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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?
Options
81
40
100
71
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Solution
81
Explanation:
Mean of three numbers = 40 and lowest number = 19 ...[Given]
Let the three observations be 19, x and y, respectively.
∴ Mean = `"Sum of all observations"/"Total number of observations"`
⇒ `40 = (19 + x + y)/3` ...[∵ Mean = 40, given]
⇒ 3 × 40 = 19 + x + y
⇒ 120 = 19 + x + y
⇒ x + y = 120 – 19
⇒ x + y = 101 ...(i)
Since, 19 is the lowest observation.
Hence, for the highest possible value of the remaining two numbers, one must be 20.
Let x = 20
From equation (i), we get
20 + y = 101
⇒ y = 101 – 20
⇒ y = 81
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