CAT Arithmetic Questions

Let the cost price of the item be C
We are given that Bina sells this at 19% loss or at (1 - 0.19)C = 0.81C at 4860
This gives us the value of C at Rs. 6000

If Bina had sold this at 17% profit, the selling price would have been $$1.17\times\ 6000\ =\ 7020$$

So Shyam bought the product at 4860 and sold it to Hari at 7020
Giving the profit made by Shyam to be $$7020-4860\ =\ 2160$$

Therefore, 2160 is the correct answer.

Total number of students is 250, and we are told that, The percentage of girls was at least 44% and at most 60%.

So the number of girls range from, $$0.44\left(250\right)\le Girls\le0.6\left(250\right)$$
$$110\le Girls\le150$$

Statement 1: 
If 50% of the boys and 80% of the girls opted for swimming, that means if the total number of Boys is B, Girls is G where B+G=250. 
Swimming is: 0.5B+0.8G

Statement 2: 
If 70%of the boys and 60% of the girls opted for running, that means
Running is 0.7B+0.6G

Total number of enrolments for swimming and running together will be
(0.7B+0.6G)+(0.5B+0.8G)=1.2B+1.4G

Using the overlapping principle, where I represents people who have enrolled only for one activity and II represents number of people who have enrolled for two activities. 
We know that, $$I+II=250=B+G$$
$$I+2II=1.2B+1.4G$$

Subtracting the two equations, 
$$II=0.2B+0.4G$$
$$II=0.2\left(B+2G\right)$$
Using B+G=250
$$II=0.2\left(250+G\right)$$

G can at-most be 150 and at least 110. 

So maximum value of II will be $$0.2\left(250+150\right)=80$$

Minimum value of II will be $$0.2\left(250+110\right)=72$$

(Note: This was a repeated question that appeared in both CAT 2023 and CAT 2024)

Let us assume the initial stock of all the fruits is S.

Let us take we have 'b' and 'a' mangoes initially.

Stock of Mangoes = 40% of S = 2S/5

The total number of fruits sold are Mangoes Sold + Apples Sold + Bananas Sold

= 2S/10 + 96 + 4a/10 = S/2 (Given)

=> S/5 + 96 + 2a/5 = S/2

=> S = $$\dfrac{\left(4a+960\right)}{3}$$

=> $$\dfrac{4a}{3}+320$$

'a' has to be a multiple of 3 for the above term to be an integer.

But 'a' has to be a multiple of 5 for 4a/10 to be an integer.

=> The smallest value of 'a' satisfying both conditions is 15.

=> $$\dfrac{4a}{3}+320=\dfrac{4\left(15\right)}{3}+320$$ = 340

Therefore, 340 is the correct answer.

Let us assign the amount of work done by Renu in one hour is R
And the amount of work done by Seema in one hour is S

We are told that a certain task with different time durations for Seema and Renu
Renu=15 days with 4 hours each day, that is total work done by Renu is $$15\times\ 4\times\ R=60R$$
Similarly for Seema, 8 days and 5 hours each day, total work done by Seema is $$40S$$
We know that $$60R=40S$$ or $$S=1.5R$$

For this task, let us assume that the amount of days that Seema decides to work is X and the hours per day that Renu decides to work is Y
We are then told that, Seema works for 2Y hours per day and Renu works for 2X days, 
Work done by them will be $$2XYR$$ and $$2XYS$$

We are told that Y=2, Making this $$4XR$$ and $$4XS$$
$$S=1.5R$$
Total work will be, $$4XR$$+$$6XR$$=$$60R$$
We get the value of $$X=6$$

X is the number of days Seema will work, which is 6. 

We are given that Sam completes a piece of work in 20 days. We are also given that Mohit is twice as fast, so he should take only 10 days. Mohit is thrice as fast as Ayna, so he would take 30 days. 

Let's take the total work to be 60 units; this would give the work done per day for Mohit, Sam, and Ayna to be 6, 3 and 2, respectively. 

On the first day Sam and Mohit work: doing 9 units
On the second day Sam and Ayan work: doing 5 units
On the third day, Mohit and Ayan work: doing 8 units

Essentially doing 22 units in a 3 days cycle. 
After two such cycles, there will be 60-44 = 16 units of work left

On day 7, Sam and Mohit would work 9 units, leaving 7 units 
On day 8, Sam and Ayan would work 5 units, leaving 2 units
And on day 9, Ayan and Mohit would complete the remaining work. 

So Sam worked for a total of 2+2+2= 6 days and on each day he did 3 units of work, completing 18 units of work.

The ratio of work done by Sam would be $$\frac{18}{60}=\frac{3}{10}$$

Therefore, Option B is the correct answer. 

Let's take Rajesh and Garima's ages to be R and G, respectively
From the given ratio, we can see that Rajesh is older than Garima, so let's take R=G+x

When Rajesh was of age G, which was x years ago, Garima was of G-x years old
Giving the ratio as $$\frac{G}{G-x}=\frac{3}{2}$$
This gives us G as 3x, which in turn gives R as 4x

We are asked the ratio when Gramia becomes 4x years old. 
By that time, Rajesh will be 5x years old. 

Giinv their ratio as $$\frac{5x}{4x}=5:4$$

Therefore, Option C is the correct answer. 

