Respuesta :
Answer:
The number of tickets that cost $ 10 is 923
The number of tickets that cost $ 20 is 1205
The number of tickets that cost $ 30 is 1111
Step-by-step explanation:
Given as :
The total number of three different tickets cost $ 10 , $ 20 and VIP seats $ 30
The total number of all tickets sold = 3239
The sale price for the tickets = $ 63720
Let The type of ticket for $ 10 = x
And The type of ticket for $ 20 = y
And The type of VIP ticket = v
According to question
The number of $ 20 tickets = The number of $ 10 tickets + 282
I.e y = x + 282
And x + y + v = 3239
Or, v = 3239 - ( x + y )
I.e v = 3239 - ( x + x +282 )
Or, v = 2957 - 2 x
Now ,
x × $ 10 + y × $ 20 = v × $ 30
Or, x × $ 10 + ( x + 282 ) × $ 20 = ( 2957 - 2 x ) × $ 30
or, 10 x + 20 x + 5640 = 88710 - 60 x
Or, 30 x + 60 x = 88710 - 5640
or, 90 x = 83070
∴ x = [tex]\frac{83070}{90}[/tex]
I.e x = 923
So, The number of tickets that cost $ 10 = x = 923
Similarly
y = x + 282
or, y = 923 + 282
I.e y = 1205
So, The number of tickets that cost $ 10 = y = 1205
And
v = 2957 - 2 x
∴ v = 2957 - 2 × 923
I.e v = 1111
So, The number of tickets that cost $ 30 = v = 1111
Hence
The number of tickets that cost $ 10 is 923
The number of tickets that cost $ 20 is 1205
The number of tickets that cost $ 30 is 1111
Answer
Answer:
1021 of the $10 tickets, 1303 of the $20 tickets and 915 of the $30 tickets.
Step-by-step explanation:
We have 3 types of tickets:
ones that cost $10, others that cost $20 and others that cost $30.
If A is the number of $10 tickets sold, B is the number of $20 tickets sold and C is the number of $30 tickets sold we have that:
A + B + C = 3239
B = A + 282
A*10 + B*20 + C*30 = 63720
We need to solve this system of equations.
From the second equation we have B isolated, we can replace it in the other equations:
A + (A + 282) + C = 3239
A*10 + (A + 282)*20 + C*30 = 63720
Now we can isolate one of the variables in the first equation and replace it in the second, lets isolate C.
C = 3239 - 2*A - 282 = 2957 - 2*A
the third equation is now:
A*10 + (A + 286)*20 + (2957 - 2*A)*30 = 63720
Now we solve it for A:
A*10 + A*20 - A*60 + 282*20 + 2957*30 = 63720
-A*30 + 94350 = 63720
A*30 = - 63720 + 94350 = 30630
A = 30639/30 = 1021
So they sold 1021 tickets of $10.
Now we can replace it in the equation:
C = 2957 - 2*A = 2957 - 2*1021 = 915
They sold 915 $30 tickets
They sold 1303 tickets of $20.
B = A + 282 = 1021 + 282 = 1303.
