Jeremy has a carton containing some golden kiwis and green kiwis. If he adds in 46 golden kiwis, 0.7 of the kiwis in the carton will be green kiwis. If he adds in 42 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 - 46 |
7 p - 42 |
1 p |
Change |
|
+ 46 |
+ 42 |
|
After |
7 u |
3 u |
7 p |
1 p |
0.7 =
710 =
710If he adds 46 golden kiwis,
Number of golden kiwis in the end
= 10 u - 7 u
= 3 u
If he adds 42 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 - 42 --- (1)
3 u - 46 = 1 p
3 u = 1 p + 46 --- (2)
Make u the same.
(1)
x3 21 u = 21 p - 126 --- (3)
(2)
x7 21 u = 7 p + 322 --- (4)
(3) = (4)
21 p - 126 = 7 p + 322
21 p - 7 p = 126 + 322
14 p = 448
1 p = 448 ÷ 14 = 32
Number of kiwis at first
= 1 p + 7 p - 42
= 8 p - 42
= 8 x 32 - 42
= 256 - 42
= 214
Answer(s): 214