Jeremy has a carton containing some bunches of red grapes and bunches of green grapes. If he adds in 39 bunches of red grapes, 0.7 of the bunches of grapes in the carton will be bunches of green grapes. If he adds in 41 bunches of green grapes,
34 of the bunches of grapes in the carton will be green bunches of grapes. How many bunches of grapes are there in the carton?
|
Bunches of green grapes |
Bunches of red grapes |
Bunches of green grapes |
Bunches of red grapes |
Before |
7 u |
3 u - 39 |
3 p - 41 |
1 p |
Change |
|
+ 39 |
+ 41 |
|
After |
7 u |
3 u |
3 p |
1 p |
0.7 =
710 =
710If he adds 39 bunches of red grapes,
Number of bunches of red grapes in the end
= 10 u - 7 u
= 3 u
If he adds 41 bunches of green grapes,
Number of bunches of red grapes 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 - 41 --- (1)
3 u - 39 = 1 p
3 u = 1 p + 39 --- (2)
Make u the same.
(1)
x3 21 u = 9 p - 123 --- (3)
(2)
x7 21 u = 7 p + 273 --- (4)
(3) = (4)
9 p - 123 = 7 p + 273
9 p - 7 p = 123 + 273
2 p = 396
1 p = 396 ÷ 2 = 198
Number of bunches of grapes at first
= 1 p + 3 p - 41
= 4 p - 41
= 4 x 198 - 41
= 792 - 41
= 751
Answer(s): 751