Thx parallel chase n other members, nice trick hidden in the prompt itself.

Is the below apporach always correct

I think that the best way to crack this one is:

1� found de MCM of 0.29 and 0.15 (it is better to use 29 and 15)
2� then if the MCM of 29 and 15 is less than 440 there is only one solution...

MCM (29x5x3) = 435 < 440 . So A is the correct answer.

I think that this is the best way to crack this kind of exercise

Well the conclusion is that you need to double check if you find this easy 2 unknowns instead of concluding that you dont have sufficient info you need to the verify if there's a possibility to get to the answer using integers..

damn.. i went for the C trap

II 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.

Can you please provide you logic in reaching the answer. Thanks.

Another way to think about this question is as follows: there must be at least one solution, since data sufficiency always gives facts that are not contradictory. In other words, there must be postive integers x and y such that 15x + 29y = 440.

To exhange 15 cent stamps for 29 cent stamps in such as way that 440 cents worth of stamps are still bought, they must be exchanged in a ratio of 29:15 , a ratio that cannot be reduced, as 29 and 15 have no prime factors in common. Thus finding another integer pair x,y would involve increasing or decreasing the number of 15 cent stamps by a multiple of 29, i,e. adding or subtracting 435 cents to their value, something that cannot be done because we have exactly 440 cents to spend. Thus there can be only one solution.

I am afraid I dont quite completely understand what kevin says here. Pls could some expert make it lucid for me? thanks

kevincanspain wrote: To exhange 15 cent stamps for 29 cent stamps in such as way that 440 cents worth of stamps are still bought, they must be exchanged in a ratio of 29:15 , a ratio that cannot be reduced, as 29 and 15 have no prime factors in common. Thus finding another integer pair x,y would involve increasing or decreasing the number of 15 cent stamps by a multiple of 29, i,e. adding or subtracting 435 cents to their value, something that cannot be done because we have exactly 440 cents to spend. Thus there can be only one solution.

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 important to note how a problem is RESTRICTED.
This problem is restricted to POSITIVE INTEGERS: Joanna cannot by 1/2 of a stamp or -2 stamps.

Statement 1: 15x + 29y = 440.
When a problem is restricted to positive integers, BE SUSPICIOUS: quite often 1 linear equation with 2 variables will provide sufficient information to determine the value of each variable.
Here, we need to determine whether more than one combination of positive integers will satisfy the equation 15x + 29y = 440.

Rephrasing the equation, we get:
29y = 440 - 15x = (multiple of 5) - (multiple 5) = multiple of 5.

Since 29y must be a multiple of 5, there are only 3 options:
Case 1: 5*29 = 145.
Case 2: 10*29 = 290.
Case 3: 15*29 = 435.

In each case, the amount remaining must be a MULTIPLE OF 15:
Case 1: 440-145=295, which is not a multiple of 15.
Case 2: 440-290=150, which is a multiple of 15.
Case 3: 440-435=5, which is not a multiple of 15.

Only Case 2 is viable:
15x = 150, implying that 10 15-cent stamps are purchased.
SUFFICIENT.

Statement 2: She bought an equal number of $0.15 stamps and $0.29 stamps.
No way to determine how many 15-cent stamps were purchased.
INSUFFICIENT.

The correct answer is A.

Nina1987 wrote:I am afraid I dont quite completely understand what kevin says here. Pls could some expert make it lucid for me? thanks

Let $F = the total revenue from the 15¢ stamps and $T = the total revenue from the 29¢ stamps.

Since the two statements cannot contradict each other, it must be possible in statement 1 that the 440¢ in total revenue is divided according to the ratio indicated in statement 2.
Statement 2 indicates that the ratio of 15¢ stamps to 29¢ stamps = 1:1.
If an equal number of each type of stamp is purchased, we get:
$F : $T = 15:29 = 150:290, for a total of 440¢ in revenue.
Here, the number of 15¢ stamps = 150/15 = 10, and the number of 29¢ stamps = 290/29 = 10.

Check whether OTHER revenue ratios are also possible.
Since the stamp values are 15¢ and 29¢, the revenue ratio can be altered only by adding a multiple of 15 and 29 to $F or $T, while subtracting this same multiple from the remaining value.
The LCM of 15 and 29 = 15*29 = 435.
If 435¢ is added to either 150¢ or 290¢, the sum will exceed 440¢.
Thus, the revenue ratio CANNOT be altered, implying that only ONE revenue ratio will satisfy statement 1:
$F = 150¢ and $T = 290¢.
Thus, the number of 15¢ stamps = 10.
SUFFICIENT.