Computer Science Canada Math problem |
| Author: | tjmoore1993 [ Mon May 11, 2009 3:01 pm ] |
| Post subject: | Math problem |
I am not sure what this is classified under but my friend didn't quite comprehend this as well and needs to figure it out. Scenario: Bobby and his friend went the the carnival. Children pay $1.50 and adults pay $4.00 to gain admission. On certain days, they had 200 guests and made a total of $550. How many adults went and how many children went. Given C = Children A = Adult I am not sure on how to set up the equation. If someone could help then thanks  |
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| Author: | DemonWasp [ Mon May 11, 2009 3:24 pm ] |
| Post subject: | RE:Math problem |
Let C be the number of children. Let A be the number of adults. You should be able to figure out an equation for profit, P, in the form: P = some function of A and C, as well as the prices 1.50 and 4.00. Your other equation is that C+A = 200. You know this because the total number of guests is C+A, and the number of guests on that day is 200. |
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| Author: | tjmoore1993 [ Mon May 11, 2009 3:44 pm ] |
| Post subject: | Re: RE:Math problem |
DemonWasp @ Mon May 11, 2009 3:24 pm wrote: Let C be the number of children. Let A be the number of adults. You should be able to figure out an equation for profit, P, in the form:
P = some function of A and C, as well as the prices 1.50 and 4.00. Your other equation is that C+A = 200. You know this because the total number of guests is C+A, and the number of guests on that day is 200. Well what we are trying to do is something like 1x + 2y = 2000 2x + 3y = 2000 It can't be done so we must do this (1x + 2y = 2000 ) * (2) 2x + 4y = 4000 -2x + 3y = 2000 -------------------- 1y = 2000 y = 2000 2x + 4(2000) = 4000 2x + 8000 = 4000 2x = 4000 - 8000 2x = -4000 x = -2000 X = -2000 Y = 2000 |
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| Author: | DanielG [ Mon May 11, 2009 5:03 pm ] |
| Post subject: | RE:Math problem |
that's one way of doing it, but not the only way, for your example of: 1x+2y=2000 2x+3y=2000 you can also solve it by doing folliwing move y to other side for first equation so x = 2000-2y substitute that back into the x of the second equation so 2(2000-2y) + 3y = 2000 solving for y gives you the y value, then it is easy to solve for x. |
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| Author: | jeffgreco13 [ Mon May 11, 2009 5:56 pm ] |
| Post subject: | RE:Math problem |
550 = 1.5C + 4A 200 = C + A ------------------ A = 200 - C 550 = 1.5C + 4(200 - C) 550 = 1.5C + 800 - 4C -250 = -2.5C C = 100 550 = 150 + 4A A = 100 |
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| Author: | jbking [ Mon May 11, 2009 6:16 pm ] |
| Post subject: | Re: Math problem |
There are a couple of other ways to solve this, to my mind if one wanted to circumvent equations: 1) Note that $1.50+$4.00 = $5.50 or 1% of what was made. Thus there were 100 of each for an initial guess that would total $550 taken in and also passes the other required condition of there being 200 guests. 2) Note that there is a difference of $2.50 as part of changing the admission. If all 200 guests were either children or adults we would have incorrect totals of $300 or $800 depending on which way one wanted to assume. The difference between those and the $550 take in is $250 which means that 100 adults or 100 children had to be in there, respectively. Granted these are sort of "tricks" to get those values of 100 for each the number of children and adults but sometimes problems are set up this way. |
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| Author: | tjmoore1993 [ Tue May 12, 2009 2:59 pm ] |
| Post subject: | RE:Math problem |
I solved the problem! The actual equation is 1.50c + 4.00a = 5050 c + a = 2200 -------------------------- (c+a = 2200) * 4.00 1.50c + 4.00a = 5050 - 4.00c + 4.00a = 8800 ---------------------------- -2.50c + 0.00a = -3750 -2.50c = -3750 c = -3750 / -2.50 c = 1500 1.50c + 4.00a = 5050 1.50(1500) + 4.00a = 5050 2250 + 4.00a = 5050 4.00a = 5050 - 2250 4.00a = 2800 a = 2800 / 4.00 a = 700 .: There were a total of 1500 children and 700 adults attending the carnival. |
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| Author: | andrew. [ Tue May 12, 2009 8:48 pm ] |
| Post subject: | RE:Math problem |
I did it like this: Let C represent the number of children and A represent the number of adults. 1.50C + 4.00A = 550 C + A = 200 A = 200 - C 1.50C + 4.00(200 - C) = 550 1.50C + 800 - 4.00C = 550 1.50C - 4.00C = 550 - 800 -2.50C = -250 C = 250/2.50 C = 100 A = 200 - C A = 200 - 100 = 100 Therefore there were 100 children and 100 adults. Edit: Just wanted to add, this is the way they teach you in grade 10. I figured it would be more helpful to your friend to see it the way they teach it in school. |
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| Author: | matt271 [ Tue May 12, 2009 11:12 pm ] |
| Post subject: | Re: Math problem |
1.50C + 4A = 550 C + A = 200 [C + A = 200] * 4 4C + 4A = 800 [4C + 4A = 800] -[1.50C + 4A = 550] 2.5C = 250 C = 250/2.5 = 2500/25 = 100 there C = 100 C + A = 200 A = 200 - C = 200 - 100 = 100 therefore A = 100 |
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| Author: | Clayton [ Wed May 13, 2009 7:54 pm ] |
| Post subject: | RE:Math problem |
When I'm solving a problem like this, I like to have as few variables as possible (makes for neater/easier modelling). So we get: Let A be the number of adults attending the carnival. Let (200 - A) be the number of children attending the carnival. 4A + 1.5(200 - A) = 550 2.5A + 300 = 550 2.5A = 250 A = 250/2.5 A = 100 .: there was 100 adults and (200 - 100) = 100 children at the carnival. |
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| Author: | jeffgreco13 [ Thu May 14, 2009 12:02 am ] |
| Post subject: | RE:Math problem |
i wonder how many more ppl will post the same answer to this question? |
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| Author: | [Gandalf] [ Thu May 14, 2009 12:07 am ] |
| Post subject: | RE:Math problem |
Haha I was thinking the same thing... Why is everyone posting their version of the solution? Isn't 5 answers enough?   |
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| Author: | jernst [ Thu May 14, 2009 8:17 am ] |
| Post subject: | Re: RE:Math problem |
[quote="[Gandalf] @ Thu May 14, 2009 12:07 am"]Haha I was thinking the same thing... Why is everyone posting their version of the solution? Isn't 5 answers enough? [/quote]
everyone wants to feel smart sometimes  |
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| Author: | Clayton [ Thu May 14, 2009 9:27 am ] |
| Post subject: | RE:Math problem |
My point was there is more than one way to solve any given problem. My method had not been posted, so I did. |
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| Author: | matt271 [ Thu May 14, 2009 8:29 pm ] |
| Post subject: | RE:Math problem |
most ppl prob read the question and none of the replies  |
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