How do I solve this mathematical word problem without using trial and error?
Timmy has almost $2.00 in allowance money that he wants to spend on candy. He has two favorite candies, one selling at $0.15 a piece and one at $0.08 a piece. If he ends up spending $1.92 on 17 pieces of candy, how many $0.15 pieces of candy did Timmy decide to buy?
Is there a formula I can apply to this? Find the answer and show me how you did it.
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11 Answers
It’s a simple linear system. Two equations, two variables.
Now think about it some more.
@bob_ is correct…
E1: a + b = 17 (two types of candy, but we know the total of both he bought is 17)
E2: 0.15 a + 0.08 b = 1.92 (we know total amount spent and how much each costs)
We want to know ‘a’, so…
Rearrange E1… b = 17 – a
Substitute into E2… 0.15 a + 0.08 (17 – a) = 1.92
Solve for a.
17 * 0.15 = 2.55
2.55 – 1.92 = 0.63
0.15 – 0.08 = 0.07
0.63 / 0.07 = 9
17 – 9 = 8
That is how I did it.
@jerv Clever. Not the most direct way, but it works.
@jerv @hiphiphopflipflapflop Thanks guys, those are both good answers. When i saw this question I immediately picked a random amount for the .15 cent candies, checked the remainder, and reached 8 by taking a few guesses and checking them.
@jerv it took me a while to extract the thought process there. Agree with @bob_ again, indirect and clever.
I have a knack and a reputation for doing things my own way.
A somewhat offbeat way of doing it.
Average price of candy = 1.92/17 = 11,3 cents (rounded)
If p is the fraction of 15 cent candies, then looking at it as a weighted average,
15 p + 8(1-p) = 11.3
7p = 3.3
p=,47 (rounded)
.47*17 = 8
It is unfortunate that weighted averages are not taught in schools. It is, after all, how people’s grades are determined.
One might argue that my method is just a complication on top of that of @hiphiphopflipflapflop
What I am doing in effect is dividing his equation by the total number 17.
0.15 a /17+ 0.08 (17 – a) /17= 1.92/17
I would argue though that average value takes on a life of its own. After computing the average price of 11.3 cents, you can compute (8+15)/2 = 11.5. That means that 11.3 is closer to 8 than 15. Without doing any other calculation, you know there will be more 8 cent candies than 15 cent ones. Further, since 11.3 is so close to 11.5, you know the numbers of the two candies will be pretty close.
8*17 = 136
15 – 8 = 7
192 – 136 = 56
56/7 = 8
@ratboy Yes, solving from the bottom and working up works just as well as going from the top down.
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