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. Average of all prime numbers between 30 to 50

37
37.8
39
39.8

Answer: Option D

Explanation:

Prime numbers between 30 and 50 are:
31, 37, 41, 43, 47

Average of prime numbers between 30 to 50 will be
\begin{aligned}
(\frac{31+37+41+43+47}{5}) = \frac{199}{5} = 39.8
\end{aligned}

2. 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

3. Average of first five multiples of 3 is

Answer: Option A

Explanation:

\begin{aligned}
Average = \frac{3(1+2+3+4+5)}{5} = \frac{45}{5} = 9
\end{aligned}

4. 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}

5. 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