A ferry can carry a maximum of 35 adults or 40 children. A ticket for an adult cost $50 and a ticket for a child cost $30. A group of adults and children took 2 such ferries at maximum load on a ferry trip. The total amount paid for the children was $1140 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 |
70 |
70 x 50 = 3500 |
0 |
0 x 30 = 0 |
3500 - 0 = 3500 |
- 7 |
|
+ 8 |
|
- 590 |
63 |
63 x 50 = 3150 |
8 |
8 x 30 = 240 |
3150 - 240 = 2910 |
42 |
|
32 |
|
1140 |
35 adults → 40 children
(÷5)7 adults → 8 children
From every decrease of 7 adults, there is an increase of 8 children and the difference in the total amounts paid between the adults and the children will be reduced by $590.
If all 2 ferries are occupied by only adults, total number of adults
= 2 x 35
= 70
Total difference
= 3500 - 1140
= $2360
Small difference
= 3500 - 2910
= $590
Number of sets
= 2360 ÷ 590
= 4
Total decrease in the number of adults
= 4 x 7
= 28
(a)
Number of adults
= 70 - 28
= 42
(b)
Number of children
= 4 x 8
= 32
Answer(s): (a) 42; (b) 32