A ferry can carry a maximum of 30 adults or 35 children. A ticket for an adult cost $55 and a ticket for a child cost $35. 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 $1500 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 55 = 4950 |
0 |
0 x 35 = 0 |
4950 - 0 = 4950 |
- 6 |
|
+ 7 |
|
- 575 |
84 |
84 x 55 = 4620 |
7 |
7 x 35 = 245 |
4620 - 245 = 4375 |
54 |
|
42 |
|
1500 |
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 $575.
If all 3 ferries are occupied by only adults, total number of adults
= 3 x 30
= 90
Total difference
= 4950 - 1500
= $3450
Small difference
= 4950 - 4375
= $575
Number of sets
= 3450 ÷ 575
= 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