BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Number of Stamps

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
Junior | Next Rank: 30 Posts
Posts: 12
Joined: Wed May 18, 2011 10:28 am

Number of Stamps

by PankajRai » Fri Jun 10, 2011 7:46 am
Please suggest the solution to this problem(OG page number 283, question number: 123), it seems OG has given wrong solution.
IT is a DS problem:

Joanna bought only $0.15 stamps and $0.29 stamps.
How many $0.15 stamps did she buy?
(1) She bought $4.40 worth of stamps.
(2) She bought an equal number of $0.15 stamps
and $0.29 stamps.


Please help ASAP!
Last edited by PankajRai on Fri Jun 10, 2011 8:11 am, edited 1 time in total.
Top
Source: — Data Sufficiency |
User avatar
Legendary Member
Posts: 1309
Joined: Mon Apr 04, 2011 5:34 am
Location: India
Thanked: 310 times
Followed by:123 members
GMAT Score:750

by cans » Fri Jun 10, 2011 7:55 am
let $0.15 stamps = x and $.29 stamps =y
to find x
a).15x + .29y = 4.4
insufficient
(one equation two variables)
b)x=y
insufficient. (no other information)
a&b) .15x + .29x = 4.4 ->x=10
Sufficient
IMO C
If my post helped you- let me know by pushing the thanks button ;)

Contact me about long distance tutoring!
[email protected]

Cans!!
Top
Legendary Member
Posts: 1448
Joined: Tue May 17, 2011 9:55 am
Location: India
Thanked: 375 times
Followed by:53 members

by Frankenstein » Fri Jun 10, 2011 8:03 am
Hi,
From(1): 0.15x+29y = 4.4 =>15x+29y = 440
So, 15x = 440-29y
x=10 and y=10 satisfies
As x and y are both integers, y can only be a multiple of 5 to satisfy this condition
for y=5, 15x = 440-29*5 = 295(not divisible by 15)
for y=15, 15x = 440- 29*15 = 5 ( not divisible by 15)
So, x=10, y=10 is a unique pair
Sufficient

From(2): x=y
Insufficient


(or)
Alternate method:
From(1): 15x+29y = 440, with 0<x<30 and 0<y<16
=>15x + (44-15)y = 440 =>15(x-y) = 44(10-y).
15 and 44 are relatively prime. So, for equality to hold, x-y = 44n and 10-y = 15n. This gives x-10 =29n.
So, x=29n+10
As 0<<x<30, only value for which equality can hold is when n=0. So x= 10 and y=10(unique set)


Hence, A
Cheers!

Things are not what they appear to be... nor are they otherwise
Top
User avatar
Junior | Next Rank: 30 Posts
Posts: 12
Joined: Wed May 18, 2011 10:28 am

by PankajRai » Fri Jun 10, 2011 8:19 am
Initially I opted for Option C, which is absolutely correct.
Within the given time frame, I did not reach upto finding the values of x and y.
Frankstien, how do you think that you need to find the values of X and Y.
Top
Legendary Member
Posts: 1084
Joined: Fri Apr 15, 2011 2:33 pm
Thanked: 158 times
Followed by:21 members

by pemdas » Fri Jun 10, 2011 8:21 am
i think the approach was unclear and would result either in waste of time to derive to "multiple of 5 theory" or guess-correct/wrong pathway

we know that 29 is prime, hence y can be either 0,1,2,3,4 ... we continue by picking only y values and dividing 440-29y by 15 (or by both 3 and 5)
y=0, 440/15
y=1, 411/15
...
y=10, 150/15



Frankenstein wrote:Hi,
From(1): 0.15x+29y = 4.4 =>15x+29y = 440
So, 15x = 440-29y
x=10 and y=10 satisfies
As x and y are both integers, y can only be a multiple of 5 to satisfy this condition
for y=5, 15x = 440-29*5 = 295(not divisible by 15)
for y=15, 15x = 440- 29*15 = 5 ( not divisible by 15)
So, x=10, y=10 is a unique pair
Sufficient

From(2): x=y
Insufficient


(or)
Alternate method:
From(1): 15x+29y = 440, with 0<x<30 and 0<y<16
=>15x + (44-15)y = 440 =>15(x-y) = 44(10-y).
15 and 44 are relatively prime. So, for them to be equal 10-y should be a multiple of 15 and (x-y) should be a multiple of 44 at the same time.
As 0<y<16, only value for which equality can hold is when 10-y = 0 and x-y=0
So, x=y=10.

Hence, A
Success doesn't come overnight!
Top
Legendary Member
Posts: 1448
Joined: Tue May 17, 2011 9:55 am
Location: India
Thanked: 375 times
Followed by:53 members

by Frankenstein » Fri Jun 10, 2011 8:24 am
PankajRai wrote:Initially I opted for Option C, which is absolutely correct.
Within the given time frame, I did not reach upto finding the values of x and y.
Frankstien, how do you think that you need to find the values of X and Y.
Hi,
Can you be more specific what you intended to ask because I have already explained how to find values of x and y.
Cheers!

