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hardest DS, help pls

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hardest DS, help pls

by tanviet » Wed Oct 07, 2009 4:15 am
this com from 15 test sets, a good source

Joana bought only 0.15 usd stamp and 0.29 usd stamp. How many 0.15 stamp did he buy

a, she bought 4.40 worth of stamp

b, she bought an equal number of 0.15 usd stamp and 0.29 usd stamp

this is a hard not easy question. pls caution
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by fiercepoint » Wed Oct 07, 2009 4:44 am
Is the answer C?

x = no of $0.15 stamps
y = no of $0.29 stamps

1) 0.15x + 0.29y = 4.40
Cannot be solved.

2) x = y
This alone doesn't help either

1&2) 0.15x + 0.29x = 4.40
Can be solved, x = 10
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by JeffB » Wed Oct 07, 2009 7:36 am
fiercepoint wrote:Is the answer C?

x = no of $0.15 stamps
y = no of $0.29 stamps

1) 0.15x + 0.29y = 4.40
Cannot be solved.

2) x = y
This alone doesn't help either

1&2) 0.15x + 0.29x = 4.40
Can be solved, x = 10
A is actually sufficient. There is only one way to have $4.40 in 15 and 29 cent stamps. (10 and 10)
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by srivas » Wed Oct 07, 2009 8:01 am
if there is only one way to have $4.40 in 15 and 29 cent stamps. (10 and 10)

then answer should be D
Gmat710,, Hyd
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by Dobrov » Mon Oct 12, 2009 2:15 pm
A is correct.

If a is the nr. of stamps that cost 0.15, and b is the nr. of stamps that cost 0.29, than we know from stml 1 that

0.15a + 0.29b = 4.4

There is only one combination that satisfies this condition: a and b = 10

Stml 2 just says that a=b, but doesnt quote the total purchase price. So a could be 1,2,3,4,5,6,7... not suff.
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by mehravikas » Mon Oct 12, 2009 5:19 pm
Should be 'A' because you can get 4.40 only in one case:

0.15 * 10 + .29 * 10 = 4.40

Statement 2 is insufficient because it does not give the total amount.
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by tanviet » Tue Oct 13, 2009 1:37 am
D is correct. this is from 15 test set

what I want to know is how do you know that A is correct , that you can know 2 solutions with 1 operation

pls help
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by glorydefined » Tue Oct 13, 2009 3:00 am
OA is "A" its an OG 12 question PS # 123
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by mehravikas » Tue Oct 13, 2009 11:14 am
D can never be correct.

Statement 1 - Try to calculate the number of .15 and .29 solutions and ensure that they add up to 4.40.
duongthang wrote:D is correct. this is from 15 test set

what I want to know is how do you know that A is correct , that you can know 2 solutions with 1 operation

pls help
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by Talkativetree » Tue Oct 13, 2009 5:15 pm
when you see that it is .29 and .15, and that they will add up to 4.40. With the formula .15x + .29y = 4.40, you can tell that y will need to be a multiple of 5, because of the 0 in the digits column of 4.40. This means that it will either be 5y, 10y, or 15y. to do a quick calculation,
5y=.29(5)=1.45
10y=2.9
15y=4.35

from this, plug those numbers quickly into the equation .15x+.29y=4.40, and y=10, x=10.

As for the answer being D, there is no way I can think of to solve the problem with only answer choice II.

Answer should be A.
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by tanviet » Wed Oct 14, 2009 11:23 pm
OK, A is correct, not D

but pls, explain why we can find out y must be multiple of 5, I am confused
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by life is a test » Thu Oct 15, 2009 11:14 pm
duongthang wrote:OK, A is correct, not D

but pls, explain why we can find out y must be multiple of 5, I am confused
15x + 29y = 440.

since the total must end in 0 it means that either both x and y need to end in 5 or 0. For 29*y to end in 5 or 0, y must be a multiple of 5 or 10.
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A is correct

by nakul_anand » Fri Oct 16, 2009 2:43 pm
Equation - 0.15x + 0.29y = 4.40

The only integer values of x and y that would result in 4.40 are x = y = 10.
Since the number of coins has to be an integer, the value of x and y is 10

Now statement I gives you the sum = 4.40

Statement II does not give you a sum. So equal number of coins could be 20,or 30, or 1000. So we need the total dollar value to calculate the number of coins.

Since this is given in statement I, statement I is sufficient.

Answer - (A)
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by Talkativetree » Sat Oct 17, 2009 2:33 pm
Okay, here's a very basic explaination of the problem. We know x and y must be integers, because we cannot have half of a stamp. After we understand that, we can go to the equation, which is .15x + .29y = 4.40

The equation is very helpful because it gives us a guideline for how to find combinations of x and y, where .15x + .29y = 4.40. NOW, looking at the equation, we see that the equation equals 4.40. Y must be a multiple of 5, because 1y=.29 and there is no value of x where .15x +.29=4.40. The same goes for 2y (.58), 3y(.87), and so forth. the only values for y that would possibly create the number 4.40, when added to .15x, are 0, 5, 10, and 15.

A really useful trick on the Gmat is to use the digits in a number to figure out the answer. so in this problem, we simply look at the 5 in .15, the 9 in .29, and 0 in 4.40, and think "what combinations of 5 and 9 can I make to get a 0). That's the underlying idea behind how I figured out we need a multiple of 5, because 5x9=45 and 45+15=60. OR 10x9=90 and 10x5=50 and 90+50=140.

Does this help?
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