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questions from OG CAT!

by ch0719 » Sat Oct 25, 2008 11:31 pm
Here are three questions from OG CAT math section that I got wrong and could not figure it out how to solve them correctly.

If anyone can get the right answers (the squares indicate the right naswers) please post your thoughts on these, thanks! :D
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by stop@800 » Sun Oct 26, 2008 1:51 pm
I:

Pay = x
saved = a
spent = x-a

saved fraction = a/x
spent fraction = (x-a)/x

As per Qn
12a = 3(x-a)

12a = 3x - 3a
a/x = 1/5
thats your answer
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by stop@800 » Sun Oct 26, 2008 2:03 pm
III:

The formula for simple interest is

Interest = Principal * Rate of interest (annual) * time (years) / 100

You can remember it as I = PRT/100

PS: The unit for R and T should be similar. Annual is not mandatory.


The question is just an application of this formula.
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by stop@800 » Sun Oct 26, 2008 2:05 pm
II
A tricky question


Qn says
IS (full/part) for x > (full/part) for z

A says
(full/part) for y < (full/part) for z
if its less for y than it has to be greater for x
so Suff


B

>50% full with x
>50% part with y

if you see the Qn
full in x is more so overall (full/part) for x will be more
as num is large

and also
part in y is more so overall (full/part) for y will be less
as denom is large

hence we can conclude that for x its greater and for y its less
so Suff

Final Answer D
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by cramya » Sun Oct 26, 2008 2:40 pm
For the Alice problem if we are not able to set up the equation like stop@800 did we can always back solve.

Let x be the take home pay
saved - x/5 not saved = 4x/5

The key is to recognize on lhs to mutliply by 12 since the problem read the amount saved at the end of the year is equal to 3 times the monthly pay not saved

i.e 12*x/5 = 3(4x/5)
12x/5 = 12x/5

Done
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by cramya » Sun Oct 26, 2008 3:01 pm
For problem 2 I have a slightly different solution(critique welcome :D )


FTX-FULL TIME EMPL OF DIV X
PTX - PART TIME EMPL OF DIV X
FTY-FULL TILE EMP FOR DIV Y
PTY-PART TIME EMP FOR DIV Y

To prove

FTX/PTX > FTY/PTY (eg:1/2 > 1/4 since 4>2 (cross multiply))

a/b > c/d if ad > bc

i.e. to prove FTX * PTY > FTY * PTX

Stmt I)

FTX+FTY / PTX+PTY > FTY / PTY

Cross mutliply

FTXPTY+FTYPTY>FTYPTX + FTYPTY (FTYPTY CANCELS)

FTXPTY > FTYPTX

Therefore FTX/PTX > FTY/PTY

SUFF

Stmt II) No proof needed

Given FTX>FTY and PTX<PTY

So
FTX/PTX ratio > FTY/PTY ratio (as the numerator increases denominator decreases the fraction value will increase)


SUFF

D)
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by stop@800 » Sun Oct 26, 2008 3:26 pm
cramya wrote:For problem 2 I have a slightly different solution(critique welcome :D )


FTX-FULL TIME EMPL OF DIV X
PTX - PART TIME EMPL OF DIV X
FTY-FULL TILE EMP FOR DIV Y
PTY-PART TIME EMP FOR DIV Y

To prove

FTX/PTX > FTY/PTY (eg:1/2 > 1/4 since 4>2 (cross multiply))

a/b > c/d if ad > bc
only if b and d are positive.

Here you are safe :) [number of empl can not be negative]

i.e. to prove FTX * PTY > FTY * PTX

Stmt I)

FTX+FTY / PTX+PTY > FTY / PTY

Cross mutliply

FTXPTY+FTYPTY>FTYPTX + FTYPTY (FTYPTY CANCELS)

FTXPTY > FTYPTX

Therefore FTX/PTX > FTY/PTY

SUFF

Stmt II) No proof needed

Given FTX>FTY and PTX<PTY

So
FTX/PTX ratio > FTY/PTY ratio (as the numerator increases denominator decreases the fraction value will increase)


SUFF

D)
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by ch0719 » Sun Oct 26, 2008 4:03 pm
thanks :)
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by getso » Wed May 12, 2010 8:31 am
cramya wrote:For problem 2 I have a slightly different solution(critique welcome :D )


FTX-FULL TIME EMPL OF DIV X
PTX - PART TIME EMPL OF DIV X
FTY-FULL TILE EMP FOR DIV Y
PTY-PART TIME EMP FOR DIV Y

To prove

FTX/PTX > FTY/PTY (eg:1/2 > 1/4 since 4>2 (cross multiply))

a/b > c/d if ad > bc

i.e. to prove FTX * PTY > FTY * PTX

Stmt I)

FTX+FTY / PTX+PTY > FTY / PTY

Cross mutliply

FTXPTY+FTYPTY>FTYPTX + FTYPTY (FTYPTY CANCELS)

FTXPTY > FTYPTX

Therefore FTX/PTX > FTY/PTY

SUFF

Stmt II) No proof needed

Given FTX>FTY and PTX<PTY

So
FTX/PTX ratio > FTY/PTY ratio (as the numerator increases denominator decreases the fraction value will increase)


SUFF

D)
I dont understand how did we deduce this FTX+FTY / PTX+PTY > FTY / PTY from statement 1...

Can anybody explain the same..

