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list of 5 different numbers

by clock60 » Tue Dec 21, 2010 2:30 pm
hi guys
can somebody help me with 2 st. spend too much time


A certain list consists of five different integers. Is the average o fthe two greatest integers in the list greater than 70?

(1) The median of the integers in the list is 70
(2) The average of the integers in the list is 70

oa is D
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by eyelikecheese » Tue Dec 21, 2010 2:38 pm
I'll attempt this:
Since there are 5 integers in the set, the middle number is going to be an integer.


1. If the median of the set is 70, there are 2 numbers under 70 and 2 numbers greater than 70. OR, the median could be 70, with each number being 70, but it is stated in the stem that each integer is different. So, we have 2 integers greater than 70, who's average must be greater than 70. SUFFICIENT

2. This could be any 5 numbers that sum up to 350. INSUFFICIENT

A
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by clock60 » Tue Dec 21, 2010 2:40 pm
hi eyelikecheese
thank you for quick reply
but oa is D and it is not a typo
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by anshumishra » Tue Dec 21, 2010 3:34 pm
clock60 wrote:hi guys
can somebody help me with 2 st. spend too much time


A certain list consists of five different integers. Is the average o fthe two greatest integers in the list greater than 70?

(1) The median of the integers in the list is 70
(2) The average of the integers in the list is 70

oa is D
Five different integers : I am writing from least to greatest: (x1,x2,x3,x4,x5)
(x4+x5)/2 > 70 ?

Statement1 :
Median - 70, how can you represent the set from lowest to greatest ?

[70-n1, 70-n2,70,70+n3,70+n4]
Clearly, 4th and 5th have to be greater than 70 - Sufficient.

Statement 2 :
(x1+x2+x3+x4+x5)/5 = 70
since, x5>x4>x3>x2>x1

x4+x5 = 350 -(x1+x2+x3)
Can it be less than 140 ever ?
Nope, As x3>x2>x1, To make the sum minimum you want to maximize (x1+x2+x3), which can never be more than (70+69+68)...because if you make x3 more than 70, x4 and x5 will automatically will become more than 70, and hence the required average will be more than 70, in any case.

- Sufficient

Hence "D".
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by clock60 » Tue Dec 21, 2010 4:04 pm
hi anshumishra
i want to think that i start to get why 2 st is sufficient due to your solving
in my own words
x1+x2+x3+x4+x5=350
if x4+x5<140, then
highest possible values for x4 and x5= 69 and 70, they can`t be higher as any adding gives 70 and 71 that is > than 140
so highest values for x1,x2,x3 are 66, 67,68 and no more as all numbers are different
so our possible sum will look like
66+67+68+69+70=340 but we need 350
thus x4+x5 must be >that 140
do you think my solving is valid?
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by anshumishra » Tue Dec 21, 2010 4:06 pm
clock60 wrote:hi anshumishra
i want to think that i start to get why 2 st is sufficient due to your solving
in my own words
x1+x2+x3+x4+x5=350
if x4+x5<140, then
highest possible values for x4 and x5= 69 and 70, they can`t be higher as any adding gives 70 and 71 that is > than 140
so highest values for x1,x2,x3 are 66, 67,68 and no more as all numbers are different
so our possible sum will look like
66+67+68+69+70=340 but we need 350
thus x4+x5 must be >that 140
do you think my solving is valid?
You are right !
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by fskilnik@GMATH » Wed Dec 22, 2010 6:09 am
clock60 wrote:A certain list consists of five different integers. Is the average o fthe two greatest integers in the list greater than 70?

(1) The median of the integers in the list is 70
(2) The average of the integers in the list is 70
Hi there,

Although anshumishra´s solution is perfect, I guess all technical "x_1, x_2, ..." formulation is really unecessary. Let´s have a look why:

(1) This sttm says that when the five numbers are put in (say) increasing order, the third one is 70. That means that the fourth and fifth are greater than 70 (they are all different), therefore the average of fourth and fifth is greater than 70.

(2) This sttm says that if all five numbers could be equal, (to sum 5*70 = 350) they would have to be equal to 70. From the fact that they are all different, the AVERAGE of the two greater ones must be above 70. I guess this is (almost) self-evident, but if you want a proof for that, it follows at the end.

Regards,
Fabio.

Proof: say the average of the three smaller numbers is m and the average of the two greater ones is M. Then from sttm (2) and the fact that m<M (remember the 5 numbers are all different) we have 5*70 = 3m+2M <3M+2M = 5M implying that M > 70.
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by clock60 » Wed Dec 22, 2010 8:23 am
hi fskilnik
i am very interested in your post and follow it scrupulously
can you please elaborate a little bit more
5*70 = 3m+2M <3M+2M = 5M

i understand that 3m+2M=350, buy next step is unclear, why do you change 3m on 3M?
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by fskilnik@GMATH » Wed Dec 22, 2010 9:14 am
clock60 wrote:hi fskilnik
i am very interested in your post and follow it scrupulously
can you please elaborate a little bit more
5*70 = 3m+2M <3M+2M = 5M

i understand that 3m+2M=350, buy next step is unclear, why do you change 3m on 3M?
Sure, clock60. I´m glad to deserve your careful attention!!

The three smaller numbers have an average (m) that is certainly less than the average of the 2 greater ones (M), therefore we are sure m<M. Multiplying the inequality by the positive number 3 we get 3m<3M and adding the same parcel (2M) to both sides of the inequality we get: 3m + 2M < 3M + 2M and we are done.

Please let me know if there is something still unclear, ok?!

Cheers,
Fabio.
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by clock60 » Wed Dec 22, 2010 11:41 am
dear fskilnik
thank you for your patience i start to get you universal solution, but what to replicate it in my own words
the mean of first three numbers -a
the mean of last two numbers -b, and according to the 2 st
3a+2b=350
a<b
3a<3b
3a+2b<3b+2b
3a+2b<5b, and as 3a+2b=350 it happens that
350<5b
b>70, what we need to prove, the mean of 2 greatest numbers>70
is it right?
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by fskilnik@GMATH » Wed Dec 22, 2010 11:48 am
clock60 wrote:dear fskilnik
thank you for your patience i start to get you universal solution, but what to replicate it in my own words
the mean of first three numbers -a
the mean of last two numbers -b, and according to the 2 st
3a+2b=350
a<b
3a<3b
3a+2b<3b+2b
3a+2b<5b, and as 3a+2b=350 it happens that
350<5b
b>70, what we need to prove, the mean of 2 greatest numbers>70
is it right?
It is not right, clock60, it is PERFECT! ;)

Important: please note the difference between proving that something is right (what we both did) and suspecting that something must be right!! I tell you that because at first I SUSPECTED the average of the two greater numbers SHOULD be greater than 70 (and this is the most important thing in terms of GMAT performance, this "hunch")... only then I used technical arguments to guarantee my "hunch" was absolutely correct... the "hunch" is AT LEAST as important as your technicall ability to prove its validity!! (My experience tells me that the ability to prove things develop your "hunch", because it increases your "mathematical maturity"... and this mathematical maturity is "sniffed" by the GMAT examiners, for sure!)
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
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