Question

the mean of three different natural number is 40 if lowest is 19 what could be the highest possible number of remaining two numbers
a) 40 b) 71. C) 81✓ d) 100
Pls give me solution of the s problem
The answer is 81

Answer 1

Answer:

Hope this helps.

Let the other two numbers be a and b, with a < b.  In order they are then:

19 < a < b

We are told the mean is 40, so

( 19 + a + b ) / 3 = 40

=> 19 + a + b = 120

=> a + b = 120 - 19 = 101

Since a > 19, we then have

-a < -19

=> b = 101 - a < 101 - 19 = 82.

So the largest that b can be is 81.

This means the numbers would be 19, 20, 81.