Mason had 50% more cards than Isaac. Evelyn had 20% fewer cards than Mason. Mason and Isaac gave Evelyn some cards in the ratio of 4:5. As a result, Evelyn had 25% more cards than before. Given that Evelyn had 240 more cards than Isaac in the end, how many cards did Mason give to Evelyn?
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Question
Answer
Issac | Mason | Evelyn | |
Start | 30u | 45u | 36u |
End | 25u = 30u – 5u | 41u = 45u- 4u | 45u = 36u + 9u |
Fill up the table above according to the reasoning below:
- Suppose Issac had 30u.
- Then Mason had 45u = 50% more than Issac.
- Evelyn had 36u = 20% fewer than Mason.
- Evelyn had 9u (25% x 36) more after Mason and Issac gave her some.
- That means Mason gave 4u and Issac gave 5u.
- Difference between Evelyn and Issac = 45u - 25u = 20u = 240.
- So u = 12.
- Mason gave 4u = 48 cards to Evelyn.
Err… how did you get the 30u, 45u… etc?
This model making thing is driving me nuts 🙂
Good question! I forgot to talk about it. Suppose we model our question using unit u. Let t = u/K so u = Kt. Then we can replace u with Kt throughout in our model. So we can solve for t first and can get back u because u = Kt. In short, we can always start with Ku for a suitably chosen scale K. The value of K is usually some multiple of the fractions appearing in the question. If K is chosen carefully, subsequent working is simpler. I actually started with Issac = 10u because it is easy to do 50% more for Mason (i.e. 15u) and 80% = 4/5 of 15u. This is what I got:
Issac | Mason | Evelyn | |
Start | 10u | 15u | 12u |
End | 15u = 12u + 3u |
At this point, I realized the 3u has to be broken up into 9 parts
because of 4:5. That suggests scaling up 3u by 3 to become 9u.
So we start with Issac = 30u and the rest follows.
There is no magic - just hard work. Do some quick exploration
(in pencil, so it is erasable) and then going back to write down
the final working (in pen).
The problem is under exam conditions, time mgt is important. So for those lacking in practice to know which whole number to use upfront, it’s better to know how to deal with fractions and move on from there to solve the problem instead of backtracking to make it into whole numbers or use the units transfer method where in this case, I could express everything in whole numbers.
I agree with you that it’s a matter of lots of practice. Without the sufficient practice to know exactly which whole number to use, to change into whole numbers to make the question look neater is not worth the time taken if one knows how to arrive at the final answer using fractions. Just my 2 cents worth for students who may not practise a lot like me since I am not a student. 🙂 However, for students who can do that, whole numbers is definitely neater to look at. So do whatever is best to ace your exams.
Just to clarify, I am not advocating that students must scale their units correctly all the time. Chief was wondering how I arrived at 30u right from the start, so I replied to him to demystify the process.
Scaling units is just another tool in the problem solver’s toolbox. Not some magic bullet. For some questions, it might not be feasible. But for students who practice enough to master it, it can save time and reduce the chance of arithmetic fumbles. This can also be said for students who practice enough to master fractions.
At the end of the day, like you said, practice is the key.
I think we are in sync on this. There is no short-cut, just master whatever suits you best.
Actually I was not going to post my solution but since Chiefkiasu asked how to get those numbers, I will submit my alternative solution (units transfer method)
Mason Issac Evelyn
At first 1.5(10p) =15p 10p 0.8(15p) = 12p
Give to Evelyn 4u 5u 1.25(12p)= 15p (so incr. by 3p)
In the end 15p-4u 10p-5u 15p
4u + 5u = 9u = 3p ==> p = 3u
Evelyn – Issac = 240
15p – (10p-5u) = 240
5p+5u = 240
p + u = 48
3u + u = 48
4u = 48
Mason gave 48 cards to Evelyn