A ferry can carry a maximum of 30 adults or 35 children. A ticket for an adult cost $40 and a ticket for a child cost $35. A group of adults and children took 4 such ferries at maximum load on a ferry trip. The total amount paid for the children was $920 less than the total amount paid for the adults.
- How many adults were there?
- How many children were there?
Number of adults |
Amount paid for adults |
Number of children |
Amount paid for children |
Difference in total amounts paid |
120 |
120 x 40 = 4800 |
0 |
0 x 35 = 0 |
4800 - 0 = 4800 |
- 6 |
|
+ 7 |
|
- 485 |
114 |
114 x 40 = 4560 |
7 |
7 x 35 = 245 |
4560 - 245 = 4315 |
72 |
|
56 |
|
920 |
30 adults → 35 children
(÷5)6 adults → 7 children
From every decrease of 6 adults, there is an increase of 7 children and the difference in the total amounts paid between the adults and the children will be reduced by $485.
If all 4 ferries are occupied by only adults, total number of adults
= 4 x 30
= 120
Total difference
= 4800 - 920
= $3880
Small difference
= 4800 - 4315
= $485
Number of sets
= 3880 ÷ 485
= 8
Total decrease in the number of adults
= 8 x 6
= 48
(a)
Number of adults
= 120 - 48
= 72
(b)
Number of children
= 8 x 7
= 56
Answer(s): (a) 72; (b) 56