Which of the following statements is/are correct ?
Select the answer using the code given below:
- 1. The average of four numbers 10, 15, 20 and 25 is 17.5
- 2. If a, b and c are three different natural numbers such that a + b + c = abc, then the average of a, b and c is 3
1 only
2 only
Both 1 and 2
Neither 1 nor 2
Answer is Wrong!
Answer is Right!
This question was previously asked in
UPSC CAPF – 2024
1. The average of four numbers 10, 15, 20 and 25:
Sum $= 10 + 15 + 20 + 25 = 70$. Number of terms $= 4$. Average $= \frac{70}{4} = 17.5$. Statement 1 is correct.
2. If a, b and c are three different natural numbers such that a + b + c = abc, then the average of a, b and c is 3.
The average of a, b, and c is $\frac{a+b+c}{3}$. Given $a+b+c = abc$, the average is $\frac{abc}{3}$. We need to find if $\frac{abc}{3}$ is always 3 for different natural numbers a, b, c satisfying $a+b+c=abc$.
We look for solutions to $a+b+c = abc$ in different natural numbers. Assume $1 \le a < b < c$. If $a=1$, $1+b+c=bc \implies bc-b-c=1 \implies (b-1)(c-1)-1=1 \implies (b-1)(c-1)=2$. Since $1 < b < c$, we have $b-1 < c-1$. The only integer factors of 2 are (1, 2). So $b-1=1$ and $c-1=2$, which gives $b=2$ and $c=3$. The set {1, 2, 3} satisfies $1+2+3=6$ and $1 \times 2 \times 3 = 6$. These are different natural numbers. For this solution set {1, 2, 3}, the average is $\frac{1+2+3}{3} = \frac{6}{3} = 2$.
Therefore, only statement 1 is correct.