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Question Detail
The average age of the mother and her six children is 12 years which is reduced by 5 years if the age of the mother is excluded. How old is the mother
- 40
- 41
- 42
- 43
Answer: Option C
1. When a student weighing 45 kgs left a class, the average weight of the remaining 59 students increased by 200g. What is the average weight of the remaining 59 students
- 55
- 56
- 57
- 58
Answer: Option C
Explanation:
Let the average weight of the 59 students be A.
So the total weight of the 59 of them will be 59*A.
The questions states that when the weight of this student who left is added, the total weight of the class = 59A + 45
When this student is also included, the average weight decreases by 0.2 kgs
\begin{aligned}
\frac{59A + 45}{60} = A - 0.2
\end{aligned}
=> 59A + 45 = 60A - 12
=> 45 + 12 = 60A - 59A
=> A = 57
2. 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}
3. The average of four consecutive odd numbers is 24. Find the largest number.
- 25
- 27
- 29
- 31
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
4. 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}
5. 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}
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