Average MCQ Questions

Share on facebook
Facebook
Share on twitter
Twitter
Share on telegram
Telegram
Share on whatsapp
WhatsApp
Share on pinterest
Pinterest
Share on reddit
Reddit
Share on tumblr
Tumblr

Practice Average MCQ Questions with Details Solution

 

Q11. A businessman purchased 30 TV sets in Rs. 300000. If the average cost of 15 TV sets is Rs. 9000, then find out the average cost of remaining TV sets.

(a) Rs. 20000

(b) Rs. 12000

(c) Rs. 11000

(d) Rs. 10000

(e) None of the above

Answer: (c) Rs. 11000

Solution: Total cost of 30 TV sets = 300000

Total cost of 15 TV sets = 9000 x 15 = 135000

Total cost of remaining 15 TV sets = 300000 – 135000 = 165000

.’. Average cost of remaining TV sets = $\frac{165000}{15}=11000$

Using Trick: Average cost of remaining TV sets = $\frac{300000-15\times 9000}{30-15}=11000$

Trick: If the average of m numbers is x and out of these ‘m’ numbers the average of n numbers is y, then the average of remaining numbers will be

Average (A) =  $\frac{mx-ny}{m-n}$

Q12. The average age of 30 girls is 13 yr. The average of first 18 girls is 15 yr. Find out the average age of remaining 12 girls.

(a) 12 yr

(b) 10 yr

(c) 16 yr

(d) 10.5 yr

(e) None of the above

Answer: (b) 10 yr

Solution: Total age of 30 girls = 30 × 13 = 390

Total age of 18 girls = 18 × 15 = 270

Total age of remaining 12 girls 390 – 270 = 120

Average age of remaining 12 girls =$\frac{120}{12}=10$

Using Trick: Average age of remaining 12 girls = $\frac{30\times 13-18\times 15}{30-18}=10$

Trick: If the average of m numbers is x and out of these ‘m’ numbers the average of n numbers is y, then the average of remaining numbers will be

Average (A) =  $\frac{mx-ny}{m-n}$

Q13. The average height of the basketball team A is 5 ft 11 inches and that of B is 6 ft 2 inches. There are 20 players in team A and 18 players in team B. The overall average height is

(a) 72.42 inches

(b) 72 inches

(c) 70.22 inches

(d) 70 inches

Answer: (a) 72.42 inches

Solution: overall average height = $\frac{20\times 71+18\times 74}{20+18}=\frac{1420+1332}{38}=72.4211$

Rules: If the given observations (x) are occurring with certain frequency (A) then,

Average = $\frac{A_1x_1+A_2x_2+…….+A_nx_n}{x_1+x_2+…..+x_n}$ , where, A1, A2, A3. ………. An are frequencies

Q14. The average of 13 results is 60. If the average of first 7 results is 59 and that of last 7 results is 61, what will be the seventh result?

(a) 90

(b) 50

(c) 75

(d) 60

(e) None of the above

Answer: (d) 60

Solution: Total of 13 results = 13 × 60 = 780

Total of first 7 results = 7 × 59 = 413

Total of last 7 results = 7 × 61 = 427

Seventh result = 413 + 427 – 780 =

In one line: Seventh result = 7 × (59+61) – 13 × 60 = 840 – 780 = 60

Q15. The average of 50 numbers is 38. If the two numbers 45 and 55 are not considered, what will be the average of remaining numbers?

(a) 36.5

(b) 37

(c) 37.5

(d) 37.52

Answer: (c) 37.5

Solution:  Total of 50 numbers = 50 × 38 = 1900

Total after eliminating 45 and 55 = 1900 – 100 = 1800

Average of remaining numbers = $\frac{1800}{48}=37.5$

In one line: Average of remaining numbers = $\frac{50\times 38-\left( 45+55 \right)}{48}=37.5$

Q16. The average of nine numbers is 50. The average of the first five numbers is 54 and that of the last three numbers is 52. Then, the sixth number is

(a) 34

(b) 24

(c) 44

(d) 30

Answer: (c) 44

Solution: Total of 9 numbers = 9 × 50 = 450

Total of first 5 numbers = 5 × 54 = 270

Total of last 3 numbers = 3 × 52 = 156

Sixth numbers = 450 – 270 – 156 = 44

Q17. A batsman makes a score of 87 runs in the 17th inning and thus increases his average by 3. Find his average after 17th inning.

(a) 36

(b) 39

(c) 42

(d) 45

Answer: (b) 39

Solution:  Let the average after 17th inning = x.

Then, average after 16th inning = (x – 3).

∴ 16(x – 3) + 87 = 17x or x = (87 – 48) = 39

Q18. A cricketer whose bowling average is 12.4 runs per wicket takes 5 wickets for 26 runs and thereby decreases his average by 0.4. The number of wickets taken by him till the last match was:

(a) 64

(b) 72

(c) 80

(d) 85

Answer: (d) 85

Solution: Let the number of wickets taken till the last match be x.

∴ $\frac{12.4\mathrm{x}+26}{\mathrm{x}+5}=12$

⇒ 12.4x + 26 = 12x + 60

⇒ 0.4x = 34

⇒ x = 85

Q19. A family consists of grandparents, parents and three grandchildren. The average age of the grandparents is 67 years, that of the parents is 35 years and that of the grandchildren is 6 years. What is the average age of the family?

(a) $28\frac{4}{7}$ Years

(b) $31\frac{5}{7}$ Years

(c) $32\frac{1}{7}$ Years

(d) None of these

Answer: (b) $31\frac{5}{7}$ Years

Solution: Required average = $\frac{2\times 67+2\times 35+3\times 6}{2+2+3}$

= $\frac{134+70+18}{7}=31\frac{5}{7}$

Q20. Of the three numbers, the first is 3 times the second and the third is 5 times the first. If the average of the three numbers is 57, the difference between the largest and the smallest number is

(a) 9

(b) 27

(c) 126

(d) 135

Answer: (c) 126

1st number = $\frac{3\times 3\times \frac{1}{5}\times 57}{3+\frac{1}{5}+\frac{3}{5}}=27$

Second number =$\frac{3\times \frac{1}{5}\times 57}{3+\frac{1}{5}+\frac{3}{5}}=9$

Third number =$\frac{3\times 3\times 57}{3+\frac{1}{5}+\frac{3}{5}}=135$

The difference between the largest and the smallest number = 135 – 9 = 126

Rule: In three numbers, if 1st number is ‘a’ times of 2nd number and ‘b’ times of 3rd number and the average of all three numbers is n,

then 1st number = $\frac{3abn}{a+b+ab}$

Second number = $\frac{3an}{a+b+ab}$

Third number =$\frac{3an}{a+b+ab}$

Pages ( 2 of 2 ): « Previous1 2

Read Important Article

Leave a Comment

error: Content is protected !!