1. Average score for Virat Kohli in a series of 10 matches is 38.9 runs. If the average for first six matches comes out to be 42 what is his average in the last 4 matches of the series?

A. 34.25 B. 35 C. 33.25 D. 34.25

Answer: Option A
Explanation: Average runs scored by Virat Kohli in 10 matches: Total runs scored/10—(1) Average runs scored by Virat Kohli in 6 matches: (Total runs in 6 matches)/6 => 42 = [latex]\frac{Runs }{6}[/latex] =>252 runs Using (1): =>38.9 = [latex]\frac{Runs }{10}[/latex] =>389 runs Runs scored in rest 4 matches: 389 - 252 runs => 137 runs Average runs scored by Virat Kohli in 4 matches: [latex]\frac{137 }{4}[/latex] = 34.25 runs
2. Average of six numbers comes out to be 3.95. Average of two of the numbers is 3.4 while the average of other two is 3.85. Determine the average of the two numbers left.

A. 4.7 B. 4.8 C. 4.3 D. 4.6

Answer: Option D
Explanation: In the given question we have taken average of 6 numbers taken 2 at a time which makes 3 numbers. => 3.95 (Average of 1st and 2nd number) => 3.85 (Average of 3rd and 4th number) => 3.4 (Average of 5th and 6th number) 3.95 = [latex]\frac{(3.4 + 3.85 + x) }{3}[/latex] x = 4.6
3. Determine a man’s present age if his present age is 125% of his age 10 years ago and 83.333% of his age 10 years from now.

A. 40 years B. 50 years C. 60 years D. 70 years

Answer: Option B
Explanation: According to the question current age of the man is: 1. 125% of his age 10 years ago 2. 83.333% of his age 10 years from now Let the current age of the man be x. Therefore, equating the two equations generated: [latex]\frac{125 }{100(x-10)}[/latex] = [latex]\frac{250 }{3(x+10)}[/latex] x = 50
4. The average runs of a cricket player of 10 innings was 20. How many runs must he make in his next innings so as to increase his average of runs by 4?

A. 12 B. 42 C. 64 D. 76

Answer: Option C
Explanation: Average after 11 innings = 24 Required number of runs = (24 * 11) – (20 * 10) = 264 – 200 = 64
5. Find the average of all the numbers between 11 and 36 which are divisible by 5.

A. 15 B. 20 C. 25 D. 30

Answer: Option C
Explanation: Average = [latex]\frac{(15 + 20 + 25 + 30 + 35) }{5}[/latex] = [latex]\frac{125 }{5}[/latex] = 25

1. The average of five consecutive odd numbers is 51. What is the difference between the highest and lowest number?

A. 3 B. 7 C. 8 D. 11

Answer: Option C
Explanation: Let the numbers be x, x + 2, x + 4, x + 6 and x + 8. Then, [latex]\frac{x + (x + 2) + (x + 4) + (x + 6) + (x + 8) }{5}[/latex] = 51 5x + 20 = 255 x = 47 So, required difference = (47 + 8) – 47 = 8
2. Find the average of first 30 natural numbers.

A. 12 B. 15.5 C. 14.5 D. 16

Answer: Option B
Explanation: Sum of first n natural numbers = [latex]\frac{n(n+1) }{2}[/latex] Hence, sum of first 30 natural numbers = [latex]\frac{30 × 31 }{2}[/latex] = 465 Therefore, required average of = [latex]\frac{465 }{30}[/latex] = 15.5
3. The average weight of three boys P, Q and R is 54 kg, while the average weight of three boys Q, S and T is 60 kg. What is the average weight of P, Q, R, S and T?

A. 66.4 kg B. 63.2 kg C. 58.8 kg D. Data Inadequate

Answer: Option D
Explanation: Total weight of (P + Q + R) = {54*3} kg = 162 kg Total weight of(Q + S + T) = (60 *3) kg = 180 kg Adding both, we get : P + 2Q + S + R + T = (162 + 180) kg = 342 kg So, to find the average weight of P, Q, R, S & T, we ought to know Q's weight, which is not given. The data is inadequate.
4. If the sum of three consecutive even numbers is 44 more than the average of these numbers, then the largest of these numbers is?

