Quant Quiz

Q1)-The average weight of 8 person's increases by 2.5 kg when a new person comes in place of one of them weighing 65 kg. What might be the weight of the new person?

76 kg

76.5 kg

85 kg

Data inadequate

Ans.- C
Total weight increased = (8 x 2.5) kg = 20 kg.
Weight of new person = (65 + 20) kg = 85 kg.

Q2)-The average of 20 numbers is zero. Of them, at the most, how many may be greater than zero?

0

1

10

19

Ans.- D
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).

Q3)-The average age of a class of 32 students is 16 yrs. if the teacher's age is also included, the average increases by one year. Find the age of the teacher

49

48

47

50

Ans.- A
Total age of students is 32X16 = 512 Years
Total age inclusive of teacher = 33X (16+1) = 561
So, Teacher's age is 561-512 = 49 Yrs
There is a shortcut for these type of problems
Teacher's age is 16+(33X1) = 49 Years

Q4)-The average weight of a class of 20 boys was calculated to be 58.4 kgs and it was later found that one weight was misread as 56 kg instead of 65 kg. What is the correct weight?

67.4kg

57.75kg

58.85kg

49.4kg

Ans.- C
Actual total weight is (20X 58.4 - 56 + 65) = 1177 Kgs
Actual average weight is 1177/20 = 58.85kgs

Q5)-The average age of father and his two sons is 27 Years. Five years ago, the average age of the two sons was 12 Years. If the difference between the ages of the two sons is four years, what is the present age of the father?

42

47

48

49

Ans.- B
The total present age of father and two sons is 3S27 = 81 yrs
The total present age of sons is (12+5) X 2 = 34Years
so, present age of father is 81 - 34 = 47 yrs

Q6)-Of three numbers, the third is twice the second and the second is 4 times the first. If their average is 78, the smallest of the three numbers is:

15

21

17

18

Ans.- D
Let first number be x.
So,2nd no. = 4x & 3rd no.=8x.
So,x+4x+8x=78 × 3 = 234.
13x = 234
x = 234/13
Hence,smallest Number x=18.

Q7)-The average age of husband, wife and their child 3 years ago was 27 years and that of wife and the child 5 years ago was 20 years. The present age of the husband is

40

35

45

55

Ans.- A
Sum of the present ages of husband, wife and child = (27 * 3 + 3 * 3) years = 90 years.
Sum of the present ages of wife and child = (20 * 2 + 5 * 2) years = 50 years.
Husband's present age = (90 - 50) years = 40 years

Q8)-Average of 10 matches is 32, How many runs one should should score to increase his average by 4 runs.

70

76

78

80

Ans.- B
Average after 11 innings should be 36
So, Required score = (11 * 36) - (10 * 32)
= 396 - 320 = 76

Q9)-The mean of five numbers is 28. If one of the numbers is excluded, the mean gets reduced by 2. Find the excluded number.

28

34

36

38

Ans.- C
Mean of 5 numbers = 28.
Sum of these 5 numbers = (28 x 5) = 140.
Mean of the remaining 4 numbers = (28 - 2) =26.
Sum of these remaining 4 numbers = (26 × 4) = 104.
Excluded number
= (sum of the given 5 numbers) - (sum of the remaining 4 numbers)
= (140 - 104)
= 36.
Hence, the excluded number is 36.

Q10)-A cricketer has a mean score of 58 runs in nine innings. Find out how many runs are to be scored by him in the tenth innings to raise the mean score to 61.

80

82

86

88

Ans.- D
Mean score of 9 innings = 58 runs.
Total score of 9 innings = (58 x 9) runs = 522 runs.
Required mean score of 10 innings = 61 runs.
Required total score of 10 innings = (61 x 10) runs = 610 runs.
Number of runs to be scored in the 10th innings
= (total score of 10 innings) - (total score of 9 innings)
= (610 -522) = 88.
Hence, the number of runs to be scored in the 10th innings = 88.