Jeremy has a carton containing some golden kiwis and green kiwis. If he adds in 44 golden kiwis, 0.7 of the kiwis in the carton will be green kiwis. If he adds in 56 green kiwis,
78 of the kiwis in the carton will be green kiwis. How many kiwis are there in the carton?
|
Green kiwis |
Golden kiwis |
Green kiwis |
Golden kiwis |
Before |
7 u |
3 u - 44 |
7 p - 56 |
1 p |
Change |
|
+ 44 |
+ 56 |
|
After |
7 u |
3 u |
7 p |
1 p |
0.7 =
710 =
710If he adds 44 golden kiwis,
Number of golden kiwis in the end
= 10 u - 7 u
= 3 u
If he adds 56 green kiwis,
Number of golden kiwis in the end
= 8 p - 7 p
= 1 p
The number of each type in both scenarios remain unchanged at first.
7 u = 7 p - 56 --- (1)
3 u - 44 = 1 p
3 u = 1 p + 44 --- (2)
Make u the same.
(1)
x3 21 u = 21 p - 168 --- (3)
(2)
x7 21 u = 7 p + 308 --- (4)
(3) = (4)
21 p - 168 = 7 p + 308
21 p - 7 p = 168 + 308
14 p = 476
1 p = 476 ÷ 14 = 34
Number of kiwis at first
= 1 p + 7 p - 56
= 8 p - 56
= 8 x 34 - 56
= 272 - 56
= 216
Answer(s): 216