Question Detail

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

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 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}

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. 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

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}