Average MCQ Questions

Solution: even number Between 11 to 63 are 12, 14, 16, … 60, 62

∴ Required Average = $\frac{\mathrm{first}  \mathrm{number}+\mathrm{last}  \mathrm{number}}{2}=\,\,\frac{12+62}{2}=37$

Rule: The average of consecutive even natural numbers.

A = $\frac{\mathrm{first} \mathrm{number}+\mathrm{last} \mathrm{number}}{2}$

Solution: Required average = $\frac{5\left( 5+1 \right) ^2}{4}=45$

Rule: The average of cubes of first ‘n’ consecutive natural numbers

A =  $\frac{n\left( n+1 \right) ^2}{4}$

Solution: Required average = $\frac{8\times 14-6\times 16}{2}=8$

Solution: Let 40 consecutive even numbers are a, a + 2, …., a+78 ( t₄₀ = a + (n – 1)d, where n= 40, d = 2)

Average = $\frac{\mathrm{first} \mathrm{number}+\mathrm{last} \mathrm{number}}{2}=\,\,\frac{\mathrm{a}+\mathrm{a}+78}{2}=\mathrm{a}+39$

∴ a + 39 = 59 ⇒ a = 20

Greatest number = a + 78 = 20 + 78 = 98

Solution: Required average = $\frac{30\times 60+10\times 56}{30+10}=59$

Rule: If the average of ‘n₁’ numbers is a₁ and the average of ‘n₂’ numbers is a₂, then average of total numbers n₁ and n₂ is

Average (A) = $\frac{n_1a_1+n_2a_2}{n_1+n_2}$

Solution: Required average = $\frac{40\times 50+35\times 60+45\times 55+42\times 45}{40+35+45+42}$

= $\frac{2000+2100+2475+1890}{162}=52.2531$

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: Sum of present ages of the six members = (17 × 6) years = 102 years.

Sum of present ages of the 5 members (excluding baby) = 5 × (17 + 3) years = 100 years.

∴ Age of the baby = 102 – 100 = 2 years

Solution: Let the number of persons, initially going for Picnic = x

∴ Sum of their ages = 16x

Now new average = $\frac{16\mathrm{x}+15\times 20}{\mathrm{x}+20}$

∴$\frac{16\mathrm{x}+15\times 20}{\mathrm{x}+20}=15.5$

⇒ 16x +300 = 15.5x + 310

⇒ 0.5x = 10

⇒x = 20

Solution:  Total age before teacher include = 36 × 14 = 504

Total age after teacher include = 37 × 15 = 555

Age of the teacher = (555 – 504) years = 51 years

Using Trick: Age of the teacher = 14 + 37 = 51

Trick: The average of n numbers is ‘a’. After adding a new number average increase by x then

The new numbers = a + x (n + 1)

Solution: Let there be x pupils in the class.

Total increase in marks = $\mathrm{x}\times \frac{1}{2}=\frac{\mathrm{x}}{2}$

$\frac{\mathrm{x}}{2}=\,\,83-63=20$

⇒ x = 40

Using Trick: The number of pupils = $\frac{83-63}{\frac{1}{2}}=40$

Trick: The average of n numbers is ‘a’. If a number ‘x’ is replaced with another new number then average increase by ‘y’.

n = $\frac{\mathrm{x}-\mathrm{new} \mathrm{number}}{\mathrm{y}}$