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Question Detail
Find the average of all numbers between 6 and 34 which are divisible by 5
- 15
- 20
- 25
- 30
Answer: Option B
Explanation:
\begin{aligned}
Average = (\frac{10+15+20+25+30}{5}) = \frac{100}{5} =20
\end{aligned}
1. Average of 10 matches is 32, How many runs one should should score to increase his average by 4 runs.
- 70
- 76
- 78
- 80
Answer: Option B
Explanation:
Average after 11 innings should be 36
So, Required score = (11 * 36) - (10 * 32)
= 396 - 320 = 76
2. 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
3. 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}
4. 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.
- 25
- 30
- 35
- 40
- 45
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}
5. Average of first five multiples of 3 is
- 9
- 11
- 13
- 15
Answer: Option A
Explanation:
\begin{aligned}
Average = \frac{3(1+2+3+4+5)}{5} = \frac{45}{5} = 9
\end{aligned}
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