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Question Detail
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}
1. 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
- 36
- 37
- 38
- 39
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
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 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
- 40
- 35
- 45
- 55
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
4. Average of five numbers is 27. If one number is excluded the average becomes 25. The excluded number is
- 35
- 45
- 55
- 65
Answer: Option A
Explanation:
Number is (5*27) - (4*25) = 135-100 = 35
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.
- 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}
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