##
__Average
Aptitude Problems with Solutions: Set 02__

__Average Aptitude Problems with Solutions: Set 02__

__Question No. 01__
A cricketer,
whose bowling average is 12.4, takes 5 wickets for 26 runs and thereby
decreases his average by 0.4. The number of wickets taken by him before his
last match is

(A) 85

(B) 82

(C) 84

(D) 83

Answer:
Option A

__Explanation:__
Let the
number of wickets taken before the last match be

*x*.
Then, (12.4

*x*+ 26)/(*x*+ 5) = 12
=> 12.4

*x*+ 26 = 12*x*+ 60
=> 0.4

*x*= 34
=>

*x*= 34/0.4 = 340/4 = 85

__Question No. 02__
The average
weight of men ‘A’, ‘B’ and ‘C’ is 84 kg. Another man ‘D’ joints the group and
the average now becomes 80 kg. If another man ‘E’, whose weight is 3 kg more
than that of ‘D’, replaces ‘A’, then the average weight of ‘B’ ‘C’ ‘D’ and ‘E’
becomes 79 kg. The weight of ‘A’ is

(A) 70 kg

(B) 72 kg

(C) 75 kg

(D) 80 kg

Answer:
Option C

__Explanation:__
A + b + C =
(84 × 3) = 252 kg

A + b + C +
D = (80 × 4) = 320 kg.

Therefore, D
= (320 - 252) kg = 68 kg.

Given, E =
(68 + 3) kg = 71 kg.

Therefore, B
+ C + D + E = (79 × 4) = 316 kg.

Now, (A + b
+ C + D) - (B + C + D + E) = (320 - 316) kg = 4 kg.

So, A - E =
4 kg.

=> A = 4
+ E = 75 kg.

__Question No. 03__
The average
weight of 16 boys in a class is 50.25 kg and that of the remaining 8 boys is
45.15 kg. Find the average weights of all the boys in the class.

(A) 47.55 kg

(B) 48 kg

(C) 48.55 kg

(D) 49.25 kg

Answer:
Option C

__Explanation:__
Required
average = {(50.25 × 16) + (45.15 × 8)} / (16 + 8)

= (804 +
361.20)/24

= 1165.20/24

= 48.55

__Question No. 04__
The average
age of a class is 15.8 years. The average age of the boys in the class is 16.4
years while that of girls is 15.4 years. What is the ratio of boys to girls in
the class?

(A) 1 : 2

(B) 3 : 4

(C) 3 : 5

(D) 2 : 3

Answer:
Option D

__Explanation:__
Let the
ratio be

*k*: 1
Then, (

*k*× 16.4) + (1 × 15.4) = (*k*+ 1) × 15.8
Or, (16.4 -
15.8)

*k*= (15.8 - 15.4)
Or,

*k*= 0.4/0.6 = 2/3
Therefore,
the required ratio =

^{2}/_{3}: 1 = 2 : 3

__Question No. 05__
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:

(A) 35 years

(B) 40 years

(C) 50 years

(D) None of
these

Answer:
Option B

__Explanation:__
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.

__Question No. 06__
The average
weight of

*A*,*B*and*C*is 45 kg. If the average weight of ‘*A*’ and ‘*B*’ be 40 kg and that of ‘*B*’ and ‘*C*’ be 43 kg, then the weight of ‘*B*’ is:
(A) 17 kg

(B) 20 kg

(C) 26 kg

(D) 31 kg

Answer:
Option D

__Explanation:__
Let A, B, C
represent their respective weights. Then, we have:

*A*+

*B*+

*C*= (45 × 3) = 135 …...... (i)

*A*+

*B*= (40 × 2) = 80 ......... (ii)

*B*+

*C*= (43 × 2) = 86 .........(iii)

Adding (ii)
and (iii), we get:

*A*+ 2*B*+*C*= 166 ........ (iv)
Subtracting
(i) from (iv), we get:

*B*= 31.
∴

*B*'s weight = 31 kg.

__Question No. 07__
Mr. '

*A*' travels to Mr. '*B*' 150 km away at an average speed of 55 km per hour and returns at 45 km per hour. His average speed for the whole journey in km per hour is
(A) 52.5

(B) 48.5

(C) 49.5

(D) 47.5

Answer:
Option C

__Explanation:__
Average
speed = 2

*xy*/(*x*+*y*)
= (2 × 55 × 45)/(55
+ 45) km/hr

= 49.5 km/hr.

__Question No. 08__
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

Answer:
Option D

__Explanation:__
Average of
20 numbers = 0.

∴ Sum
of 20 numbers (0 × 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*).

__Question No. 09__
A pupil's
marks were wrongly entered as 83 instead of 63. Due to that the average marks
for the class got increased by half (1/2). The number of pupils in the class
is:

(A) 10

(B) 20

(C) 40

(D) 73

Answer:
Option C

__Explanation:__
Let there
be

*x*pupils in the class
Total
increase in marks = (

*x*× ½) =*x*/2
∴

*x*/2= (83 - 63)
=>

*x*/2= 20
=>

*x*= 40

__Question No. 10__
A family
consists of two grandparents, two parents and three grandchildren. The average
age of the grandparents is 67 years, that of the parents is 35 years and that
of the grandchildren is 6 years. What is the average age of the family?

(A) 28

^{4}/_{7}
(B) 31

^{5}/_{7}
(C) 32

^{1}/_{7}
(D) None of
these

Answer:
Option B

__Explanation:__
Required
average = {(67 × 2) + (35 × 2) + (6 × 3)} / (2 + 2 + 3)

= (134 + 70 +
18)/7

= 222/7 = 31

^{5}/_{7}years.
## 0 blogger:

## Post a Comment