Hi!
Normally for this type of question we can rely on our old friend, "number of equations vs number of unknowns". However, there are some cases, such as this one, in which you have more info that you originally thought and you don't actually need as many equations as you may think.
The key to solving this question is noting that stamps are indivisible - in other words, Joanna has to by an integer number of each denomination. Always be on the lookout for objects that are indivisible!
Let's call the two types of stamps c (for cheap) and e (for expensive). From (1), we know that:
15c + 29e = 440
Now, without knowing that c and e have to be integers, we'd quickly say that (1) is insufficient, since we have 2 variables and only 1 equation. However, since c and e have to be integers, it's possible that there's only one pair of values that satisfy the equation. So, we need a quick way to determine if there's one unique solution.
A great way to solve quickly is to pay attention to the units digit of the sum. Since 440 ends in a 0, and since 15c will end in either 5 or 0 every time, 29e must also end in 5 or 0. Focusing on values of e that are multiples of 5 should make the process fairly quick.
e=5... 29e = 145. 440-145=295. Is 295 divisible by 15? NO (since 15 goes into 300)
e=10... 29e = 290. 440-290=150. Is 150 divisible by 15? YES
e=15... 20e = 435. 440-435=5. Is 5 divisible by 15? NO
So the only possible value for e is 10 (which means there's also a unique value for c). (1) is sufficient alone... choose (A)!
(I skipped over looking at (2) - we can pick any equal values for c and e to make (2) true.)
kheba wrote:Joanna bought only $0.15 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.
OA A
I'd like to see what are the efficient ways of solving it. I took more than 2 minutes for this problem.
Thanks
