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M03 #16 PS: Average [#permalink]
03 Oct 2008, 00:17

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The average of 4 different positive integers is 125 and the largest of these integers is 150, what is the least possible value of the smallest of the 4 integers?

Re: PS: Average m03#16 [#permalink]
03 Oct 2008, 02:19

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According to question the largest of 4 positive different integers is 150, so b cannot be 300.

sum of 4 integers a,b,c and d is 125*4 =500 given that a=150 => b+c+d = 350 considering b and c as 150 each(assumption) d = 350-150-150=50. Thus smallest integer has to be more than 50 which removes 1,2 and 12 as choices. So remaining choices are 53 and 100.

The 4 integers needs to be different and we need to find least value for smallest integer.

The answer will be 53(considering b as 149 and c as 148 to minimise value of d). _________________

the numerator and denominator in B are actually the same.

Small note/clarification: In statement 2, b = 0 but c is not = 0. If c = 0, b/c = infinite, which is not the case in 2. So the numerator and denominator in 2 are not the same. _________________

Re: M03 #16 PS: Average [#permalink]
14 May 2010, 05:40

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My approach:

We know the average of 4 numbers is 125

Therefore Sum is 500 Largest Number is 150

Now to minimize the smallest integer i maximize the largest integers.

As a result 1st Integer = 150 (given) 2nd Integer = 149 (Next biggest integer value i can assign) 3rd Integer = 148 4rt (smallest) = 500 - (150 + 149 +148) = 53-------------Choice D

Re: M03 #16 PS: Average [#permalink]
14 May 2010, 08:00

the above was my approach too.... I had initial choice of 50 but since it has to be 4 different integers, that made me change the other 2 to 149,148 =>53=>D

Re: M03 #16 PS: Average [#permalink]
18 May 2011, 19:39

amitdgr wrote:

The average of 4 different positive integers is 125 and the largest of these integers is 150, what is the least possible value of the smallest of the 4 integers?

The smallest integer d cant be 1 ? what if b= 300 and c = 49,

150 + 300 + 49 + 1 = 500

the average still remains 125. What am i missing here ?

It's given that all four numbers are different. As average is given to be 125, one number must be less than 125 and this number will be least when other three numbers will be maximum. So, three greatest numbers which we can assume will be 150,149, and 148. Now we have to find out the number which should be added to these three numbers and divided by 4 and it should give 125.

=> 125=150+149+148+least number/4 =>500-447=Least number =>53= Least number

Re: M03 #16 PS: Average [#permalink]
25 May 2011, 07:21

will go with 53.

Key here is "4 different positive integers"

Sum of all "4 different positive integers" = \(4 * 125 = 500\)

sum of least 3 positive integers = \(500 - (largest integer) = 500-150 = 350\)

other possible integers has to be lesser than 150. so we could choose 149 (as large as possible to find out the least possible integer among the 4 different integers)

similarly, the next has to be 148

least possible 4th integer = \(350 - (149+148) = 53\)

Re: M03 #16 PS: Average [#permalink]
22 May 2013, 04:06

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Expert's post

The average (arithmetic mean) of 4 different positive integers is 125 and the largest of these integers is 150, what is the least possible value of the smallest of the 4 integers?

A. 1 B. 2 C. 12 D. 53 E. 100

Suppose the 4 integers in ascending order are \(x\), \(y\), \(z\) and 150.

Since the average of 4 different positive integers is 125 then their sum is 125*4=500. So, \(x+y+z+150=500\) --> \(x+y+z=350\).

We want to minimize \(x\), so we should maximize \(y\) and \(z\), since given that all integers are distinct then \(y_{max}=148\) and \(z_{max}=149\) --> \(x+148+149=350\) --> \(x=53\).

Re: M03 #16 PS: Average [#permalink]
22 May 2013, 10:24

125 = 150+x+y+z/4 hence 350=x+y+z since 150 is the largest no. of he four x,y,z cant be more than 150 even if it is 150 350 = 150+150+z the smallest no. has to be more than 50 hence the ans is d

gmatclubot

Re: M03 #16 PS: Average
[#permalink]
22 May 2013, 10:24