Let us assume the amount invested to be $$P$$, and the rate of interest to be $$r$$. 
Value of the investment after 3 years will be, $$P\left(1+r\right)^3$$
Value of the investment after 5 years will be, $$P\left(1+r\right)^5$$

$$\left(1+r\right)^2=\frac{36}{25}$$
$$\left(1+r\right)^2=1.44$$
$$r=0.2$$

We need to find the value of n for which $$4000\left(1+r\right)^n>20000$$
$$\left(1+r\right)^n>5$$
$$\left(1.2\right)^n>5$$
We see that, 
$$1.2^8=4.2999$$
$$1.2^9=5.15$$

Hence it takes 9 years to grow to over 20,000. 

We are told that, the incomes of Kamal, Amal and Vimal are in the ratio 8 ∶ 6 ∶ 5
Lets assume them to be 80X, 60X, 50X respectively.

Money that each one of them pays towards rent, 
Kamal 15% which comes to 12X
Amal 12% which comes to 7.2X
Vimal 18% which comes to 9X
Total Rental expenditure will be, 28.2X

Their incomes increase by 10%, 12% and 15% respectively, 
Kamal 10% which comes to 88X
Amal 12% which comes to 67.2X
Vimal 15% which comes to 57.5X
Total income will be 212.7X

We are told the rent is the same, so it will be 28.2X

Percentage of income going towards rent in October will be, $$\frac{28.2X}{212.7X}=0.13258$$

Hence, the answer is 13.26%

We can divide the list into 4 elements: the first 10 as a, the next 10 as b, the next 10 as c, and the last 10 as d 

From the relations we are given, we can form the equations: $$\frac{a+b+c}{3}=40,000$$
$$\frac{b+c+d}{3}=60,000$$ and $$\frac{a+d}{2}=50,000$$

Adding the first two equations, we get $$a+2\left(b+c\right)+d=300,000$$
We can substitute the value of a+d as 100,000 to get b+c as 100,000

Using this value in the first and second equation would give a and d as 20,000 and 80,000, respectively. 

We are told that the average of the first 10 employees increases by 100%, that is, it changes from 20,000 to 40,000
The average of the last 10 increases by 200%; that is, it changes from 80,000 to 240,000

The total of all the four elements would be 40,000+100,000+240.000 = 380,000

Giving the average to be $$\frac{380,000}{4}=95,000$$

Therefore, Option A is the correct answer. 

Let us assume that the distance is D, speed of the train is S and time taken by the train is t. 

t is nothing but $$\frac{D}{S}$$

Statement 1: Had the speed been 6 km per hour more, it would have needed 4 hours less
$$\frac{D}{S+6}=t-4$$

$$\frac{D}{S+6}=\frac{D}{S}-4$$

$$4=\frac{D}{S}-\frac{D}{S+6}$$

$$\frac{S+6-S}{S\left(S+6\right)}=\frac{4}{D}$$

$$\frac{6}{S\left(S+6\right)}=\frac{4}{D}$$

$$D=\frac{2S\left(S+6\right)}{3}$$

Statement 2:  Had the speed been 6 km per hour less, it would have needed 6 hours more
$$\frac{D}{S-6}=t+6$$

$$D\left[\frac{1}{S-6}-\frac{1}{S}\right]=6$$

$$\frac{S-S+6}{S\left(S-6\right)}=6$$

$$D=S\left(S-6\right)$$

Equating the two equations for distance, 
$$S\left(S-6\right)=\frac{2S\left(S+6\right)}{3}$$

$$3S-18=2S+12$$

$$S=30$$

Hence the speed is 30 kmph

We know that the distance $$D=S\left(S-6\right)$$ = 30*24  =720 km

We can take the work done by Amal, Vimal and Sunil to be A, V and S, respectively. 

Let's take the total work they did to be T.

We are given the equations:
150A + 150V = T      ...(1)
100V + 100S = T      ...(2)

75A + 135V + 45S = T    ...(3)

Adding (1) and (2), we get: 150A + 250V + 100S = 2T   ...(4)

And multiplying (3) with 2 we get: 150A + 270V + 100S = 2T  ...(5)

Subtracting (5) from (4), we get 10S = 20V or simply S = 2V

Using this in (2), we get the total work T to be 300V, and using that result in (1), we get A=V

Therefore, the work done by A, V and S equals V, V, and 2V units per day. 

Now, in the question, we are given work cycles. To simplify the calculations, we should consider a time duration that is the LCM of the period taken by the three agents, which in this case would be 6

In 6 days, Amal will work for 6 days, doing 6V units of work.
Vimal will work for 3 days, doing 3V units of work. 
Sunil will work for 2 days, doing 4V units of work. 

So, in one 6-day cycle, 13V units of work will be done. 
Dividing the total work (300V) by 13V we can see that in 23 cycles 299V units of work will be done. 

These 23 cycles will be $$23\times\ 6\ =\ 138$$ days
The remaining 1V units of work will be done the next day. 

Therefore, a total of 139 days will be required. 

Let's take the amount invested by Sunil to be X. 

The amount received by Anil at the end of 6 years would be $$22000\left(1+\frac{4}{2\times\ 100}\right)^{6\times\ 2}=22000\left(1.02\right)^{12}$$

The amount received by Sunil at the end of 5 years would be $$X\left(1.02\right)^{10}$$
In the 6th year, Sunil invests this at a simple interest of 10%, giving him an interest of $$X\left(1.02\right)^{10}\times\ 0.1$$
Giving the total amount with him at the end of 6 years to be $$X\left(1.02\right)^{10}\times\ \left(1+0.1\right)$$

Equating the final amount with Sunil and Anil, we get:
$$X\left(1.02\right)^{10}\times\ \left(1.1\right)=22000\left(1.02\right)^{12}$$
$$X=\frac{22000\left(1.02\right)^2}{1.1}=20808$$

Therefore, Option D is the correct answer. 

