Jeremy has a carton containing some yellow pears and green pears. If he adds in 20 yellow pears, 0.6 of the pears in the carton will be green pears. If he adds in 60 green pears,
67 of the pears in the carton will be green pears. How many pears are there in the carton?
|
Green pears |
Yellow pears |
Green pears |
Yellow pears |
Before |
3 u |
2 u - 20 |
6 p - 60 |
1 p |
Change |
|
+ 20 |
+ 60 |
|
After |
3 u |
2 u |
6 p |
1 p |
0.6 =
610 =
35If he adds 20 yellow pears,
Number of yellow pears in the end
= 5 u - 3 u
= 2 u
If he adds 60 green pears,
Number of yellow pears in the end
= 7 p - 6 p
= 1 p
The number of each type in both scenarios remain unchanged at first.
3 u = 6 p - 60 --- (1)
2 u - 20 = 1 p
2 u = 1 p + 20 --- (2)
Make u the same.
(1)
x2 6 u = 12 p - 120 --- (3)
(2)
x3 6 u = 3 p + 60 --- (4)
(3) = (4)
12 p - 120 = 3 p + 60
12 p - 3 p = 120 + 60
9 p = 180
1 p = 180 ÷ 9 = 20
Number of pears at first
= 1 p + 6 p - 60
= 7 p - 60
= 7 x 20 - 60
= 140 - 60
= 80
Answer(s): 80