Question

9. The mean of three
different natural numbers is 40. If lowest is 19, what could be highestpossible number of remaining two numbers? ,
(b) 71
(c) 81
(d) 100
(a) 40

with explanation
​

Answer 1

Step-by-step explanation:

mean = 40

Let the two unknown numbers be x and y, such that x > y.

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

19 + x + y = 120

x + y = 111

from the options the highest possible number would be 100.

as,

100 + y = 111

y = 11

here,

100 > 11

so,

x > y

Answer is 100