Let us say the cost price of an item is X
It is said that it is marked to make a profit of 20%. 
That means it is marked at 1.2X

Ravi gets a 10% discount on the marked price, 
$$0.9\left(1.2X\right)=1.08X$$

Saves 15 rupees, so 1.2X-1.08X
0.12X=15
X=125

Profit made by Gopi is 0.08(125)=10 rupees. 

Let's take the scheduled time taken by the bus to be t
From the first statement (bus travelling at 60 kmph), we can get the total distance travelled by bus to 60(t+3.5)

The second scenario gives us that the bus covered two-thirds of the distance in one-third of the time, meaning that the remaining one-third distance was covered in two-thirds of the time, giving us the relation $$\frac{1}{3}st$$ covered in $$\frac{2}{3}t$$ giving the speed to be $$\frac{s}{2}$$ which is given as 40 km/h, thereby giving the usual speed of the bus to be 80 km/hr

Now the first relation we get 60(t+3.5)=80t
Giving us t=10.5 hours

Thus, the bus usually takes 10.5 hours on its journey. 
Staring at 9:00, it will complete the journey at 7:30 pm

Therefore, Option A is the correct answer. 

Let us assume the amount of Milk in the container to be X and the amount of water in the container to be Y. 
We are told that X+Y=300. 

Now, an operation is given where "an amount of this solution was taken out from the container which was twice the volume of water that was earlier poured into it, and water was poured to refill the container again" 
Volume of the water initially is Y. If twice that amount is taken out, the percentage of the solution that is taken out will be, $$\frac{2Y}{X+Y}$$
That means the quantity of milk that will remain in the solution will be, $$X\left(1-\frac{2Y}{X+Y}\right)$$
This value is given to be 72%, 72% of 300 will be 216

$$X\left(1-\frac{2Y}{X+Y}\right)=216$$
$$X\left(\frac{X+Y-2Y}{X+Y}\right)=216$$
Writing $$X=300-Y$$
$$\left(300-Y\right)\left(300-2Y\right)=64800$$

Expanding this we have, 
$$2Y^2-900Y+25200=0$$
Factorising this equation we have, 
$$2\left(Y-30\right)\left(Y-420\right)=0$$

Y is either 30 or 420. 

Given that the capacity of the container itself is 300, Y has to be 30. 

Hence the amount of water initially is 30 Litres. 

Let us fix the Cost Price of the product to be X, and the Selling Price of the product to be 1.4X, since it is given that it is fixed to have a profit of 40%. 
If the CP has been 40% less, making the CP 0.6X, 
And the selling price is 5 rupees less, making it 1.4X-5
Profit will be 50%, 
So, $$1.5\left(0.6X\right)=1.4X-5$$
$$0.9X=1.4X-5$$
$$0.5X=5$$
$$X=10$$

Original selling price will be 14. 

Let us say the capacity of the glass is X, and it is completely filled with milk, 
If two-thirds of its content is poured out and replaced with water, the remaining fraction of the milk will be one third. 
And this is said to be done three more times, that means a total of 4 times. 
So the contents of Milk initially being X,
And after 4 times the contents will be, $$X\left(1-\frac{2}{3}\right)^4=\frac{X}{81}$$

Since the total contents is X, and the milk contents is X/81, the water contents will be 80X/81. 

Ratio of milk to water will be $$\frac{X}{81}:\ \frac{80X}{81}$$

Answer is $$1:80$$

The number of apples and mangoes are in the ratio 5 : 2.
Let us write the number of apples as 5X and number of mangoes as 2X
This means oranges will be 187-7X. 

After selling the remaining fruits, 
Apples: 5X-75
Mangoes: 2X-26
Oranges: (187-7X)/2

Unsold Apples to Unsold oranges is 3:2

$$\frac{2\left(5x-75\right)}{187-7x}=\frac{3}{2}$$

$$20x-300=561-21x$$
$$41x=861$$
$$x=21$$

Total number of unsold fruits will be, 
Apples: 30
Mangoes: 16
Oranges: 20

Total is 66. 

We can use the formula for when two people moving towards each other and meeting in a straight line. $$t^2=t_1\times\ t_2$$
Where $$t$$ is time taken for them to meet each other. $$t_1$$ is the time taken by person 1 to reach the destination after meeting and 
$$t_2$$ is the time taken by person 2 to reach the destination after meeting

We are told they meet each other in 1 and a half hours, that is 90 minutes. 
And if Sunil takes X minutes to reach A, Anil will take X+75 minutes
Since it is given that, Anil reaches B exactly 1 hour 15 minutes after Sunil reaches A

$$90^2=x\left(x+75\right)$$
$$8100=x^2+75x$$
$$x^2+75x-8100=0$$
$$\dfrac{\left(-75\pm\sqrt{5625+32400}\right)}{2}$$
$$\dfrac{\left(-75+195\right)}{2}$$
$$x=60$$

Total time of travel of Anil is, $$90\ \min\ +60\ \min\ +75\ \min$$
Total of 225 minutes. 
Time in hours will be 3.75 hours. 
Speed is $$\dfrac{45}{3.75}=12\ \dfrac{km}{hr}$$

Let us assume the four numbers to be a, b, c and d in ascending order. 

