Problems on Average - Solved Examples

Problems on Average - Solved Examples

1. In the first 10 overs of a cricket game, the run rate was only 3.2. What should be the run rate in the remaining 40 overs to reach the target of 282 runs?
A. 6.25	B. 5.5
C. 7.4	D. 5

answer with explanation

Answer: Option A
Explanation:
Runs scored in the first 10 overs = 10 × 3.2 = 32
Total runs = 282

Remaining runs to be scored = 282 - 32 = 250
Remaining overs = 40

Run rate needed = $\frac{250}{40}=6.25$

2. A grocer has a sale of Rs. 6435, Rs. 6927, Rs. 6855, Rs. 7230 and Rs. 6562 for 5 consecutive months. How much sale must he have in the sixth month so that he gets an average sale of Rs. 6500?
A. 4800	B. 4991
C. 5004	D. 5000

answer with explanation

Answer: Option B
Explanation:
Let the sale in the sixth month $=x$

Then $\frac{6435+6927+6855+7230+6562+x}{6}$ $=6500$

=> $6435+6927+6855+7230+6562+x$ $=6\times 6500$

=> $34009+x=39000$
=> $x=39000-34009=4991$

3. The average of 20 numbers is zero. Of them, How many of them may be greater than zero, at the most?
A. 1	B. 20
C. 0	D. 19

answer with explanation

Answer: Option D
Explanation:
Average of 20 numbers = 0

=> $\frac{\text{Sum of 20 numbers}}{20}=0$

=> Sum of 20 numbers = 0

Hence at the most, there can be 19 positive numbers.
(Such that if the sum of these 19 positive numbers is x, 20th number will be -x)

4. The captain of a cricket team of 11 members is 26 years old and the wicket keeper is 3 years older. If the ages of these two are excluded, the average age of the remaining players is one year less than the average age of the whole team. Find out the average age of the team.
A. 23 years	B. 20 years
C. 24 years	D. 21 years

answer with explanation

Answer: Option A
Explanation:
Number of members in the team = 11

Let the average age of of the team = $x$

=> $\frac{\text{Sum of ages of all 11 members}}{11}=x$

=> Sum of the ages of all 11 members = $11x$

Age of the captain = 26
Age of the wicket keeper = 26 + 3 = 29

Sum of the ages of 9 members of the team excluding captain and wicket keeper
$=11x-26-29=11x-55$

Average age of 9 members of the team excluding captain and wicket keeper
$=\frac{11x-55}{9}$

Given that $\frac{11x-55}{9}=(x-1)$

$\Rightarrow 11x-55=9(x-1)\Rightarrow 11x-55=9x-9\Rightarrow 2x=46\Rightarrow x=\frac{46}{2}=23\text{years}$

5. The average monthly income of A and B is Rs. 5050. The average monthly income of B and C is Rs. 6250 and the average monthly income of A and C is Rs. 5200. What is the monthly income of A?
A. 2000	B. 3000
C. 4000	D. 5000

answer with explanation

Answer: Option C
Explanation:
Let monthly income of A = a
monthly income of B = b
monthly income of C = c

a + b = 2 × 5050 .... (Equation 1)
b + c = 2 × 6250 .... (Equation 2)
a + c = 2 × 5200 .... (Equation 3)

(Equation 1) + (Equation 3) - (Equation 2)
=> a + b + a + c - (b + c) = (2 × 5050) + (2 × 5200) - (2 × 6250)
=> 2a = 2(5050 + 5200 - 6250)
=> a = 4000

i.e., Monthly income of A = 4000

6. A car owner buys diesel at Rs.7.50, Rs. 8 and Rs. 8.50 per litre for three successive years. What approximately is the average cost per litre of diesel if he spends Rs. 4000 each year?
A. Rs. 8	B. Rs. 7.98
C. Rs. 6.2	D. Rs. 8.1

answer with explanation

Answer: Option B
Explanation:
Total Cost = 4000 × 3

Total diesel used = $\frac{4000}{7.5}+\frac{4000}{8}+\frac{4000}{8.5}$

Average cost per litre of diesel
$=\frac{4000\times 3}{(\frac{4000}{7.5}+\frac{4000}{8}+\frac{4000}{8.5})}=\frac{3}{(\frac{1}{7.5}+\frac{1}{8}+\frac{1}{8.5})}$

It is important how you proceed from this stage. Remember time is very important and if we solve this in the normal method, it may take lot of time. Instead, we can find out the approximate value easily and select the right answer from the given choices.

