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students are made to stand

by sanju09 » Wed Apr 04, 2012 4:08 am
70 students are made to stand in rows such that each row has the same number of students. The same students are rearranged now to stand in 4 more rows than before such that each row still has same number of students. If each row now has 2 fewer students than before, how many rows were there before the rearrangement?
(A) 5
(B) 6
(C) 7
(D) 10
(E) 14



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by neelgandham » Wed Apr 04, 2012 5:47 am
Initial arrangement
Let the number of rows be R and the number of students in each row be S
R*S = 70
Final arrangement
The number of rows is now R+4 and the number of students in each row = S-2
(R+4)*(S-2) = 70

So R and R+4 are multiples of 70

(A)5 is a multiple of 70, 5+4 = 9 is not a multiple of 70, so incorrect
(B)6 is not a multiple of 70, so incorrect
(C)7 is a multiple of 70, 7+4 = 11 is not a multiple of 70, so incorrect
[spoiler](D)10[/spoiler] is a multiple of 70, 10+4 = 14 is a multiple of 70, Bingo!
(E)Blah Blah
Last edited by neelgandham on Thu Apr 05, 2012 4:47 am, edited 1 time in total.
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by sanju09 » Thu Apr 05, 2012 4:44 am
neelgandham wrote:Initial arrangement
Let the number of rows be R and the number of students in each row be S
R*S = 70
Final arrangement
The number of rows is now R+4 and the number of students in each row = S-2
(R+4)*(S-2) = 70

So R and R+4 are multiples of 70

(A)5 is a multiple of 70, 5+4 = 9 is not a multiple of 70, so incorrect
(B)6 is not a multiple of 70, so incorrect
(C)7 is a multiple of 70, 7+4 = 11 is not a multiple of 70, so incorrect
[spoiler](D)10[/spoiler] is a multiple of 70, 10+4 = 14 is not a multiple of 70, WHY Bingo?
(E)Blah Blah
Nice approach, but?
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by neelgandham » Thu Apr 05, 2012 4:46 am
My bad! It is just that I love copy+paste so much I tend to... I will edit it now :)
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by gmatmath » Fri Apr 06, 2012 1:20 am
initially there are 70 students.
This can be arranged in 2 ways:
1) 10 students of 7 rows each or
2) 7 students of 10 rows each.

If we take choice 1, adding 4 more rows,makes it 11 rows and reduce 2 students from each row ,we get 8 students in each row. Since there are only 70 students, we will have 2 options which do not satisfy the criteria.

a) either we will have 8 rows of 8 students and the 9th row with 6 students or
b) 10 rows of 6 students and the 11th row having 10 students.

Both these arrangements violate the fact that all rows should have equal number of students. This does not satisfy our criteria of total= 70 students,and each row with equal number of students.

In the second choice, we have 10 rows with 7 students each. Here, if we reduce 2 students from each row and add 4 more rows such that all rows have equal students, we would get,5 students in 14 rows. This works out to be 14 x 5 = 70 students as the total number of students.

Hence, the number of rows originally were 10, which is option D.
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by Birottam Dutta » Fri Apr 06, 2012 7:56 am
Let there be A rows of B students each. Therefore,

Total number of students= A*B= 70.

Now, 70= 2*5*7 (reducing to prime factors)

Then, following combination of rows and students can exist:

(2*5 rows) * (7 students in each row) => 10 rows * 7 students in each row----- (1)
(7 rows) * (2*5 students in each row) => 7 rows * 10 students in each row----- (2)
(2*7 rows) * (5 students in each row) => 14 rows * 5 students in each row----- (3)
(5 rows) * (2*7 students in each row) => 5 rows * 14 students in each row----- (4)
(5*7 rows) * (2 students in each row) => 35 rows * 2 students in each row----- (5)
(2 rows) * (5*7 students in each row) => 2 rows * 35 students in each row----- (6)

Now, considering the condition in the problem, the rearrangement of the students results in 4 additional rows and 2 students lesser in each row than the previous combination.

This condition is met only by (1) and (3), i.e., initially there were 10 rows of 7 students each, and in the latter arrangement 14 rows of 5 students each.

So, answer is 7.

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by sanju09 » Sat Apr 07, 2012 5:01 am
Birottam Dutta wrote:Let there be A rows of B students each. Therefore,

Total number of students= A*B= 70.

Now, 70= 2*5*7 (reducing to prime factors)

Then, following combination of rows and students can exist:

(2*5 rows) * (7 students in each row) => 10 rows * 7 students in each row----- (1)
(7 rows) * (2*5 students in each row) => 7 rows * 10 students in each row----- (2)
(2*7 rows) * (5 students in each row) => 14 rows * 5 students in each row----- (3)
(5 rows) * (2*7 students in each row) => 5 rows * 14 students in each row----- (4)
(5*7 rows) * (2 students in each row) => 35 rows * 2 students in each row----- (5)
(2 rows) * (5*7 students in each row) => 2 rows * 35 students in each row----- (6)

Now, considering the condition in the problem, the rearrangement of the students results in 4 additional rows and 2 students lesser in each row than the previous combination.

This condition is met only by (1) and (3), i.e., initially there were 10 rows of 7 students each, and in the latter arrangement 14 rows of 5 students each.

So, answer is 7.

Regards,

Birottam Dutta
The question is, "how many rows were there before the rearrangement?", hence if 7 is the answer then it won't allow 7 + 4 = 11 to be a multiple of 70, as it must be so. Please recheck your work.
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by Brent@GMATPrepNow » Sat Apr 07, 2012 5:29 am
sanju09 wrote:70 students are made to stand in rows such that each row has the same number of students. The same students are rearranged now to stand in 4 more rows than before such that each row still has same number of students. If each row now has 2 fewer students than before, how many rows were there before the rearrangement?
(A) 5
(B) 6
(C) 7
(D) 10
(E) 14
I think the fastest approach here is to check the answer choices.
A) 5 rows (in the original arrangement) means 14 students per row. The new arrangement must have 9 rows. Since 70 is not divisible by 9, we'll stop here and eliminate A.
B) 6 rows. Since 70 is not divisible by 6, we'll stop here and eliminate B.
C) 7 rows (in the original arrangement) means 10 students per row. The new arrangement must have 11 rows. Since 70 is not divisible by 11, we'll stop here and eliminate C.
D) 10 rows (in the original arrangement) means 7 students per row. The new arrangement must have 14 rows, which means 5 students per row. Since this meets the given criteria, we can stop here and conclude that the answer is D.

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