filled till 10pm = 8x-7y (1) .

Here A operates for 8 hours and B operates for 7 hours .

As per condition 2

We get Volume filled till 6pm = 4x-2y (2)

Here A operates for 4 hours and B operates for 2 hours .

Now equating (1) and (2)

we get 8x-7y =4x-2y

so we get 4x =5y

y =4x/5

So volume of tank = $8x-7\times\ \frac{4x}{5}=\frac{12x}{5}$

So time taken by A alone to fill the tank = $\frac{\frac{12x}{5}}{x}=\frac{12}{5}hrs\ $

= 144 minutes

**Question 124: **One day, Rahul started a work at 9 AM and Gautam joined him two hours later. They then worked together and completed the work at 5 PM the same day. If both had started at 9 AM and worked together, the work would have been completed 30 minutes earlier. Working alone, the time Rahul would have taken, in hours, to complete the work is

a) 11.5

b) 10

c) 12.5

d) 12

**124) Answer (B)**

**Solution:**

Let Rahul work at a units/hr and Gautam at b units/hour

Now as per the condition :

8a+6b =7.5a+7.5b

so we get 0.5a=1.5b

or a=3b

Therefore total work = 8a +6b = 8a +2a =10a

Now Rahul alone takes 10a/10 = 10 hours.

**Question 125: **In a tournament, a team has played 40 matches so far and won 30% of them. If they win 60% of the remaining matches, their overall win percentage will be 50%. Suppose they win 90% of the remaining matches, then the total number of matches won by the team in the tournament will be

a) 80

b) 78

c) 84

d) 86

**125) Answer (C)**

**Solution:**

Initially number of matches = 40

Now matches won = 12

Now let remaining matches be x

Now number of matches won = 0.6x

Now as per the condition :

$\frac{\left(12+0.6x\right)}{40+x}=\frac{1}{2}$

24 +1.2x=40+x

0.2x=16

x=80

Now when they won 90% of remaining = 80(0.9) =72

So total won = 84

**Question 126: **The total of male and female populations in a city increased by 25% from 1970 to 1980. During the same period, the male population increased by 40% while the female population increased by 20%. From 1980 to 1990, the female population increased by 25%. In 1990, if the female population is twice the male population, then the percentage increase in the total of male and female populations in the city from 1970 to 1990 is

a) 68.25

b) 68.75

c) 68.50

d) 69.25

**126) Answer (B)**

**Solution:**

Let us solve this question by assuming values(multiples of 100) and not variables(x).

Since we know that the female population was twice the male population in 1990, let us assume their respective values as 200 and 100.

**Note that while assuming numbers, some of the population values might come out as a fraction(which is not possible, since the population needs to be a natural number). However, this would not affect our answer, since the calculations are in ratios and percentages and not real values of the population in any given year.**

Now, we know that the female population became 1.25 times itself in 1990 from what it was in 1980.

Hence, the female population in 1980 = 200/1.25 = 160

Also, the female population became 1.2 times itself in 1980 from what it was in 1970.

Hence, the female population in 1970 = 160/1.2 = 1600/12 = 400/3

Let the male population in 1970 be x. Hence, the male population in 1980 is 1.4x.

Now, the total population in 1980 = 1.25 times the total population in 1970.

Hence, 1.25 (x + 400/3) = 1.4x + 160

Hence, x = 400/9.

Population change = 300 – 400/9 – 400/3 = 300 – 1600/9 = 1100/9

percentage change = $\frac{\frac{1100}{9}}{\frac{1600}{9}}\times\ 100\ =\ \frac{1100}{16}\%=68.75\%$

**Question 127: **The arithmetic mean of scores of 25 students in an examination is 50. Five of these students top the examination with the same score. If the scores of the other students are distinct integers with the lowest being 30, then the maximum possible score of the toppers is

**127) Answer: 92**

**Solution:**

Let sum of marks of students be x

Now therefore x = 25*50 =1250

Now to maximize the marks of the toppers

We will minimize the marks of 20 students

so their scores will be (30,31,32…..49 )

let score of toppers be y

so we get 5y +$\frac{20}{2}\left(79\right)$=1250

we get 5y +790=1250

5y=460

y=92

So scores of toppers = 92

**Question 128: **One part of a hostel’s monthly expenses is fixed, and the other part is proportional to the number of its boarders. The hostel collects ₹ 1600 per month from each boarder. When the number of boarders is 50, the profit of the hostel is ₹ 200 per boarder, and when the number of boarders is 75, the profit of the hostel is ₹ 250 per boarder. When the number of boarders is 80, the total profit of the hostel, in INR, will be

a) 20200

b) 20500

c) 20800

d) 20000

**128) Answer (B)**

**Solution:**

Profit per boarder = Total profit / Number of boarders.

