Test: Ratio & Proportion (February 5) - 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 = 1
• p² + 2p =  p² + 3p - 4
• -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.

Let the incomes be 4x, 5x, 6x and the spending be 6y, 7y, 8y and savings are (4x–6y), (5x–7y) & (6x–8y)
Sheldon saves 1/4^th of his income.

Therefore:

⇒ 4x – 6y = 4x / 4
⇒ 4x – 6y = x
⇒ 3x = 6y
⇒ x / y = 2
∴ y = x / 2

Ratio of Sheldon’s Leonard’s & Howard’s savings:

= 4x – 6y : 5x – 7y : 6x – 8y
= x : 5x – 7y : 6x – 8y
= x : 5x – 7x / 2 : 6x – 8x / 2
= x : 3x / 2 : 2x
= 2 : 3 : 4 

Given:

A : B = 3 : 4

B : C = 5 : 6

C : D = 8 : 9

Sum to divided among them = Rs. 12,384

Concept used:

Ratio Proportion

Calculation:

A : B = 3 : 4 = 15 : 20

B : C = 5 : 6 = 20 : 24

C : D = 8 : 9 = 24 : 27

A : B : C : D = 15 : 20 : 24 : 27

Share of C = 24/(15 + 20 + 24 + 27) × 12384 = Rs. 3456

∴ The share of C is Rs. 3456.

• 
• 1248 + M = 312 x 5
• M = 1560 - 1248 = 312

Let's denote the total weight of persons A, B, and C by S.

When person D joins:

• The new total weight becomes S + D.
• The new average weight becomes (S + D) / 4.
• We are told that this new average is x kg less than the original average (which is S/3). That gives us the equation:

  (S + D) / 4 = (S / 3) - x

When instead person E joins:

• The new total weight becomes S + E.
• The new average weight becomes (S + E) / 4.
• In this case, the average increases by 2x kg compared to the original average, so:

  (S + E) / 4 = (S / 3) + 2x

We are also told that E weighs 12 kg more than D, which means:

  E = D + 12

Step 1. Subtract the two equations

Subtract the equation for D from the equation for E:

  [(S + E) / 4] – [(S + D) / 4] = [(S / 3) + 2x] – [(S / 3) - x]

Simplify the left side:

  (E - D) / 4

Simplify the right side:

  (S / 3 cancels) and we get 2x + x = 3x

So we have:

  (E - D) / 4 = 3x

Multiply both sides by 4:

  E - D = 12x

Step 2. Substitute the known difference

We are given that E - D = 12 kg. So:

  12 = 12x

Divide both sides by 12:

  x = 1

Conclusion

The value of x is 1 kg.

Answer: 1

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

SP = 1.50 Rs. Profit = 25%.
CP = 1.50 × (100/125) = 1.20 Rs.

► If new values of x, y, z are x′, y′ and z′, and respectively then x′ :  y′ = 4 : 5, y′ :  z′ = 3 : 4

⇒ x′ :  y′ :  z′ = 12 : 15 : 20
⇒ x + y + z = 5000
⇒ x′ + 50 + y′ + 100 + z′ + 150 = 5000 x′ + y′ + z′ = 4700
⇒ 12k + 15k + 20k = 4700 k = 100

► x = 1200 + 50 = 1250
► y = 1500 + 100 = 1600 z = 2000 + 150 = 2150
► x + y = 1250 + 1600 = 2850