Average Test-2 (Finding average of specified numbers)

Answer: Option A

Explanation

According to the formula,

The average of 1 to n odd numbers =$\frac{\mathrm{Last} \mathrm{odd} \mathrm{number}+1}{2}$

Here, The last odd number= 99

∴ Average of 1 to 99 odd numbers =$\frac{99+1}{2}=\frac{100}{2}=50$

Again, According to the formula,

The average of 1 to n even numbers=$\frac{\mathrm{Last} \mathrm{even} \mathrm{number}+2}{2}$

Here, Last even number = 98

∴ Average of 1 to 44 even numbers =$\frac{98+2}{2}=\frac{100}{2}=50$

Hence, Option A is correct.

Answer: Option C

Explanation

According to the formula,

The average of squares of first 'n' natural numbers =$\frac{(n+1)(2n+1)}{6}$

Here, n=25

∴ Required average 

$$\begin{gathered} =\frac{\left(25+1\right)\left(2\times 25+1\right)}{6} \\ =\frac{26\times 51}{6} \\ =\frac{1326}{6}= 221 \end{gathered}$$

Hence, Option C is correct.

Answer: Option D

Explanation

According to the formula,

The average of the cubes of first 'n' natural numbers= $\frac{\mathrm{n}{\left(\mathrm{n}+1\right)}^2}{4}$

where, n=9

∴ Required average =$\frac{9x{\left(9+1\right)}^2}{4}=\frac{9\times 100}{4}$

= 9×25=225

Hence, Option D is correct.

Answer: Option B

Explanation

According to the formula,

The average of 'n' multiples of a number =$\frac{\mathrm{Number}\times \left(\mathrm{n}+1\right)}{2}$

∴ n=9, number=5

∴ Required average =$\frac{5\times \left(9+1\right)}{2}=\frac{50}{2}=25$

Hence, Option B is correct.

Answer: Option C

Explanation

According to the formula,

The average of 1st 'n' natural numbers= $\frac{\mathrm{n}+1}{2}$

∴ Average of 1st 91 natural numbers

$$\begin{gathered} = \frac{91+1}{2} \\ =\frac{92}{2} \\ =46 \end{gathered}$$

Hence, Option C is correct.