Let the number of boarders be n.

Profit/boarder = 1600 – (Total cost/n)

Let the total cost be a + bn, where a = fixed, and b is the variable additional cost per boarder.

Profit/boarder = 1600 – (a + bn)/n

Profit/boarder = 1600 – a/n – b

1600 – a/50 – b = 200

1600 – a/75 – b = 250

Solving, we get a = 7500, and b = 1250

Hence, total profit with 80 people = 80 ( 1600 – 7500/80 – 1250) = 80 (350 – 7500/80) = 28000 – 7500 = Rs. 20500

**Question 129: **A tea shop offers tea in cups of three different sizes. The product of the prices, in INR, of three different sizes is equal to 800. The prices of the smallest size and the medium size are in the ratio 2 : 5. If the shop owner decides to increase the prices of the smallest and the medium ones by INR 6 keeping the price of the largest size unchanged, the product then changes to 3200. The sum of the original prices of three different sizes, in INR, is

**129) Answer: 34**

**Solution:**

Let price of smallest cup be 2x and medium be 5x and large be y

Now by condition 1

we get $2x\ \times\ \ 5x\ \times\ y\ =800$

we get $x^2y\ =80$ (1)

Now as per second condition ;

$\left(2x+6\right)\times\ \left(5x+6\right)\ y\ =3200$ (2)

Now dividing (2) and (1)

we get $\frac{\left(\left(2x+6\right)\times\ \left(5x+6\right)\right)}{x^2}=40$

we get $10x^2+42x+36\ =\ 40x^2$

we get $\ 30x^2-42x-36=0$

$5x^2-7x-6=0$

we get x=2

So 2x=4 and 5x=10

Now substituting in (1) we get y =20

Now therefore sum = 4+10+20 =34

**Question 130: **Mira and Amal walk along a circular track, starting from the same point at the same time. If they walk in the same direction, then in 45 minutes, Amal completes exactly 3 more rounds than Mira. If they walk in opposite directions, then they meet for the first time exactly after 3 minutes. The number of rounds Mira walks in one hour is

**130) Answer: 8**

**Solution:**

Considering the distance travelled by Mira in one minute = M,

The distance traveled by Amal in one minute = A.

Given if they walk in the opposite direction it takes 3 minutes for both of them to meet. Hence 3*(A+M) = C. (1)

C is the circumference of the circle.

Similarly, it is mentioned that if both of them walk in the same direction Amal completes 3 more rounds than Mira :

Hence 45*(A-M) = 3C. (2)

Multiplying (1)*15 we have :

45A + 45M = 15C.

45A – 45M = 3C.

Adding the two we have A = $\frac{18C}{90}$

Subtracting the two M = $\frac{12C}{90}$

Since Mira travels $\frac{12C}{90}$ in one minute, in one hour she travels :$\frac{12C}{90}\cdot60\ =\ 8C$

Hence a total of 8 rounds.

Alternatively,

Let the length of track be L

and velocity of Mira be a and Amal be b

Now when they meet after 45 minutes Amal completes 3 more rounds than Mira

so we can say they met for the 3rd time moving in the same direction

so we can say they met for the first time after 15 minutes

So we know Time to meet = Relative distance /Relative velocity

so we get $\frac{15}{60}=\frac{L}{a-b}$ (1)

Now When they move in opposite direction

They meet after 3 minutes

so we get $\frac{3}{60}=\frac{L}{a+b}$ (2)

Dividing (1) and (2)

we get $\frac{\left(a+b\right)}{\left(a-b\right)}=5$

or 4a =6b

or a = 3b/2

Now substituting in (1)

we get :

$\frac{L}{b}\times\ 2=\ \frac{15}{60}$

so $\frac{L}{b}\ =\frac{1}{8}$

So we can say 1 round is covered in $\frac{1}{8}$ hours

so in 1-hour total rounds covered = 8.

