Q. 135.0( 2 Votes )

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

Answer :

Let the three numbers be a, b and c.

According to the problem, a = 19


And


Or


Which gives 19+b+c = 120


So b+c = 101


Consider b = 100, which gives c = 1 which is not possible because the lowest number allowed is 19.


Consider b = 40 which gives c = 61


Consider b = 71, which gives c = 30


Consider b = 81, which gives c = 20


Among these b = 81 is the highest number.
Therefore, the correct answer is 81.

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