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Arranging the children

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by 2010gmat » Thu Oct 22, 2009 8:11 pm
A group of 810 children arranged themselves in N rows for a group photograph. In each row there were 3 more children than the row ahead. Which of the following can be the value of N?

a.) 6, b.) 8, c.) 5, d.) 10, e.) 12

i could solve this question by imputing the values, any other way to solve such type of questions?
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Re: Arranging the children

by uttam.albela » Thu Oct 22, 2009 9:41 pm
2010gmat wrote:A group of 810 children arranged themselves in N rows for a group photograph. In each row there were 3 more children than the row ahead. Which of the following can be the value of N?

a.) 6, b.) 8, c.) 5, d.) 10, e.) 12

i could solve this question by imputing the values, any other way to solve such type of questions?
As the answer choices are given it is very easy to solve:

Assume first row has x children.
next row x+3 and so on. So this is Arithmetic Progression.

Sum of all children in N rows= N * (x + (N-1)*3/2) = 810

x + (N-1)*3/2 must be an integer as N * integer = 810

x is an integer as it is number of children in first row. We can not have 1/2 or 1/3 children.
So, (N-1)*3/2 must be an integer.

now N-1 must be even.

N must be odd.

and we have only one answer option with odd number.
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by 2010gmat » Thu Oct 22, 2009 10:31 pm
How come sum of children = N * (x + (N-1)*3/2)?


(N-1)*3/2 ?? or it should be (N-1)*3

and this question had one add and other even choices...so i also started by putting 5 in the equation which i had arrived at....but in case there is a 2-3 split between even and odd choices...

then how to go about it??
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by uttam.albela » Thu Oct 22, 2009 11:15 pm
2010gmat wrote:How come sum of children = N * (x + (N-1)*3/2)?


(N-1)*3/2 ?? or it should be (N-1)*3 : the formula for AP series is n (a + (n-1)*d/2) where n=number of terms
a first term
d common difference

and this question had one add and other even choices...so i also started by putting 5 in the equation which i had arrived at....but in case there is a 2-3 split between even and odd choices...

then how to go about it??


(N-1)*3/2 ?? or it should be (N-1)*3 : the formula for AP series is n (a + (n-1)*d/2) where n=number of terms
a first term
d common difference
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by life is a test » Fri Oct 23, 2009 1:05 am
2010gmat wrote:How come sum of children = N * (x + (N-1)*3/2)?


(N-1)*3/2 ?? or it should be (N-1)*3

and this question had one add and other even choices...so i also started by putting 5 in the equation which i had arrived at....but in case there is a 2-3 split between even and odd choices...
then how to go about it??
sum of AP = n *(2a1 + [n-1]d) / 2 --> N * 2x+ 3[N-1] /2 --> 3 [N-1]/2 * N * 2x

For 3[N-1]/2 to be an int, N-1 must be a multiple of 2 which means it must be even hence N must be odd.

uttam.albela - did you mistype x instead of 2x in the formula since it should be 2 times the first term (a1) as in the above formula?

uttam.albela
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by uttam.albela » Fri Oct 23, 2009 2:25 am
life is a test wrote:
2010gmat wrote:How come sum of children = N * (x + (N-1)*3/2)? Divide by 2 is inside the bracket. It is same as what you have written but after division of terms withing outside bracket by 2.


(N-1)*3/2 ?? or it should be (N-1)*3

and this question had one add and other even choices...so i also started by putting 5 in the equation which i had arrived at....but in case there is a 2-3 split between even and odd choices...
then how to go about it??
sum of AP = n *(2a1 + [n-1]d) / 2 --> N * 2x+ 3[N-1] /2 --> 3 [N-1]/2 * N * 2x

For 3[N-1]/2 to be an int, N-1 must be a multiple of 2 which means it must be even hence N must be odd.

uttam.albela - did you mistype x instead of 2x in the formula since it should be 2 times the first term (a1) as in the above formula?

uttam.albela
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