In this case, answer
$=\frac{3}{(\frac{1}{7.5}+\frac{1}{8}+\frac{1}{8.5})}\approx \frac{3}{(\frac{1}{8}+\frac{1}{8}+\frac{1}{8})}\approx \frac{3}{\left(\frac{3}{8}\right)}\approx 8$

We got that answer as approximately equal to 8. From the given choices, the answer can be 8 or 7.98 or 8.1 . But which one from these?

It is easy to figure out. We approximated the denominator,$(\frac{1}{7.5}+\frac{1}{8}+\frac{1}{8.5})$ to $\frac{3}{8}$. However
$\frac{1}{7.5}+\frac{1}{8.5}=\frac{1}{8-0.5}+\frac{1}{8+0.5}=\frac{8+0.5+8-0.5}{(8-0.5)(8+0.5)}=\frac{16}{(8^2-{0.5}^2)}[\because (a-b\left)\right(a+b)=a^2-b^2]=\frac{16}{64-0.25}$

i.e., $\frac{1}{7.5}+\frac{1}{8.5}=\frac{16}{(64-0.25)}$
We know that $\frac{1}{8}+\frac{1}{8}=\frac{1}{4}=\frac{16}{64}$

=> $\frac{1}{7.5}+\frac{1}{8.5}>\frac{1}{8}+\frac{1}{8}$

Early we had approximated the denominator to $\frac{3}{8}$. However, from the above mentioned equations, now we know that actual denominator is slightly greater than $\frac{3}{8}$. It means answer is slightly lesser than 8. Hence we can pick the choice 7.98 as the answer

Try to remember such relations between numbers which can save lot of time in calculations.

7. In Kiran's opinion, his weight is greater than 65 kg but less than 72 kg. His brother does not agree with Kiran and he thinks that Kiran's weight is greater than 60 kg but less than 70 kg. His mother's view is that his weight cannot be greater than 68 kg. If all are them are correct in their estimation, what is the average of different probable weights of Kiran?
A. 70 kg	B. 69 kg
C. 61 kg	D. 67 kg

answer with explanation

Answer: Option D
Explanation:
Let Kiran's weight = x. Then

According to Kiran, 65 < x < 72 ----(equation 1)

According to brother, 60 < x < 70 ----(equation 2)

According to mother, x $\le$ 68 ----(equation 3)

Given that equation 1,equation 2 and equation 3 are correct. By combining these equations,we can write as

$65<x\le 68$

i.e., x = 66 or 67 or 68

Average of different probable weights of Kiran = $\frac{66+67+68}{3}=67$

8. 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. 48.55	B. 42.25
C. 50	D. 51.25

answer with explanation

Answer: Option A
Explanation:
Average weight of 16 boys = 50.25
Total weight of 16 boys = 50.25 × 16

Average weight of remaining 8 boys = 45.15
Total weight of remaining 8 boys = 45.15 × 8

Total weight of all boys in the class = (50.25 × 16)+ (45.15 × 8)
Total boys = 16 + 8 = 24

Average weight of all the boys = $\frac{(50.25\times 16)+(45.15\times 8)}{24}$
$=\frac{(50.25\times 2)+(45.15\times 1)}{3}=(16.75\times 2)+15.05=33.5+15.05=48.55$

9. A library has an average of 510 visitors on Sundays and 240 on other days. What is the average number of visitors per day in a month of 30 days beginning with a Sunday?
A. 290	B. 304
C. 285	D. 270

answer with explanation

Answer: Option C
Explanation:
In a month of 30 days beginning with a Sunday, there will be 4 complete weeks and another two days which will be Sunday and Monday.

Hence there will be 5 Sundays and 25 other days in a month of 30 days beginning with a Sunday

Average visitors on Sundays = 510
Total visitors of 5 Sundays = 510 × 5

Average visitors on other days = 240
Total visitors of other 25 days = 240 × 25

Total visitors = (510 × 5) + (240 × 25)
Total days = 30

Average number of visitors per day
$=\frac{(510\times 5)+(240\times 25)}{30}=\frac{(51\times 5)+(24\times 25)}{3}=(17\times 5)+(8\times 25)=85+200=285$

10. A student's mark was wrongly entered as 83 instead of 63. Due to that the average marks for the class got increased by $\frac{1}{2}$. What is the number of students in the class?
A. 45	B. 40
C. 35	D. 30

answer with explanation

Answer: Option B
Explanation:
Let the total number of students = $x$

The average marks increased by $\frac{1}{2}$ due to an increase of 83 - 63 = 20 marks.

But total increase in the marks = $\frac{1}{2}\times x=\frac{x}{2}$

Hence we can write as
$\frac{x}{2}=20\Rightarrow x=20\times 2=40$