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Question

The mean of the three numbers is 40. All three numbers are different natural numbers. If the lowest is 19, what could be the highest possible number of the remaining two numbers?


A

81

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B

40

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C

100

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D

71

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Solution

The correct option is A

81


The explanation for the correct option.

Let us consider that the remaining two numbers are aandb.

Since the mean of the three numbers is 40 and the lowest number is 19.

a+b+193=40a+b+19=120a+b=120-19a+b=101

Let one number b is 20 because the lowest number is 19.

a+20=101b=101-20b=81

Thus, the highest possible number is 81.

Hence, the option Ais correct.


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