Average MCQ Questions

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) =  $\frac{mx-ny}{m-n}$

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) =  $\frac{mx-ny}{m-n}$

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

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

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 = $\frac{50\times 38-\left( 45+55 \right)}{48}=37.5$

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

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

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

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

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

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