Jeremy has a carton containing some yellow pears and green pears. If he adds in 42 yellow pears, 0.7 of the pears in the carton will be green pears. If he adds in 50 green pears,
34 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 - 42 |
3 p - 50 |
1 p |
Change |
|
+ 42 |
+ 50 |
|
After |
7 u |
3 u |
3 p |
1 p |
0.7 =
710 =
710If he adds 42 yellow pears,
Number of yellow pears in the end
= 10 u - 7 u
= 3 u
If he adds 50 green pears,
Number of yellow pears in the end
= 4 p - 3 p
= 1 p
The number of each type in both scenarios remain unchanged at first.
7 u = 3 p - 50 --- (1)
3 u - 42 = 1 p
3 u = 1 p + 42 --- (2)
Make u the same.
(1)
x3 21 u = 9 p - 150 --- (3)
(2)
x7 21 u = 7 p + 294 --- (4)
(3) = (4)
9 p - 150 = 7 p + 294
9 p - 7 p = 150 + 294
2 p = 444
1 p = 444 ÷ 2 = 222
Number of pears at first
= 1 p + 3 p - 50
= 4 p - 50
= 4 x 222 - 50
= 888 - 50
= 838
Answer(s): 838