At a local festival, the entire town showed up, a total of 500 people. The
event raised $3,000. Tickets were priced as follows:
$7.48 per ticket for men
$7.12 per ticket for women
$0.45 per ticket for children.
How many children were there?
2007-10-15 08:59:26 · 7 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
You must show or tell how you got your answer!
2007-10-15 08:59:51 · update #1
Begin like sahsjing /1st answer above/:
748m + 712w + 45c = 300 000
m + w + c = 500, multiply by 712:
712m + 712w + 712c = 356 000 and subtract the above:
667c - 36m = 56 000, take a congruence by modulo 36:
667c ≡ 56000 (mod 36), that leads to
19c ≡ 20 (mod 36) or c ≡ 20 (mod 36), so
c = 20, 56, 92, 128 etc. /arithmetic progression, difference 36/, but the first 2 values lead to negative
m = (667c - 56000)/36, so the least possible is
c = 92, then m = (667*92 - 56000)/36 = 149 and w = 259.
On the other hand c = 128 produces m > 500 so no other values for c are possible.
Finally: 149 men, 259 women, 92 children is the ONLY solution to this nice Diophantine system.
P.S.(EDIT - after having read other answers) rhsaunders: "the number of kids' tickets must be a multiple of 20" - no, that's wrong, it must be of the form 20 + 36k /k=0,1,2,.../, if You have tried these values instead of 20, 40, 60 etc., You would have reached for k=2 the correct answer c = 92.
2007-10-15 09:29:06 · answer #1 · answered by Duke 7 · 2⤊ 0⤋
This is a problem in what is called Diphantine equations. There are often a multiplicity of solutions for the things. The key to solving this one is the rather oddball pricing: it requires some art to get the numbers to add up to exactly $3,000. In particular, the number of kids' tickets must be a multiple of 20, and the numbers of other tickets must obey M+W = 5k, where k is some integer. Also, we can conclude that the number of children present was greater than zero, as the numbers don't work otherwise. Also, a division shows that the number of women's tickets cannot exceed 421. Start with a bit of a table, showing number of kid's tickets along with possible number of women's tickets:
Kid's tix Women's tix
20...........420
40...........418
60...........417
80...........416
100.........415
We see that there must be at least 100 kid's tickets, as otherwise there would not be 500 sold. From this point, one can proceed to extend the list, seeing if a valid number of men's tickets can be determined using the k-constraint. No obvious way of simplifying this is apparent, so just grind out the numbers and see what happens.
2007-10-15 09:34:06 · answer #2 · answered by Anonymous · 0⤊ 0⤋
complex subject. look into into yahoo. just that can help!
2014-11-05 15:08:55 · answer #3 · answered by Anonymous · 0⤊ 0⤋
a million. hi-jinx 2. he's down contained in the dumps 3. a number of options: You pass as a lot because the decrease Up your decrease you're less than the decrease Your higher decrease 4. massive way 5. immediately jacket 6. start up on the starting up of the tale 7. turn to the left of him 8. Getting (get in) switched off line
2016-10-21 05:24:07 · answer #4 · answered by Anonymous · 0⤊ 0⤋
Sorry, Sounds more like homework than a brain teaser.
2007-10-15 09:08:11 · answer #5 · answered by jon_mac_usa_007 7 · 0⤊ 1⤋
7.48m+7.12w+0.45c = 3000
m+w+c = 500, where m, w, and c are positive integers
2007-10-15 09:03:02 · answer #6 · answered by sahsjing 7 · 0⤊ 0⤋
are you cheating on your homework?
2007-10-15 09:10:04 · answer #7 · answered by naruto_fan JC 3 · 0⤊ 1⤋