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

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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 runs

(b) 174 runs

(c) 165 runs

(d) 172 runs

The correct answers to the above question in:

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

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Question : 1

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 : 2

The average of runs of a cricket player of 10 innings was 32. How many runs must he make in his next inning so as to increase his average of runs by 4 ?

a) 70

b) 2

c) 76

d) 4

Answer: (c)

Let the batsman make x runs. Total runs in 10 innings = 10 × 32 = 320

∴ ${320+x}/11$ = 32+4

⇒ 320 + x = 36 × 11

⇒ x = 396 – 320 = 76

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 = 11, t = 4

Required Run = T + Nt

[Here N is taken as (n + 1)]

= 32 + 11 × 4

= 32 + 44 = 76

Question : 3

The bowling average of a cricketer was 12.4. He improves his bowling average by 0.2 points when he takes 5 wickets for 26 runs in his last match. The number of wickets taken by him before the last match was

a) 150

b) 200

c) 125

d) 175

Answer: (d)

Let the number of wickets taken by the cricketer before the last match = x

According to the question,

${12.4x+26}/{x+5}$ =12.2

⇒ 12.4x + 26 = 12.2x + 61

⇒ 0.2x = 61 – 26 = 35

⇒ x = $35/{0.2}$=$350/2$=175

Question : 4

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 : 5

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

Answer: (d)

Question : 6

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

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