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Joanna bought only $0.15 stamps and $0.29 stamps. How many

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Senior Manager
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Joanna bought only $0.15 stamps and $0.29 stamps. How many [#permalink] New post 15 Sep 2007, 14:45
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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.
VP
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Re: DS : Set 6, Q30 - Joanna and her stamps [#permalink] New post 15 Sep 2007, 15:00
gluon wrote:
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.


it is combined 1 and 2.

she bought equal number of stamps so it can be denoted by x
.15x+.29x=4.40
.44x=4.4
x=10

so she bought 10 of .15 and 10 of .29
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 [#permalink] New post 15 Sep 2007, 18:38
Isn't (1) by itself sufficient to answer the question ? The only positive integer values of x and y that satifsy 0.15x+0.29y=4.40 are x=y=10.
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 [#permalink] New post 15 Sep 2007, 23:26
coldweather999 wrote:
Isn't (1) by itself sufficient to answer the question ? The only positive integer values of x and y that satifsy 0.15x+0.29y=4.40 are x=y=10.


Hey coolwaether999.. you are cool (right). Only 1 combination works that way.. So A is alone suff. How do you figure out these kind of solution ? I am sure, trial and error will consume lot of time ... do you just hear your ear on that ..?

But the OA that I have says the the correct ans as D. I am just not able to figure out how B alone suff..
can anyone shed some light..
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Re: DS : Set 6, Q30 - Joanna and her stamps [#permalink] New post 16 Sep 2007, 00:33
gluon wrote:
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.



S1)

$4.40--> 440 We know that 15(x) +29(y)=440. Since 440 has a 0 on the end of it we have to have 29(y) equal to a number that will give us 0 on the end. Since the other number is 15 we must multiply 29 by a number that will give us a 0 as the units digit.

29*10=290. 15*10=150 150+290=440. The only number that satisfys this condition for 29 is 10. Since 20*29 is too great.

Suff.

S2)

This has to be insuff. Ok so the number is equal... so what... we aren't given any constraints as in S1. I really don't see how this can be suff since we can have an infinite number of values.

Could have 5 stamps of .15 thus 5 stamps of .29. could have 15 stamps of .15 thus 15 of .29. These give us different values.

Gluon wut do u have as the OA?
Manager
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 [#permalink] New post 16 Sep 2007, 01:15
Yes. D is the answer.
Both the statements by themselves are sufficient.
It took some time, but trial and error seemed to be the only way for me to arrive at making St. 1 suff.
St. B is simple algebra.

Amit05 wrote:
coldweather999 wrote:
Isn't (1) by itself sufficient to answer the question ? The only positive integer values of x and y that satifsy 0.15x+0.29y=4.40 are x=y=10.


Hey coolwaether999.. you are cool (right). Only 1 combination works that way.. So A is alone suff. How do you figure out these kind of solution ? I am sure, trial and error will consume lot of time ... do you just hear your ear on that ..?

But the OA that I have says the the correct ans as D. I am just not able to figure out how B alone suff..
can anyone shed some light..
VP
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Re: DS : Set 6, Q30 - Joanna and her stamps [#permalink] New post 16 Sep 2007, 02:45
gluon wrote:
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.


how can 1 be alone sufficient. We cannot simply say she bought x numer of stamps of each
how do we know that she bought equal number of stamps from 1


how can 2 be alone suffuicient. we do not know how much she spend.


combined should be fine.
Senior Manager
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 [#permalink] New post 16 Sep 2007, 06:22
The OA is A.

I got it wrong while practising too and then the OA stunned me. I hate trick problems!!

Ravshonbek... I agree with you that we should not assume that x = y for (1). But if you look closely we dont need to.

This is a very tricky problem. We know that 0.15x + 0.29y = 4.40

If this was a generic equation then we could have x = -1 and y = (4.55/0.29)

But, since Joanna is buying stamps it is implicit that x and y are positive integers. Joanna cannot buy -1 or (4.55/0.29) stamps. The only positive integers that satisfy this equation are x = 10, y = 10. Hence, (1) is sufficient.

If there are some other positive integers that satisfy this equation then we would be correct about (1) being insufficient.

Last edited by gluon on 16 Sep 2007, 06:40, edited 2 times in total.
Manager
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Status: Post MBA, working in the area of Development Finance
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Kudos [?]: 2 [0], given: 1

 [#permalink] New post 16 Sep 2007, 06:30
Speed kills! Careless error!
St. by 2 itself is not suff. There could be innumerable combinations that could suffice the eqn. [Let the no. of stamps of $0.15 denomination be x and $0.29 deno. be y
Then x*$0.15+y*$0.29=?]

So, the trial and error will give the ans. A.
Only the combination of 10 each of the two stamps suffice st.1 Remember the no. of stamps has to be +ve integer.
Am not sure how B is suff.!

Artemov wrote:
Yes. D is the answer.
Both the statements by themselves are sufficient.
It took some time, but trial and error seemed to be the only way for me to arrive at making St. 1 suff.
St. B is simple algebra.

Amit05 wrote:
coldweather999 wrote:
Isn't (1) by itself sufficient to answer the question ? The only positive integer values of x and y that satifsy 0.15x+0.29y=4.40 are x=y=10.


Hey coolwaether999.. you are cool (right). Only 1 combination works that way.. So A is alone suff. How do you figure out these kind of solution ? I am sure, trial and error will consume lot of time ... do you just hear your ear on that ..?

But the OA that I have says the the correct ans as D. I am just not able to figure out how B alone suff..
can anyone shed some light..
Senior Manager
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 [#permalink] New post 16 Sep 2007, 07:10
I got into trap too, I have to read questions once again after finding the solution.

Ans: A
  [#permalink] 16 Sep 2007, 07:10
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