Jeremy has a carton containing some yellow pears and green pears. If he adds in 21 yellow pears, 0.7 of the pears in the carton will be green pears. If he adds in 49 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 - 21 |
7 p - 49 |
1 p |
Change |
|
+ 21 |
+ 49 |
|
After |
7 u |
3 u |
7 p |
1 p |
0.7 =
710 =
710If he adds 21 yellow pears,
Number of yellow pears in the end
= 10 u - 7 u
= 3 u
If he adds 49 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 - 49 --- (1)
3 u - 21 = 1 p
3 u = 1 p + 21 --- (2)
Make u the same.
(1)
x3 21 u = 21 p - 147 --- (3)
(2)
x7 21 u = 7 p + 147 --- (4)
(3) = (4)
21 p - 147 = 7 p + 147
21 p - 7 p = 147 + 147
14 p = 294
1 p = 294 ÷ 14 = 21
Number of pears at first
= 1 p + 7 p - 49
= 8 p - 49
= 8 x 21 - 49
= 168 - 49
= 119
Answer(s): 119