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# Ratio and Proportion Questions for CAT

Ratio and Proportion is a very important topic in the CAT Quantitative Ability (QA) Section. Ratio and Proportion questions are interesting to solve, and not very difficult if you learn the basics and practice them well. Questions on Ratio and Proportion are asked frequently in the CAT. You can expect a few questions from Ratio and Proportion in the latest format of the CAT Quant section. This article will look into some important Ratio and Proportion questions for CAT. If you want to practice these important Ratio and Proportion questions, you can download the PDF below, which is completely Free.

Question 1:Â Let A and B be two solid spheres such that the surface area of B is 300% higher than the surface area of A. The volume of A is found to be k% lower than the volume of B. The value of k must be
[CAT 2003 leaked]

a)Â 85.5

b)Â 92.5

c)Â 90.5

d)Â 87.5

Solution:

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%

Question 2:Â At the end of year 1998, Shepard bought nine dozen goats. Henceforth, every year he added p% of the goats at the beginning of the year and sold q% of the goats at the end of the year where p > 0 and q > 0. If Shepard had nine dozen goats at the end of year 2002, after making the sales for that year, which of the following is true?
[CAT 2003 leaked]

a)Â p = q

b)Â p < q

c)Â p > q

d)Â p = q/2

Solution:

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.

Question 3:Â A milkman mixes 20 litres of water with 80 litres of milk. After selling one-fourth of this mixture, he adds water to replenish the quantity that he had sold. What is the current proportion of water to milk?
[CAT 2004]

a)Â 2 : 3

b)Â 1 : 2

c)Â 1 : 3

d)Â 3 : 4

Solution:

After selling 1/4th of the mixture, the remaining quantity of water is 15 liters and milk is 60 liters. So the milkman would add 25 liters of water to the mixture. The total amount of water now is 40 liters and milk is 60 liters. Therefore, the required ratio is 2:3.

Question 4:Â A student took five papers in an examination, where the full marks were the same for each paper. His marks in these papers were in the proportion of 6 : 7 : 8 : 9 : 10. In all papers together, the candidate obtained 60% of the total marks. Then the number of papers in which he got more than 50% marks is

a)Â 2

b)Â 3

c)Â 4

d)Â 5

Solution:

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

Question 5:Â Fresh grapes contain 90% water by weight while dried grapes contain 20% water by weight and the remaining proportion being pulp. What is the weight of dry grapes available from 20 kg of fresh grapes?

a)Â 2 kg

b)Â 2.4 kg

c)Â 2.5 kg

d)Â None of these

Solution:

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

Question 6:Â Amol was asked to calculate the arithmetic mean of 10 positive integers, each of which had 2 digits. By mistake, he interchanged the 2 digits, say a and b, in one of these 10 integers. As a result, his answer for the arithmetic mean was 1.8 more than what it should have been. Then |b – a| equals

a)Â 1

b)Â 2

c)Â 3

d)Â None of these

Solution:

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

Question 7:Â Two liquids A and B are in the ratio 5 : 1 in container 1 and 1 : 3 in container 2. In what ratio should the contents of the two containers be mixed so as to obtain a mixture of A and B in the ratio 1 : 1?

a)Â 2 : 3

b)Â 4 : 3

c)Â 3 : 2

d)Â 3 : 4

Solution:

Fraction of A in contained 1 = $\frac{5}{6}$

Fraction of A in contained 2Â = $\frac{1}{4}$

Let the ratio of liquid required from containers 1 and 2 be x:1-x

x($\frac{5}{6}$) + (1-x)($\frac{1}{4}$) =Â $\frac{1}{2}$

$\frac{7x}{12}$ = $\frac{1}{4}$

=> x = $\frac{3}{7}$

=> Ratio = 3:4

Question 8:Â Fresh grapes contain 90% water while dry grapes contain 20% water. What is the weight of dry grapes obtained from 20 kg fresh grapes?

a)Â 2 kg

b)Â 2.5 kg

c)Â 2.4 kg

d)Â 10 kg

Solution:

Let the total weight of freshÂ 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.

