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

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Joined: 24 Oct 2016
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GMAT 1: 670 Q46 V36
In a certain lottery drawing, five balls are selected from a tumbler i  [#permalink]

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02 Aug 2018, 09:32
1
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Difficulty:

45% (medium)

Question Stats:

68% (02:14) correct 32% (02:40) wrong based on 74 sessions

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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

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In a certain lottery drawing, five balls are selected from a tumbler i  [#permalink]

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02 Aug 2018, 09:54
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

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).

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Re: In a certain lottery drawing, five balls are selected from a tumbler i  [#permalink]

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02 Aug 2018, 10:07
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
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Re: In a certain lottery drawing, five balls are selected from a tumbler i  [#permalink]

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02 Aug 2018, 10:10
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Akash720 wrote:
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

Typo. They should be 61 and 62. The least integers greater than 60.
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Re: In a certain lottery drawing, five balls are selected from a tumbler i  [#permalink]

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02 Aug 2018, 20:51
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)
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In a certain lottery drawing, five balls are selected from a tumbler i  [#permalink]

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02 Aug 2018, 23:58
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

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.

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In a certain lottery drawing, five balls are selected from a tumbler i &nbs [#permalink] 02 Aug 2018, 23:58
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