## The average weight of a group of boys is 30 kg. After a boy of weight 35 kg joins the group, the average weight of the group goes up by 1 kg. Find the number of boys in the group originally?

A. 4
B. 6
C. 7
D. 9

### Solution

Let the number of boys in the group originally be x.
Total weight of the boys = 30x
After the boy weighing 35 kg joins the group, total weight of boys = 30x + 35
So 30x + 35 + 31(x + 1) = > x = 4.

## The ratio of speeds of a car, a train and a bus is 5:9:4. The average speed of the car, the bus and the train is 72 km/hr together. What is the average speed of the car and the train together?

A. 60 km/hr
B. 72 km/hr
C. 84 km/hr
D. 96 km/hr

### Solution

Let the speeds of the car, train, and bus be 5x, 9x and 4x km/hr respectively.
Average speed = 5x + 9x + 4x/3 = 18x /3 = 6x km/hr.
Also, 6x = 72 => x = 12 km/hr

Therefore, the average speed of the car and train together is = 5x+9x/2 = 7x = 7 ×12 = 84 km/hr

## A cricketer has a certain average for 10 innings, In the eleventh inning, he scored 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

### Solution

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.

## Murtaza average expenditure for January to June is Rs. 4200 and he spends Rs.1200 in January and Rs.1500 in July. The average expenditure for the months of February to July is:

A. Rs. 4060
B. Rs. 4250
C. Rs. 4520
D. Rs. 4870

### Solution

Murtaza total expenditure for Jan – June  = 4200 × 6 = 25200
Expenditure for Feb – June = 25200-1200 = 24000
Expenditure for the months of Feb – July = 24000 + 1500 =25500
The average expenditure = 25500 ÷ 6 = 4250

## In a men’s hostel, there were 95 students. To accommodate 30 more students the average is decreased by rupees 5. But total expenditure increased by Rs.500. Find the total expenditure of the hostel now?

A. Rs. 4420
B. Rs. 4680
C. Rs. 4960
D. Rs. 5480

### Solution

Let the expenditure on the 95 students be ‘x’

Then,

95x + 500 = 120(x – 5)
95x + 500 = 120x – 600
120x – 95x = 1100
25x = 1100
=> x = 44

Therefore, total expenditure of the hostel becomes
95 x 44 + 500 = 4680.

## The average age of 7 boys is increased by 1 year when one of them whose age is replaced by a girl. What is the age of the girl?

A. 21 years
B. 27 years
C. 29 years
D. 33 years

### Solution

Let the avg age of 7 boys be ‘p’ years
Let the age of the girl be ‘q’ years
From the given data,
The age of 7 boys = 7p years
Now the new average = (p + 1) when 22 yrs is replaced by q
Now the age of all 7 will become = 7(p + 1) yrs
Hence, 7p – 22 + q = 7(p + 1) yrs
7p – 22 + q = 7p + 7

q = 22 + 7 = 29

Therefore, the age of girl = q = 29 years.

## 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. 12
B. 14
C. 15
D. 16

### Solution

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.

## The average of marks in 3 subjects is 224. The first subject marks are twice the second and the second subject marks is twice the third. Find the second subject marks?

A. 96
B. 192
C. 384
D. 206

### Solution

Let the third subject marks be ‘x’
=> Second subject marks = 2x
=> Third subject marks = 4x
Given avg = 224
x + 2x + 4x = 224 x 3
=> 7x = 224 x 3
=> x = 96
Hence, Second subject marks = 2x = 2 x 96 = 192.

## Out of the five integral numbers C is the average of A and D. B is greater than C and less than D.Also B is the average of A and E . The middle most number in the sequence?

A. A
B. B
C. C
D. D

### Solution

D _  C  _ A          ——–(1)

D > B > C           ——–(2)

from (1) and (2)

D > B > C > A   ———(3)

Again      E _ B _ A

But B > A, from (3)

So          E > D > B > C > A      [ Since B is the average of E and A so it is equidistant from both E and A]

## It costs Rs. p each to send the first thousand messages and Rs. q to send each subsequent one . If r is greater than 1,000, how many Rupees will it cost to send r messages?

A. 1000 (r – p) + pq
B. 1000 p + qr
C. 1000 (r – q) + pr
D. 1000(p – q) + qr

Correct Answer: D. 1000(p – q) + qr

### Solution

We need to find the total cost to send r messages, r > 1000.
The first 1000 messages will cost Rs.p each (Or)
The total cost of first 1000 messages = Rs.1000p
The remaining (r – 1000) messages will cost Rs.q each (Or)
The cost of the (r – 1000) = Rs.(r – 1000)y

Therefore, total cost = 1000p + rq – 1000q

= 1000(p – q) + qr