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Question

Question 13
The mean of three numbers is 40. All the three numbers are different natural numbers. If lowest is 19, what could be the highest possible number of remaining two numbers?
(a) 81
(b) 40
(c) 100
(d) 71

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Solution

Mean of three numbers = 40 and lowest number= 19 [given]
Let the three observations be 19, x and y, respectively.
Mean=Sum of all observationsTotal number of observations 40=19+x+y3 [ mean=40, given] 3×40=19+x+y
120 = 19 + x + y
x + y = 120 - 19
x + y = 101
Since, 19 is the lowest observation.
Hence, for the highest possible value of remaining two numbers, one must be 20.
Let x = 20
From Eq. (i), we get
20 + y = 101
y = 101 - 20
y = 81

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