**Question 131: **If a certain weight of an alloy of silver and copper is mixed with 3 kg of pure silver, the resulting alloy will have 90% silver by weight. If the same weight of the initial alloy is mixed with 2 kg of another alloy which has 90% silver by weight, the resulting alloy will have 84% silver by weight. Then, the weight of the initial alloy, in kg, is

a) 3.5

b) 2.5

c) 3

d) 4

**131) Answer (C)**

**Solution:**

Let the alloy contain x Kg silver and y kg copper

Now when mixed with 3Kg Pure silver

we get $\frac{\left(x+3\right)}{x+y+3}=\frac{9}{10}$

we get 10x+30 =9x+9y+27

9y-x=3 (1)

Now as per condition 2

silver in 2nd alloy = 2(0.9) =1.8

so we get$\frac{\left(x+1.8\right)}{x+y+2}=\frac{21}{25}$

we get 21y-4x =3 (2)

solving (1) and (2) we get y= 0.6 and x =2.4

so x+y = 3

**Question 132: **Bank A offers 6% interest rate per annum compounded half-yearly. Bank B and Bank C offer simple interest but the annual interest rate offered by Bank C is twice that of Bank B. Raju invests a certain amount in Bank B for a certain period and Rupa invests ₹ 10,000 in Bank C for twice that period. The interest that would accrue to Raju during that period is equal to the interest that would have accrued had he invested the same amount in Bank A for one year. The interest accrued, in INR, to Rupa is

a) 3436

b) 2436

c) 2346

d) 1436

**132) Answer (B)**

**Solution:**

Bank A: 6% p.a. 1/2 yearly (CI)

Bank B: x% p.a (SI)

Bank C: 2x% p.a (SI)

Let Raju invest Rs P in bank B for t years. Hence, Rupa invests Rs 10,000 in bank C for 2t years.

Now,

$P\left(\frac{x}{100}\right)t\ =\ P\left(1+\frac{3}{100}\right)^2-P$

$\left(\frac{x}{100}\right)t\ =\ 1.0609-1$

$\left(\frac{x}{100}\right)t\ =\ 0.0609$

We need to calculate

SI = $10000\times\ 2t\times\ \left(\frac{2x}{100}\right)=40000\left(\frac{x}{100}\right)t=40000\times\ 0.0609=2436$

**Question 133: **Anil can paint a house in 12 days while Barun can paint it in 16 days. Anil, Barun, and Chandu undertake to paint the house for ₹ 24000 and the three of them together complete the painting in 6 days. If Chandu is paid in proportion to the work done by him, then the amount in INR received by him is

**133) Answer: 3000**

**Solution:**

Now Anil Paints in 12 Days

Barun paints in 16 Days

Now together Arun , Barun and Chandu painted in 6 Days

Now let total work be W

Now each worked for 6 days

So Anil’s work = 0.5W

Barun’s work = $\frac{6W}{16}=\frac{3W}{8}$

Therefore Charu’s work = $\frac{W}{2}-\frac{3W}{8}=\ \frac{W}{8}$

Therefore proportion of charu =$\frac{24000}{8}=\ 3,000$

**Question 134: **In a village, the ratio of number of males to females is 5 : 4. The ratio of number of literate males to literate females is 2 : 3. The ratio of the number of illiterate males to illiterate females is 4 : 3. If 3600 males in the village are literate, then the total number of females in the village is

**134) Answer: 43200**

**Solution:**

In the question, it is given that the ratio of number of literate males to literate females is 2 : 3.

Given, the number of literate males = 3600

The number of literate females = $\frac{3600}{2}\times3$ = 5400

Males : Females = 5 : 4

Let the number of males be 5y and the number of females be 4y

Illiterate males = 5y – 3600

Illiterate females = 4y – 5400

It is given,

$\ \frac{\ 5y-3600}{4y-5400}=\frac{4}{3}$

15y – 10800 = 16y – 21600

y = 10800

Number of females = 4y = 4*10800 = 43200

**Question 135: **The average weight of students in a class increases by 600 gm when some new students join the class. If the average weight of the new students is 3 kg more than the average weight of the original students, then the ratio of the number of original students to the number of new students is

a) 1 : 4

b) 1 : 2

c) 4 : 1

d) 3 : 1

**135) Answer (C)**

**Solution:**

Let the original number of students be ‘n’ whose average weight is ‘x’

Let the number of students added be ‘m’ and the average weight will be x + 3

We need to find the value of n : m

It is given, average weight of students in a class increased by 0.6 after new students are added.