Average of first two numbers is 1 more than the first number
$$\frac{\left(a+b\right)}{2}=a+1$$
$$b-a=2$$
$$b=a+2$$

Average of first three numbers is 2 more than average of first two numbers
$$\frac{\left(a+b+c\right)}{3}=\frac{\left(a+b\right)}{2}+2$$
$$2c=a+b+12$$
Substituting the value for b
$$2c=a+a+2+12$$
$$2c=2a+14$$
$$c=a+7$$

Average of first four numbers is 3 more than average of first three numbers.
$$\frac{\left(a+b+c+d\right)}{4}=\frac{\left(a+b+c\right)}{3}+3$$
$$3d=a+b+c+36$$
Substituting the value of b and c
$$3d=a+a+2+a+7+36$$
$$3d=3a+45$$
$$d=a+15$$

d is the largest and a is the smallest and we know that d=a+15

Hence the difference between the smallest and the largest values is 15. 

We are told that, 10000 is deposited in bank A for a certain number of years at a simple interest of 5% per annum. 
Let us say that the number of years is x
Total value of the deposit after x years is, $$10000\left(1+x\left(0.05\right)\right)$$

On maturity, the total amount received is deposited in bank B for another 5 years at a simple interest of 6% per annum
Here we know the years and the interest rate, 
$$10000\left(1+x\left(0.05\right)\right)\left(1+5\left(0.06\right)\right)$$
$$10000\left(1+\left(0.05\right)x\right)\left(1.3\right)$$

Interest received from Bank A is $$\left(x\left(0.05\right)\right)10000$$
Interest received from Bank B is $$0.3\left(10000\left(1+x\left(0.05\right)\right)\right)$$

This ratio is given to be 10:13. 

$$\dfrac{x\left(0.05\right)}{0.3\left(1+x\left(0.05\right)\right)}=\dfrac{10}{13}$$

$$0.65x=3+0.15x$$
$$0.5x=3$$
$$x=6$$

Hence the number of years the money was invested in Bank A is 6 years. 

We are told that Rajesh manages 20 hectares and Vimal manages 30 hectares

For Vimal, we know the distribution of the land between Wheat and Mustard, 5:3
So, wheat area will be, $$\frac{5}{8}\left(30\right)$$
Mustard area will be, $$\frac{3}{8}\left(30\right)$$

Similarly, let us assume that the distribution of crops between Wheat and Mustard to be k:1
Wheat will be, $$\frac{k}{k+1}\left(20\right)$$
Mustard will be, $$\frac{1}{k+1}\left(20\right)$$

We are told that total area of Wheat and Mustard is in the ratio 11:9

Adding them up we get, 
$$\dfrac{\left(\frac{150}{8}+\frac{20k}{k+1}\right)}{\left(\frac{90}{8}+\frac{20}{k+1}\right)}=\dfrac{11}{9}$$

$$\dfrac{\left(\frac{15}{8}+\frac{2k}{k+1}\right)}{\left(\frac{9}{8}+\frac{2}{k+1}\right)}=\dfrac{11}{9}$$

$$\frac{135}{8}+\frac{18k}{k+1}=\frac{99}{8}+\frac{22}{k+1}$$

$$\frac{36}{8}=\frac{22-18k}{k+1}$$

$$44-36k=9k+9$$

$$45k=35$$

$$k=\frac{7}{9}$$

Hence the ratio of distribution of area between Wheat and Mustard for Rajesh is $$\dfrac{7}{9}$$

Let us say the quantity of grains is X
For the first customer he sells $$\frac{X}{2}+3$$
Remaining is $$\frac{X}{2}-3$$

Second customer he sells: $$\frac{X}{4}-\frac{3}{2}+3=\frac{X}{4}+\frac{3}{2}$$
Remaining will be $$\frac{X}{4}-\frac{9}{2}$$

Third customer he sells: $$\frac{X}{8}-\frac{9}{4}+3$$ = $$\frac{X}{8}+\frac{3}{4}$$
Remaining will be $$\frac{X}{8}-\frac{21}{4}$$
Now, this is said to be 0,
$$\frac{X}{8}-\frac{21}{4}=0$$

X=42

Let's start from the step when there was 50% concentration. 
Let's take there to ee 2T solution: T acid and T water.

Adding 15 litres of acid increases the acid concentration to  80%, giving the equation $$\frac{T+15}{2T+15}=\frac{4}{5}$$
Solving this would give us T=5

This means that there were 5 litres of acid and 5 litres of water after mixing 2 litres of water. 

Therefore, there would be 5 litres of acid and 3 litres of water before adding the water. 
We are asked the ratio of water to acid, which would be 3:5

Therefore, Option B is the correct answer. 

We are told that there was two successive increments in the salary, with the second increment percentage twice the first one. Total Increment was 187.5%. 

Drawing up the equation
$$\left(1+z\right)\left(1+2z\right)=1.875$$
$$1+3z+2z^2=1.875$$
$$2z^2+3z-0.875$$
$$z=\frac{\left(-3\pm\sqrt{9+8\left(0.875\right)}\right)}{4}$$
$$z=\frac{\left(-3\pm\ 4\right)}{4}$$
$$z=0.25$$

Answer is 25%. 

It is given that a merchant purchases a cloth at a rate of Rs.100 per meter and receives 5 cm length of cloth free for every 100 cm length of cloth purchased by him.

Hence, the cost price of 105 cm clothes is 100 rupees. 

It is also known that he marked the price of 100 cm clothes as 110 rupees, and gave a 5% discount, and he cheated his customers by giving 95 cm length of cloth for every 100 cm length of cloth purchased by the customers.

