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’)