In a certain lottery drawing, five balls are selected from a tumbler i

Question

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

A) 43

B) 48

C) 51

D) 53

E) 56

Bunuel · Answer

Say the five numbers drawn are a, b, c, d, e, where: a < b < c < d < e (each ball is printed with a  different  two-digit positive integer).

The average (arithmetic mean) of the five numbers drawn is 56 and the median is 60: median = c = 60 and a + b + 60 + d + e = 56*5 = 280.

a + b + d + e = 220.

What is the  greatest value  that the  lowest number  selected could be? The lowest number is a. To maximize a, we need to minimize other numbers. The lowest values of d and e are 61 and 62.

a + b + 61 + 62 = 220.
a + b = 97.

a = 48 and b = 49 (since a < b).

Answer: B.

Akash720 · Answer

Hi  Bunuel

Kindly, help me out here. I did not understand the part where you chose d and e as 51,52
Since a<b<c<d<e and c = 60 how are d and e less than 60?

Posted from my mobile device

Bunuel · Answer

Typo. They should be 61 and 62. The least integers greater than 60.

PKN · Answer

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)

SSM700plus · Answer

All the 5 terms can be estimated as  a<b<c<d<e . Median = 60. Average = 56.

It implies c = 60. 
d & e - can be more, but not less than 60.
a & b - should be less than 60.

Consider the arrangement as  a<b<c=60<d=61<e=62

In order to maintain Average as 56, a & b should add up to 56+56-15= 97.

a + b = 97.

a =48 & b=49 is one only largest value possible.

Clearly C,D & E choice is not possible. So eliminate it.

A & B is possible, but A is out because Question is to find  the greatest value that the lowest number selected could be . So we have to find largest possible value of a.

Answer is B.

BrushMyQuant · Answer

We have five different 2 digit numbers whose average is 56 and median is 60

Since the Median is 60 => the middle number is 60
Average of the number is 56 => Sum = 5*56 = 280
Now we have to find the greatest possible value of lowest number=> numbers greater than 60 should be smallest
=> x , y ,60,61,62 should be other numbers (where x and y are the two numbers smaller than 60, x being the smallest)
=> x + y + 60 + 61 + 62 = 280
=> x + y = 97
Now, for x to be maximum, y should be minimum
Minimum value of y will be x + 1 (as y cannot be less than x and x has to be greatest)
x + x + 1 = 97
=> x = 48, y = 49

So, answer will be B
Hope it helps!

SaquibHGMATWhiz · Answer

Solution:

Sum of 5 chosen numbers  =5	imes 56=280  
 Median i.e., 3rd (in a sorted manner) chosen number is 60 
 To be able to maximize the lowest number, we have to minimize the rest of them
  __, __, 60, __, __   
 4th and 5th numbers have to be greater than 60. So, the minimum values possible are 61 and 62 
  __, __, 60, 61, 62   
 Sum of 1st and 2nd number  =280-60-61-62=97  
 Greatest possible value of 1st number =  internal value of  \frac{97}{2} = 48

48 , 49, 60, 61, 62

Hence the right answer is    Option B

bumpbot · Answer

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In a certain lottery drawing, five balls are selected from a tumbler i

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