(A) 5

(B) 6

(C) 7

(D) 10

(E) 14

[spoiler]made up

*by*Sanjeev K Saxena

*for*Avenues Abroad[/spoiler]

- sanju09
- GMAT Instructor
**Posts:**3650**Joined:**21 Jan 2009**Location:**India**Thanked**: 267 times**Followed by:**80 members**GMAT Score:**760

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

[spoiler]made up*by* Sanjeev K Saxena *for* Avenues Abroad[/spoiler]

(A) 5

(B) 6

(C) 7

(D) 10

(E) 14

[spoiler]made up

The mind is everything. What you think you become. -Lord Buddha

Sanjeev K Saxena

Quantitative Instructor

The Princeton Review - Manya Abroad

Lucknow-226001

www.manyagroup.com

Sanjeev K Saxena

Quantitative Instructor

The Princeton Review - Manya Abroad

Lucknow-226001

www.manyagroup.com

- neelgandham
- Community Manager
**Posts:**1060**Joined:**13 May 2011**Location:**Utrecht, The Netherlands**Thanked**: 318 times**Followed by:**52 members

Let the number of rows be R and the number of students in each row be S

R*S = 70

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.

Anil Gandham

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- sanju09
- GMAT Instructor
**Posts:**3650**Joined:**21 Jan 2009**Location:**India**Thanked**: 267 times**Followed by:**80 members**GMAT Score:**760

Nice approach, but?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, WHYBingo?

(E)Blah Blah

The mind is everything. What you think you become. -Lord Buddha

Sanjeev K Saxena

Quantitative Instructor

The Princeton Review - Manya Abroad

Lucknow-226001

www.manyagroup.com

Sanjeev K Saxena

Quantitative Instructor

The Princeton Review - Manya Abroad

Lucknow-226001

www.manyagroup.com

- neelgandham
- Community Manager
**Posts:**1060**Joined:**13 May 2011**Location:**Utrecht, The Netherlands**Thanked**: 318 times**Followed by:**52 members

My bad! It is just that I love copy+paste so much I tend to... I will edit it now

Anil Gandham

Welcome to BEATtheGMAT | Photography | Getting Started | BTG Community rules | MBA Watch

Check out GMAT Prep Now's online course at https://www.gmatprepnow.com/

Welcome to BEATtheGMAT | Photography | Getting Started | BTG Community rules | MBA Watch

Check out GMAT Prep Now's online course at https://www.gmatprepnow.com/

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

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

- Birottam Dutta
- Master | Next Rank: 500 Posts
**Posts:**342**Joined:**08 Jul 2009**Thanked**: 214 times**Followed by:**19 members**GMAT Score:**740

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

- sanju09
- GMAT Instructor
**Posts:**3650**Joined:**21 Jan 2009**Location:**India**Thanked**: 267 times**Followed by:**80 members**GMAT Score:**760

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.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 mind is everything. What you think you become. -Lord Buddha

Sanjeev K Saxena

Quantitative Instructor

The Princeton Review - Manya Abroad

Lucknow-226001

www.manyagroup.com

Sanjeev K Saxena

Quantitative Instructor

The Princeton Review - Manya Abroad

Lucknow-226001

www.manyagroup.com

- [email protected]
- GMAT Instructor
**Posts:**15420**Joined:**08 Dec 2008**Location:**Vancouver, BC**Thanked**: 5254 times**Followed by:**1266 members**GMAT Score:**770

I think the fastest approach here is to check the answer choices.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

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.

Cheers,

Brent