Thanks,
Shobha
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by clock60 » Wed May 12, 2010 2:01 pm
as for Q2
___________ division x_______ division y________company z
part time_____ x1_____________ y1_____________x1+y1
full time______ x2_____________ y2_____________ x2+y2

and now the quesion is: is x2/x1>(x2+y2)/(x1+y1). or assuming that all numbers are+ve, and after little transforming is
x2/x1>y2/y1

(1) not very difficult
y2/y1<(x2+y2)/(x1+y1) again assuming that all numbers +ve cross multiply and cancel

y2*x1+y2*y1<x2*y1+y2*y1 here we can cancel y1*y2 and left with
x2*y1>y2*x1 and so

x2/x1>y2/y1- sufficient

(2) as for 2 very hard to me because of wording and transforming into math language
final version looks like
x2>1/2(x2+y2)
2*x2>x2+y2
x2>y2

the same for division y

y1>1/2(x1+y1)
2*y1>x1+y1
y1>x1

so we left with two inequalities

x2>y2
y1>x1
multiply each other
x2*y1>y2*x1
x2/x1>y2/y1

so both sufficient
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by deepakteja1988 » Mon Jul 11, 2011 10:28 pm
More than ½ of full-time employees of company Z are employees of div X, and more than ½ of part-time employees of company Z are employees of div Y.

From the first part of the line,

It should be -> (x2 + x1) > 1/2(x2 + y2) as
employees of X are (x2 + x1) and full time employees of Z are (x2 + y2)

and from the second part of the line,

(y1 + y2) > 1/2(x1 + y1) as
employees of Y are (y1 + y2) and part time employees of Z are (y1 + y2).

But I see in the above post as something different. Please correct me if am wrong. Am having tough time in interepreting the second stmt.
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by GMATGuruNY » Tue Jul 12, 2011 3:23 am
For question 1 -- about Alice's take-home pay -- check here:

https://bt.www.beatthegmat.com/alice-s-t ... 72398.html

Question 2:
Each employee of Company Z is an employee of either Division X or Division Y, but not both. If each division has some part-time employees, is the ratio of the number of full-time employees to the number of part-time employees greater for Division X than for Company Z?

(1) The ratio of the number of full-time employees to the number of part-time employees is less for Division Y than for Company Z.

(2) More than half of the full-time employees of Company Z are employees of Division X, and more than half of the part-time employees of Company Z are employees of Division Y.
This is a mixture problem.
X is being combined with Y to form Z.
Unless all the ratios are equal, the ratio for Z must be between the ratios for X and Y.

Statement 1: The ratio of the number of full-time employees to the number of part-time employees is less for Division Y than for Company Z.
Ratio for Z > Ratio for Y.
Since the ratio for Z is between the ratios for X and Y, Ratio for X > Ratio for Z > Ratio for Y.
Thus, Ratio for X > Ratio for Z.
Sufficient.

Statement 2: More than half of the full-time employees of Company Z are employees of Division X, and more than half of the part-time employees of Company Z are employees of Division Y.

Let F = full-time employees and P = part-time employees.
Follow the portion in red:

Ratio for X = (more than half of F)/(less than half of P)
Ratio for Y = (less than half of F)/(more than half of P)

To compare ratios, we cross-multiply.
The NUMERATOR USED IN THE GREATER PRODUCT belongs to the greater ratio.

Cross-multiplying, we get:
(more than half of F)(more than half of P) vs. (less than half of F)(less than half of P).
The product on the left -- which includes MORE THAN HALF OF BOTH GROUPS -- clearly is greater.
Since the NUMERATOR OF X -- (more than half of F) -- is used in the greater product, RATIO for X > RATIO for Y.

Since the Ratio for Z is between X and Y, Ratio for X > Ratio for Z > Ratio for Y.
Thus, Ratio for X > Ratio for Z.
Sufficient.

The correct answer is D.
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by nonameee » Sat Feb 04, 2012 3:04 am
Since (more than half of F) is the numerator of X, Ratio for X > Ratio for Y.
Mitch, could you please explain how you deducted the above?

How do you know that (more than half of F) is greater than (less than half of P)?

Thank you.
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by GMATGuruNY » Sat Feb 04, 2012 3:46 am
nonameee wrote:
Since (more than half of F) is the numerator of X, Ratio for X > Ratio for Y.
Mitch, could you please explain how you deducted the above?

How do you know that (more than half of F) is greater than (less than half of P)?

Thank you.
I did not deduce a relationship between F and P; I deduced only a relationship between the RATIO for X and the RATIO for Y. Follow the portion in red:

Ratio for X = (more than half of F)/(less than half of P)
Ratio for Y = (less than half of F)/(more than half of P)

To compare ratios, we cross-multiply.
The NUMERATOR USED IN THE GREATER PRODUCT belongs to the greater ratio.

Cross-multiplying, we get:
(more than half of F)(more than half of P) vs. (less than half of F)(less than half of P).
The product on the left -- which includes MORE THAN HALF OF BOTH GROUPS -- clearly is greater.

Since the NUMERATOR OF X -- (more than half of F) -- is used in the greater product, RATIO for X > RATIO for Y.
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I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
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by geet_ge » Sat Feb 04, 2012 4:59 am
stop@800 wrote:I:I am not quite clear why on the left hand side we are multiplying by 12 in equation, I am assuming pay to be annual then y=3(x-y)? Please help me understand.

Pay = x
saved = a
spent = x-a

saved fraction = a/x
spent fraction = (x-a)/x

As per Qn
12a = 3(x-a)

12a = 3x - 3a
a/x = 1/5
thats your answer
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