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 $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 $1960 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 |
140 |
140 x 50 = 7000 |
0 |
0 x 35 = 0 |
7000 - 0 = 7000 |
- 7 |
|
+ 8 |
|
- 630 |
133 |
133 x 50 = 6650 |
8 |
8 x 35 = 280 |
6650 - 280 = 6370 |
84 |
|
64 |
|
1960 |
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 $630.
If all 4 ferries are occupied by only adults, total number of adults
= 4 x 35
= 140
Total difference
= 7000 - 1960
= $5040
Small difference
= 7000 - 6370
= $630
Number of sets
= 5040 ÷ 630
= 8
Total decrease in the number of adults
= 8 x 7
= 56
(a)
Number of adults
= 140 - 56
= 84
(b)
Number of children
= 8 x 8
= 64
Answer(s): (a) 84; (b) 64