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61) The average marks obtained by 40 students of a class is 86. If the 5 highest marks are removed, the average reduces by one mark. The average marks of the top 5 students is

(A) 92

(B) 96

(D) 97

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Correct Answer - Option (C)

Explanation :
The average marks obtained by 40 students of a class = 86 If the 5 highest marks are removed, the average reduces by one mark , then Sum of marks of top 5 students = 40 × 86 – 35 × 85 Sum of marks of top 5 students = 3440 – 2975 = 465 ∴ Their average = 465 ÷ 5 = 93

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62) The average marks obtained by 22 candidates in an examination are 45. The average marks of the first 10 candidates are 55 and those of the last eleven are 40. The number of marks obtained by the eleventh candidate is

(A) 45

(B) 0

(D) 47.5

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Correct Answer - Option (B)

Explanation :
Given that , The average marks obtained by 22 candidates in an examination = 45 The average marks of the first 10 candidates = 55 and The average marks of the last eleven = 40 ∴ Marks obtained by eleventh candidate = 22 × 45 – (10 × 55 + 11 × 40) Marks obtained by eleventh candidate = 990 – (550 + 440) Marks obtained by eleventh candidate= 990 – 990 = 0

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63) In an examination, the average of marks was found to be 50. For deducting marks for computational errors, the marks of 100 candidates had to be changed from 90 to 60 each and so the average of marks came down to 45. The total number of candidates, who appeared at the examination, was

(A) 600

(B) 300

(D) 150

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Correct Answer - Option (A)

Explanation :
Here , The average of marks was found to be 50 Let total number of candidates be n. According to question, ∴ 50n – 30 × 100 = 45n ⇒ 50n - 45n = 30 × 100 ⇒ 5n = 3000 ⇒ n = 3000 ÷ 5 = 600

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64) In an exam, the average marks obtained by the students was found to be 60. After omission of computational errors, the average marks of 100 candidates had to be changed from 60 to 30 and the average with respect to all the examinees came down to 45 marks. The total number of candidates who took the exam, was

(A) 200

(B) 210

(D) 180

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Correct Answer - Option (A)

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65) The average of 18 observations is recorded as 124. Later it was found that two observations with values 64 and 28 were entered wrongly as 46 and 82. Find the correct average of the 18 observations.

(A) 111(7/9)

(B) 122

(D) 137(3/9)

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Correct Answer - Option (B)

No Explanation found. Add Explanation and get +2 points.

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66) A batsman in his 12th innings makes a score of 63 runs and there by increases his average scores by 2. What is his average after the 12th innings?

(A) 39

(B) 13

(D) 87

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Correct Answer - Option (A)

Explanation :
A batsman in his 12th innings makes a score = 63 runs And there by increases his average scores = 2 Extra runs = 12 × 2 = 24 ∴ Required average = 63 – 24 = 39

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67) In a 20 over match, the required run rate to win is 7.2. If the run rate is 6 at the end of the 15th over, the required run rate to win the match is

(A) 1.2

(B) 13.2

(D) 12

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Correct Answer - Option (C)

Explanation :
Here , In a 20 over match, the required run rate to win = 7.2 Total runs = 20 × 7.2 = 144 Total runs in 15 overs = 15 × 6 = 90 Runs to be scored in the next 5 overs = 144 – 90 = 54 ∴ Required run-rate = 54 ÷ 5 = 10.8

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68) A cricketer has a mean score of 60 runs in 10 innings. Find out how many runs are to be scored in the eleventh innings to raise the mean score to 62?

(A) 83

(B) 82

(D) 81

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Correct Answer - Option (B)

Explanation :
Given that , The mean score of a cricketer in 10 innings = 60 runs The mean score of a cricketer in eleventh innings = 62 runs Required runs = 60 + 11 × 2 = 82 runs Second method to solve this question with the help of given formula : Here, T = 60, N = (10 + 1) t = 62 – 60 = 2 Required Runs = T + Nt Required Runs = 60 + 11 × 2 = 82

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69) 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) 165 runs

(D) 172 runs

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Correct Answer - Option (C)

Explanation :
Let the highest score be n. ∴ Lowest score = n – 172 According to question , ∴ n + n – 172 = 40 × 50 – 38 × 48 ⇒ 2n – 172 = 2000 – 1824 = 176 ⇒ 2n = 176 + 172 = 348 ∴ n = 348 ÷ 2 = 174

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70) The average of six numbers is 20. If one number is removed, the average becomes 15. What is the number removed ?

(A) 5

(B) 35

(D) 45

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Correct Answer - Option (D)

Explanation :
Given that ,The average of six numbers = 20 ⇒ sum of six numbers = 6 × 20 = 120 The average of 5 numbers = 15 ⇒ sum of six numbers = 5 × 5 = 75 ∴ Required number = sum of six numbers – sum of five numbers Required number = 120 – 75 = 45

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