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(#M40015921) MBA question averages, percentages, profit loss and interest Keep an EYE

The average of five natural numbers is 150. A particular number among the five exceeds another by 100. The rest three numbers lie between these two numbers and they are equal. How many different values can the largest number among the five take?

1) 59
2) 19
3) 21
4) 42

Asked In MBA (11 years ago)

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Aniruddh Shah  (8 years ago)

Let the 5 natural numbers in increasing order be: a, b, b, b, a+100
Note that a < b < a+100

(a+b+b+b+a+100)5=150(a+b+b+b+a+100)5=150
2a+3b=6502a+3b=650

Since a < b, let's find the point where a = b
2a + 3a = 650
a = 130 = b
But b must be greater than a. If we increase b by 1, we need to decrease a by 3 to keep the average same. But that decreases the largest number too, so increase b by another 1.
So a = 127, b = 132 give us the numbers as 127, 132, 132, 132, 227

Since b < a+100, let's find the point where b = a+100
2a + 3(a+100) = 650
a = 70, b = 170
But b must be less than a+100. If we decrease b by 1, we need to increase a by 3 to keep the average same. But that increases the largest number too, so decrease b by another 1.
So a = 73, b = 168 give us the numbers as 73, 168, 168, 168, 173

Values of a will be: 73, 76, 79, ....127 (Difference of 3 to make b a natural number)
This is an AP.
Last term = First term + (n - 1)*Common difference
127 = 73 + (n - 1)*3
n = 19
So the last term (a+100) will also take 19 distinct values.

Answer (B)

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