Jeremy has a carton containing some green pears and yellow pears. If he adds in 16 green pears, 0.7 of the pears in the carton will be yellow pears. If he adds in 46 yellow pears,
34 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 - 16 |
3 p - 46 |
1 p |
Change |
|
+ 16 |
+ 46 |
|
After |
7 u |
3 u |
3 p |
1 p |
0.7 =
710 =
710If he adds 16 green pears,
Number of green pears in the end
= 10 u - 7 u
= 3 u
If he adds 46 yellow pears,
Number of green 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 - 46 --- (1)
3 u - 16 = 1 p
3 u = 1 p + 16 --- (2)
Make u the same.
(1)
x3 21 u = 9 p - 138 --- (3)
(2)
x7 21 u = 7 p + 112 --- (4)
(3) = (4)
9 p - 138 = 7 p + 112
9 p - 7 p = 138 + 112
2 p = 250
1 p = 250 ÷ 2 = 125
Number of pears at first
= 1 p + 3 p - 46
= 4 p - 46
= 4 x 125 - 46
= 500 - 46
= 454
Answer(s): 454