model 9 average on cricket/exam Section-Wise Topic Notes With Detailed Explanation And Example Questions

MOST IMPORTANT quantitative aptitude - 9 EXERCISES

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Questions : The batting average for 40 innings of a cricket player is 50 runs. His highest score exceeds his lowest score by 172 runs. If these two innings are excluded, the average of the remaining 38 innings is 48 runs. The highest score of the player is

(a) 170

(b) 174

(c) 165

(d) 172

The correct answers to the above question in:

Answer: (b)

Let the highest score of cricketer be x runs.

∴ His lowest score = (x – 172) runs

According to the question, 38 × 48 + x + x – 172 = 40 × 50

⇒ 1824 – 172 + 2x = 2000

⇒ 1652 + 2x = 2000

⇒ 2x = 2000 – 1652 = 348

∴ x = 348 2 = 174 runs

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Question : 1

A cricketer, whose bowling average was 12.4 runs/wicket takes 5 wickets for 22 runs in a match, thereby decreases his average by 0.4. The number of wickets, taken by him before this match was :

a) 87

b) 105

c) 78

d) 95

Answer: (d)

Let the number of wickets before the last match be x.

According to the question,

12.4x + 22 = (x + 5 ) × 12

⇒ 12.4x + 22 = 12x + 60

⇒ 12.4x – 12x = 60 – 22

⇒ 0.4x = 38

⇒ x = $38/{0.4}$ =$380/4$ = 95

Question : 2

The average of runs scored by a player in 10 innings is 50. How many runs should he score in the 11th innings so that his average is increased by 2 runs ?

a) 72 runs

b) 54 runs

c) 80 runs

d) 60 runs

Answer: (a)

Let the number of runs scored in 11th innings be x.

∴ 10 × 50 + x = 11 × 52

⇒ 500 + x = 572

⇒ x = 572 – 500 = 72 runs

Aliter : Using Rule 18,

If in the group of N persons, a new person comes at the place of a person of 'T’ years, so that average age,

increases by 't’ years

Then, the age of the new person = T + N.t.

Here, T = 50, N = (10 + 1) = 11, t = 2

Required Runs = T + Nt

= 50 + 11 × 2 = 72

Question : 3

The average run of a player is 32 out of 10 innings. How many runs must he make in the next innings so as to increase his average by 6 ?

a) 40

b) 98

c) 38

d) 6

Answer: (b)

Runs scored in the next innings = x (let)

According to the question,

10 × 32 + x = 11 × 38

⇒ 320 + x = 418

⇒ x = 418 – 320 = 98

Aliter : Using Rule 18,

If in the group of N persons, a new person comes at the place of a person of 'T’ years, so that average age,

increases by 't’ years

Then, the age of the new person = T + N.t.

Here, T = 32, N= (10 + 1) = 11, t = 6

Required Runs = T + Nt

= 32 + 11 × 6

= 32 + 66 = 98

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