Therefore,

$\ \frac{\ nx+m\left(x+3\right)}{n+m}=x+0.6$

$\ \ nx+mx+3m=mx+nx+0.6n+0.6m$

$2.4m=0.6n$

$4m=n$

$\frac{n}{m}=\frac{4}{1}$

The answer is option C.

**Question 136: **Trains A and B start traveling at the same time towards each other with constant speeds from stations X and Y, respectively. Train A reaches station Y in 10 minutes while train B takes 9 minutes to reach station X after meeting train A. Then the total time taken, in minutes, by train B to travel from station Y to station X is

a) 6

b) 15

c) 10

d) 12

**136) Answer (B)**

**Solution:**

M – First meeting point

Let the speeds of trains A and B be ‘a’ and ‘b’, respectively.

$\frac{x}{a}=\ \frac{\ D-x}{b}$

It is given,

$\frac{D}{a}=10$ and $\frac{x}{b}=9$

$\frac{x}{\frac{D}{10}}=\ \frac{\ D-x}{\frac{x}{9}}$

$\frac{10x}{D}=\ \frac{\ 9D-9x}{x}$

$10x^2=\ \ 9D^2-9Dx$

$10x^2+9Dx-9D^2=\ 0$

Solving, we get $x=\frac{3D}{5}$

$\frac{x}{b}=9$

$\frac{3D}{b\times5}=9$

$\frac{D}{b}=15$

The total time taken by train B to travel from station Y to station X is 15 minutes.

The answer is option B

**Question 137: **Ankita buys 4 kg cashews, 14 kg peanuts and 6 kg almonds when the cost of 7 kg cashews is the same as that of 30 kg peanuts or 9 kg almonds. She mixes all the three nuts and marks a price for the mixture in order to make a profit of ₹1752. She sells 4 kg of the mixture at this marked price and the remaining at a 20% discount on the marked price, thus making a total profit of ₹744. Then the amount, in rupees, that she had spent in buying almonds is

a) 1680

b) 1176

c) 2520

d) 1440

**137) Answer (A)**

**Solution:**

It is given,

7C = 30P = 9A and Ankita bought 4C, 14P and 6A.

Let 7C = 30P = 9A = 630k

C = 90k, P = 21k, and A = 70k

Cost price of 4C, 14P and 6A = 4(90k)+14(21k)+6(70k) = 1074k

Marked up price = 1074k + 1752

S.P = $\frac{1}{6}\left(1074k+1752\right)+\left(\frac{4}{5}\right)\left(\frac{5}{6}\right)\left(1074k+1752\right)$ = $\frac{5}{6}\left(1074k+1752\right)$

S.P – C.P = profit

$1460-\frac{1074k}{6}=744$

$\frac{1074k}{6}=716$

k = 4

Money spent on buying almonds = 420k = 420*4 = Rs 1680

The answer is option A.

**Question 138: **Amal buys 110 kg of syrup and 120 kg of juice, syrup being 20% less costly than juice, per kg. He sells 10 kg of syrup at 10% profit and 20 kg of juice at 20% profit. Mixing the remaining juice and syrup, Amal sells the mixture at ₹ 308.32 per kg and makes an overall profit of 64%. Then, Amal’s cost price for syrup, in rupees per kg, is

**138) Answer: 160**

**Solution:**

Total syrup – 110 kg

Total juice – 120 kg

It is given, cost price of syrup is 20% less than the cost price of juice.

Let the cost price of juice per kg be 10CP

Cost price of syrup per kg is 8CP

10kg syrup -> cost price = 80CP

It is given, 10kg syrup is sold at 10% profit. This implies selling price = 1.1*80CP = 88CP

20kg juice -> cost price = 200CP

It is given, 20kg juice is sold at 20% profit. This implies selling price = 1.2*200CP = 240CP

It is given, Mixing the remaining juice and syrup, Amal sells the mixture at ₹ 308.32 per kg

Selling price of the remaining mixture = 308.32*200 = Rs 61664

Total S.P = 61664 + 328CP

Total C.P = 880CP + 1200CP = 2080CP

Overall profit = 64%

$61664+328CP=\frac{164}{100}\left(2080CP\right)$

Solving, we get CP = 20

Cost price for syrup per kg = 8CP = 8*20 = Rs 160

**Question 139: **The average of three integers is 13. When a natural number n is included, the average of these four integers remains an odd integer. The minimum possible value of n is

a) 3

b) 4

c) 5

d) 1

**139) Answer (C)**

**Solution:**

It is given that average of three numbers is 13.

