Tickets at a football match sell at k3, k5 and k7.50. The number of k5 tickets sold was three times more than the number of k3 tickets sold and 500 less than the number of k7.50 tickets sold. If the total gate receipts were k42225, how many of each ticket were sold?
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let a = k3 , b = k5 , c = k7.50
b = 3 a
c = b + 500
3 a + 5 b + 7.5 c = 42225
substituting ... 3(b / 3) + 5 b + 7.5(b + 500) = 42225
solve for b , then substitute back to find a and c
To solve this problem, let's assign variables to represent the number of each type of ticket sold.
Let's assume:
- The number of k3 tickets sold is represented by x.
- The number of k5 tickets sold is represented by 3x (three times more than k3 tickets).
- The number of k7.50 tickets sold is represented by 3x + 500 (500 less than k7.50 tickets).
We can set up an equation to represent the total gate receipts:
(k3 ticket value * number of k3 tickets) + (k5 ticket value * number of k5 tickets) + (k7.50 ticket value * number of k7.50 tickets) = total gate receipts
Plugging in the given values, the equation becomes:
(k3 * x) + (k5 * 3x) + (k7.50 * (3x + 500)) = 42225
Now let's find the numerical values of the ticket prices:
k3 = k3 = 3; k5 = 5; k7.50 = 7.50
Substituting the values, the equation becomes:
(3 * x) + (5 * 3x) + (7.50 * (3x + 500)) = 42225
Now we can solve for x:
3x + 15x + 7.50(3x + 500) = 42225
3x + 15x + 22.5x + 3750 = 42225
40.5x + 3750 = 42225
40.5x = 38475
x ≈ 951 (rounded to the nearest whole number)
Therefore, approximately 951 k3 tickets were sold.
The number of k5 tickets sold is three times more, so 3x ≈ 3 * 951 ≈ 2853.
Approximately 2853 k5 tickets were sold.
The number of k7.50 tickets is 3x + 500, so 3x + 500 ≈ 3 * 951 + 500 ≈ 3786.
Approximately 3786 k7.50 tickets were sold.
So, the approximate number of each ticket sold is:
k3 tickets: 951
k5 tickets: 2853
k7.50 tickets: 3786
Note: The values are rounded to the nearest whole number.