CAT Percentages 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 $$0.2\left(250+110\right)=72$$

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%

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 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.

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

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%

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

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

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

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

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

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

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

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

Each transfer can be captured through the following table.

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

=14%

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

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 volume of the first and the second solution be 100 and 300.
When they are mixed, quantity of ethanol in the mixture 
= (20 + 300S)
Let this solution be mixed with equal volume i.e. 400 of third solution in which the strength of ethanol is 20%.
So, the quantity of ethanol in the final solution 
= (20 + 300S + 80) = (300S + 100)
It is given that, 31.25% of 800 = (300S + 100)
or, 300S + 100 = 250
or S = $$\frac{1}{2}$$ = 50%
Hence, 50 is the correct answer.

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%

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%

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%

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%

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

Fresh grapes contain 90% water so water in 20kg of fresh pulp = (90/100)x20= 18kg. 

In 20kg fresh grapes, the weight of water is 18kg and the weight of pulp is 2kg.

The concept that we apply in this question is that the weight of pulp will remain the same in both dry and fresh grapes. If this grape is dried, the water content will change but pulp content will remain the same.

Suppose the weight of the dry grapes be D.

80% of the weight of dry grapes = weight of the pulp = 2 kg

(80/100) x D =2 kg.

D = 2.5 kg

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

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 the total weight of fresh grapes be 100 gm.

=> Fresh grapes have 90 gm of water and 10 gm of fruit.

When these grapes are dried, the amount of fruit does not change.

=> 10 grams will become 80% of the content in dry grapes

=> Weight of dry grapes = $$\frac{10}{0.8}$$ = 12.5 gm

So, the weight of fresh grapes reduces to 1/8th of its original weight.

=> 20 kg of fresh grapes give 2.5 kg of dry grapes.