Respuesta :
Answer:
There were 220 tickets of $8, 130 tickets of $10 and 70 tickets of $12 were sold.
Step-by-step explanation:
Given,
Total number of tickets sold = 420
Total money collected = $3900
We need to find the number of each type of ticket.
Solution,
Let the number of $8 ticket be 'x'.
Also let the number of $10 ticket be 'y'.
And also let the number of $12 ticket be 'z'.
So the total number of tickets is equal to the sum of the number each type of ticket.
framing in equation form, we get;
[tex]x+y+z=420\ \ \ \ \ equation\ 1[/tex]
Also given the combined number of $8 and $10 tickets sold was 5 times the number of $12 tickets sold.
So we can frame it as;
[tex]x+y=5z\ \ \ \ equation\ 2[/tex]
Now substituting the value of equation 2 in equation 1, we get;
[tex]5z+z=420\\\\6z=420[/tex]
On dividing both side by '6' using division property, we get;
[tex]\frac{6z}{6}=\frac{420}{6}\\\\z=70[/tex]
Now we get from equation 1;
[tex]x+y+z=420\\\\x+y+70=420\\\\x+y=420-70[/tex]
[tex]x+y=350[/tex] ⇒ equation 3
Also we can say that;
Total money collected is equal to 8 multiplied by number of $8 ticket plus 10 multiplied by number of $10 ticket plus 12 multiplied by number of $12 ticket.
framing in equation form we get;
[tex]8x+10y+12z=3900[/tex]
Now we will substitute value of z in above equation we get;
[tex]8x+10y+12\times70=3900\\\\8x+10y+840=3900\\\\8x+10y=3900-840\\\\8x+10y = 3060[/tex]
Now Dividing by 10 we get;
[tex]0.8x+y=306[/tex] ⇒ equation 4
subtracting equation 4 from equation 3 we get;
[tex]x+y-(0.8x+y)=350-306\\\\x+y-0.8x-y=44\\\\0.2x=44[/tex]
Dividing both side by 0.2 we get;
[tex]\frac{0.2x}{0.2}=\frac{44}{0.2}\\\\x=220[/tex]
Now substituting the value of x in equation 3 we get.
[tex]x+y=350\\\\220+y=350\\\\y=350-220\\\\y= 130[/tex]
Hence There were 220 tickets of $8, 130 tickets of $10 and 70 tickets of $12 were sold.
