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Download Mains Level Average Questions with Detailed Explanation for SBI, IBPS and RBI Grade B 2021

Directions : Read the following questions carefully and choose the right answer.
1
The average temperature of Delhi in the first four days of June 2017 was 52 degrees. The average for the second, third, fourth and fifth days was 55 degrees. If the temperatures of the first and fifth days were in the ratio 5 : 7, then what was the temperature of Delhi on 5th June 2017?
» Explain it
D
Sum of temperature on 1st, 2nd, 3rd, and 4th days = (52 × 4) = 208----------(I)
 
Sum of temperature on 2nd, 3rd, 4th, and 5th days = (55 × 4) = 220---------(II)
 
Subtracting equation (I) from (II)
 
Temperature on 5th June 2017 –  temperature on 1st June 2017 = 12 degrees. 
 
Let the temperature on 1st and 5th June 2017 be 5x and 7x degrees respectively. 
 
Then, 7x – 5x = 12
 
2x = 12
 
X = 6
 
Since temperature on the 5th June 2017 in Delhi = 7x = 7 × 6 = 42 degrees. 
 
Hence, option (D) is correct.
2
The average weekly salary per head of all employees (supervisors and labourers) is Rs. 100. The average weekly salary per head of all the supervisors is Rs. 600 while the average weekly salary per head of all the labourers is Rs. 75. Find the number of supervisors in the factory if there are 840 labourers in it.    

A
46
» Explain it
B
Using Allegation method,

Average salary of       Average salary of
Labourers (Rs. 75)       Supervisors (Rs. 600)
  \   /  
    Mean (Rs. 100)    
  /   \  
500       25

Required ratio is 500 : 25 or 20 : 1

No. of Labourers  =  20
No. of Supervisors 1

⇒  840  =  20
No. of supervisors 1

Hence, no. of supervisors = 42

Therefore, option (B) is correct.
3
The average score (runs/match) of Dravid before the start of World Cup was 35. In the world cup of 6 matches, his total score was 355 runs. Find the total number of matches played by him till date, if his new average after the series is 37.5.

A
55
» Explain it
D
Let the total number of matches Dravid has played, till the start of world cup be X
 
So, total runs scored by Dravid till the beginning of the world cup be 35 × X

After the world cup, the new average will be  35 × X + 355
X + 6

According to question,
 
35 × X + 355  = 37.5
X + 6

⇒ 37.5 × X + 225 = 35 × X + 355 
 
⇒ 2.5 × x = 130
 
 ⇒ X = 52
 
Hence, the total number of matches played by Dravid till date = X + 6 = 52 + 6 = 58

Therefore, option (D) is correct.
4
City Pretoria is 7800 km directly to the east of city Canberra . A plane takes off from Canberra at 8:00am one day and lands at Pretoria at 3:00 am the next day. Five hours later it takes off from Pretoria and lands at 2:00 pm the same day at Canberra. There is a steady wind blowing at 25 km/hr from east to west. With respect to the wind, the plane cruises at the same uniform speed in either direction, but the speed with respect to the ground is different. What is the time difference between Canberra and Pretoria?
» Explain it
D
Canberra
8:00 am (let Mon)
2:00 pm (Tue)		 Distance = 7800km Pretoria
3:00 am (Tue)
8:00 am (Tue)

Let the plane speed with respect to the wind be u km/hr
 
The wind speed is 25 km/hr
 
Also, let Pretoria be t hours ahead of Canberra
 
Now, 

7800  = 19 – t ………. (i)
u – 25

7800  = 6 + t ………….(ii)
u + 25
 
Adding (i) and (ii), we get
 
  7800  +  7800  = 25
u – 25 u + 25
⇒ 7800(2u) = 25(u2 – 625)
 
⇒ 312(2u) = (u2 – 625)
 
⇒ u2 – 624u – 625 = 0
 
⇒ (u – 625) (u + 1) = 0
 
As, u > 0 then 
 
u – 625 = 0 ⇒ u = 625
 
Substituting in (i), we get

7800  = 19 – t ⇒ 13 = 19 – t ⇒ t = 6
600

Thus, is the time difference between Canberra and Pretoria is 6 hours.

Hence, option (D) is correct.
5
In a school, there are five friends- A, B, C, D and E. The weight of A is equal to the average weight of B, C and D and the weight of B is equal to the average weight of A, C and D. The average weight of A and C is equal to the average weight of C and D. E is 30 kg heavier than C and the average weight of B and D is 60 kg. What is the average weight (in kgs) of A, B, C, D and E?
» Explain it
D
Let the weight of four friends A, B, C, D and E are a, b, c, d and e kgs respectively

From the question,

3a = b + c + d ----------- (i)

3b = a + c + d ----------(ii)

Subtracting equation (i) and (ii)

3a – 3b = b – a

b = a ---- (iii)

Adding equation (i) and (ii)

3a + 3b = a + b + 2c + 2d

a + b = c + d ---- (iv)

The average weight of A and C is equal to the average weight of C and D

It means, a + c = c + d

a = d

Put a = d in equation (iv)

We get b = c

It means, b = a = c = d

The average weight of B and D is 60 kgs

b + d = 60 × 2 = 120 kgs

So, b = d = 60 kgs = a = c

E is 30 kg heavier than C

e = (60 + 30) kgs

By solving, e = 90 kg

The sum of the weight of A, B, C, D and E = 60 + 60 + 60 + 60 + 90 = 330 kgs

The average =  330  = 66 kgs
5

Hence, option D is correct.