Tickets for a certain show cost $14, $21, or, for VIP seats, $41. If ten times as many $14 tickets were sold as VIP tickets, and the number of $14 tickets sold was 54 more than the sum of the number of $21 tickets and VIP tickets, sales of all three kinds would total $48,446. How many of each kind of ticket would have been sold?

Number of $14 tickets sold = 1340

Number of $21 tickets sold = 1152

Number of VIP tickets sold = 134

Step-by-step explanation:

step 1 :

Let,

The equation is 14a + 21b + 41c = 48446     ----------(1)

Given,

Ten times as many $14 tickets were sold as VIP tickets.

⇒  10c = a  ---------(2)

step 2:

The number of $14 tickets sold was 54 more than the sum of the number of $21 tickets and VIP tickets.

⇒ a = 54 + (b+c)

⇒ a - 54 = b+c

Substitute a = 10c in the above eqn,

⇒ 10c - 54 = (b+c)

⇒ b = 9c - 54  -----------(3)

step 3 :

substitute (2) and (3) in eqn (1)

 14(10c) + 21(9c - 54) + 41c = 48446

Solve for the c value,

140c + 189c - 1134 + 41c = 48446

370c = 48446  + 1134

c = 49580 / 370

c = 134

step 4 :

Substitute c=134 in eqn (2),

10c = a

10 134 = a

a = 1340

Substitute c=134 in eqn (3),

b = 9(134) - 54

b = 1206 - 54

b = 1152

step 5 :

∴ a = number of $14 tickets = 1340

b = number of $21 tickets = 1152

c = number of VIP tickets = 134

the answer is 200

step-by-step explanation:

the way the question is worded, it seems to imply the $264 interest is simple interest, not compound interest.

the simple interest for 2 years is $264; the simple interest for 1 year is $132.

$132 interest on a $600 investment means an interest rate of (132/600)*100 = 22%.

the value of $200 at 22% interest compounded annually for three years is

200%281.22%29%5e3

the amount of interest is that value, minus the original $200.

answer:

c 60% on edg