<p>Hey everyone! I answered a question in the Math sections in one of my practice test, but the solution took so long time, I was wondering if I got it the long way.. here's the question..
Tickets for a community play cost $2.00 for a child and $4.00 for an adult. If 200 tickets were sold for a total of $700, what was the ratio of the number of the children's tickets sold to the number of adults' tickets sold?</p>
<p>Solution:
(2x+4y=700) - (x+y=200) = 500
x/3 + 3y/y=500/3</p>
<p>x/3 + y = 500/3
y= 500/3 - x/3
y=1500-3x/9
x+(1500-3x/9)=200
6x+1500/9=200
1800=6x+1500
300=6x
x=50</p>
<p>50+Y=200
Y=150
50/150 = 1/3 (which is the correct answer)</p>
<p>I’d have used substitution.</p>
<p>Say that x = 200-y (or that y = 200-x), and substitute into the other equation.</p>
<p>It shouldn’t take very long.</p>
<p>whoa! yeah, it shouldn’t have taken that long… thanks, Sikorsky.</p>
<p>
</p>
<p>Method #1 - Algebraically Solve for X and Y.</p>
<p>This problem has two variables, thus you need to equations to solve.</p>
<p>Equation #1: 2x + 4y = 700
Equation #2: x + y = 200</p>
<p>Substitute x or y.</p>
<p>If you ever are stumped on a problem like this don’t be afraid to resorting to “plug and chug” method if time permits.</p>
<hr>
<p>Method #2 - “Plug and Chug”</p>
<p>4 * 50 = $200
2 * 150 = $300
Total = $500</p>
<p>4 * 75 = $300<br>
2 * 125 = $250
Total = $550 <– if you increase 4 by 25 and decrease 2 by 25 it goes up by $50. Now you need to gain another $150 so increase the amount of $4 tickets by 75 while decreasing the amount of $2 tickets by 75.</p>
<p>4 * 150 = $600
2 * 50 = $100
Total = $700</p>
<p>1 (children’s) : 3 (adults’)</p>