Jeremy has a carton containing some green kiwis and golden kiwis. If he adds in 15 green kiwis, 0.7 of the kiwis in the carton will be golden kiwis. If he adds in 50 golden kiwis,
45 of the kiwis in the carton will be green kiwis. How many kiwis are there in the carton?
|
Golden kiwis |
Green kiwis |
Golden kiwis |
Green kiwis |
Before |
7 u |
3 u - 15 |
4 p - 50 |
1 p |
Change |
|
+ 15 |
+ 50 |
|
After |
7 u |
3 u |
4 p |
1 p |
0.7 =
710 =
710If he adds 15 green kiwis,
Number of green kiwis in the end
= 10 u - 7 u
= 3 u
If he adds 50 golden kiwis,
Number of green kiwis in the end
= 5 p - 4 p
= 1 p
The number of each type in both scenarios remain unchanged at first.
7 u = 4 p - 50 --- (1)
3 u - 15 = 1 p
3 u = 1 p + 15 --- (2)
Make u the same.
(1)
x3 21 u = 12 p - 150 --- (3)
(2)
x7 21 u = 7 p + 105 --- (4)
(3) = (4)
12 p - 150 = 7 p + 105
12 p - 7 p = 150 + 105
5 p = 255
1 p = 255 ÷ 5 = 51
Number of kiwis at first
= 1 p + 4 p - 50
= 5 p - 50
= 5 x 51 - 50
= 255 - 50
= 205
Answer(s): 205