Things are not what they appear to be... nor are they otherwise
Top
User avatar
Junior | Next Rank: 30 Posts
Posts: 12
Joined: Wed May 18, 2011 10:28 am

by PankajRai » Fri Jun 10, 2011 8:36 am
Frankenstein, Thanks for the help.
My question is : Why did you think that you need to find the values of Y from the Stmt 1 itself( This thought did not come to my mind, by seeing the Statement 1.), it was very clear and easy to take both the stmts and go for option C?
Top
Legendary Member
Posts: 1448
Joined: Tue May 17, 2011 9:55 am
Location: India
Thanked: 375 times
Followed by:53 members

by Frankenstein » Fri Jun 10, 2011 8:42 am
PankajRai wrote:Frankenstein, Thanks for the help.
My question is : Why did you think that you need to find the values of Y from the Stmt 1 itself( This thought did not come to my mind, by seeing the Statement 1.), it was very clear and easy to take both the stmts and go for option C?
Hi,
One reason is clearly 15+29 makes 44. So, just wanna check if it is the only pair possible or not.
Cheers!

Things are not what they appear to be... nor are they otherwise
Top
User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pm
Location: New York, NY
Thanked: 13060 times
Followed by:1906 members
GMAT Score:790

by GMATGuruNY » Fri Jun 10, 2011 8:57 am
PankajRai wrote:Frankenstein, Thanks for the help.
My question is : Why did you think that you need to find the values of Y from the Stmt 1 itself( This thought did not come to my mind, by seeing the Statement 1.), it was very clear and easy to take both the stmts and go for option C?
When a problem is restricted to positive integers, the odds increase that 1 linear equation with 2 variables will provide sufficient information.
The DS above is restricted to positive integers; Joanna cannot buy 1/2 a stamp or -3 stamps.
Thus, given the information in statement 1, we need to determine whether more than one combination of positive integers will satisfy the equation 15x + 29y = 440.
Since only combination of positive integers works (x=10 and y=10), statement 1 is sufficient.
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.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Top

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 2623
Joined: Mon Jun 02, 2008 3:17 am
Location: Montreal
Thanked: 1090 times
Followed by:355 members
GMAT Score:780

by Ian Stewart » Fri Jun 10, 2011 10:07 am
PankajRai wrote: it was very clear and easy to take both the stmts and go for option C?
This should set off some kind of 'warning bell' in your head. If it's so easy to see that both statements together answer the question, then why would the GMAT bother asking the question? Everyone would get it right, in which case the question serves no purpose. When it is "too obvious" that you can answer a question using both statements together, you should be absolutely certain you've considered whether either statement might be sufficient alone. Sometimes, as in the question above, one statement can surprisingly turn out to give you sufficient information. The obvious answer is often the wrong answer on the GMAT.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

ianstewartgmat.com
Top
Master | Next Rank: 500 Posts
Posts: 116
Joined: Tue May 31, 2011 7:52 pm
Location: Bangalore, India
Thanked: 2 times
Followed by:2 members

by Sanjay2706 » Sat Jun 11, 2011 5:33 am
IMO C is the answer.
Top
Senior | Next Rank: 100 Posts
Posts: 39
Joined: Wed Jan 19, 2011 7:57 am
Location: bangalore

by sandeep1306 » Sat Jun 11, 2011 6:01 am
GMATGuruNY wrote:
PankajRai wrote:Frankenstein, Thanks for the help.
My question is : Why did you think that you need to find the values of Y from the Stmt 1 itself( This thought did not come to my mind, by seeing the Statement 1.), it was very clear and easy to take both the stmts and go for option C?
When a problem is restricted to positive integers, the odds increase that 1 linear equation with 2 variables will provide sufficient information.
The DS above is restricted to positive integers; Joanna cannot buy 1/2 a stamp or -3 stamps.
Thus, given the information in statement 1, we need to determine whether more than one combination of positive integers will satisfy the equation 15x + 29y = 440.
Since only combination of positive integers works (x=10 and y=10), statement 1 is sufficient.


IMHO, this is the most plausible explanation that GMATGuruNY has given.Of course the solution is correct but it is easy to fall for the trap laid out by GMAC, especially if you get this kind of question when there are just two minutes to solve it.If you are skeptical for every easy C and E, i think you are going to waste a lot of time. what do others opine??


Thanks,
Sandeep
Thanks,
Sandeep
Top
Senior | Next Rank: 100 Posts
Posts: 42
Joined: Fri Jul 03, 2009 9:52 am
Location: San Jose, CA
Thanked: 2 times

by rveeraga » Sat Jun 11, 2011 1:47 pm
At first look, one may think that there could be several combination of multiples for .15 and .29 to result in 4.4. But, in Data Sufficiency, statements 1 and 2 do not contradict each other. So, by using the information in statement 2, not directly but indirectly, you can take it as only equal number of multiples for .15 and .29 will result in 4.4. It is easy to identify 10 as one equal multiple from the numbers. When you identify equal multiple 9 results in less than 4.4 and 11 results in more than 4.4, you can conclude there is only one possible combination of multiples for .15 and .29 to result in 4.4.
Top
Post new topic Post Reply
• Page 1 of 1