Sum = 3*13 = 39

It is given, $\ \frac{\ 39+n}{4}$ is a odd number.

Minimum value $\ \frac{\ 39+n}{4}$ can take such that n is a natural number is 11

$\ \frac{\ 39+n}{4}=11$

n = 5

The answer is option C.

**Question 140: **A mixture contains lemon juice and sugar syrup in equal proportion. If a new mixture is created by adding this mixture and sugar syrup in the ratio 1 : 3, then the ratio of lemon juice and sugar syrup in the new mixture is

a) 1 : 4

b) 1 : 5

c) 1 : 6

d) 1 : 7

**140) Answer (D)**

**Solution:**

Lemon juice : sugar syrup in the mixture is 1:1, i.e. 50% Lemon juice and 50% sugar syrup.

In sugar syrup, 100% is sugar syrup.

These two are mixed in the ratio 1:3.

Lemon juice = $\ \frac{\ 1\left(50\%\right)}{1+3}$

Sugar syrup = $\ \frac{\ 1\left(50\%\right)+3\left(100\%\right)}{1+3}=\frac{350}{4}$

Required ratio = 50:350 = 1:7

The answer is option D.

**Question 141: **Alex invested his savings in two parts. The simple interest earned on the first part at 15% per annum for 4 years is the same as the simple interest earned on the second part at 12% per annum for 3 years. Then, the percentage of his savings invested in the first part is

a) 37.5%

b) 62.5%

c) 60%

d) 40%

**141) Answer (A)**

**Solution:**

Let the savings invested in first part and second part be ‘x’ and ‘y’, respectively.

It is given,

$\ \frac{\ x\times15\times4}{100}=\ \frac{\ y\times12\times3}{100}$

60x = 36y

5x = 3y

Required percentage = $\frac{x}{x+y}\times100=\frac{3}{3+5}\times100=37.5\%$

The answer is option A.

**Question 142: **Pinky is standing in a queue at a ticket counter. Suppose the ratio of the number of persons standing ahead of Pinky to the number of persons standing behind her in the queue is 3 : 5. If the total number of persons in the queue is less than 300, then the maximum possible number of persons standing ahead of Pinky is

**142) Answer: 111**

**Solution:**

Let the number of persons standing ahead and behind of Pinky be 3a and 5a.

Total number of persons = 3a + 5a + 1(including pinky) = 8a + 1

8a + 1 < 300

8a < 299

a < 37.375

Maximum value a can take is 37.

The maximum possible number of persons standing ahead of Pinky = 3a = 3*37 = 111

**Question 143: **Mr. Pinto invests one-fifth of his capital at 6%, one-third at 10% and the remaining at 1%, each rate being simple interest per annum. Then, the minimum number of years required for the cumulative interest income from these investments to equal or exceed his initial capital is

**143) Answer: 20**

**Solution:**

Let the total investment me 15x and the no. of years required be T years

$\frac{\left(3x\times6\times T\right)}{100}+\frac{\left(5x\times10\times T\right)}{100}+\frac{\left(7x\times1\times T\right)}{100}\ge15x$

or, $\frac{75xT}{100}\ge15x$

or, $T\ge20$

So minimum value of T is 20 years

**Question 144: **Manu earns ₹4000 per month and wants to save an average of ₹550 per month in a year. In the first nine months, his monthly expense was ₹3500, and he foresees that, tenth month onward, his monthly expense will increase to ₹3700. In order to meet his yearly savings target, his monthly earnings, in rupees, from the tenth month onward should be

a) 4400

b) 4200

c) 4300

d) 4350

**144) Answer (A)**

**Solution:**

Savings target in a year = 550*12 = Rs 6600

Saving in first 9 months = 9(4000-3500) = Rs 4500

Saving for remaining 3 months should be 6600-4500, i.e. Rs 2100

Savings for each month in last 3 months = $\frac{2100}{3}$ = Rs 700

It is given, monthly expenses in last 3 months = Rs 3700

This implies, his monthly earnings from 10th month should be 3700+700, i.e. Rs 4400

The answer is option A.

