Averages, Mixtures and Alligation — Practice

Question 1

Find the average of 10, 20, 30 and 40.
(a) 25 (b) 30 (c) 20 (d) 35

Accepted Answer

(a) 25.

Average = (10 + 20 + 30 + 40) ÷ 4 = 100 ÷ 4 = 25.

Question 2

What is the average of the first five natural numbers (1, 2, 3, 4, 5)?
(a) 3 (b) 2.5 (c) 3.5 (d) 4

Accepted Answer

(a) 3.

Sum = 1 + 2 + 3 + 4 + 5 = 15; average = 15 ÷ 5 = 3.

Question 3

The average of 5 numbers is 20. If the number 30 is removed, what is the average of the remaining 4 numbers?
(a) 17.5 (b) 20 (c) 18 (d) 16

Accepted Answer

(a) 17.5.

Sum = 20 × 5 = 100. After removing 30, sum = 70; new average = 70 ÷ 4 = 17.5.

Question 4

The average age of 4 people is 25 years. A fifth person, aged 35, joins them. What is the new average age?
(a) 27 (b) 26 (c) 28 (d) 30

Accepted Answer

(a) 27.

Sum = 25 × 4 = 100. New sum = 100 + 35 = 135; new average = 135 ÷ 5 = 27 years.

Question 5

A shopkeeper mixes 3 kg of rice costing ₹40 per kg with 2 kg costing ₹50 per kg. What is the average cost per kg of the mixture?
(a) ₹44 (b) ₹45 (c) ₹46 (d) ₹43

Accepted Answer

(a) ₹44.

Weighted average = (3 × 40 + 2 × 50) ÷ (3 + 2) = (120 + 100) ÷ 5 = 220 ÷ 5 = ₹44. It is below the simple mean ₹45 because the cheaper rice weighs more.

Question 6

A 40-litre mixture contains milk and water in the ratio 5 : 3. How much water does it contain?
(a) 15 litres (b) 12 litres (c) 25 litres (d) 8 litres

Accepted Answer

(a) 15 litres.

Total parts = 5 + 3 = 8. Water = (3/8) × 40 = 15 litres (milk = 25 litres).

Question 7

In what ratio must rice at ₹30 per kg be mixed with rice at ₹45 per kg to obtain a mixture worth ₹40 per kg?
(a) 1 : 2 (b) 2 : 1 (c) 1 : 3 (d) 3 : 2

Accepted Answer

(a) 1 : 2.

By alligation, cheaper : dearer = (45 − 40) : (40 − 30) = 5 : 10 = 1 : 2. So the ₹30 rice and ₹45 rice are mixed in the ratio 1 : 2.

Question 8

The average of 11 results is 50. The average of the first 6 is 49 and that of the last 6 is 52. What is the 6th result?
(a) 56 (b) 50 (c) 54 (d) 48

Accepted Answer

(a) 56.

Total of 11 = 50 × 11 = 550. First 6 = 49 × 6 = 294; last 6 = 52 × 6 = 312. Their sum, 294 + 312 = 606, counts the 6th result twice, so the 6th result = 606 − 550 = 56.

Practice — Averages, Mixtures and Alligation

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