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Question Detail
Find the average of first 10 multiples of 7
- 35.5
- 37.5
- 38.5
- 40.5
Answer: Option C
Explanation:
\begin{aligned}
= \frac {7(1+2+3+...+10)}{10}
\end{aligned}
\begin{aligned}
= \frac {7(10(10+1))}{10 \times 2}
\end{aligned}
\begin{aligned}
= \frac {7(110)}{10 \times 2} = 38.5
\end{aligned}
1. If the average marks of three batches of 55, 60 and 45 students respectively is 50, 55, 60, then the average marks of all the students is
- 54.48
- 54.68
- 54.60
- 54.58
Answer: Option B
Explanation:
\begin{aligned}
\frac{(55 \times 50) +(60 \times 55) +(45 \times 60) }{55 + 60 + 45}
\end{aligned}
\begin{aligned}
\frac{8750}{160} = 54.68
\end{aligned}
2. A library has an average of 510 visitors on Sundays and 240 on other day. The average number of visitors in a month of 30 days starting with sunday is
- 280
- 285
- 290
- 295
Answer: Option B
Explanation:
As the month begin with sunday, so there will be five sundays in the month. So result will be:
\begin{aligned}
= (\frac{510 \times 5 + 240 \times 25}{30})
= (\frac{8550}{30}) = 285
\end{aligned}
3. The average age of the mother and her six children is 12 years which is reduced by 5 years if the age of the mother is excluded. How old is the mother
- 40
- 41
- 42
- 43
Answer: Option C
4. The average age of husband, wife and their child 3 years ago was 27 years and that of wife and the child 5 years ago was 20 years. The present age of the husband is
- 40
- 35
- 45
- 55
Answer: Option A
Explanation:
Sum of the present ages of husband, wife and child = (27 * 3 + 3 * 3) years = 90 years.
Sum of the present ages of wife and child = (20 * 2 + 5 * 2) years = 50 years.
Husband's present age = (90 - 50) years = 40 years
5. Average of 10 matches is 32, How many runs one should should score to increase his average by 4 runs.
- 70
- 76
- 78
- 80
Answer: Option B
Explanation:
Average after 11 innings should be 36
So, Required score = (11 * 36) - (10 * 32)
= 396 - 320 = 76
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