Hence, the selling price of 95 cm clothes is 110*(19/20) rupees.

Therefore, the selling price of 105 cm clothes is 115.5 rupees.

Hence, the profit is 15.5%

The correct option is C

Given that in the final mixture, the ratio of coffee and cocoa is 16:9

Let us assume coffee is 16 units and cocoa is 9 units.

=> Initially, there are 25 units of coffee and 0 units of cocoa

Let's say x units of the mixture is removed and replaced with cocoa

=> Now, we have (25-x) coffee and x units of cocoa. => Mixture P

Now, if x units of the mixture is removed:

Amount of coffee present = (25-x) - $$\dfrac{\left(25-x\right)}{25}\times\ x$$

=> $$\left(25-x\right)\left(1-\dfrac{x}{25}\right)=16$$

=> $$\left(25-x\right)^2=16\times\ 25$$

=> 25 - x = 20 => x = 5.

In mixture P, cocoa = x = 5

In mixture Q, cocoa = 9 units.

=> Required ratio = 5:9

Let the time taken by A to fill the tank alone be x hours, which implies the time taken by B to empty the tank alone is (x-1) hours (B is the drainage pipe), and the time taken by C to fill the tank is y hours.

It is given that when pipes A, B, and C are turned on together, the empty tank is filled in two hours. 

Hence, $$\frac{1}{x}-\frac{1}{x-1}+\frac{1}{y}=\frac{1}{2}$$ .... Eq(1)

It is given that if pipes B and C are turned on together when the tank is empty and Pipe B is turned off after one hour, then Pipe C takes another one hour and 15 minutes to fill the remaining tank.

Hence, B worked for 1 hour, and C worked for 2 hours 15 minutes, which is equal to $$\frac{9}{4}$$ hours.

In 1 hour, B worked $$-\frac{1}{x-1}$$ units, and in $$\frac{9}{4}$$ hours, C worked $$\frac{9}{4y}$$ units.

Hence, $$\frac{9}{4y}-\frac{1}{x-1}=1$$ .... Eq(2)

Solving both equations, we get $$y=\frac{3}{2}$$, and $$x=3$$

Hence, the time taken by C is $$\frac{3}{2}$$ hours, which is equal to $$90$$ minutes.

The correct option is A

It is given that Anil borrows Rs 2 lakhs at an interest rate of 8% per annum, compounded half-yearly. It is also known that he repays Rs 10320 at the end of the first year and closes the loan by paying the outstanding amount at the end of the third year.

The total amount at the end of the first year is: $$200000\times\ \frac{104}{100}\times\ \frac{104}{100}=216320$$

He repays 10320 rupees at the end of the first year, which implies the amount that remains unpaid at the end of the first year is 206000 rupees.

This unpaid amount will accrue interest for another two years.

Hence, the final amount at the end of three years is $$206000\times\ \frac{104}{100}\times\ \frac{104}{100}\times\ \frac{104}{100}\times\ \frac{104}{100}=240990.86$$

Hence, the accrued interest in these two years is (240990.86-206000) = 34990.86 rupees.

Hence, the total interest accrued over the three years = (34990.86+16320) = 51311 rupees.

The correct option is B

Let the work done by Rahul, Rakshita, and Gurmeet be a, b, and c units per day, respectively, and the total units of work are W.

Hence, we can say that 7(a+b+c) < W ( Rahul, Rakshita, and Gurmeet, working together, would have taken more than 7 days to finish a job).

Similarly, we can say that 15(a+c) > W ( Rahul and Gurmeet, working together would have taken less than 15 days to finish the job)

Now, comparing these two inequalities, we get: 7(a+b+c) < W < 15(a+c)

It is also known that they all worked together for 6 days, followed by Rakshita, who worked alone for 3 more days to finish the job. Therefore, the total units of work done is: W = 6(a+b+c)+3b

Hence, we can say that 7(a+b+c) < 6(a+b+c)+3b < 15(a+c)

Therefore, (a+b+c) < 3b => a+c < 2b, and 9b < 9(a+c) => b < a+c

=> a+b+c < 3b => 7(a+b+c) < 21b , and 15b < 15(a+c)

Hence, The number of days required for b must be in between 15 and 21 (both exclusive).

Hence, the correct option is D

The given time is 8:48 AM.

Angle made by hours hand w.r.t 12 is 8 * 30 (30 degrees in 1 hour) + 0.5 * 48 (0.5 degree in 1 minute) = 240 + 24 = 264 degrees.

Angle made by minutes hands w.r.t 12 is 48 * 6 = 288 degrees.

=> The angle between them is 288 - 264 = 24 degrees.

This should further increase by 12 degrees (50% of 24)

After m minutes, the further increase in angle = (6 - 0.5)*m = $$\dfrac{11}{2}m=12$$ => m = $$\dfrac{24}{11}$$

Let us assume the initial selling prices of A and B is p.

Given, she made profit of 20% on A => 1.2 * c = p => c = 5p/6 => cost of A is $$\dfrac{5}{6}p$$

Given, she made a loss of 10% on B => 0.9 * c = p => c = 10p/9 => cost of B is $$\dfrac{10}{9}p$$

Now, she sold them at a price such that a 10% profit is made on B

=> Selling price = s = 11/10 * 10/9 p => $$\dfrac{11}{9}p$$

=> Profit % on A = $$\dfrac{\left(\dfrac{11}{9}-\dfrac{5}{6}\right)}{\left(\dfrac{5}{6}\right)}\times\ 100$$ = 46.66% = nearly 47%

It is given that the speed of Ravi is 40 kmph, which is equal to $$\frac{100}{9}$$ m/s. It is also known that the speed of Ashok is 50 kmph, which is equal to $$\frac{125}{9}$$ m/s.

