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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The following question based on average topic of quantitative aptitude

Questions : A batsman has a certain average of runs for 12 innings. In the 13th innings he scores 96 runs thereby increasing his average by 5 runs. What will be his average after 13th innings?

(a) 32

(b) 42

(c) 28

(d) 36

The correct answers to the above question in:

Answer: (d)

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Question : 1

A cricket batsman had a certain average of runs for his 11 innings. In the 12th innings, he made a score of 90 runs and thereby his average of runs was decreased by 5. His average of runs after 12th innings is :

a) 150

b) 140

c) 155

d) 145

Answer: (d)

Let the batsman’s average in 11 innings be x runs.

∴ ${11x + 90}/12$ = x-5

⇒ 11x + 90 = 12x – 60

⇒ x = 150

∴ Required average = 150 – 5 = 145

Question : 2

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 runs

b) 174 runs

c) 165 runs

d) 172 runs

Answer: (b)

Let the highest score be x.

∴ Lowest score = x – 172

∴ x + x – 172 = 40 × 50 – 38 × 48

⇒ 2x – 172 = 2000 – 1824 = 176

⇒ 2x = 176 + 172 = 348

∴ x = $348/2$ = 174

Question : 3

A cricket player after playing 10 tests scored 100 runs in the 11th test. As a result, the average of his runs is increased by 5. The present average of runs is

a) 40

b) 55

c) 45

d) 50

Answer: (d)

If the average in 10 tests be x, then,

x × 10 + 100 = (x + 5) × 11

⇒ 11x – 10x = 100 – 55

⇒ x = 45

∴ Required average = 50

Question : 4

A cricketer whose bowling average is 24.85, runs per wicket, takes 5 wickets for 52 runs and thereby decreases his average by 0.85. The number of wickets taken by him till the last match was :

a) 72

b) 96

c) 64

d) 80

Answer: (d)

Let the no. of wickets taken till the last match be n.

∴ Total runs at 24.85 runs per wicket = 24.85n

Total runs after the current match = 24.85n + 52

Total no. of wickets after the current match = n + 5

Bowling Average after the current match

⇒ ${24.8n+52}/{n+5}$ = 24.85 – 0.85

∴ ${24.8n+52}/{n+5}$ = 24

or 24.85n + 52 = 24n + 120

or 0.85n = 120 – 52

or n = $68/{0.85}$ = 80

Question : 5

The average of runs scored by a cricketer in his 99 innings is 99. How many runs will he have to score in his 100th innings so that his average of runs in 100 innings may be 100?

a) 99

b) 101

c) 100

d) 199

Answer: (d)

Number of runs scored in 100th innings

= 100 × 100 – 99 × 99

= 10000 – 9801 = 199

OR

Increase in average = 1 run

∴ Runs scored in 100th innings = 100 + 99 = 199

Question : 6

A cricketer had a certain average of runs for his 64 innings. In his 65th innings, he is bowled out for no score on his part. This brings down his average by 2 runs. His new average of runs is

a) 128

b) 68

c) 130

d) 70

Answer: (a)

Let the cricketer’s average of runs for his 64 innings be x runs.

∴ Total number of runs in 64 innings = 64x

According to the question,

${64x+0}/65$ = x-2

⇒ 64x = 65x – 130

⇒ x = 130

∴ New average of runs = x – 2

= 130– 2 = 128

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