<p>An amusement park charges $8 for an adults ticket, and $6 for a childrens ticket. On a certain day, a total of 150 tickets were sold for a total cost of $1020. How many more children's tickets were sold than adult's tickets?</p>
<p>a) 30
b) 50
c) 60
d) 90
e) 120</p>
<p>im having trouble translating and setting this problem up correctly. help would be very much appreciated!</p>
<p>Let y = number of adult tickets and z = number of children tickets
8y + 6z = 1020
y + z = 150
6y + 6z = 900
2y = 120
y = 60
60 + z = 150
z = 90
90 - 60 = 30 = A</p>
<p>ahh thanks a lot fusion! i see it now.</p>
<p>I agree that A is the answer too!
I just do not understand how Fusion solved the problem.</p>
<p>a+c=150</p>
<h1>8a+6c=1020</h1>
<p>I think you should just try to see which answer works.
c-a=30
That means a = 60, but that does work since 8<em>60 + 6</em>90 = 1020.</p>
<p>Fusion used the substitution method. :)
Just reiterating.</p>
<p>Psycho_Paroxysm, does mine work too?</p>
<p>there is a simpler technique. say the number of adult tickets is x and the number of childrens is 150-x.</p>
<p>8x+6(150-x)=1020
x=60</p>
<p>so you get 90 childrens tickets and 60 adult tickets. 90-60=30</p>
<p>Shadow, yes your method works too. :) However, the other method is more foolproof because yours involves some guessing. It's not everyday that anyone can come up with the value of 60 and 90 right away from the equation c-a=30.</p>
<p>I do it the risky method. Look at the answers and start with A (dont bother with the actualy algebra lol!! :) ) Then I multiply 90x6= total money of children's tickets. Then get the remainder and multiply 60x8= total money of adults. Add them up and see if the sum is $1020. If it does not work then move on to B. While I do all this I cross out E because I can subconcioesly (dont know how to spell it) see that it does not work. Anyone else use the unorthadox method described above?</p>
<p>i agree with coolkid1992's method
thats howi did it at least
it seems quicker, or easier</p>