If anyone can get the right answers (the squares indicate the right naswers) please post your thoughts on these, thanks!
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questions from OG CAT!
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!
If anyone can get the right answers (the squares indicate the right naswers) please post your thoughts on these, thanks!
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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.
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.
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
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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cramya
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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
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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cramya
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For problem 2 I have a slightly different solution(critique welcome )
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)
)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)
only if b and d are positive.cramya wrote:For problem 2 I have a slightly different solution(critique welcome
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
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)
I dont understand how did we deduce this FTX+FTY / PTX+PTY > FTY / PTY from statement 1...cramya wrote:For problem 2 I have a slightly different solution(critique welcome
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)
Can anybody explain the same..
Thanks,
Shobha
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clock60
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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
___________ 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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deepakteja1988
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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.
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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GMATGuruNY
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For question 1 -- about Alice's take-home pay -- check here:
https://bt.www.beatthegmat.com/alice-s-t ... 72398.html
Question 2:
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.
https://bt.www.beatthegmat.com/alice-s-t ... 72398.html
Question 2:
This is a mixture problem.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.
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.
Last edited by GMATGuruNY on Sat Feb 04, 2012 3:49 am, edited 1 time in total.
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Mitch, could you please explain how you deducted the above?Since (more than half of F) is the numerator of X, Ratio for X > Ratio for Y.
How do you know that (more than half of F) is greater than (less than half of P)?
Thank you.
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GMATGuruNY
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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:nonameee wrote:Mitch, could you please explain how you deducted the above?Since (more than half of F) is the numerator of X, Ratio for X > Ratio for Y.
How do you know that (more than half of F) is greater than (less than half of P)?
Thank you.
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.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
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I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
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Followed here and elsewhere by over 1900 test-takers.
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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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
