2
Q:

# 8 years ago there were 5 members in the Arthur's family and then the average age of the family  was 36 years. Mean while Arthur got married and gave birth to a child. Still the average age of his family is same now. What is the age of his wife at the time of his child's birth was.If the difference between the age of her  child and herself was 26 years.

 A) 25 years B) 26 years C) 20 years D) can't be determined

Explanation:

Since we know that the difference b/w the age of any two persons remains always constant, while the ratio of their ages gets changed as the time changes.

so, if the age of his child be x (presently)

Then the age of  wife be x + 26 (presently)

Thus the total age = x + ( x + 26) = 32  [$\inline&space;\because$ 252-220 =32]

$\inline&space;\Rightarrow$ x = 3

$\inline&space;\therefore$ The age of her child is 3 years and her self  is 29 years. Hence her age at the time of the birth of her child was 26 years.

Q:

On a school’s Annual day sweets were to be equally distributed amongst 112 children. But on that particular day, 32 children were absent. Thus the remaining children got 6 extra sweets. How many sweets was each child originally supposed to get ?

 A) 14 B) 16 C) 12 D) 15

Explanation:

Let 'K' be the total number of sweets.
Given total number of students = 112
If sweets are distributed among 112 children,
Let number of sweets each student gets = 'L'

=> K/112 = L ....(1)
But on that day students absent = 32 => remaining = 112 - 32 = 80
Then, each student gets '6' sweets extra.

=> K/80 = L + 6 ....(2)
from (1) K = 112L substitute in (2), we get
112L = 80L + 480
32L = 480
L = 15

Therefore, 15 sweets were each student originally supposed to get.

3 15
Q:

A batsman scores 26 runs and increases his average from 14 to 15. Find the runs to be made if he wants top increasing the average to 19 in the same match ?

 A) 74 B) 79 C) 72 D) 60

Explanation:

Number of runs scored more to increse the ratio by 1 is 26 - 14 = 12
To raise the average by one (from 14 to 15), he scored 12 more than the existing average.
Therefore, to raise the average by five (from 14 to 19), he should score 12 x 5 = 60 more than the existing average. Thus he should score 14 + 60 = 74.

4 56
Q:

There are 44 students in a hostel, due to the administration,  15 new students has joined. The expense of the mess increase by Rs. 33 per day. While the average expenditure per head diminished by Rs. 3, what was the original expenditure of the mess ?

 A) Rs. 404 B) Rs. 514 C) Rs. 340 D) Rs. 616

Explanation:

Let the average expenditure per head be Rs. p
Now, the expenditure of the mess for old students is Rs. 44p
After joining of 15 more students, the average expenditure per head is decreased by Rs. 3 => p-3
Here, given the expenditure of the mess for (44+15 = 59) students is increased by Rs. 33
Therefore, 59(p-3) = 44p + 33
59p - 177 = 44p + 33
15p = 210
=> p = 14
Thus, the expenditure of the mess for old students is Rs. 44p = 44 x 14 = Rs. 616.

3 17
Q:

Sripad has scored average of 65 marks in three objects. In no subjects has he secured less than 58 marks. He has secured more marks in Maths than other two subjects. What could be his maximum score in Maths ?

 A) 79 B) 77 C) 76 D) 73

Explanation:

Assuming Sripad has scored the least marks in subject other than science,
Then the marks he could secure in other two are 58 each.
Since the average mark of all the 3 subject is 65.
i.e (58+58+x)/3 = 65
116 + x = 195
x = 79 marks.

Therefore, the maximum marks he can score in maths is 79.

4 17
Q:

Three maths classes: A, B and C take an algebra test. The average score of class A is 83. The average score of class B is 76. The average score of class C is 85. The average score of class A and B is 79 and average score of class B and C is 81. What is the average score of classes A, B, C ?

 A) 81 B) 78 C) 80.5 D) 81.5

Explanation:

Let the number of students in classes A, B and C be P, Q and R respectively.
Then, total score of A = 83P, total score of B = 76Q, total score of C = 85R.
Also given that,
(83P + 76Q) / (P + Q) = 79
=>4P = 3Q.
(76Q + 85R)/(Q + R) = 81
=>4R = 5Q,
=>Q = 4P/3 and R = 5P/3
Therefore, average score of A, B, C = ( 83P + 76Q + 85R ) / (P + Q + R) = 978/12 = 81.5