CAT Averages, Ratio and Proportion Questions

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'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. 

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. 

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. 

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 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

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 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%

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 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.

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.

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.

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

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

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 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

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

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

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 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 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

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

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

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

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)*114

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

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

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

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%

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.

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.

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. 

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. 

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. 

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%

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.

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

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

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%

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

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%

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

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 $$(A+9.5) - A = 9.5$$

The odd numbers in the set are 3, 5, 7, ...2n+1

Sum of the odd numbers = 3+5+7+...+(2n+1) = $$n^2 + 2n$$

Average of odd numbers = $$n^2 + 2n$$/n = n+2

Sum of even numbers = 2 + 4 + 6 + ... + 2n = 2(1+2+3+...+n) = 2*n*(n+1)/2 = n(n+1)

Average of even numbers = n(n+1)/n = n+1

So, difference between the averages of even and odd numbers = 1

Ten years ago, the total age of the family is 231 years.

Seven years ago, (Just before the death of the first person), the total age of the family would have been 231+8*3 = 231+24 = 255.
This is because, in 3 years, every person in the family would have aged by 3 years,
Total change in age = 231+24 = 255

After the death of one member, the total age is 255-60 = 195 years.

Since a child takes birth in the same year, the number of members remain the same i.e. (7+1) = 8

Four years ago, (i.e. 6 years after start date) one of the member of age 60 dies, 
therefore, total age of the family is 195+24-60 = 159 years.

Since a child takes birth in the same year, the number of members remain the same i.e. (7+1) = 8

After 4 more years, the current total age of the family is = 8x4 + 159 = 191 years  

The average age is 191/8 = 23.875 years = 24 years (approx)


Alternatively,
Since the number of members is always the same throughout
The 2 older members dropped their age by 60
So, after 10yrs, total age = 231 + 8*10 - 2*60 = 191
Average age = 191/8 = 23.875 $$\simeq$$ 24

Surface area of sphere A (of radius a) is $$4\pi*a^2$$
Surface area of sphere B (of radius b) is $$4\pi*b^2$$
=> $$4\pi*a^2$$/$$4\pi*b^2$$ = 1/4 => a:b = 1:2
Their volumes would be in the ratio 1:8
Therefore, k = 7/8 * 100% = 87.5%

By the end of the year 2002, Shepard bought 4 times and sold 4 times. He is left with the initial number of goats that he had in 1998. If the percentage of goats bought is equal to or lesser than the percentage of goats sold, then there would be a net decrease in the total number of goats. For the number of goats to remain the same, p has to be greater than q, because p% is being calculated in a lesser number and q% is being calculated on a greater number. Hence, p > q.

Let the actual average be n. So, the new average is n + 1.8
Actual total = 10n
New total = 10n + 18

Let the number which was miswritten = ab(a is the tenth's digit and b is the units digit) = 10a+b

and reversed number ba = 10b+a

So, 10b + a - (10a + b) = 18
=> 9(b-a) = 18
=> b-a = 2

Out of 200 of the seating capacity, 180 seats are filled out of which 108 are males and 72 are females. Remaining 20 seats are vacant. According to given condition seating capacity in both the planes is 120 . Considering flight A - we can find that 100 passenger in waiting hall will be taking fight A . So 80 people remain in in the waiting hall who will be taking flight B . Now for every 20 people taking flight B we have a air hostess in flight A . So in total there are 4 air hostess in flight A. Flight B having 120 as seating capacity, 40 remain vacant. So required ratio 40:4 = 10:1 .

Let there be total 100 people whom the college will ask for donation. Out of these 60 people have already given average donation of 600 Rs. Thus total amount generated by 60 people is 36000. This is 75% of total amount required . so the amount remaining is 12000 which should be generated from remaining 40 people. So average amount needed is 12000/40 = 300

Let x , y and z be no. of students in class X, Y ,Z respectively.

