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. 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. Average weight of 10 people increased by 1.5 kg when one person of 45 kg is replaced by a new man. Then weight of the new man is

Answer: Option C

Explanation:

Total weight increased is 1.5 * 10 = 15.
So weight of new person is 45+15 = 60

3. The average of six numbers is X and the average of three of these is Y.If the average of the remaining three is z, then

x = y + z
2x = y + z
x = 2y + z
x = y + 2z

Answer: Option B

Explanation:

X =((3y+3z)/6)
or
2X= y + z

4. 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}

5. 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}