Question

The mean of three distinct natural numbers is 40. If the smallest number is 19, then what could be largest possible number of remaining two numbers?

• A. 81
• B. 140
• C. 100
• D. 71

Answer 1

The correct option is A 81

Let the highest number be $y$, the middle number be $x$ and $z$ be the smallest number
Smallest number is given as 19
We know that, Mean $=\frac{\text{(sum of observations)}}{\text{(number of observations)}}$
Mean $=\frac{(x+y+z)}{3}$
Given that, Mean = 40 and the lowest number = 19.
$40=\frac{(x+y+19)}{3}$
$\Rightarrow 120=x+y+19$
$\Rightarrow x+y=101$
$\Rightarrow y=101-x$

Therefore, for $y$ to be maximum, $x$ should be minimum
Also $x>19$ [As $x$ is not the smallest]
$\Rightarrow x=20$
Therefore, maximum value $y=101-20=81$