From 1st condition we have

83*x+76*y = 79*x+79*y which give 4x = 3y.

Next we have 76*y + 85*z = 81(y+z) which give 4z = 5y .

Now overall average of all the classes can be given as $$\frac{83x+76y+85z}{x+y+z}$$

Substitute the relations in above equation we get, 

$$\frac{83x+76y+85z}{x+y+z}$$  = (83*3/4 + 76 + 85*5/4)/(3/4 + 1 + 5/4) = 978/12 = 81.5

Since the number starts from 1 if there are n numbers then initial average = $$\dfrac{n+1}{2}$$.

Average of N natural number can be either an integer {ab} or {ab.50} type. For example average of first 10 number = 5.5 whereas the average of first 11 natural numbers is 6. 
Even if we erased the largest number change in average will be always less than 0.5. 
Here we are given the average is 602/17 or 35$$\frac{7}{17}$$ Hence we can say that average must have been 35.5 or 35 before. 
Case 1: If the average was 35.5 before the erasing process. 
We know that average of 1st N natural number = $$\dfrac{N+1}{2}$$
35.5 = $$\dfrac{N+1}{2}$$
N = 70. 
Sum of these 70 numbers = 70*71/2 = 35*71 = 2485.
Sum of the 69 numbers which we are left with after removing a number = (602/17)*69 = 2443.41. Which is not possible as the sum of natural numbers will always be an integer. Hence, we can say that case is not possible. 

Case 1: If the average was 35 before the erasing process. 
We know that average of 1st N natural number = $$\dfrac{N+1}{2}$$
35 = $$\dfrac{N+1}{2}$$
N = 69. 
Sum of these 69 numbers = 69*70/2 = 35*69 = 2415.
Sum of the 68 numbers which we are left with after removing a number = (602/17)*68 = 2408. 
Hence, we can say that the erased number = 2415 - 2408 = 7. 

Let the marks in the five papers be 6k, 7k, 8k, 9k and 10k respectively.
So, the total marks in all the 5 papers put together is 40k. This is equal to 60% of the total maximum marks. So, the total maximum marks is 5/3 * 40k
So, the maximum marks in each paper is 5/3 * 40k / 5 = 40k/3 = 13.33k
50% of the maximum marks is 6.67k
So, the number of papers in which the student scored more than 50% is 4

The first five numbers could be n-2, n-1, n, n+1, n+2. The next two number would then be, n+3 and n+4, in which case, the average of all the 7 numbers would be $$\frac{(5n+2n+7)}{7}$$ = n+1

Let the individual weights be a,b,c,d,e in increasing order such that e is max and a is min. Adding all the addition of weight together we get 4*(a+b+c+d+e) = 1156 so a+b+c+d+e = 289 . Out of these a+b will be lowest sum and d+e will be the max . so a+b=110 and d+e=121 so we get value of c as 58 . now c have the 3rd highest weight so addition of c and e must give the second largest total i.e 120 . hence e = 120-58 = 62

Let the fixed income be x and the number of boarders be y.

x + 25y = 17500

x + 50y = 30000

=> y = 500 and x = 5000

x + 100y = 5000 + 50000 = 55000

Average expense = $$\frac{55000}{100}$$ = Rs.550.

Let the number of employees be 100.

=> 40 men and 60 women.

Number of men getting more than 25000 = 30

Number of people getting more than 25000 = 45

Number of women getting more than 25000 = 45 - 30 = 15

Fraction of women getting less than 25000 = $$\frac{45}{60}$$ = $$\frac{3}{4}$$

let's say x is the initial number of bacterias :
So in 2nd generation no. of bacterias = $$\frac{8x}{2} = 4x$$
In 3rd generation, it will be = 16x
4th gen. = 64x
5th gen. = 256x
6th gen. = 1024x
7th gen. = 4096x 
Hence x = 1 million

