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A group of 630 children is arranged in rows for a photograph

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A group of 630 children is arranged in rows for a photograph

by GMATinsight » Sat May 20, 2017 7:20 pm
A group of 630 children is arranged in rows for a photograph session. Each row contains 3 fewer children than the row in front of it. What number of rows from the following numbers is not a possible number of rows.

A) 3
B) 4
C) 5
D) 6
E) 7

ANS: B
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by Matt@VeritasPrep » Fri May 26, 2017 2:47 pm
Since nobody else answered this ...

Suppose r is our smallest row. The numbers of children in each row are thus r, r + 3, r + 6, ..., r + 3*(n - 1), where n is the number of rows.

If we try answer B, we find that

r + r + 3 + r + 6 + r + 9 = 630, which gives us a solution.

If we try answer D, we find that

r + (r + 3) + (r + 6) + (r + 9) + (r + 12) + (r + 15) = 630, which does NOT give us an integer solution.
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by Matt@VeritasPrep » Fri May 26, 2017 2:53 pm
A neat arithmetic way of working this out:

Since our rows are evenly spaced, we know that the mean and the median number in each row must be the same.

If we take the total and divide by the number of rows, that number will be both the mean and the median.

For A, this gives 630/3 = 210, so the middle row is 210, and the others are 207 and 213.

For B, this gives 630/4 = 157.5. Since this is the median of an evenly spaced set, the two terms that generate it are equidistant from it, so 156 and 159. Our other rows are 153 and 162, so we're done!

For C, this gives 630/5 = 106, with the other rows 100, 103, 109, and 112.

For D, however, this gives 630/6 = 105. Since we have an even number of terms in the set, the two terms that average out to 105 must be equidistant from it ... but these are 103.5 and 106.5, which are not integers.

E is also simple enough, 630/7 = 90, and the rest of the set is 81, 84, 87, 93, 96, and 99.
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by Matt@VeritasPrep » Fri May 26, 2017 2:59 pm
If we wanted to do this algebraically, we could! Suppose r = the number of rows and n = the number of students in the least populated row.

We know that n + (n + 3) + (n + 6) + ... + (n + 3*(r - 1)) = 630. Simplifying this, we get

n*r + 3 + 6 + ... + 3*(r - 1), or

n*r + 3*(1 + 2 + ... + r - 1), or

n*r + 3*r*(r - 1)*(1/2), or

nr + 1.5r*(r-1), or

r * (n + 1.5r - 1.5)

Setting that equal to 630, we get

r * (n + 1.5r - 1.5) = 630

If we try r = 4, we get 4 * (n + 4.5) = 630, or n = 153. If we try r = 6, we get 6 * (n + 7.5) = 630, or n = 97.5.

Since n must be an integer, we know that r = 6 is not a solution.
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by Matt@VeritasPrep » Fri May 26, 2017 3:04 pm
We could go a little further to find all possible values of r.

Once we have the equation n r + 1.5 r² - 1.5 r - 630 = 0, we can isolate r in terms of n:

r = (1/6) * (±√(4n² - 12n + 15129) - 2n + 3)

Any positive integer n that makes that beastly equation in ( )s come out to a positive integer divisible by 6 will give us a valid value of r.
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by Jay@ManhattanReview » Tue May 30, 2017 4:36 am
GMATinsight wrote:A group of 630 children is arranged in rows for a photograph session. Each row contains 3 fewer children than the row in front of it. What number of rows from the following numbers is not a possible number of rows.

A) 3
B) 4
C) 5
D) 6
E) 7

ANS: B
The correct answer is D (6 number of rows), not 4.

Let us start with option A.

There are 3 rows.

Say, the first row has x, the second row has (x-3), and the third row has (x-6) number of children.

Thus, x + (x-3) + (x-6) = 630

3x - 9 = 630

3x = 630 + 9

3x = 639

x = 213 (A positive integer); thus, 3 number of rows is possible.

Option B:

For 3 number of rows, we had 3x = 639, thus for 4 rows, we would have 4x = 639 + (4-1)*3

4x = 639 + 3*3 = 639 + 9 = 648

x = 648/4 = 162 (A positive integer); thus, 4 number of rows is possible.

Option C:

For 4 number of rows, we had 4x = 648, thus for 5 rows, we would have 5x = 648 + (5-1)*3

5x = 648 + 4*3 = 648 + 12 = 660

x = 660/5 = 132 (A positive integer); thus, 5 number of rows is possible.

Option D:

For 5 number of rows, we had 5x = 660, thus for 6 rows, we would have 6x = 660 + (6-1)*3

6x = 660 + 5*3 = 660 + 15 = 675

x = 675/6 = A decimal number; thus, 6 number of rows is not possible.

The correct answer: D

Hope this helps!

-Jay
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Re: A group of 630 children is arranged in rows for a photograph

by Scott@TargetTestPrep » Tue Dec 08, 2020 1:19 pm
GMATinsight wrote: ↑
Sat May 20, 2017 7:20 pm
 A group of 630 children is arranged in rows for a photograph session. Each row contains 3 fewer children than the row in front of it. What number of rows from the following numbers is not a possible number of rows.

A) 3
B) 4
C) 5
D) 6
E) 7

ANS: B
Solution:

Since each row contains three fewer children than the row in front of it, the number of children form an evenly spaced set. In other words, if there are an odd number of rows, the middle row has the median, or average, number of children. Of course, that number has to be an integer because the number of children has to be a whole number. Since 630 is divisible by 3, 5, and 7, we see the middle row will indeed have an integer number of children. If any of these numbers of rows is not possible, the number of children in the middle row will not be an integer. However, since they are, they can’t be the correct answer and we are only left with either choice B or D as the correct answer.

Let’s examine choice B first. We can let x = the number of children in the first row. Therefore, the number of children in the second, third and fourth rows are x - 3, x - 6, and x - 9, respectively. We can create the equation:

x + x - 3 + x - 6 + x - 9 = 630

4x - 18 = 630

4x = 648

x = 162

Since x is an integer, we see that the number of rows can be 4. Since we are looking for the number of rows that is not possible, we are left with choice D as the correct answer.

Answer: D

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