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!
Number of Stamps
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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
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
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Cans!!
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Cans!!
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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
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!
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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
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
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Hi,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.
Can you be more specific what you intended to ask because I have already explained how to find values of x and y.
Cheers!
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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?
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?
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Hi,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?
One reason is clearly 15+29 makes 44. So, just wanna check if it is the only pair possible or not.
Cheers!
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When a problem is restricted to positive integers, the odds increase that 1 linear equation with 2 variables will provide sufficient information.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?
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.
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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.PankajRai wrote: it was very clear and easy to take both the stmts and go for option C?
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GMATGuruNY wrote:When a problem is restricted to positive integers, the odds increase that 1 linear equation with 2 variables will provide sufficient information.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?
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
Sandeep
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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.