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Number

Expert replies
by Nycgrl » Sat Sep 13, 2008 3:36 pm
Which of the following procedures is always equivalent to adding 5 given numbers and then dividing the sum by 5 ?
I. Multiplying the 5 numbers and then finding the 5th root of the product.
II. Adding the 5 numbers, doubling the sum, and then moving the decimal point one place to the left.
III. Ordering the 5 numbers numerically and then selecting the middle number.
(A) None
(B) I only
(C) II only
(D) III only
(E) I and III
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Source: — Problem Solving |

Re: Number

by Sunny22uk » Sat Sep 13, 2008 6:20 pm
Nycgrl wrote:Which of the following procedures is always equivalent to adding 5 given numbers and then dividing the sum by 5 ?
I. Multiplying the 5 numbers and then finding the 5th root of the product.
II. Adding the 5 numbers, doubling the sum, and then moving the decimal point one place to the left.
III. Ordering the 5 numbers numerically and then selecting the middle number.
(A) None
(B) I only
(C) II only
(D) III only
(E) I and III
It took some time to solve this problem, I did it by picking numbers.
"Which of the following procedures is always equivalent to adding 5 given numbers and then dividing the sum by 5 ?" means Arithmetic mean of 5 numbers
lets say 1,5,7,8,10 are the numbers
Statement 1 is not true
Statement 2 is true
Statement 3 is not true
Is the OA C?
You cannot discover new oceans unless you have the courage to loose sight of the shore.
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by Nycgrl » Sat Sep 13, 2008 9:12 pm
Yes OA is C

Even I solved it by picking number. I just wanted to check if there is any other method to do it?
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by jawanindia2007 » Sat Sep 13, 2008 10:16 pm
my take is.......

option 2 says.........doubling the sum in the numerator......that amounts to multiplying by 2 and the second part says moving the decimal point one place to the left........this amounts to multiplying the denominator by 2 again....as the denominator already has 5....

also no choice has 2 in it apart from ..c.
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by vcb » Sat Sep 13, 2008 10:55 pm
Yeah, I started out the same way, picking numbers, and options I and III didnt work out. For II, what we would be doing is, adding the five numbers, multiplying the sum by 2 and then dividing the whole thing by 10.
x*2/10 = x/5, where x is the sum of the five numbers...so it has to be II. Or, C.
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by gmatguy16 » Sun Sep 14, 2008 8:57 am
can someone please elaborate with example what does by mean "moving decimal point one place to the left"?
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by vishubn » Sun Sep 14, 2008 9:10 am
Yaa...
my take on the question and possible explanation of moving decimal ppint to left by one position

So the question stem is basically way top fond out the average... soo
qith the option a I,II,III ... fifth root option is crazy .. i elliminated on the fist go ...

And then multiilpyig the five numbers and option is is increasing by the order all the time so u left with only opion of II only ,,....

decimal pint shift.....


eg---1,2,3,4,5 for simplicity ... are the numbers

1+2+3+4+5/=15/5
=3

11 option says double the sum of five numbers .... 15+15=30
which is also equal to 30.0 so shifting one decimal pint to left ....point shifts from 30.0 to 3.0 so which is the answer ....

Hope it helps

Vishu
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by cramya » Sun Sep 14, 2008 6:19 pm
Picking numbers is ok. I think we just need to make sure we pick atleast 2 different sets to confirm our choice(disprove the remaining choices).

For example if we pick 2,3,4,5,6 then 2 and 3 would both work.
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by 4meonly » Tue Sep 16, 2008 7:45 am
This is a very old way to divide a number by 5
1. Multiply by 2
2. Move decimal point
Generally, this means to multiply numerator and denumerator by 2

60/5 = 60*2/5*2 = 12
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