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 $25. A group of adults and children took 3 such ferries at maximum load on a ferry trip. The total amount paid for the children was $1110 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 |
90 |
90 x 40 = 3600 |
0 |
0 x 25 = 0 |
3600 - 0 = 3600 |
- 6 |
|
+ 7 |
|
- 415 |
84 |
84 x 40 = 3360 |
7 |
7 x 25 = 175 |
3360 - 175 = 3185 |
54 |
|
42 |
|
1110 |
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 $415.
If all 3 ferries are occupied by only adults, total number of adults
= 3 x 30
= 90
Total difference
= 3600 - 1110
= $2490
Small difference
= 3600 - 3185
= $415
Number of sets
= 2490 ÷ 415
= 6
Total decrease in the number of adults
= 6 x 6
= 36
(a)
Number of adults
= 90 - 36
= 54
(b)
Number of children
= 6 x 7
= 42
Answer(s): (a) 54; (b) 42