# QUESTIONS ON AVERAGE (PART-2)

## QUESTIONS ON AVERAGE (PART-2)

#### QUERY 11

**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 before the last match was?**

A) 64

B) 72

C) 80

D) 96

**MAHA GUPTA**

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

So, his runs scored when his number of wickets is x = 24.85x

His number of wickets after = x + 5

And runs scored finally = 24.85x + 52

His final average per wicket after those last 5 wickets = 24.85 – 0.85 = 24

Therefore, 24.85x + 52 = (x +5)*24

By solving

x = 80 (option ‘C’)

TRICK

Take answer options one by one and check when the given conditions satisfy. For example we take 80 first as it’s a multiple of 10 and calculations therefore are easier.

His runs before the last match = 80*24.85 = 1988

His total runs including those 5 wickets = 1988 + 52 = 2040 —- (i)

His total wickets = 80 + 5 = 85

His average after = 24.85 – 0.85 = 24

His total runs including those 5 wickets = 85*24 = 2040 —- (ii)

We see figures at (i) and (ii) are equal, hence our correct answer option.

#### QUERY 12

**The batting average of a cricketer is 50 runs per innings. In the next 5 innings, he scores 400 runs and thereby increases his batting average by 5 runs. What is total number of innings played by him till the last match?**

A) 25

B) 30

C) 40

D) 50

**MAHA GUPTA**

Let the number of innings played by him till the last match = x

Hence, his number of innings before = x – 5

So, his runs scored when his number of total innings is x = (x – 5)*50

And runs scored finally = (x – 5)*50 + 400

His final average per innings after those last 5 innings = 50 + 5 = 55

Thus his runs = 55x

Therefore, (x – 5)*50 + 400 = 55x

By solving

x = 30 (option ‘B’)

ANOTHER METHOD

Average runs in the last 5 innings = 400/5 = 80 runs

We see this average is more than the old average by 80 – 50 = 30

Hence, his excess runs in the last 5 innings = 30*5 = 150

Excess of average after those 5 innings = 5 (given)

Therefore, number of total innings played by him till the last match = 150/5 = 30 (option ‘B’)

ALLEGATION METHOD

Average runs of the last 5 innings = 400/5 = 80 runs

Old average = 50 runs

New average = 50 + 5 = 55 runs

**OLD AVERAGE AVERAGE OF LAST 5 INNINGS**

50 | 80 | |||

55 | ||||

55 – 80 = 25 | 55 – 50 = 5 |

We see INNINGS PLAYED BEFORE : INNINGS PLAYED AFTER = 25 : 5 = 5 : 1

Innings played after = 1 ratio = 1*5 = 5

Hence, innings played before = 5*5 = 25

Therefore, number of total innings played by him till the last match = 25+ 5 = 30 (option ‘B’)

TRICK

Take answer options one by one and check when the given conditions satisfy. For example we take 25 first.

His runs including those last 5 innings = (25 – 5)*50 + 400 = 1400 —- (i)

His final average = 50 + 5 = 55

Hence, his runs including those last 5 innings = 25*55 = 1375 —- (ii)

We see figures at (i) and (ii) are not equal, hence our incorrect answer option. Likewise we can check with other answer options.

#### QUERY 13

**The batting average of a cricketer is 72 runs per innings. In the next 4 innings, he scores 80 runs and thereby decreases his batting average by 2 runs. What is total number of innings played by him till the last match?**

A) 72

B) 88

C) 92

D) 104

**MAHA GUPTA**

Let the number of innings played by him till the last match = x

Hence, his number of innings before = x – 4

So, his runs scored when his number of total innings is x = (x – 4)*72

And runs scored finally = (x – 4)*72 + 80

His final average per innings after those last 5 innings = 72 – 2 = 70

Thus his runs = 70x

Therefore, (x – 4)*72 + 80 = 70x

By solving

x = 104 (option ‘D’)

ANOTHER METHOD

Average runs in the last 4 innings = 80/4 = 20 runs

We see this average is less than the old average by 72 – 20= 52

Hence, his short runs in the last 4 innings = 52*4 = 208

Short of average after those 4 innings = 2 (given)

Therefore, number of total innings played by him till the last match = 208/2 = 104 (option ‘D’)

ALLEGATION METHOD

Average runs of the last 4 innings = 80/4 = 20 runs

Old average = 72 runs

New average = 72 – 2 = 70 runs

**OLD AVERAGE AVERAGE OF LAST 4 INNINGS**

72 | 20 | |||

70 | ||||

70 – 20 = 50 | 70 – 72 = 2 |

We see INNINGS PLAYED BEFORE : INNINGS PLAYED AFTER = 50 : 2 = 25 : 1

Innings played after = 1 ratio = 1*4 = 4

Hence, innings played before = 25*4 = 100

Therefore, number of total innings played by him till the last match = 100 + 4 = 104 (option ‘D’)

TRICK

Take answer options one by one and check when the given conditions satisfy. For example we take 104 first.

His runs including those last 4 innings = (104 – 4)*72 + 80 = 7280 —- (i)

His final average = 72 – 2 = 70

Hence, his runs including those last 4 innings = 104*70 = 7280 —- (ii)

We see figures at (i) and (ii) are equal, hence our correct answer option.

#### QUERY 14

**A cricket player has an average score of 30 runs for 42 innings played by him. In an innings his highest score exceeds his lowest score by 100 runs. If these 2 innings are excluded, his average of the remaining 40 innings is 28 runs. His highest score in an innings is?**

A) 125

B) 120

C) 110

D) 100

**THE LOST SOUL ……**

Total score of 42 innings = 30 x 42 = 1260

Score of 40 innings 40 x 28 = 1120

Therefore the difference = 140

But his highest score exceeds his lowest score by 100

Thus his highest will be 120 and the lowest as 20 (option ‘B’)

Alternate Method

**RONNIE BANSAL
**His score of the two innings that were excluded = 2 x the old average + reduction in the average x remaining innings

= (2 x 30) + (2 x 40)

= 60 + 80 = 140

But his highest score exceeds his lowest score by 100

Thus his highest will be 120 and the lowest as 20 (option ‘B’)

#### QUERY 15

**A batsman has certain average runs for his 11 innings. In his 12th innings he scores zero runs. Thereby his average decreases by 3 runs. Find his average after his 12th innings.**

A) 33

B) 30

C) 32

D) 45

**MAHA GUPTA**

Let his average after 12th innings = x

Therefore his average after 11th innings = x + 3

And his total score after 11th innings = 11(x +3)

Hence his score after 12th innings = His total score of 11 innings + his score of 12th innings = 11(x +3) + 0

But his total score after 12th innings also = 12*x = 12x

Hence, 11(x +3) + 0 = 12x

=> x = 33

So, his average after his 12th innings = 33 (option ‘A’)