The mean of three numbers is 40. All the three numbers are different natural numbers. If lowest is 19, what could be highest possible number of remaining two numbers?
A. 81
B. 40
C. 100
D. 71

Given:

The mean of three numbers is 40.

All the three numbers are different natural numbers and the lowest number is 19.
To do:

We have to find the possible highest number from among the given options. 

Solution:

Let the highest number be $x$ and the other number be $y$.

This implies,

$40=\frac{x+y+19}{3}$

$40\times3=x+y+19$

$x+y=120-19$

$x+y=101$

Given that 19 is the lowest number. This implies y can be 20 or greater than 20.

From the given options,

If $x=81$ then $y=101-81=20$.

If $x=40$ then $y=101-40=61$ which is not possible as $y>x$ here.

If $x=100$ then $y=101-100=1$ which is not possible as $y$ should be greater than 19.

If $x=71$ then $y=101-71=30$.

Therefore, from among the given options, $x=81$ is the correct answer.

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Updated on: 10-Oct-2022

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