Average - Aptitude MCQ Questions and Solutions with Explanations

Average
$$ = \frac{{11 + 22 + 33 + 44 + 55 + 66 + 77 + 88 + 99}}{9}$$
$$ = \frac{{\left( {11 + 99} \right) + \left( {22 + 88} \right) + \left( {33 + 77} \right) + \left( {44 + 66} \right) + 55}}{9}$$
$$\eqalign{ & = \frac{{4 \times 110 + 55}}{9} \cr & = \frac{{495}}{9} \cr & = 55 \cr} $$

Let the number be x
Then,
$$\eqalign{ & \Rightarrow \frac{{x + {x^2}}}{2} = 5x \cr & \Rightarrow {x^2} - 9x = 0 \cr & \Rightarrow x\left( {x - 9} \right) = 0 \cr & \Rightarrow x = 0{\text{ or }}x = 9 \cr} $$
So, the number is 9

According to the question,
Increased age of 11 players
= 11 × 3
= 33 years
Current age of 3 players who are replaced
= 3 × 33
= 99 years
∴ Age of 3 newcomers
= 99 - 33
= 66 years
∴ Average age = $$\frac{66}{3}$$  = 22 years

Sum of five number
= 27 × 5 = 135
Sum of four number
= 25 × 4 = 100
Excluded number
= 135 - 100 = 35

According to the question,
Required Average
$$\eqalign{ & = \frac{{6 \times 50 + 51 \times 2 + 55 \times 2}}{{10}} \cr & = \frac{{300 + 212}}{{10}} \cr & = \frac{{512}}{{10}} \cr & = 51.2{\text{ Kg}} \cr} $$

As we know that average of square of "n" natural number is
$$\eqalign{ & = \frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{{6n}} \cr & = \frac{{\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} \cr} $$
According to the question,
Average of square of first ten natural number is
$$\eqalign{ & = \frac{{\left( {10 + 1} \right)\left( {20 + 1} \right)}}{6} \cr & = \frac{{11 \times 21}}{6} \cr & = 38.5 \cr} $$

According to the question,
Average of 11 numbers is = 50
sum of 11 numbers is = 50 × 11 = 550

∴ VI number
= 312 + 294 - 550
= 56

According to the question,
$$\eqalign{ & = \frac{{13 \times 70 + 15 \times 60 + 12 \times 65}}{{40}} \cr & = \frac{{910 + 900 + 780}}{{40}} \cr & = \frac{{2590}}{{40}} \cr & = 64.75 \cr} $$

Let the total number of worker = x
According to the question,
⇒ 12 × 400 + (x - 12) × 56 = 60x
⇒ 4800 + 56x - 672 = 60x
⇒ 4128 = 4x
⇒ x = $$\frac{4128}{4}$$
= x = 1032

According to the question,
$$\eqalign{ & \Rightarrow \frac{{x + y + z}}{3} = 45 \cr & \Rightarrow x + y + z = 135.....(i) \cr & \Rightarrow x = \frac{{y + z}}{2} + 9 \cr & \Rightarrow 2x - y - z = 18.....(ii) \cr & x + y + z = 135 \cr & \underline {2x - y - z = 18} \cr & 3x = 153 \cr & x = 51 \cr} $$
From (i)
y + z = 135 - 51 = 84.....(iii)
Also,
$$\eqalign{ & \Rightarrow \frac{{y + z}}{2} = y + 2 \cr & \Rightarrow y + z = 2y + 4 \cr & \Rightarrow z - y = 4 \cr & + y + z = 84 \cr & \underline { - y + z = 4} \cr & \,\,\,\,\,\,\,\,\,2z = 88 \cr & \,\,\,\,\,\,\,\,\,\,\,\,z = 44 \cr} $$
Required difference
= 51 - 44 = 7