Jeremy has a carton containing some yellow pears and green pears. If he adds in 44 yellow pears, 0.7 of the pears in the carton will be green pears. If he adds in 56 green pears,
78 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 |
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 yellow pears,
Number of yellow pears in the end
= 10 u - 7 u
= 3 u
If he adds 56 green pears,
Number of yellow pears 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 pears at first
= 1 p + 7 p - 56
= 8 p - 56
= 8 x 34 - 56
= 272 - 56
= 216
Answer(s): 216