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# Ratio And Proportion Questions For CAT Exam Set-3 PDF

Download Ratio And Proportion Questions For CAT Exam Set-3 PDF. Practice important Ratio And Proportion Questions with detailed answers and explanations. These questions are based on previous CAT question papers.

Question 1:Â A, B and C are 3 varieties of beer which contains starch, hops and water in varying proportions. A contains x% starch and y% hops. B contains y% starch and y% water. C contains x% hops and y% water. When A, B and C are mixed in the ratio 2:3:5, then the final mixture has 38% starch and 28% hops. Find the sum of x and y?

a) 45
b) 50
c) 60
d) 70

Question 2:Â The ratio of a two digit natural number to the number formed by reversing its digit is 1:3, how may such two digit numbers are possible?

a) 2
b) 3
c) 5
d) None of these

Question 3:Â The present ages of Ram and Shyam are in the ratio 6:7. 14 years ago the ratio of the ages of the two was 4:5. What will be the ratio of their ages 21 years from now?

a) 9:10
b) 8:9
c) 7:9
d) 10:11

Question 4:Â Three friends played a card game between themselves and started with money in the ratio 5:3:4. By the end of their game their money was in the ratio 2:4:3, in the same order as before. If the money lost by one of the friends is Rs. 21, then what is the amount of money that friend had in the beginning of the game?

a) Rs. 45
b) Rs. 38
c) Rs. 27
d) cannot be determined

Question 5:Â Ramu is a crooked milkman who mixes water in the milk. He keeps two cans of the mixture with him, one for his premium customers which contains 85% milk and the other for the regular customers which has 70% milk.
It is known that the ratio of the number of premium customers and regular customers is 17: 12. If he sells the milk 10% above the cost price to premium customers and 5% above the cost price to regular customers, then find his overall profit percentage (Each customer buys equal quantity of milk and assume the water to be free)

a) 33%
b) 41%
c) 47%
d) 37%

Given concentration of starch in final mixture is 38%,then
(2x/10) + (3y/10) + (5(100 -(x+y))/10) = 38
3x + 2y = 120 – (1)
Also, given concentration of hops in final mixture is 28%, then
(2y/10) + (3(100-2y)/10) + (5x/10) = 28
-5x + 4y = 20 – (2)
Solving (1) and (2) we get x = 20 and y = 30
âˆ´ x + y = 50

Let the two digit number be ab.
==> (10a+b)/(10b+a) = 1/3 ==> 30a+3b = 10b+a ==> 29a = 7b
Since 29 and 7 are prime numbers, â€˜aâ€™ should be a multiple of 7 and b should be a multiple of 29.
But a and b are single digit natural numbers.
Therefore no such number exists.

Let the present age of Ram be 6x and that of Shyam be 7x.
So $\frac{6x-14}{7x-14} = \frac{4}{5}$
=> 30x – 70 = 28x – 56
=> 2x = 14 => x =7
Thus the present ages of the two are 42 and 49 respectively.
20 years from now, the ratio of ages of the two will be 63 and 70 respectively.
So the required ratio will be 9:10

Let the total amount of money with the 3 friends initially be k. Then the amount of money with each of the 3 friends initially is 5k/12, 3k/12 and 4k/12 and that with the friends in the same order after the game finished is 2k/9, 4k/9 and 3k/9.
The person who had 4k/12 finished with 3k/9 at the end i.e. he neither won or lost his amount. The person with 5k/12 finished with 2k/9 which means he lost money. Thus, the third friend would definitely have won money from the game.
Therefore, the money lost by the person with 5k/12 initially = 5k/12 – 2k/9 = 21
On solving we get k = 108
Thus, the initial amount of money the person who lost Rs.21 had = 5k/12 = Rs. 45

Profit percentage = (.25*17y + .35*12y)*100/(.85y*17+.70y*12)Â = $\frac{4.25y + 4.2y}{14.45y + 8.4y}\times100$
=> $\frac{8.45y}{22.85y}\times100$ = 36.97% ~ 37%