Let's say number of rubbies are x and emeralds are y.
So 0.3x + 0.4y = 12
And total wealth = 4x+5y
Now putting value of x from eq.1 to eq.2 
i.e. total wealth = 4(12-0.4y)/0.3 + 5y
Now for maximizing total wealth y should be equal to zero.
Hence x = 40

Let's say number of coins are 2.5x , 3x and 4x
So total amount will be = 2.5x + 3x(0.5) + 4x(0.25) = 210
So x = 42
And number of 1 rs. coins = 2.5x = 105

Let's say he scored marks as $$10x,9x,8x,7x,6x$$ or total of $$40x$$ which is 60% of total maximum marks(T).

$$\frac{T \times 60}{100}=40x$$

So T (total maximum marks)=$$\frac{400x}{6}$$

Or Individual max. marks = $$\frac{T}{5}=\frac{80x}{6}$$

Passing marks =50% of individual max. marks =$$\frac{40x}{6}=6.66x$$

Hence he scored more than passing marks in four subjects as $$10x,9x,8x$$ and $$7x$$ and failed in one subject as scoring $$6x$$ marks which is less than passing marks of $$6.66x$$

Total marks = 80 x 10 = 800
Total marks except highest and lowest marks = 81 x 8 = 648
So Summation of highest marks and lowest marks will be = 800 - 648 = 152
When highest marks is 92, lowest marks will be = 152-92 = 60

Value of coin = $$k (2r)^2 t$$ (where k is proportionality constant, 2r is diameter and t is thickness)
So (value of first coin) = 4 (value of second coin)

$$k (2r_1)^2 t_1 = 4 \times (k(2r_2)^2 t_2)$$

or $$\frac{t_1}{t_2} = \frac{9}{4}$$  (As ratio of diameters 2r will be 9:4)

 

As data regarding weights of people is not given, hence we can't determine the avg. weight of people in group D 

As data regarding weights is not available, hence we can't say about the increment or decrement in group A an B
But avg. weight for overall class will remain same as no student went out or came inside to make any changes in total weight or total number of students.

Except option C, all are correct because if everyone has same weight than avg. weight of all groups remain same.
So avg. weight of D will be equal to avg. weight of A.

Let's say the amount of money Gopal has is 100x rs. With this money he can buy either 50 oranges or 40 mangoes.

Hence, the cost of 1 orange = $$\dfrac{100x}{50}$$ = Rs. 2x

Similarly, the cost of 1 mango = $$\dfrac{100x}{40}$$ = Rs. 2.5x

Taxi fare = 0.1*100x = 10x

Remaining money with Gopal = 90x
Cost incurred in buying 20 mangoes = 2.5*20 = 50x

Hence, the amount of money left with Gopal = 100x - 10x - 50x = 40x

Therefore, we can say that with this much money he can but a maximum of $$\dfrac{40x}{2x}$$ = 20 oranges. 

Let's, say students contributed x rs.
So teachers contributed = $$\frac{3x}{2}$$
And external benefactor contributed = $$\frac{9x}{2}$$
Summation of all contribution = 4200 = $$7x$$
or $$x=600$$
Hence, teachers contributed = 900

Let's say total products maufactured by M1, M2 and M3 are 100.
So M1 produced 40, M2 produced 30 and M3 produced 30
Defective pieces for M1 = $$\frac{120}{100}$$
Defective pieces for M2 = $$\frac{30}{100}$$
Defective pieces for M3 = $$\frac{150}{100}$$
So total defective pieces are $$\frac{150+30+120}{100}$$ = $$\frac{300}{100}$$ = 3% of total products.

Assume one person is born every day. In 100 years, there will be 25 leap years. So 25*1 additional people will be born on these days.

So, total people born will be = $$365 \times 100 \times 1 + 25 \times 1$$
And people born on 29th february = $$25 \times 1$$ 
Hence percentage will be = $$\frac{25 \times 1}{365 \times 100 \times 1 + 25 \times 1} \times 100$$ = 0.0684