Jeremy has a carton containing some green pears and yellow pears. If he adds in 36 green pears, 0.7 of the pears in the carton will be yellow pears. If he adds in 56 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 |
7 u |
3 u - 36 |
4 p - 56 |
1 p |
Change |
|
+ 36 |
+ 56 |
|
After |
7 u |
3 u |
4 p |
1 p |
0.7 =
710 =
710If he adds 36 green pears,
Number of green pears in the end
= 10 u - 7 u
= 3 u
If he adds 56 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.
7 u = 4 p - 56 --- (1)
3 u - 36 = 1 p
3 u = 1 p + 36 --- (2)
Make u the same.
(1)
x3 21 u = 12 p - 168 --- (3)
(2)
x7 21 u = 7 p + 252 --- (4)
(3) = (4)
12 p - 168 = 7 p + 252
12 p - 7 p = 168 + 252
5 p = 420
1 p = 420 ÷ 5 = 84
Number of pears at first
= 1 p + 4 p - 56
= 5 p - 56
= 5 x 84 - 56
= 420 - 56
= 364
Answer(s): 364