Jeremy has a carton containing some green pears and yellow pears. If he adds in 27 green pears, 0.6 of the pears in the carton will be yellow pears. If he adds in 52 yellow pears,
45 of the pears in the carton will be green pears. How many pears are there in the carton?
|
Yellow pears |
Green pears |
Yellow pears |
Green pears |
Before |
3 u |
2 u - 27 |
4 p - 52 |
1 p |
Change |
|
+ 27 |
+ 52 |
|
After |
3 u |
2 u |
4 p |
1 p |
0.6 =
610 =
35If he adds 27 green pears,
Number of green pears in the end
= 5 u - 3 u
= 2 u
If he adds 52 yellow pears,
Number of green pears in the end
= 5 p - 4 p
= 1 p
The number of each type in both scenarios remain unchanged at first.
3 u = 4 p - 52 --- (1)
2 u - 27 = 1 p
2 u = 1 p + 27 --- (2)
Make u the same.
(1)
x2 6 u = 8 p - 104 --- (3)
(2)
x3 6 u = 3 p + 81 --- (4)
(3) = (4)
8 p - 104 = 3 p + 81
8 p - 3 p = 104 + 81
5 p = 185
1 p = 185 ÷ 5 = 37
Number of pears at first
= 1 p + 4 p - 52
= 5 p - 52
= 5 x 37 - 52
= 185 - 52
= 133
Answer(s): 133