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10 adults & 10 children's
The formula to determine the total money is nA + mC = 1317.50 where n is the number of adult tickets sold and A is the price of the adult tickets and m OS the number of child tickets and C is the price of a child ticket. Substituting in the actual prices we get... eq 1. n*7.50 + m * 3.50 = 1317.50 eq 2. n+m = 205 rearranging eq 2 we get n = 205 - m substituting this into eq 1 we get.. (205 -m) * 7.5 + m * 3.5 = 1317.50 Simplifying we get 1537.5 - 7.5m + 3.5m = 1317.50 1537.5 - 1317.5 = 4m 220 = 4m 55 = m substituting into eqn 2 we get n + 55 = 205 n = 150. Therefor 150 Adult tickets were sold 55 Child tickets were sold
Let's assume the cost of a child's ticket is x. Then, the cost of an adult's ticket would be 2x. The total cost of two adults' tickets and three children's tickets is 2(2x) + 3(x) = 4x + 3x = 7x. We know that 7x = 28, so x = 4. Therefore, the cost of a child's ticket is $4, and the cost of an adult's ticket is $8.
It costs 15.00 for two adults and two children.
Ok Let Children=C Let Adult=A Set up two equations: 3C+9A=$1905 C+A=419 tickets Solve the second equation for C or A, I'm choosing A A=419-C Sub it into the other equation 3C+9(419-C)=1905 3C+3771-9C=1905 Distribute 3771-6C=1905 Combine like terms -6C=(-1866) Subtract 3771 from both sides C=311 Divide by -6 Take 311 and plug it into the other equation "C+A=419" to find A=108 Answer Sold 108 adult tickets and 311 child tickets
adult:£7.50 child:£5.00
7 Adult tickets 13 Child tickets
10 adults & 10 children's
1-Day Park Hopper$940 total ($97 each for adult tickets, $87 each for child tickets)1-Day 1-Park$690 total ($72 each for adult tickets, $62 each for child tickets)
Adult tickets are $9.50 Senior tickets are $7.00 Child tickets are $7.50
The tickets are $5 adult and $2 child. Here is the algebraic substitution. A = adult ticket money C = child ticket money 120C + 80A = 640 C = A-3 (three dollars less) 120 (A-3) + 80A = 640 120 A - 360 + 80A = 640 200 A = 1000 A = 5 C = A - 3 = 2 Checking your answer: 120(2) + 80 (5) = 240 + 400 = 640
At Bella Terra there about $11.25 per adult and $8.25 per child
One-day One-park$426.60 ($71.10 each) for 6 adult tickets$367.20 ($61.20 each) for 6 child ticketsOne-day Park Hopper$582.60 ($97.10 each) for 6 adult tickets$523.20 ($87.20 each) for 6 child tickets
The formula to determine the total money is nA + mC = 1317.50 where n is the number of adult tickets sold and A is the price of the adult tickets and m OS the number of child tickets and C is the price of a child ticket. Substituting in the actual prices we get... eq 1. n*7.50 + m * 3.50 = 1317.50 eq 2. n+m = 205 rearranging eq 2 we get n = 205 - m substituting this into eq 1 we get.. (205 -m) * 7.5 + m * 3.5 = 1317.50 Simplifying we get 1537.5 - 7.5m + 3.5m = 1317.50 1537.5 - 1317.5 = 4m 220 = 4m 55 = m substituting into eqn 2 we get n + 55 = 205 n = 150. Therefor 150 Adult tickets were sold 55 Child tickets were sold
[100 + 54] 910 + 291.60 = 1,201.60 (too much) [90 + 64] 819 + 345.60 = 1,164.60 (too much) [80 + 74] 728 + 399.60 = 1,127.60 (too little) so answer is between 65 and 73 child tickets 1,135.00 - 1,127.60 = 7.40 9.10 - 5.40 = 3.70 7.40 / 3.70 = 2 so adjusting [80 + 74] by that 2 ... [82 + 72] 746.20 + 388.80 = 1,135.00 Answer: 72 child tickets (and 82 adult tickets)
For 1 park (Adventure Island) an adult ticket is $41.95 and for a child ticket it is $37.95, but if you want two parks (Adventure Island and Busch Gardens) it'll cost $89.95 for an adult and $79.95 for a child.
Let's assume the cost of a child's ticket is x. Then, the cost of an adult's ticket would be 2x. The total cost of two adults' tickets and three children's tickets is 2(2x) + 3(x) = 4x + 3x = 7x. We know that 7x = 28, so x = 4. Therefore, the cost of a child's ticket is $4, and the cost of an adult's ticket is $8.