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GMAT Statistics

by abhi75 » Fri Jun 13, 2008 6:37 am
A certain list consists of 21 different numbers. If n is in the list and n is 4 times the average (arithmetic mean) of the other 20 numbers in the list, then n is what fraction of the sum of the 21 numbers in the list?

a) 1/20
b) 1/6
c) 1/5
d) 4/21
e) 5/21

I got this one correct but I am sure there are better ways to crack this than my method. Can you please suggest any alternative method to solve it more efficiently.

Thanks.

My method

According to MGMAT in statistics you should always express the average in terms of sum.

let the average of first 20 number be a
Its average a = sum/20
Therefore sum = 20a

Now lets b be the average of n and the other 20 numbers which gives us
sum + n/21 = b
==> sum + n = 21b (Substitute n = 4a in this eqn and sum=20a)
==> 20a + 4a = 21b
==> 24a = 21b
==> 8a = 7b
==> b = 8/7a

Now n/n+ sum = 4a/21b (Substitute b = 8/7a in this eqn)
==> 4a/21(8/7a) = 4a/24a
==> 1/6

Therefore OA is B. It took me about 4 to 5 mins to do this right. I have seen this pattern in GMAT problems that you have derive eqns isolate variables and then substitute that variable in the other eqns.

Also can someone please tell me what score range this question is. I believe its in 700-800 question range.

Thanks,
-A
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by durgesh79 » Fri Jun 13, 2008 6:44 am
It took me less than 1 min and I dint look at the answer before

let S is the sum of 20 numbers, we have to finu out n/(n+S)

n = 4(S/20) = S/5

S/n = 5

add 1 on both sides

(S+n)/n = 6
n/(S+n) = 1/6
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by abhi75 » Fri Jun 13, 2008 6:55 am
Thanks Durgesh. Thats really a good way of solving it.

-A
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by kindofbluenote » Fri Jun 13, 2008 8:09 am
Another method would be to assume that the first 20 integers are consecutive 1-20. The sum of consecutive integers is found by multiplying the average by 20.

The average is the first integer in the set plus the last divided by 2 (20+1/2=10.5) so the sum is (20*10.5=210).

The 21st integer is given as four times the average, so 10.5*4=42. The ratio of the 21st integer to the sum of all 21 is 42/(210+42) =42/252=1/6.
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by AleksandrM » Fri Jun 13, 2008 9:47 am
durgesh,
durgesh79 wrote:add 1 on both sides

(S+n)/n = 6
n/(S+n) = 1/6
This part is not entirely clear to me. Can you elaborate. How do you end up with (S+n)/n = 6 after adding 1. I understand the 6, but what about the other side?

Thanks.
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by durgesh79 » Fri Jun 13, 2008 9:57 am
AleksandrM wrote:durgesh,
durgesh79 wrote:add 1 on both sides

(S+n)/n = 6
n/(S+n) = 1/6
This part is not entirely clear to me. Can you elaborate. How do you end up with (S+n)/n = 6 after adding 1. I understand the 6, but what about the other side?

Thanks.
one way of understanding
S/n + 1
= S/n + n/n ---------- 1 can be written as n/n
= (S+n)/n

Another way ... normal fraction sum
S/n + 1
take LCM of n and 1 in denominator, which is n
(S + n)/n
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by rahul.s » Fri Jan 22, 2010 4:48 am
kindofbluenote wrote:Another method would be to assume that the first 20 integers are consecutive 1-20. The sum of consecutive integers is found by multiplying the average by 20.

The average is the first integer in the set plus the last divided by 2 (20+1/2=10.5) so the sum is (20*10.5=210).

The 21st integer is given as four times the average, so 10.5*4=42. The ratio of the 21st integer to the sum of all 21 is 42/(210+42) =42/252=1/6.
Brilliant approach! Thank you :)
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by kirsten H » Wed Jan 27, 2010 3:34 am

I solve it as below

At first, n=S20/20 X 4
n=S20/5 i.e. S20=5n

And then, suppose if n=10, substitue that figures instead of n. S20=50
hence S21=60

Lastly, Qs is <n is what fraction of the sum of 21 numbers in digit>,
The answer is 10/60 = 1/6

Hope this helps.
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by sanju09 » Wed Jan 27, 2010 4:42 am
AleksandrM wrote:durgesh,
durgesh79 wrote:add 1 on both sides

(S+n)/n = 6
n/(S+n) = 1/6
This part is not entirely clear to me. Can you elaborate. How do you end up with (S+n)/n = 6 after adding 1. I understand the 6, but what about the other side?