It is known that the distance between Ravi and Ashok is 225 meters, and the relative speed of Ravi and Ashok is $$\frac{125}{9}+\frac{100}{9}=25$$ m/s 

Hence, they will meet each other in $$\frac{225}{25}=9$$ seconds. The distance traveled by Ravi in these 9 seconds is $$\frac{100}{9}\times\ 9=100$$ meters.

Since Vijay was already 54 meters behind Ravi when they were starting, Vijay must travel (100+54) = 154 meters in these 9 seconds.

Hence, the speed of Vijay is $$\frac{154}{9}$$ m/s, which is equal to $$\frac{154}{9}\times\ \frac{18}{5}\ =\frac{308}{5}=61.6$$ kmph.

The correct option is C

Let the volume of mixture A be 200 ml, which implies the quantity of cocoa in the mixture is 120 ml, and the quantity of sugar In the mixture 80 ml.

Similarly, let the volume of the mixture be 300 ml, which implies the quantity of coffee, and sugar in the mixture is 210, and 90 ml, respectively.

Now we combine mixture A, and B in the ratio of 2:3 (if 200 ml mixture A, then 300 ml of mixture B).

Hence, the volume of the mixture C is (200+300) = 500 ml, and the quantity of the sugar is (90+80) = 170 ml. 

Now he mixes C with an equal amount of milk to make a drink, which implies the quantity of the final mixture is (500+500) = 1000 ml.

The quantity of sugar in the final mixture is 170 ml.

Hence, the percentage is 17% 

The correct option is A

Given the total number of kilometres travelled, including the trip = is 26862 Km, and the duration of the trip is 8 hrs.

If avg. speed of the car during the trip is 's' => the km travelled till just before the trip is 26862 - 8s, which should also be a palindrome.

=> From the options if s = 110 => The reading will be 26862 - 110*8 = 25982 (Not a palindrome)

=> If s = 100 => The reading will be 26862 - 100*8 = 26062 => It is a palindrome.

=> s = 100 is the correct option.

The cost price of the sunglass for Meenu when he purchased it for the first time was 1000 rupees, and he sold it to Kanu at 20% profit. Hence, the selling price of the sunglass is 1200 rupees, which Kanu purchased. Hence, the profit made by Meenu is (1200-1000) = 200 rupees.

Hence, the cost price of the same sunglass for Kanu is 1200 rupees, and now he sold it to Meenu at a 20% loss. Hence, the selling price of the sunglass now is (1200*0.8) = 960 rupees.

The cost price of the same sunglass for Meenu when he purchased it for the second time was 960 rupees. Now Meenu sold it Tanu, at a certain price such that the total profit of Meenu becomes 500 rupees.

Hence, on the second transaction (selling it to Tanu), Meenu made a profit of (500-200) = 300 rupees.

Hence, the profit made by Minu in the second transaction is (300/960)*100% = 31.25%

The correct option is C

it is given that the price of a precious stone is directly proportional to the square of its weight. Let the price be denoted by C and the weight is denoted by W.

Hence, $$C\ ∝\ W^2$$ => $$C\ =kw^2$$ (where k is the proportional constant)

Now, Sita has a precious stone weighing 18 units.

Therefore, $$C\ =kw^2=k\cdot18^2\ =\ 324$$

If she breaks it into four pieces with each piece having a distinct integer weight, then the difference between the highest and lowest possible values of the total price of the four pieces will be 288000.

To get the lowest possible value of C, we will get the weight of the four-piece as close as possible (3,4,5,6). To get the highest value we will try to take three pieces as low as possible, and one is as high as possible (1, 2, 3, 12).

Hence, the maximum cost = $$k(12^2+1^2+2^2+3^2) = 158k$$, and the minimum cost is $$k(3^2+4^2+5^2+6^2) = 86k$$

Hence, the difference is $$(158k - 86k) = 72k$$, which is equal to 288000.

=> $$72k = 288000$$

=> $$k = 4000$$

Hence, the price of the original stone is $$324k = 324\times\ 4000\ =\ 1296000$$

The correct option is D

Let us assume that A, B, C, D, and E weights are a, b, c, d, and e.

1st condition

$$\frac{\left(a+b+c\right)}{3}-\frac{\left(a+b+c+d\right)}{4}=x$$

2nd condition

$$\frac{\left(a+b+c+e\right)}{4}-\frac{\left(a+b+c\right)}{3}=2x$$

Adding both the equations, we get:

$$\frac{\left(e-d\right)}{4}=3x$$

=> $$\frac{\left(e-d\right)}{4}=3x$$ => e - d = 12x

Given that 12x = 12 => x = 1.

Let us assume the speed of the 1st boat is b, the 2nd boat is s, and the river's speed is r.

Let 'd' be the distance between A and B.

=> d = 2(b+r) and d = 3(b-r)

=> b + r = d/2 and b - r = d/3 => r = d/12 (subtracting both equations).

