Average
- 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
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Let the number of wickets taken by the cricketer before the last match = p
According to the question,12.4p + 26 = 12.2 p + 5
⇒ 12.4p + 26 = 12.2p + 61
⇒ 0.2p = 61 – 26 = 35
Correct Option: C
Let the number of wickets taken by the cricketer before the last match = p
According to the question,12.4p + 26 = 12.2 p + 5
⇒ 12.4p + 26 = 12.2p + 61
⇒ 0.2p = 61 – 26 = 35
.⇒ p = 35 = 350 = 175 0.2 2
- 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 :
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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⇒ 24.85n + 52 = 24.85 - 0.85 n + 5 ∴ 24.85n + 52 = 24 n + 5
Correct Option: C
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⇒ 24.85n + 52 = 24.85 - 0.85 n + 5 ∴ 24.85n + 52 = 24 n + 5
⇒ 24.85n + 52 = 24n + 120
⇒ 0.85n = 120 – 52⇒ n = 68 = 80 0.85
- The average of 100 items was found to be 30. If at the time of calculation, two items were wrongly taken as 32 and 12 instead of 23 and 11, then the correct average is :
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Given that , The average of 100 items = 30
If at the time of calculation, two items were wrongly taken as 32 and 12 instead of 23 and 11 , then
Correct sum of 100 items = 30 × 100 – 32 – 12 + 23 + 11Correct Option: C
Given that , The average of 100 items = 30
If at the time of calculation, two items were wrongly taken as 32 and 12 instead of 23 and 11 , then
Correct sum of 100 items = 30 × 100 – 32 – 12 + 23 + 11
Correct sum of 100 items = 3000 – 44 + 34 = 2990
∴ Required average = 2990 ÷ 100 = 29.9
- The average marks secured by 36 students was 52. But it was discovered that an item 64 was misread as 46. What is the correct mean of marks ?
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Here ,The average marks secured by 36 students = 52
Difference of correct and incorrect marks = 64 – 46 = 18∴ Correct mean = 52 + 18 36
Correct mean = 52.5
Second method to solve this question with the help of given formula :
Here, n = 36, m = 52 , a = 64, b = 46New mean = m + (a - b) n New mean = 52 + (64 - 46) 36
Correct Option: D
Here ,The average marks secured by 36 students = 52
Difference of correct and incorrect marks = 64 – 46 = 18∴ Correct mean = 52 + 18 36
Correct mean = 52.5
Second method to solve this question with the help of given formula :
Here, n = 36, m = 52 , a = 64, b = 46New mean = m + (a - b) n New mean = 52 + (64 - 46) 36 New mean = 52 + 18 36 New mean = 52 + 1 = 52.5 2
- The average of 50 numbers is 38. If two numbers, namely 45 and 55 are discarded, the average of the remaining numbers is
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Here , The average of 50 numbers = 38
Sum of 50 numbers = 50 × 38 = 1900
If two numbers, namely 45 and 55 are discarded , then Sum of 48 numbers = 1900 – 45 – 55 = 1800Correct Option: A
Here , The average of 50 numbers = 38
Sum of 50 numbers = 50 × 38 = 1900
If two numbers, namely 45 and 55 are discarded , then Sum of 48 numbers = 1900 – 45 – 55 = 1800
∴ Required average = 1800 ÷ 48 = 37.5
