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Question

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
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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
Let the highest number be y and the middle number be x.
We know that,
Mean=sum of observationsnumber of observations
Mean=x+y+z3

Given that, Mean =40 and the lowest number =19.

40=x+y+193
x+y+19=120
x+y=12019=101
y=101x
Therefore, for y to be maximum, x should be minimum
Also x>19 [As x is not the smallest]
x=20

Therefore, maximum value y=10120=81

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