Hi vinviper1,
Doing this question quickly and efficiently requires that you be a very smart test taker.
First, notice that Kaye must begin with a number of stamps that is a multiple of 5, and end with a number of stamps that is a multiple of 7. So we can quickly eliminate a lot of possibilities (for example, she can't possibly start with 60 stamps, because after giving away 10, she'll be left with 50, which is not divisible by 7).
Thus, we can determine that she must have started with a number of stamps that's 10 more than a multiple of 7. So our only choices are 45, 80, 115, 150, etc., since all those numbers are divisible by 5, and after subtracting out 10, are divisible by 7.
Now, we notice that the answer choices are pretty big, so we want them to have a large spread between the number of stamps, and we'll probably want to go towards the higher numbers. Try plugging in 150. If Kaye started with 150, then Alberto started with 90 (according to the 5:3 ratio). After Kaye gives away 10 stamps to Alberto, she'll be left with 140 while Alberto will have 100 (a ratio of 7:5 as desired).
Therefore, the difference between Kaye's ending 140 stamps and Alberto's 100 stamps is 40, the correct answer.
I'm sure there's some sort of equation to do this too, but in my opinion, thinking about problems logically will save you more time and energy on the test in the long run.
Hope this helps!
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Thanks Jim. Developing that skill is indeed a big help. I did it in equation format which makes sense...but it is "multi stepped" Tough to commit to when you are on Q 31.
I think that those question can be very quickly solve with equation. At question 30 with timing pressure and stress from the test, having a standardised way of thinking about it can be easy. (anyway, that is just my opinion). I think the 'think in a clever way' approach is not that easy to apply in a stressful situation.
Equation is a 5-8 steps approach.
Before exchange
K/A = 5/3 <-> A = 3K / 5
After exachange
K-10 / A+10 = 7/5
If you combine the two:
(K - 10) / ((3K/5)+10) = 7/5
5K - 50 = (21K/5) + 70
(25K - 250)/5 = (21K + 350)/5
4K = 600
K = 150
So before exchange: K = 150 and A = 450/5 = 90
After exchange: K = 140 and A = 100
Equation is a 5-8 steps approach.
Before exchange
K/A = 5/3 <-> A = 3K / 5
After exachange
K-10 / A+10 = 7/5
If you combine the two:
(K - 10) / ((3K/5)+10) = 7/5
5K - 50 = (21K/5) + 70
(25K - 250)/5 = (21K + 350)/5
4K = 600
K = 150
So before exchange: K = 150 and A = 450/5 = 90
After exchange: K = 140 and A = 100