**Question 145: **Two ships meet mid-ocean, and then, one ship goes south and the other ship goes west, both travelling at constant speeds. Two hours later, they are 60 km apart. If the speed of one of the ships is 6 km per hour more than the other one, then the speed, in km per hour, of the slower ship is

a) 20

b) 12

c) 18

d) 24

**145) Answer (C)**

**Solution:**

Let the speeds of two ships be ‘x’ and ‘x+6’ km per hour

Distance covered in 2 hours will be 2x and 2x+12

It is given,

$\left(2x\right)^2+\left(2x+12\right)^2=60^2$

$\left(x\right)^2+\left(x+6\right)^2=30^2$

$2x^2+12x+36=900$

$x^2+6x+18=450$

$x^2+6x-432=0$

Solving, we get x = 18

The speed of slower ship is 18 kmph

The answer is option C.

**Question 146: **In an election, there were four candidates and 80% of the registered voters casted their votes. One of the candidates received 30% of the casted votes while the other three candidates received the remaining casted votes in the proportion 1 : 2 : 3. If the winner of the election received 2512 votes more than the candidate with the second highest votes, then the number of registered voters was

a) 50240

b) 40192

c) 60288

d) 62800

**146) Answer (D)**

**Solution:**

Let the number of registered votes be 100x

The number of votes casted = 80x

Votes received by one of the candidates = $\frac{30}{100}\times80x$ = 24x

Remaining votes = 80x – 24x = 56x

Votes received by other three candidates is $\frac{56x}{6},\frac{2\times56x}{6},\ \frac{\ 3\times56x}{6}$

It is given,

28x – 24x = 2512

4x = 2512

x = 628

The number of registered votes = 100x = 62800

The answer is option D.

**Question 147: **Working alone, the times taken by Anu, Tanu and Manu to complete any job are in the ratio 5 : 8 : 10. They accept a job which they can finish in 4 days if they all work together for 8 hours per day. However, Anu and Tanu work together for the first 6 days, working 6 hours 40 minutes per day. Then, the number of hours that Manu will take to complete the remaining job working alone is

**147) Answer: 6**

**Solution:**

Let the time taken by Anu, Tanu and Manu be 5x, 8x and 10x hours.

Total work = LCM(5x, 8x, 10x) = 40x

Anu can complete 8 units in one hour

Tanu can complete 5 units in one hour

Manu can complete 4 units in one hour

It is given, three of them together can complete in 32 hours.

32(8 + 5 + 4) = 40x

x = $\frac{68}{5}$

It is given,

Anu and Tanu work together for the first 6 days, working 6 hours 40 minutes per day, i.e. 36 + 4 = 40 hours

40(8 + 5) + y(4) = 40x

4y = 24

y = 6

Manu alone will complete the remaining work in 6 hours.

**Question 148: **There are two containers of the same volume, first container half-filled with sugar syrup and the second container half-filled with milk. Half the content of the first container is transferred to the second container, and then the half of this mixture is transferred back to the first container. Next, half the content of the first container is transferred back to the second container. Then the ratio of sugar syrup and milk in the second container is

a) 4 : 5

b) 6 : 5

c) 5 : 4

d) 5 : 6

**148) Answer (D)**

**Solution:**

Step 1: Half the content of the first container is transferred to the second container

Step 2: Half of the mixture of second container is transferred back to the first container

Step 3: Half the content of the first container is transferred back to the second container

Sugar syrup : Milk in second container = 62.5 : 75 = 5 : 6

The answer is option D.

**Question 149: **If a and b are non-negative real numbers such that a+ 2b = 6, then the average of the maximum and minimum possible values of (a+ b) is

a) 3

b) 4

c) 3.5

d) 4.5

**149) Answer (D)**

**Solution:**

a + 2b = 6

From the above equation, we can say that maximum value b can take is 3 and minimum value b can take is 0.

a + b + b = 6

a + b = 6 – b

a + b is maximum when b is minimum, i.e. b = 0

Maximum value of a + b = 6 – 0 = 6

a + b is minimum when b is maximum, i.e. b = 3

Minimum value of a + b = 6 – 3 = 3

Average = $\ \frac{\ 6+3}{2}$ = 4.5

The answer is option D.

