dabaobao wrote:

In a certain lottery drawing, five balls are selected from a tumbler in which each ball is printed with a different two-digit positive integer. If the average (arithmetic mean) of the five numbers drawn is 56 and the median is 60, what is the greatest value that the lowest number selected could be?

A) 43

B) 48

C) 51

D) 53

E) 56

Let the 5 numbers are:

--, --,

60, --, -- (ordering pattern:- Lowest to highest)

We need the highest value of the lowest number(1st number). So, the 4th and 5th numbers are to be minimum. Therefore, we can allocate 61 and 62 in the 4th and 5th place respectively.(Since all the 2-digit integers are distinct).

For the 1st place to be a highest value less than median, the 1st and 2nd place numbers must be consecutive.( say n and n+1).

Given, mean=56

Or, \(\frac{n+n+1+60+61+62}{5}=56\)

Or, 2n+184=56*5=280

Or, 2n=96

Or, n=48

Ans. (B)

_________________

Regards,

PKN

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