Question Detail

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

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

1. 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}

2. The average score of a cricketer for ten matches is 38.9 runs. If the average for first six matches is 42, then average for last four matches is

33.25
32.25
34.25
34.50

Answer: Option C

Explanation:

\begin{aligned}
= \frac{(38.9 \times 10)-(42 \times 6)}{4}
\end{aligned}

\begin{aligned}
= \frac{(1216 - 750)}{4} = 34.25
\end{aligned}

3. 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

Answer: Option A

Explanation:

4. The average weight of a class of 24 students is 35 kg. If the weight of the teacher be included, the average rises by 400 g. The weight of the teacher is

Answer: Option A

5. Marks of a student were wrongly entered in computer as 83, actual marks of that student were 63. Due to this mistake average marks of whole class got increased by half (1/2). Find the total number of students in that class.

Answer: Option D

Explanation:

Suppose total number of students are = X
\begin{aligned}
Total increase = x*\frac{1}{2} = \frac{x}{2} \\
=> \frac{x}{2} = 83-63 = 20 \\
=> x = 40
\end{aligned}