A. Child's tickets= $4, adult's ticket= $8
B. Child's tickets= $2, adult's ticket= $4
C. Child's tickets= $5.60, adult's ticket= $11.20
D. Child's tickets= $7, adults ticket= $14
please need ur help plzzz help me!!
2007-07-27 05:04:22 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Ok -
Let t = cost of child's ticket and x = cost of adult ticket. Now your answers all have the relationship that 2t = x.
Then:
2*x+3t =28
2(2t)+3t =28
7t = 28
t=4
x =8
Hence the answer is A
2007-07-27 05:10:16 · answer #1 · answered by nyphdinmd 7 · 0⤊ 0⤋
So I just started with A and put in the numbers to see which one worked. If there are two adults at $8 a ticket then that makes $16 (8 x 2). And for three children at $4 a ticket makes $12 (4 x 3). Then add the two together and you get $28. End of problem.
2007-07-27 05:14:26 · answer #2 · answered by eri 2 · 0⤊ 0⤋
A.(4 x 3) + (8 x 2) = 12 + 16 = $28
2007-07-27 05:11:48 · answer #3 · answered by Kenneth Koh 5 · 0⤊ 0⤋
Just multiply the 2 adult admissions by their cost and the 3 child admissions by their price for each possibility. For A you get 16+12 which does = 28 so "A" is the answer.
2007-07-27 05:13:02 · answer #4 · answered by Rich Z 7 · 0⤊ 0⤋
2 x $8 = $16
3 x $4 = $12
$16 + $12 = $28
It's 'A'
2007-07-27 05:11:32 · answer #5 · answered by Pedantic Scorpion 3 · 0⤊ 0⤋
2*A+3*C=28
from A:total=28
B:total=18
C:total=39.2
D:total=49
hence ans is A.
2007-07-27 05:31:26 · answer #6 · answered by sarvotham s 1 · 0⤊ 0⤋