**Question 150: **Five students, including Amit, appear for an examination in which possible marks are integers between 0 and 50, both inclusive. The average marks for all the students is 38 and exactly three students got more than 32. If no two students got the same marks and Amit got the least marks among the five students, then the difference between the highest and lowest possible marks of Amit is

a) 22

b) 21

c) 24

d) 20

**150) Answer (D)**

**Solution:**

The average marks for all the students is 38.

Sum = 5*38 = 190

To find the minimum marks scored by Amit, we need to maximise the score of remaining students.

Maximum scores sum of remaining students = 50 + 49 + 48 + 32 = 179

Minimum possible score of Amit = 190 – 179 = 11

It is given, Amit scored least. This implies maximum possible score of Amit is 31.

Difference = 31 – 11 = 20

The answer is option D.

**Question 151: **A group of N people worked on a project. They finished 35% of the project by working 7 hours a day for 10 days. Thereafter, 10 people left the group and the remaining people finished the rest of the project in 14 days by working 10 hours a day. Then the value of N is

a) 150

b) 23

c) 36

d) 140

**151) Answer (D)**

**Solution:**

Let the unit of work done by 1 man in 1 hour and 1 day be 1 MDH unit (Man Day Hour).

Thus, in 7 hours per day for 10 days, the work done by N people =$N\times\ 7\times\ 10$ MDH units.

Since this is equal to 35% of the total work,

35% of the total work = $N\times\ 7\times\ 10$ MDH units.

Total work = $\frac{\left(N\times\ 100\times\ 7\times\ 10\right)}{35}=200\times N$ MDH units.

The work left = $200N-70N=130N$ MDH units.

Now, 10 people left the job. So, the number of people left = (N-10)

Since (N-10) people completed the rest of work in 14 days by working 10 hours a day,

$(N-10)\times\ 14\times\ 10=130N$

$10N=1400$

N = 140

Thus, the correct option is D.

**Question 152: **A glass contains 500 cc of milk and a cup contains 500 cc of water. From the glass, 150 cc of milk is transferred to the cup and mixed thoroughly. Next, 150 cc of this mixture is transferred from the cup to the glass. Now, the amount of water in the glass and the amount of milk in the cup are in the ratio

a) 1 : 1

b) 10 : 13

c) 3 : 10

d) 10 : 3

**152) Answer (A)**

**Solution:**

Initially: a glass 500cc milk and a cup 500cc water

Step 1: 150 cc of milk is transferred to the cup from glass

After step 1: Glass – 350 cc milk, Cup – 150 cc milk and 500 cc water

Step 2: 150 cc of this mixture is transferred from the cup to the glass

After step 2:

Glass – 350 cc milk + 150 cc mixture with milk:water ratio 3:10

Cup – 500 cc mixture with milk:water ratio 3:10

water in glass : milk in cup = $\frac{10}{13}\times150\ :\ \frac{3}{13}\times500=1:1$

The answer is option A.

**Question 153: **Nitu has an initial capital of ₹20,000. Out of this, she invests ₹8,000 at 5.5% in bank A, ₹5,000 at 5.6% in bank B and the remaining amount at x% in bank C, each rate being simple interest per annum. Her combined annual interest income from these investments is equal to 5% of the initial capital. If she had invested her entire initial capital in bank C alone, then her annual interest income, in rupees, would have been

a) 700

b) 800

c) 900

d) 1000

**153) Answer (B)**

**Solution:**

It is given,

$\ \frac{\ 5.5\times1\times8000}{100}+\ \frac{\ 5.6\times1\times5000}{100}+\ \frac{\ x\times1\times7000}{100}=\frac{5}{100}\times20000$

$440+280+70x=1000$

x = 4%

Interest = $\ \frac{\ 20000\times4\times1}{100}$ = Rs 800

The answer is option B.

**Question 154: **Two cars travel from different locations at constant speeds. To meet each other after starting at the same time, they take 1.5 hours if they travel towards each other, but 10.5 hours if they travel in the same direction. If the speed of the slower car is 60 km/hr, then the distance traveled, in km, by the slower car when it meets the other car while traveling towards each other, is

a) 100

b) 90

c) 120

d) 150

**154) Answer (B)**

**Solution:**

Both the cars take 1.5 hrs to meet when they travel towards each other.

