Jeremy has a carton containing some green pears and yellow pears. If he adds in 35 green pears, 0.7 of the pears in the carton will be yellow pears. If he adds in 50 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 - 35 |
4 p - 50 |
1 p |
Change |
|
+ 35 |
+ 50 |
|
After |
7 u |
3 u |
4 p |
1 p |
0.7 =
710 =
710If he adds 35 green pears,
Number of green pears in the end
= 10 u - 7 u
= 3 u
If he adds 50 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 - 50 --- (1)
3 u - 35 = 1 p
3 u = 1 p + 35 --- (2)
Make u the same.
(1)
x3 21 u = 12 p - 150 --- (3)
(2)
x7 21 u = 7 p + 245 --- (4)
(3) = (4)
12 p - 150 = 7 p + 245
12 p - 7 p = 150 + 245
5 p = 395
1 p = 395 ÷ 5 = 79
Number of pears at first
= 1 p + 4 p - 50
= 5 p - 50
= 5 x 79 - 50
= 395 - 50
= 345
Answer(s): 345