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Complicated OG Problem

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by [email protected] » Sun Mar 17, 2013 9:37 pm
Want to understand if there is a simplere way to solve these problems from OG and Quat. Rev.

Thanks
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135 OG.jpg
Is there any simpler way to solve this problem? What concepts is it trying to test
QReview 135.jpg
Quantitative Review Problem Solving 175
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by Anju@Gurome » Sun Mar 17, 2013 9:46 pm
Question #135:
The question can be rephrased as "If 3 < m < 13 < n, is n/m an integer?"

Statement 1: The information in statement 1 implies that 3n/m is an integer. Now we have find whether n/m is an integer.
Given that 3 < m < 13 < n, if n = 36 and m = 6, then n/m is an integer.
On the other hand if n = 40 and m = 6, then n/m is not an integer.

Not sufficient

Statement 2: According to the statement, 13n/m is an integer.
3 < m < 13 < n implies that m lies between 3 and 13 but is not 13, so 13n/m can be integer only if n/m is an integer.

Sufficient

The correct answer is B.
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Backup Methods : General guide on plugging, estimation etc.
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by GMATGuruNY » Mon Mar 18, 2013 3:20 am
A square wooden plaque has a brass inlay in the center, leaving a wooden strip of uniform width around the brass square. If the ratio of the brass area to the wooden area is 25 to 39, which of the following could be the width, in inches of the wooden strip?

I. 1
II. 3
III. 4

a. I only
b. II only
c. I and II only
d. I and III only
e. I, II, and III
There is no math to be done here; just use common sense. Why couldn't the width of the strip be any value? We could choose a width for the strip, then shrink or expand the inlay until the ratio of inlay area:strip area = 25:39.

The correct answer is E.
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by GMATGuruNY » Mon Mar 18, 2013 3:23 am
A school administrator will assign each student in a group of n students to one of m classrooms. If 3 < m < 13 < n, is it possible to assign each of the n students to one of the m classrooms so that each classroom has the same number of students assigned to it?

(1) It is possible to assign each of 3n students to one of m classrooms so that each classroom
has the same number of students assigned to it.

(2) It is possible to assign each of 13n students to one of m classrooms so that each classroom has the same number of students assigned to it.
To assign the same number of students to each classroom, the number of students (n) must be divisible by the number of classrooms (m).

Question rephrased: Is n/m an integer?

Statement 1: It is possible to assign each of 3n students to one of m classrooms so that each classroom has the same number of students assigned to it.
In other words, the number of students here (3n) is divisible by the number of classrooms (m), implying that 3n/m is an integer.
Since we need to determine whether m will always divide into n, plug in EXTREME values for m.

m=4:
It's possible that m=4 and n=16, with the result that 3n/m = (3*16)/4 = 12.
In this case, then n/m = 16/4 = 4, which is an integer.

m=12:
It's possible that m=12 and n=16, with the result that 3n/m = (3*16)/12 = 4.
In this case, n/m = 16/12 = 4/3, which is NOT an integer.
INSUFFICIENT.

Statement 2: It is possible to assign each of 13n students to one of m classrooms so that each classroom has the same number of students assigned to it.
In other words, the number of students here (13n) is divisible by the number of classrooms (m), implying that 13n/m is an integer.
It is not possible that m divides into 13, since the only factors of 13 are 1 and 13, and m must be BETWEEN 3 and 13.
Thus, for 13n/m to be an integer, m must divide into n, implying that n/m is an integer.
SUFFICIENT.

The correct answer is B.
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by [email protected] » Thu Mar 21, 2013 9:06 pm
GMATGuruNY wrote:
A square wooden plaque has a brass inlay in the center, leaving a wooden strip of uniform width around the brass square. If the ratio of the brass area to the wooden area is 25 to 39, which of the following could be the width, in inches of the wooden strip?

I. 1
II. 3
III. 4

a. I only
b. II only
c. I and II only
d. I and III only
e. I, II, and III
There is no math to be done here; just use common sense. Why couldn't the width of the strip be any value? We could choose a width for the strip, then shrink or expand the inlay until the ratio of inlay area:strip area = 25:39.

But the doubt I have is that to reach this defined ratio the width will be fixed wont it, for example if in a given scenario, the inlay width is 3 the resultants ratio might not necessarily end as 25:39??[/u]
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by rintoo22 » Fri Mar 22, 2013 1:52 am
A square wooden plaque has a brass inlay in the center, leaving a wooden strip of uniform width around the brass square. If the ratio of the brass area to the wooden area is 25 to 39, which of the following could be the width, in inches of the wooden strip?

I. 1
II. 3
III. 4

a. I only
b. II only
c. I and II only
d. I and III only
e. I, II, and III

Area (Brass Inlay)/ Area (Wooden Strip) = 25/39
Area (Wooden plaque) / Area (Brass Inlay) = 64/25

Side (Brass Inlay) : Y(5)
Side (Wooden plaque) : X(8)

Y/X = 5/8
Y = 5/8 * X ------(1)

Width of Wooden Strip : (X-Y)/2 => W=(X-Y)/2 => Put the value of Y from eq 1.
Thus W=(3/16)*X

Therefore, for any positive value of W, A brass Inlay with side of length (16/3)*W will have an strip of W inches, and thus II. 3 width is possible and not I. 1 and III. 4.

Thanks
Ritesh
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