Ratio & Proportion Questions for CAT with Answers PDF

Let x be the number of boys and y be the number of girls in Ritu's class.
Based on the chocolate distribution,
3x + 4y = 124
Had she given 2 chocolates to each boy and girl:
2x + 2y = 124 - 52 = 72
Solving both equations, we get
x = 20 and y = 16
Ratio of girls to boys = y/x = 16/20 = 4 : 5

Glasses A and B have same volume, let it be 1 L.
Glass A has 2/3 L of wine and 1/3 L of water.
Glass B has 4/7 L of wine and 3/7 L of water.
Total water in the mixture = (1/3 + 3/7) L = 16/21 L
Total wine in the mixture = (2/3 + 4/7) L = 26/21 L
Ratio = (16/21)/(26/21) = 16/26 = 8/13

Suppose total number of students in a certain dormitory = y
Then, total number of first-year students = y/2
Total number of first-year students who have not declared a major = (2/5)y
So, fraction of first-year students who have declared a major = (1/10)y
Total number of fraction of second-year students who have declared a major = (3/10)y
Total number of fraction of second-year students who have not declared a major = (1/5)y
Then, required fraction = (1/5)

Since he scored 96 marks in section A, he must have attempted 24 questions in that section and all should have been right.
So,
Number of questions attempted in section A = 24
Number of questions attempted in section B = 20
Number of questions attempted in section C = 16
And,
Number of correct attempted questions in section A = 24
Number of correct attempted questions in section B = 18
Number of correct attempted questions in section C = 12
Thus, in section C, he will get:
12 × 4 - 4 = 44 marks

Cost ∝ Weight²
Let original weight = 10x
Then, original cost = 100x²k (where k is proportionality constant)
After breakage weight = 1x + 2x + 3x + 4x
After breakage cost = kx²(1 + 4 + 9 + 16) = 30x²k
Now, (100 - 30)x²k = 70,000
Or, 70x²k = 70,000
Or, x²k = 1,000
Hence, original cost = 100x²k
= 100 × 1,000 = Rs. 1,00,000

Suppose after the reduction of Rs. 25, the shares of A, B and C are x, 3x and 2x, respectively.
Total money = 6x (after reduction)
Initially, share (A) + share (B) + share (C) = 735 … (1)
After reduction,
share (A) - 25 + share (B) - 25 + share (C) - 25 = 6x
⇒ share (A) + share (B) + share (C) - 75 = 6x ... (2)
From (1) and (2), we get
⇒ 735 - 75 = 6x
⇒ x = 660/6 = 110
Now, after deduction, C has, 2x = 2 × 110 = Rs. 220.
Before deduction, C had, 220 + 25 = Rs. 245

Share of men = (12/43) × 430 = $ 120
Share of women = (15/43) × 430 = $ 150
Share of children = (16/43) × 430 = $ 160
Ratio of individual share = 6 : 5 : 4
So, individual shares be 6n, 5n and 4n, respectively.
Number of members = 45


n = 2
So, individual shares are $12, $10 and $8, respectively.

In P, quantity of Supari = 9/13
And quantity of Saunf = 4/13
In Q, quantity of Supari = 9/26
And quantity of Saunf = 17/26
When we add these two mouth fresheners:
Total quantity of Supari = 
Total quantity of Saunf = 
So, ratio of Saunf and Supari in R = 25 : 27

Let B be the cost of a banana and A be the cost of an apple.
Let x be the total cost of 8 bananas and 6 apples.
According to the question,
8B + 6A = x
8.8B + 5.4A = x
⇒ 0.8B = 0.6A
B/A = 3/4

Let G be the number of goals scored and p be the number of international players.
G α p²
∴ G = kp²
When 4 players play, G₁ = k × 4² = 16k
When 6 players play, G₂ = k × 6² = 36k
G₂ - G₁ = 5 goals
∴ 5 = k (36 - 16) or k = ¼
When 8 players play,
G = (1/4) × 8² = 16

Total number of cars = 30
Let the number of cars sold by Rex be z.
Then, number of cars sold by Jagan = 30 - z
Let the price of each car sold by Jagan be Rs. y.
Let the price of each car sold by Rex be Rs. x.
According to the question:
xz = (30 - z)y --- (1)
And x(30 - z) = 32 lakh --- (2)
zy = 24.5 lakh --- (3)

From (2), x =  lakh

From (3), y =  lakh

Putting the values of x and y in (1), we get
32z² = 24.5(30 - z)²

⇒ 32z² = 24.5(900 + z² - 60z)

⇒ 7.5z² = 22,050 - 1470z

⇒ z² + 196z - 2940 = 0

⇒ z = 14, z = -210

Since number of cars sold cannot be negative, z = 14.

⇒ Number of cars sold by Rex = 14

Price of each car sold by Rex =  = Rs. 2 lakh

Force of attraction F is given by: F ∝ 
or, F = K
Here, K is a constant.
When the masses and the distance between them are doubled, the force of attraction F₁ will be
F₁ = K = K
Hence, the force of attraction remains unchanged.

 Now, The Rice Tent Company sold 200 tents, out of which one-quarter were large.
So, they sold 50 large tents and 150 small tents last year.
Let L be the price of a large tent and S be the price of a small tent.
Then, their income from large tents was 50L and from small tents was 150S.
Their total income last year was 50L + 150S.
From the given information:
50L = 
150L = 50L + 150S
100L = 150S

Therefore, the ratio of the price of a large tent to the price of a small tent is 3 : 2.

The merchant had nine score pets at the end of year 2008.
When he added M% of 9 scores and sold N% of the total value in the year 2009:
Number of pets at the end of year 2009 = 
Since the same process is followed for the years 2010, 2011 and 2012;
At the end of the year 2012, number of pets = 
It is given that the merchant had nine score pets at the end of the year 2012.
So,

After simplifying, we get


That is, the ratio M : N must be greater than one.
Hence, option (c) is correct.

Let the amount with Robin initially = 4m
Let the amount with Robert initially = 3m
Now, amount with Robin after he gives Rs.p to Robert = 4m - p
Amount with Robert after the above said transaction = 3m + p
Now, according to question, 4m - p = 3m + p
Or, m = 2p
Now, money with Robin = 4m - p = 4 × 2p - p = 7p
Money with Robert = 3m + p = 3 × 2p + p = 7p
Now, money left with Robert after he spends Rs.q = 7p - q
Now, according to question, 

Or, 21p = 28p - 4q
Or, 4q = 7p
Or,
 
Hence, answer option c is correct.

Ratio of students who passed to those who failed = 5 : 2
Let the number of students who passed be 5x and those who failed be 2x.
Total number of students = 5x + 2x = 7x
When 25 more students appeared, then total students = 7x + 25
When students increased, 5 less passed = (5x - 5)
Now, number of students who passed = 5x - 5
Number of students who failed = (7x + 25) - (5x - 5) = 7x + 25 - 5x + 5 = 2x + 30
New ratio of passed to failed = 3 : 2
According to the question,
  = 
⇒ 2(5x - 5) = 3(2x + 30)
⇒ 10x - 10 = 6x + 90
⇒ 10x - 6x = 90 + 10
4x = 100
x = 25
Hence, number of students who appeared = 7 × 25 = 175

By taking the value of p = 4 from Option (b), the required ratio of 1: 1 is achieved

Total distances covered under each mode = 32, 4 and 12 km respectively.
Total charges = 32 × 24 + 4 × 3 + 12 × 12 = ₹ 924.

(20 + 60 + 30P)/(2 + 3 + P) = 23 → 80 + 30P = 
115 + 23P or P = 5