Thanks.
Please revise componendo and dividendo theory in ratio and proportions.

durgesh79 wrote
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Never ask who's the third otherwise durgesh would yell...Yahoo! it's you!!
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by bigmonkey31 » Wed Jan 27, 2010 7:24 am
This is a similar method to durgesh....

So you know that there's 21 numbers in the set.... 2, 5, 7, n, 10, 11, .............. 100 (you get the point).

The average of the other 20 numbers = [Sum(of all 21 numbers) - n] / 20
Thus, 4 time the average of these 20 nubmers = 4 * [Sum(of all 21 numbers) - n] / 20

And since you know that 'n' is defined by '4 time the average of these 20 numbers' set this equation up:
n = 4 * [Sum(of all 21 numbers) - n] / 20

SOLVE:
n = 1/5 [Sum(of all 21 numbers) - n]
n = 1/5*Sum(21 numbers) - 1/5*n
(6/5)*n = 1/5*sum(21 numbers)
n = 1/6 * Sum(21 numbers)

--> 'n' is '1/6' of the 'Sum of all 21 numbers'

Answer choice B
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by acoustic2 » Wed Feb 10, 2010 5:21 pm
I'm sure everyone has posted this in one way or another, but here is a really simple way to look at it:

a=average
s=sum
x=answer

n=4a (this is stated in the problem)
s=20a (relationship between the sum and the average)
x=n/(s+n) (the answer is going to equal n divided by the total sum which is n+s)

Now we have 3 simple equations and we can plug the first two into the third to get:

x=(4a)/(20a+4a)
x=(4/24)(a/a)
x=1/6

Easy money.
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by chrisinkiwiland » Thu Feb 11, 2010 4:15 pm
this might be an easier approach still:

suppose all the numbers apart from n are 1, then the sum is 20 and the average is 1. therefore n needs to be 4 and 4/24 is equal to 1/6.

I know it says all the numbers in the set should be different but it makes no difference as we are only concerned with the average.
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by sanju09 » Fri Feb 12, 2010 12:26 am
abhi75 wrote:A certain list consists of 21 different numbers. If n is in the list and n is 4 times the average (arithmetic mean) of the other 20 numbers in the list, then n is what fraction of the sum of the 21 numbers in the list?

a) 1/20
b) 1/6
c) 1/5
d) 4/21
e) 5/21

I got this one correct but I am sure there are better ways to crack this than my method. Can you please suggest any alternative method to solve it more efficiently.

Thanks.

My method

According to MGMAT in statistics you should always express the average in terms of sum.

let the average of first 20 number be a
Its average a = sum/20
Therefore sum = 20a

Now lets b be the average of n and the other 20 numbers which gives us
sum + n/21 = b
==> sum + n = 21b (Substitute n = 4a in this eqn and sum=20a)
==> 20a + 4a = 21b
==> 24a = 21b
==> 8a = 7b
==> b = 8/7a

Now n/n+ sum = 4a/21b (Substitute b = 8/7a in this eqn)
==> 4a/21(8/7a) = 4a/24a
==> 1/6

Therefore OA is B. It took me about 4 to 5 mins to do this right. I have seen this pattern in GMAT problems that you have derive eqns isolate variables and then substitute that variable in the other eqns.

Also can someone please tell me what score range this question is. I believe its in 700-800 question range.

Thanks,
-A
If x is the AM of 21 different numbers of the list, then their sum is 21 x. When n is taken out of the set, the AM of the remaining 20 different numbers of the list is given by (21 x - n)/20, and we are asked to answer n/ (21 x), with the following fact

n = 4 {(21 x - n)/20} or 5 n = 21 x - n or n/ (21 x) = 1/6.

[spoiler]B[/spoiler]
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by komal » Tue Feb 16, 2010 12:05 pm
abhi75 wrote:A certain list consists of 21 different numbers. If n is in the list and n is 4 times the average (arithmetic mean) of the other 20 numbers in the list, then n is what fraction of the sum of the 21 numbers in the list?

a) 1/20
b) 1/6
c) 1/5
d) 4/21
e) 5/21

let x be the sum of the other 20 numbers
n=4*x/20=x/5

the sum of 21 numbers is x/5+x=6x/5

n / sum of 21 numbers is:
(x/5)/(6x/5)=5x/30x=1/6
answer B
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by abhi332 » Wed Feb 17, 2010 6:40 am
you can also try with back solving

if you consider a option 1/6

sum of 20 positive number will be >= 20 (assuming all numbers are positive)

multiplying denominator and numerator of 1/6 by 20

20/120 = 20/(20 + 100) ---> n/(n+s)

s= 100 therefore avg is 100/20 = 5 which when multiplied by 4 gives 20
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