Now, it is given that

$$\dfrac{d}{s+r}+\dfrac{d}{s-r}=6$$

=> $$\dfrac{d}{s+\dfrac{d}{12}}+\dfrac{d}{s-\dfrac{d}{12}}=6$$

=> $$2ds=6\left(s^2-\dfrac{d^2}{144}\right)$$

=> $$144s^2-48ds-d^2=0$$

Solving the quadratic equation, we get:

$$s=d\left(\dfrac{\left(48+\sqrt{\ 48^2+4\left(144\right)}\right)}{2\times\ 144}\right)$$

$$s=d\left(\dfrac{1}{6}+\dfrac{\sqrt{\ 5}}{12}\right)$$

=> Required value of $$\dfrac{d}{s+r}$$

= $$\dfrac{d}{\dfrac{d}{6}+\dfrac{\sqrt{5}d}{12}+\dfrac{d}{12}}$$

= $$\dfrac{12}{3+\sqrt{\ 5}}=\dfrac{\left(12\right)\left(3-\sqrt{\ 5}\right)}{4}$$

= $$3\left(3-\sqrt{\ 5}\right)$$

Let the number of total employees in the company be 100x, and the total salary of all the employees be 100y.

It is given that 20% of the employees work in the manufacturing department, and the total salary obtained by all the manufacturing employees is one-sixth of the total salary obtained by all the employees in the company.

Hence, the total number of employees in the manufacturing department is 20x, and the total salary received by them is (100y/6)

Average salary in the manufacturing department = (100y/6*20x) = 5y/6x

Similarly, the total number of employees in the nonmanufacturing department is 80x, and the total salary received by them is (500y/6)

Hence, the average salary in the nonmanufacturing department = (500y/6*80x) = 25y/24x

Hence, the ratio is:- (5y/6x): (25y/24x) 

=> 120: 150 = 4:5

The correct option is B

Given, the salaries of Sita, Gita and Mita are initially in the ratio 5 : 6 : 7, respectively, Let us assume their salaries are 5p, 6p and 7p.

They get salary hikes of 20%, 25% and 20%, respectively.

=> Their salaries are 6/5 * 5p, 5/4 * 6p and 6/5 * 7p => 6p, 7.5p, 8.4p

Now, Sita and Mita get salary hikes of 40% and 25%, respectively

=> Sita's salary = 1.4 * 6p = 8.4p and Mita's salary = 1.25 * 8.4p = 10.5p

Let Gita's salary be 'g' after hike

=> 3g = 8.4p + g + 10.5p => 2g = 18.9p => g = 9.45p

=> Hike % = $$\dfrac{\left(9.45-7.5\right)}{7.5}\times\ 100$$ = 26%

Let us assume the speeds of Arvind and Surbhi are 'a' and 's', respectively.

Let us say they meet after 't' hours

=> Arvind travelled s*t distance in 6 hrs and Surbhi travelled a*t in 24 hrs

=> s*t = a*6 and a*t = s*24 => $$t^2=6\times\ 24$$ => t = 12

Given a = 54 => s*12 = 54*6 => s = 27.

=> Total distance between A and B is (s+a)*t = (54+27)*12 = 81*12 = 972 Kms.

Anil invested 22000 for 6 years at 4% interest compounded half-yearly

=> Amount =$$22000\left(1.02\right)^{12}$$

Let Sunil invest 'S' rupees for 5 years at 4% C.I. half-yearly and 10% S.I. for 1 additional year

=> Amount = $$S\left(1.02\right)^{10}\left(1.1\right)$$

Given that the both amounts are equal

=> $$22000\left(1.02\right)^{12}=S\left(1.02\right)^{10}\left(1.1\right)$$

=> $$S=\dfrac{22000\left(1.02\right)^2}{1.1}=20808$$

Let the number of white shirts be m, and the number of blue shirts be n. Hence, the total cost of the shirts = (1000m+1125n), and the number of shirts is (m+n)

The average price of the shirts is $$\ \frac{\ 1000m+1125n}{m+n}$$.

It is given that he set a fixed market price which was 25% higher than the average cost of all the shirts. He sold all the shirts at a discount of 10%.

Hence, the average selling price of the shirts = $$\left(\ \frac{\ 1000m+1125n}{m+n}\right)\times\ \frac{5}{4}\times\ \frac{9}{10}=\frac{9}{8}\left(\ \frac{\ 1000m+1125n}{m+n}\right)$$

The average profit of the shirts = $$\frac{9}{8}\left(\ \frac{\ 1000m+1125n}{m+n}\right)-\frac{\ 1000m+1125n}{m+n}=\frac{1}{8}\left(\frac{\ 1000m+1125n}{m+n}\right)$$

The total profit of the shirts = $$\frac{1}{8}\left(\frac{\ 1000m+1125n}{m+n}\right)\times\ \left(m+n\right)\ =\ \frac{1}{8}\left(1000m+1125n\right)$$

Now, $$=>\frac{1}{8}\left(1000m+1125n\right)=51000$$

$$=>1000m+1125n=51000\times\ 8=408000$$

Now to get the maximum number of shirts, we need to minimize n (since the coefficient of n is greater than the coefficient of m), but it can't be zero. Therefore, m has to be maximum.

$$m\ =\ \ \frac{\ 408000-1125n}{1000}$$

The maximum value of m such that m, and both are integers is m = 399, and n = 8 (by inspection)

Hence, the maximum number of shirts = m+n = 399+8 = 407

Let's assume that after n iteration, the volume of the milk will be less than 50%, which is less than 20 liters.