Question 9:Â The value of each of a set of coins varies as the square of its diameter, if its thickness remains constant, and it varies as the thickness, if the diameter remains constant. If the diameter of two coins are in the ratio 4 : 3, what should be the ratio of their thickness’ be if the value of the first is four times that of the second?

a)Â 16 : 9

b)Â 9 : 4

c)Â 9 : 16

d)Â 4 : 9

Solution:

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

Question 10:Â One bacterium splits into eight bacteria of the next generation. But due to environmental condition only 50% survives and remaining 50% dies after producing next generation. If the seventh generation number is 4,096 million, what is the number in first generation?

a)Â 1 million

b)Â 2 million

c)Â 4 million

d)Â 8 million

Solution:

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

Question 11:Â You can collect as many rubies and emeralds as you can. Each ruby is worth Rs. 4 crore and each emerald is worth Rs. 5 crore. Each ruby weighs 0.3 kg. And each emerald weighs 0.4 kg. Your bag can carry at the most 12 kg. What should you collect to get the maximum wealth?

a)Â 20 rubies and 15 emeralds

b)Â 40 rubies

c)Â 28 rubies and 9 emeralds

d)Â None of these

Solution:

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

Question 12:Â There are two containers: the first contains 500 ml of alcohol, while the second contains 500 ml of water. Three cups of alcohol from the first container is taken out and is mixed well in the second container. Then three cups of this mixture is taken out and is mixed in the first container. Let A denote the proportion of water in the first container and B denote the proportion of alcohol in the second container. Then,

a)Â A > B

b)Â A < B

c)Â A = B

d)Â Cannot be determined

Solution:

Let the volume of the cup be V.
Hence, after removing three cups of alcohol from the first container,

Volume of alcohol in the first container is 500-3V
Volume of water in the second container is 500 and volume of alcohol in the second container is 3V.
So, in each cup, the amount of water contained is $\frac{500}{500+3V}*V$

Hence, after adding back 3 cups of the mixture, amount of water in the first container is $0+\frac{1500V}{500+3V}$
Amount of alcohol contained in the second container is $3V – \frac{9V^2}{500+3V} = \frac{1500V}{500+3V}$

So, the required proportion of water in the first container and alcohol in the second container are equal.

Question 13:Â A stall sells popcorn and chips in packets of three sizes: large, super, and jumbo. The numbers of large, super, and jumbo packets in its stock are in the ratio 7 : 17 : 16 for popcorn and 6 : 15 : 14 for chips. If the total number of popcorn packets in its stock is the same as that of chips packets, then the numbers of jumbo popcorn packets and jumbo chips packets are in the ratio

a)Â 1 : 1

b)Â 8 : 7

c)Â 4 : 3

d)Â 6 : 5

Solution:

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

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

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

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

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

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

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

Question 14:Â Bottle 1 contains a mixture of milk and water in 7: 2 ratio and Bottle 2 contains a mixture of milk and water in 9: 4 ratio. In what ratio of volumes should the liquids in Bottle 1 and Bottle 2 be combined to obtain a mixture of milk and water in 3:1 ratio?

a)Â 27:14

b)Â 27:13

c)Â 27:16

d)Â 27:18

Solution:

The ratio of milk and water in Bottle 1 is 7:2 and the ratio of milk and water in Bottle 2 is 9:4
Therefore, the proportion of milk in Bottle 1 is $\frac{7}{9}$ and the proportion of milk in Bottle 2 is $\frac{9}{13}$

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

Hence, the total volume of milk is $\frac{7X}{9}+\frac{9}{13}$
The total volume of water is $\frac{2X}{9}+\frac{4}{13}$
They are in the ratio 3:1

Hence,Â $\frac{7X}{9}+\frac{9}{13} = 3*(\frac{2X}{9}+\frac{4}{13})$
Therefore, $91X+81=78X+108$

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

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

a)Â 13600

b)Â 13000

c)Â 13400

d)Â 14000

Solution:

Let the total number of shirts be x.
Hence number of non defective shirts = x – 15% of x = 0.85x
Number
of shirts left for export = No of non defective shirts – number of
shirts sold in domestic market
= No of non defective shirts – 20% of No of non defective shirts
= 80% of No of non defective shirts

Hence 8840 = 0.8 * (0.85x) .
Solving for x we get, x = 13000