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# M03 #16 PS: Average

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03 Oct 2008, 01:17
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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?

(A) 1
(B) 2
(C) 12
(D) 53
(E) 100

[Reveal] Spoiler: OA
D

Source: GMAT Club Tests - hardest GMAT questions

sum of 4 integers a,b,c and d is 125*4 =500

given that a=150 => b+c+d = 350

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 ?
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03 Oct 2008, 03:19
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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).
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03 Oct 2008, 04:06
whoops I overlooked the greatest number is 150 bit. Great explanation !! thanks +1
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13 Jan 2009, 18:50
vksunder wrote:
a,b and c are integers. Is abc = 0?

1. a^2 = 2a
2. b/c = (a+b)^2 / (a^2+2ab+b^2) - 1

1: a^2 = 2a
a = 0 or 2.

2. b/c = [(a+b)^2 / (a^2+2ab+b^2)] - 1
b/c = [(a+b)^2 - (a^2+2ab+b^2)] / (a^2+2ab+b^2)
b/c = 0

b = 0. so suff.
B.
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13 Jan 2009, 23:30
GMAT TIGER wrote:
vksunder wrote:
a,b and c are integers. Is abc = 0?

1. a^2 = 2a
2. b/c = (a+b)^2 / (a^2+2ab+b^2) - 1

1: a^2 = 2a
a = 0 or 2.

2. b/c = [(a+b)^2 / (a^2+2ab+b^2)] - 1
b/c = [(a+b)^2 - (a^2+2ab+b^2)] / (a^2+2ab+b^2)
b/c = 0

b = 0. so suff.
B.

Precisely. The main trick is to realize that the numerator and denominator in B are actually the same.
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14 Jan 2009, 18:51
chicagocubsrule wrote:
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.
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20 Jan 2009, 02:25
GMAT TIGER wrote:
vksunder wrote:
a,b and c are integers. Is abc = 0?

1. a^2 = 2a
2. b/c = (a+b)^2 / (a^2+2ab+b^2) - 1

1: a^2 = 2a
a = 0 or 2.

2. b/c = [(a+b)^2 / (a^2+2ab+b^2)] - 1
b/c = [(a+b)^2 - (a^2+2ab+b^2)] / (a^2+2ab+b^2)
b/c = 0

b = 0. so suff.
B.

Agree with the above.

Do we have OA?
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Re: M03 #16 PS: Average [#permalink]

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14 May 2010, 06:40
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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
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Re: M03 #16 PS: Average [#permalink]

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14 May 2010, 09:00
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

keep it simple!
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Re: M03 #16 PS: Average [#permalink]

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14 May 2010, 20:24
easy one ANS:D
A+148+149+150=500

So A=53
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Re: M03 #16 PS: Average [#permalink]

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18 May 2011, 05:25
Total Sum = 500

=> Sum of remaining 3 = 350

To make the smallest one have least value, the 2 bigger ones should have maximum value

So the 2nd largest = 149

and 3rd largest = 148

=> Smallest one = 350 - 297 = 53

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Re: M03 #16 PS: Average [#permalink]

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18 May 2011, 08:32

Average =125*4 =500

4 Different Numbers largest is 150

+149+148+53 =500
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Re: M03 #16 PS: Average [#permalink]

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18 May 2011, 20:39
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?

(A) 1
(B) 2
(C) 12
(D) 53
(E) 100

[Reveal] Spoiler: OA
D

Source: GMAT Club Tests - hardest GMAT questions

sum of 4 integers a,b,c and d is 125*4 =500

given that a=150 => b+c+d = 350

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
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Re: M03 #16 PS: Average [#permalink]

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21 May 2011, 12:38
125 = (a+b+c+d)/4
500 = a+b+c+d
500 = 150 + 149 + 148 + d
d = 53
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Re: M03 #16 PS: Average [#permalink]

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23 May 2011, 09:43
I also used numbers to solve the question ..
I hate algebra ..
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Re: M03 #16 PS: Average [#permalink]

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25 May 2011, 08:21
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 = $$350 - (149+148) = 53$$
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Re: M03 #16 PS: Average [#permalink]

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22 May 2013, 05:06
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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 $$z_{max}=149$$ --> $$x+148+149=350$$ --> $$x=53$$.

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Re: M03 #16 PS: Average [#permalink]

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22 May 2013, 11:24
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
Re: M03 #16 PS: Average   [#permalink] 22 May 2013, 11:24
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