Practice Test: Ratio & Proportion- 1 - CAT MCQ

Work done will be directly proportional to number of men and days.
So according to the question:

[(p)(p + 2)] / [(p + 4)(p - 1)] = 1/1 
p² + 2p = p² + 4p - p - 4
p = 4

Given :

The ratio of the income of X and Y is 4 : 3.

The ratio of monthly expenses of X and Y is 3 : 2. 

X and Y save 6000 rupees each month.

Concept used :

Savings = Income - expense

Calculations :

Let the ratio of monthly income of X and Y be 4a and 3a respectively. 

Let the ratio of monthly expenses of X and Y be 3b and 2b respectively. 

Savings of X = 4a - 3b

4a - 3b = 6000      ....(1) 

Savings of Y = 3a - 2b 

3a - 2b = 6000      ....(2) 

Solving equation 1 and 2 

We get a = 6000 and b = 6000

Total monthly income of X and Y = 4a + 3a = 7a 

⇒ 7 × 6000 

⇒ 42000 rupees 

∴ Option 2 is the correct answer.

C₅ – C₁ = 19.
∴ 100x - 81x = 19
​​​​​​​⇒ x = 1

∴ Total profit = 81x + 90x + 72x + 95x + 100x = 438x = 438 crore.

Hence, option (a).

Total number of popcorn packets in stock = total number of chips packets = T

Popcorn:
Large : Super : Jumbo = 7 : 17 : 16
⇒ Jumbo popcorn packets = 16/40T
Chips:
Large : Super : Jumbo = 6 : 15 : 14
⇒ Jumbo popcorn packets = 14/35T
∴ Required ratio = 16/40T : 14/35T = 1 : 1.
​​​​​​​Hence, option (a).

Given:
₹ 785 in the denomination of ₹ 2, ₹ 5 and ₹ 10 coins
The coins are in the ratio of 6 : 9 : 10
Calculation:
Let the number of coins of ₹ 2, ₹ 5 and ₹ 10 be 6x, 9x, and 10x respectively
⇒ (2 × 6x) + (5 × 9x) + (10 × 10x) = 785
⇒ 157x = 785
∴ x = 5
Number of coins of ₹ 5 = 9x = 9 × 5 = 45
∴ 45 coins of ₹ 5 are in the bag

Sita takes 1/3rd of the total mints which implies that the total number of mints in the bowl should be a multiple of 3. None of the options is a multiple of 3.
Alternatively
This problem can be best solved by working backwards.
Number of mints before Eswari = (17 − 2) × 2 = 30
∴ Number of mints before a Fatima 
∴ Number of mints before Sita 
Hence, option (d).

Let x and y be mass of two alloys mixed.
In first alloy:

Gold = x × 1 / (1 + 2) = x/3
Silver = x × 2 / (1 + 2) = 2x/3

In second alloy:

Gold = y × 2 / (2 + 3) = 2y/5
Silver = y × 3 / (2 + 3) = 3y/5

In resulting alloy: 

Gold / Silver = 3 / 5
(x/3+2y/5) / (2x/3+3y/5) = 3 / 5
(x/3+2y/5) × 5 = (2x/3+3y/5) × 3
5x/3 + 2y = 2x + 9y/5
5x/3 - 2x = 9y/5 - 2y
-x/3 = -y/5
x / y = 3 / 5

Therefore, two alloys should be taken in ratio of 3 : 5.

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 D