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
(C) 10.8
(D) 12
Solution:
According to the question, 20 over match required run rate = 7.2 Total runs are = 7.2 × 20 = 144 runs If the run rate is 6 at the end of the 15th over ∴ Required runs = 144 - 90 = 54 runs Required run rate = = 10.8
122.
The average of 20 numbers is zero. Of them, at the most, how many may be greater than zero?
(A) 0
(B) 1
(C) 10
(D) 19
Solution:
Average of 20 numbers = 0 ∴ Sum of 20 numbers (0 x 20) = 0 It is quite possible that 19 of these numbers may be positive and if their sum is a then 20th number is (-a)
123.
The average of 50 numbers is 30. If two numbers, 35 and 40 are discarded, then the average of the remaining numbers is nearly:
(A) 28.32
(B) 29.68
(C) 28.78
(D) 29.27
Solution:
Total sum of 48 numbers, = (50 × 30) – (35 +40) = 1500 – 75 = 1425 Average = = 29.68
124.
Average of 7 consecutive odd numbers is 31. If the previous and next odd number to these 7 odd numbers are also included, then what will be the new average?
(A) 33
(B) 31
(C) 35
(D) 29
Solution:
∵ Average of 7 consecutive odd numbers = 31 ∴ Mid odd number = fourth term = 31 So, seven consecutive number According to the question, Previous number will be = 25 - 2 = 23 And next odd number will be = 37 + 2 = 39 Average ∴ Average will remain same because 2 subtracted in previous and 2 added in the next number.
125.
The following table shows the number of working hours and the number of employees employed in a small scale industry
No. of working hours
No. of employees
3 - 5
7
5 - 7
10
7 - 9
18
9 - 11
57
11 - 13
14
13 - 15
8
The average number of working hours of an employee is
(A) 8.5
(B) 9.5
(C) 10.5
(D) None of these
Solution:
We have : Mean working hours 4 6 8 10 12 14 No. of employees 7 10 18 57 14 8 Sum of working hours of all the employees = (4 × 7 + 6 × 10 + 8 × 18 + 10 × 57 + 12 × 14 + 14 × 8) = (28 + 60 + 144 + 570 + 168 + 112) = 1082 Total number of employees = (7 + 10 + 18 + 57 + 14 + 8) = 114 ∴ Average number of working hours = = 9.49 9.5
126.
The average of first five multiples of 3 is:
(A) 9
(B) 10
(C) 8
(D) 11
Solution:
First five multiples of three are: 3, 6, 9, 12, 15. Alternatively, Average =
127.
If the arithmetic mean of seventy-five numbers us calculated, it is 35. If each number is increased by 5, then mean of new numbers is:
(A) 30
(B) 40
(C) 70
(D) 90
Solution:
Arithmetic mean of 75 members = 35 Sum of 75 numbers = (75 × 35) = 2625 Total increase = (75 × 5) = 375 Increased sum= (2625 + 375) = 3000 Increased average = = 40
128.
The average of 25 numbers is 64. The average of the first 13 numbers and that of the last 13 numbers are 62.8 and 72.2, respectively. If the 12th number is 61, and if the 12th and 13th numbers are excluded, then what is the average of the
remaining number (correct to one decimal place)?
(A) 59.2
(B) 61.5
(C) 60.2
(D) 62.2
Solution:
Average of 25 numbers = 64 Sum of 25 numbers = 64 × 25 = 1600 Average of first 13 numbers = 62.8 Sum of first 13 numbers = 62.8 × 13 = 816.4 Average of last 13 numbers = 72.2 Sum of last 13 numbers = 72.2 × 13 = 938.6 13th number = 816.4 + 938.4 - 1600 = 1754.8 - 1600 = 154.8 12th number = 61 If we remove 12th and 13th number then, sum of remaining numbers = 1600 - 61 - 154.8 = 1384.2 Average
129.
A batsman in his 12th innings makes a score of 63 runs and there by increases his average score by 2. What is his average after the 12th innings ?
(A) 13
(B) 39
(C) 41
(D) 87
Solution:
Let the average score till his 11 innings = x According to the question, ⇒ = x + 2 ⇒ 11x + 63 = 12x + 24 ⇒ x = 39 12th innings average = 39 + 2 = 41
130.
The average weight of 15 students in a class increases by 1.5 kg when one of the students weighing 40 kg is replaced by a new student. What is the weight (in kg) of the new student ?
(A) 64.5 kg
(B) 56 kg
(C) 60 kg
(D) 62.5 kg
Solution:
Let the weight of 1 student = x kg The weight to 15 student = 15x kg Let the weight of new comer = y kg ∴ According to the question, ⇒ 15x - 40 + y = 15 (x + 1.5) ⇒ 15x - 40 + y = 15x + 22.5 ⇒ y = 62.5 kg