Average - Aptitude MCQ Questions and Solutions with Explanations

Let Ajit's average be x for 9 innings. So, Ajit scored 9x run in 9 innings.
In the 10^th inning, he scored 100 runs then average became (x+8). And he scored (x + 8) × 10 runs in 10 innings.
Now,
$$\eqalign{ & \Rightarrow 9x + 100 = 10 \times \left( {x + 8} \right) \cr & {\text{or}},\,9x + 100 = 10x + 80 \cr & {\text{or}},\,x = 100 - 80 \cr & {\text{or}},\,x = 20 \cr & {\text{New}}\,{\text{average}} = \left( {x + 8} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 28\,{\text{runs}} \cr} $$

Average temperature for Wednesday, Thursday and Friday = 40° C
Total temperature = 3 × 40 = 120° C
Average temperature for Thursday, Friday and Saturday = 41° C
Total temperature = 41 × 3 = 123° C
Temperature on Saturday = 42° C
Now,
(Thursday + Friday + Saturday) - (Wednesday + Thursday + Friday) = 123 - 120;
Saturday - Wednesday = 3
Wednesday = 42 - 3 = 39° C

$$\eqalign{ & {\text{Required}}\,{\text{average}} \cr & = {\frac{{{\text{total}}\,{\text{sum}}\,{\text{of}}\,{\text{multiple}}\,{\text{of}}\,9}}{5}} \cr & = {\frac{{9 + 18 + 27 + 36 + 45}}{5}} \cr & = 27 \cr} $$
Note that, average of 9 and 45 is also 27.
And average of 18 and 36 is also 27.

$$\eqalign{ & {\text{Average speed}}, \cr & = \frac{{ {2 \times x \times y} }}{{ {x + y} }} \cr & = \frac{{ {2 \times 100 \times 150} }}{{ {100 + 150} }} \cr & = \frac{{ {200 \times 150} }}{{250}} \cr & = 120\,{\text{km/hr}} \cr} $$

1^st Method:
Average of 1^st n natural number is given by
$$ = \frac{{ {\frac{{ {{\text{n}} \times \left( {{\text{n}} + 1} \right)} }}{2}} }}{{\text{n}}}$$
Average of 1^st 97 natural number is given by
$$\eqalign{ & {\frac{{ {\frac{{ {97 \times \left( {97 + 1} \right)} }}{2}} }}{{97}}} \cr & = 49 \cr} $$

2^st Method:
These numbers are in AP series so average.
$$\eqalign{ & = \frac{{ {{\text{sum of corresponding term}}} }}{2} \cr & = \frac{{ {1 + 97} }}{2} = 49 \cr & {\text{or, }}\frac{{ {2 + 96} }}{2} = 49 \cr & {\text{or, }}\frac{{ {3 + 95} }}{2} = 49 \cr & {\text{And so on}}{\text{.}} \cr} $$

Sum of ages of three boys = 45 years
Now, (3x + 5x + 7x) = 45
⇒ 15x = 45
⇒ x = 3
So, age of youngest boy = 3x
= 3 × 3
= 9 years

Let N be the no. of persons in the group.
Required number of person is given by;
Member in group × aged increased = difference of replacement
N × 5 = 38 - 18
Or, 5N = 20
Or, N = 4

Member in group × age increased = difference of replacement
Or, 8 × 1 = new comer - man going out
Or, new comer = 8 + 60
Or, new comer = 68 kg.

Sum of 1^st 12 results = 12 × 14
Sum of last 12 results = 12 × 17
13^th result = x (let)
Now,
12 × 14 + 12 × 17 + x = 25 × 18
Or, x = 78

$$\eqalign{ & {\text{Average}}\,{\text{speed}} \cr & = \frac{{{\text{total}}\,{\text{distance}}\,{\text{covered}}}}{{{\text{total}}\,{\text{time}}}} \cr & {\text{Total}}\,{\text{Distance}} \cr & = {16 + 20 + 10} \cr & = 46\,{\text{km}} \cr & {\text{Time}}\,{\text{taken}} \cr & = {\frac{{16}}{{20}}} + {\frac{{20}}{{40}}} + {\frac{{10}}{{15}}} \cr & = {\frac{{4}}{{5}}} + {\frac{{1}}{{2}}} + {\frac{{2}}{{3}}} \cr & = {\frac{{24 + 15 + 20}}{{30}}} \cr & = \frac{{59}}{{30}} \cr & {\text{Average}}\,{\text{speed}} \cr & = \frac{{46 \times 30}}{{59}} \cr & = 23\frac{{23}}{{59}}\,{\text{km/hr}} \cr} $$