It is given, speed of slower car is 60 km/hr

Therefore, distance covered by slower car before they meet = 60*1.5 = 90 km

The answer is option B.

**Question 155: **In an examination, the average marks of students in sections A and B are 32 and 60, respectively. The number of students in section A is 10 less than that in section B. If the average marks of all the students across both the sections combined is an integer, then the difference between the maximum and minimum possible number of students in section A is

**155) Answer: 63**

**Solution:**

Let the number of students in section A and B be a and b, respectively.

It is given, a = b – 10

$\ \ \frac{\ 32a+60b}{a+b}$ is an integer

$\ \ \frac{\ 32a+60\left(a+10\right)}{a+a+10}=k$

$\ \ \frac{\ 46a+300}{a+5}=k$

$k=\ \frac{\ 46\left(a+5\right)}{a+5}+\frac{70}{a+5}$

$k=\ \ 46+\frac{70}{a+5}$

a can take values 2, 5, 9, 30, 65

Difference = 65 – 2 = 63

**Question 156: **Moody takes 30 seconds to finish riding an escalator if he walks on it at his normal speed in the same direction. He takes 20 seconds to finish riding the escalator if he walks at twice his normal speed in the same direction. If Moody decides to stand still on the escalator, then the time, in seconds, needed to finish riding the escalator is

**156) Answer: 60**

**Solution:**

Let the speed of Moody be ‘x’ steps/sec and that of the escalator be ‘y’ steps/sec.

In 30 seconds, Moody will finish riding the escalator when going in the same direction.

Thus, total steps = 30(x+y)

If Moody’s speed becomes twice, the time becomes 20 seconds.

Thus, total steps = 20(2x+y)

Or 30x + 30y = 40x + 20y

Or x = y

So, total steps = 60y.

Time taken by only escalator= 60y/y = 60s.

**Question 157: **Two ships are approaching a port along straight routes at constant speeds. Initially, the two ships and the port formed an equilateral triangle with sides of length 24 km. When the slower ship travelled 8 km, the triangle formed by the new positions of the two ships and the port became right-angled. When the faster ship reaches the port, the distance, in km, between the other ship and the port will be

a) 4

b) 12

c) 8

d) 6

**157) Answer (B)**

**Solution:**

Let S be the slower ship and F be the faster ship.

It is given that when S travelled 8 km, the positions of ships with the port is forming a right triangle.

Since one of the angles is 60(since one vertex is still part of the equilateral triangle),

the other two vertexes will have angles of 30 and 90.

The distance between O and S = 24 – 8 = 16

In triangle OFS, $\cos60^0\ =\ \frac{OF}{OS}$

Thus, OF = 8.

Thus in the time, S covered 8 km, F will cover 24 – 8 = 16 km.

Thus, the ratio of their speeds is 2:1,

Thus, when F covers 24 km, S will cover 12 km.

The correct option is B.

**Question 158: **Bob can finish a job in 40 days, if he works alone. Alex is twice as fast as Bob and thrice as fast as Cole in the same job. Suppose Alex and Bob work together on the first day, Bob and Cole work together on the second day, Cole and Alex work together on the third day, and then, they continue the work by repeating this three – day roster, with Alex and Bob working together on the fourth day, and so on. Then, the total number of days Alex would have worked when the job gets finished, is

**158) Answer: 11**

**Solution:**

Let the efficiency of Bob be 3 units/day. So, Alex’s efficiency will be 6 units/day, and Cole’s will be 2 units/day.

Since Bob can finish the job in 40 days, the total work will be 40*3 = 120 units.

Since Alex and Bob work on the first day, the total work done = 3 + 6 = 9 units.

Similarly, for days 2 and 3, it will be 5 and 8 units, respectively.

Thus, in the first 3 days, the total work done = 9 + 5 + 8 = 22 units.

The work done in the first 15 days = 22*5 = 110 units.

Thus, the work will be finished on the 17th day(since 9 + 5 = 14 units are greater than the remaining work).

Since Alex works on two days of every 3 days, he will work for 10 days out of the first 15 days.

Then he will also work on the 16th day.

The total number of days = 11.