Initially, the amount of milk is 40 liters, after the first iteration, the volume of milk is$$40\cdot\frac{9}{10}$$

After the second iteration, the volume of milk is $$40\times\left(\frac{9}{10}\right)^{^2}$$

Similarly, after the n iterations, the volume of milk is $$40\times\left(\frac{9}{10}\right)^{^n}$$

Now,

$$40\times\left(\frac{9}{10}\right)^{^n}\ \le\ 20$$

=> $$\left(\frac{9}{10}\right)^{^n}\ \le\ \frac{1}{2}$$

=> $$n\ge\ 7$$

Hence, the correct answer is 7

Let 'g' and 's' be the efficiencies of Gautam and Suhani. Let W is the total amount of work.

=> g + s = W/20 (1 day work) ----(1)

Also Gautam doing only 60% => 3g/5 and Suhani doing 150% => 3s/2

=> 3g/5 + 3s/2 = W/20 (1 day work)

=> $$g+s=\dfrac{3g}{5}+\dfrac{3s}{2}$$

=> $$\dfrac{s}{g}=\dfrac{4}{5}$$ => Gautam is the more efficient person.

Now, from the 1st equation

=> $$g+\dfrac{4g}{5}=\dfrac{W}{20}$$

=> $$\dfrac{9}{5}g=\dfrac{W}{20}$$

=> $$g=\dfrac{W}{36}$$

=> Gautam takes 36 days to finish the complete work.

Let us assume the efficiencies of Amal, Sunil, and Kamal are a, s, and k, respectively.

Given that they are in H.P.

=> $$\dfrac{2}{s}=\dfrac{1}{a}+\dfrac{1}{k}$$ ---(1)

Also, given that Kamal takes twice as much time as Amal to do the same amount of job

=> a = 2k

Given that when Amal and Sunil work for 4 days and 9 days, respectively, Kamal needs to work for 16 days to finish the remaining job.

=> If W is the total work => 4a + 9s + 16k = W.

from (1)$$\dfrac{2}{s}=\dfrac{1}{a}+\dfrac{2}{a}$$ => $$a=\dfrac{3}{2}s$$ and $$k=\dfrac{3}{4}s$$

=> $$4\left(\dfrac{3s}{2}\right)+9s+16\left(\dfrac{3s}{4}\right)=W$$

=> $$6s+9s+12s=W$$

=> $$27s=W\ =>\ s\ =\ \dfrac{W}{27}$$

=> Sunil will take 27 days to finish the work when working alone.

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

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

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.

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.

M - First meeting point

Let the speeds of trains A and B be 'a' and 'b', respectively.

$$\dfrac{x}{a}=\ \dfrac{\ D-x}{b}$$

It is given,

$$\dfrac{D}{a}=10$$ and $$\dfrac{x}{b}=9$$

$$\dfrac{x}{\dfrac{D}{10}}=\ \dfrac{\ D-x}{\dfrac{x}{9}}$$

$$\dfrac{10x}{D}=\ \dfrac{\ 9D-9x}{x}$$

$$10x^2=\ \ 9D^2-9Dx$$

$$10x^2+9Dx-9D^2=\ 0$$

Solving, we get $$x=\dfrac{3D}{5}$$

$$\dfrac{x}{b}=9$$

$$\dfrac{3D}{b\times5}=9$$

$$\dfrac{D}{b}=15$$ 

The total time taken by train B to travel from station Y to station X is 15 minutes.

The answer is option B

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.

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.

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.

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.

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.

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.

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

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.

Let the number of students in section A and B be a and b, respectively.

It is given, a = b - 10

$$\ \ \dfrac{\ 32a+60b}{a+b}$$ is an integer

$$\ \ \dfrac{\ 32a+60\left(a+10\right)}{a+a+10}=k$$

$$\ \ \dfrac{\ 46a+300}{a+5}=k$$

$$k=\ \dfrac{\ 46\left(a+5\right)}{a+5}+\dfrac{70}{a+5}$$

$$k=\ \ 46+\dfrac{70}{a+5}$$

a can take values 2, 5, 9, 30, 65

Difference = 65 - 2 = 63

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.

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 = $$\ \dfrac{\ 1\left(50\%\right)}{1+3}$$

Sugar syrup = $$\ \dfrac{\ 1\left(50\%\right)+3\left(100\%\right)}{1+3}=\dfrac{350}{4}$$

Required ratio = 50:350 = 1:7

The answer is option D.

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.

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.

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.

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.

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.

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.

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

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.

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.

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, $$x=190\ m$$

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.

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

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.

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.

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

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

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 = $$\frac{\frac{1100}{9}}{\frac{1600}{9}}\times\ 100\ =\ \frac{1100}{16}\%=68.75\%$$

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

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

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

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

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

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
 

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 $$P\left(1+r\right)^2$$

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

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

$$X=1.075\times\ 866.72=931.72$$

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.

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

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

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

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 = $$10000\times\ 2t\times\ \left(\frac{2x}{100}\right)=40000\left(\frac{x}{100}\right)t=40000\times\ 0.0609=2436$$

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.

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

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

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 =$$\frac{24000}{8}=\ 3,000$$

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.  

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 = $$\left(\frac{25}{11}-1\right)\times\ 100\approx\ 127\%$$

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

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

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 = $$\frac{5}{\frac{2}{\frac{7}{2}}}=\frac{5}{7}$$

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%

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

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

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. 

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

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

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

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

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.

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

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

Given,

x+y=50000...(i)

1.2x+0.9y=50000(1.02)=51000...(ii)

(ii)-0.9(i) gives

0.3x=6000=> x=20000.

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.

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

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. 

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 $$\frac{\left(150-120\right)}{120}\cdot100\ =\ 25\%$$

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'. 

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.

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

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%.

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.

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.