169+ Problems on Average With Explanation

Q:

The average expenditure of the hotel when there are 10 guests is Rs. 60 per guests and the average expenditure is Rs.40 when there are 20 Guests. If it is known that there are some fixed expenses irrespective of the number of guests then the average expenditure per guest whe there are 40 guests in the hotel.

A) Rs. 30	B) Rs. 25
C) Rs. 20	D) can't be determined

Answer & Explanation Answer: A) Rs. 30

Explanation:

Let the fixed expenditure of the hotel be Rs.x and the variable expenditure ( which is dependent on the guest ) is Rs.y , then

x + 10y = 600 ---------(1)

x + 20y = 800 ----------(2)

From (1) & (2)

10y = 200

=> y = Rs. 20 and x= 400

Hence the total expenditure when there are 40 guests = 400 + 40 x 20 = 1200

Therefore, average expenditure = 1200/40 = Rs. 30

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12 9320

Q:

The average weight of 45 passengers in a bus is 52 kg. 5 of them whose average weight is 48 kg leave the bus and other 5 passengers whose average weight is 54 kg join the bus at the same stop. What is the new average weight of the bus?

A) 54.21 kgs	B) 51.07 kgs
C) 52.66 kgs	D) 53.45 kgs

Answer & Explanation Answer: C) 52.66 kgs

Explanation:

Given total number of passengers in the bus = 45

First average weight of 45 passengers = 52 kgs

Average weight of 5 passengers who leave bus = 48

Average weight of 5 passengers who joined the bus = 54

Therefore, the net average weight of the bus is given by

$= \frac{45 x 52 - (5 x 48) + (5 x 54)}{45} = \frac{2370}{45} = \frac{158}{3} = 52.66 kgs .$

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31 9263

Q:

A cicketer bas a certain average for 10 innings, In the eleventh inning, he scorod 108 runs, thereby increasing his average by 6 runs. His new average is :

A) 44 runs	B) 46 runs
C) 48 runs	D) 49 runs

Answer & Explanation Answer: C) 48 runs

Explanation:

Let average for 10 innings be x. Then,

(10x + 108)/11 = x + 6 => 11x + 66 = 10x + 108 => x = 42.

New average = (x + 6) = 48 runs.

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3 9014

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

Answer & Explanation Answer: 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.

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10 8828

Q:

When all the students in a school are made to stand in rows of 54, 30 such rows are formed. If the students are made to stand in rows of 45, how many such rows will be formed ?

A) 22	B) 36
C) 32	D) 26

Answer & Explanation Answer: B) 36

Explanation:

Total number of students = 54 x 30
When arranged in rows of 45, number of rows formed are,
$\frac{54 x 30}{45}$ = 36.

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9 8256

Q:

The average of a non-zero number and its square is 11 times the number. The number is ?

A) 22	B) 21
C) 11	D) 17

Answer & Explanation Answer: A) 22

Explanation:

Let the number be 'x'.

$\frac{x + x^{2}}{2}$ = 11x
$x^{2}$ = 22x
=> x = 22.

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8 8102

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

Answer & Explanation Answer: 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

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12 8066

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

Answer & Explanation Answer: 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.

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