A. 20 B. 24 C. 22 D. None of these

Answer: Option B
Explanation: Let the smallest of these number be x. The other two numbers are (x + 2) and (x + 4). x + (x + 2) + (x + 4) = [latex]\frac{(X + (X+2) + (x+4)) }{3}[/latex] + 44 3x + 3*(x + 2) + 3*(x + 4) = x + (x + 2) + (x + 4) + 132 9x + 18 = 3x + 138 6x = 120 x = 20 Therefore, the largest number is 24.
5. If from a group of 5 people, an old member is replaced by a new one, the average age is same as it was 3 years ago. What is the difference between the ages of the old member and the new one?

A. 12 B. 15 C. 11 E. 18

Answer: Option B
Explanation: The present average age is (x + 3) when the old member is present and it equals to x when an old member is replaced by a new one. The difference in the ages of the old member and the new one is equal to the difference in the total age before and after the replacement = 15 years.

1. The batting average of a particular batsman is 60 runs in 46 innings. If the difference in his highest and lowest score is 150 runs and his average excluding these two innings is 58 runs, find his highest score.

A. 179 B. 208 C. 210 D. None of these

Answer: Option A
Explanation: Total runs scored by the batsman = 60*46 = 2760 runs Now excluding the two innings the runs scored = 58*44 = 2552 runs Hence the runs scored in the two innings = 2760 – 2552 = 208 runs. Let the highest score be x, hence the lowest score = x – 150 x + (x - 150) = 208 2x = 358 x = 179 runs
2. Vijay says that his weight is between 75kg and 85 kg. But his younger sister disagrees with Vijay and says that his weight is greater than 70kg but less than 80 kg. Vijay’s mother says that his weight cannot be greater than 78 kg. If all the three predictions are correct, what is the average of different probable weights of Vijay?

A. 75 B. 80 C. 78.5 D. 76.5

Answer: Option D
Explanation: Vijay’s opinion: - 75 < Vijay’s age < 85 Sister’s opinion: - 72 < Vijay’s age < 82 Mother’s opinion: - Vijay’s age < 78 As all the above three conditions are true, the values satisfying them are 76 & 77. Average value = [latex]\frac{(76 + 77)}{2}[/latex] = 76.5
3. The average marks obtained by 100 candidates in an examination is 50. Find how many candidates have passed in the examination, if the average marks of candidates who passed is 70 and that of failed candidates is 20.

A. 90 B. 75 C. 60 D. 58

Answer: Option C
Explanation: Let the number of students who passed the examination be X Number of students failed = (100 - X) Total marks of students who have passed = 70X Total marks of 100 students = 100 * 50 = 5000 Total marks of students who have failed = 20(100 – X) 20(100 – X) + 70X = 5000 2000 – 20 X + 70X = 5000 50 X = 3000 X = 60
4. Three numbers are in the ratio 4 : 5 : 6 and their average is 30 . The largest number is:

A. 28 B. 32 C. 36 D. 42

Answer: Option C
Explanation: Let the numbers be 4x, 5x and 6x. Therefore, [latex]\frac{(4x+5x+6x)}{3}[/latex] = 30 15x = 90 x = 6 Largest number = 6x = 36.
5. A grocer has a sale of Rs. 5420, Rs. 5660, Rs. 6200, Rs. 6350 and Rs. 6500 for 5 consecutive months. Find the sale he should have in the sixth month, so that he gets an average sale of Rs. 6000?

A. Rs. 5870 B. Rs. 5991 C. Rs. 6020 D. Rs. 6850

Answer: Option C
Explanation: Total sale for 5 months = Rs. (5420 + 5660 + 6200 + 6350 + 6500) = Rs. 30,130 Therefore, required sale = Rs. [(6000 * 6) – 30,130] = Rs. (36000 – 30,130) = Rs. 5870