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. Average of five numbers is 27. If one number is excluded the average becomes 25. The excluded number is

Answer: Option A

Explanation:

Number is (5*27) - (4*25) = 135-100 = 35

2. Average of 10 numbers is zero. At most how many numbers may be greater than zero

Answer: Option D

3. When a student weighing 45 kgs left a class, the average weight of the remaining 59 students increased by 200g. What is the average weight of the remaining 59 students

Answer: Option C

Explanation:

Let the average weight of the 59 students be A.
So the total weight of the 59 of them will be 59*A.

The questions states that when the weight of this student who left is added, the total weight of the class = 59A + 45
When this student is also included, the average weight decreases by 0.2 kgs

\begin{aligned}
\frac{59A + 45}{60} = A - 0.2
\end{aligned}

=> 59A + 45 = 60A - 12
=> 45 + 12 = 60A - 59A
=> A = 57

4. Average of 10 matches is 32, How many runs one should should score to increase his average by 4 runs.

Answer: Option B

Explanation:

Average after 11 innings should be 36

So, Required score = (11 * 36) - (10 * 32)
= 396 - 320 = 76

5. 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}