Question Detail

A motorist travel to a place 150 km away at an avearge speed of 50 km/hr and returns ar 30 km/hr.His average speed for the whole jouney in km/hr is

36.5 km/hr
37.5 km/hr
35.5 km/hr
34.5 km/hr

Answer: Option B

Explanation:

Average speed will be

\begin{aligned}

\frac{2xy}{x+y} km/hr

= \frac{2(50)(30)}{50+30} km/hr
\end{aligned}

= 37.5 km/hr

1. The average of four consecutive odd numbers is 24. Find the largest number.

Answer: Option B

Explanation:

Let the numbers are x, x+2, x+4, x+6, then
\begin{aligned}
=> \frac{x+(x+2)+(x+4)+(x+6)}{4} = 24
\end{aligned}

\begin{aligned}
=> \frac{4x+12)}{4} = 24
\end{aligned}

\begin{aligned}
=> x+3 = 24 => x = 21
\end{aligned}

So largest number is 21 + 6 = 27

2. A batsman makes a score of 87 runs in the 17th match and thus increases his average by 3. Find his average after 17th match

Answer: Option D

Explanation:

Let the average after 17th match is x
then the average before 17th match is x-3

so 16(x-3) + 87 = 17x
=> x = 87 - 48 = 39

3. Reeya obtained 65, 67, 76, 82 and 85 out of 100 in different subjects, What will be the average.

Answer: Option B

Explanation:

\begin{aligned}
( \frac {65+67+76+82+85}{5} ) = 75
\end{aligned}

4. 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}

5. 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