Zachary had 290 balls kept in 2 separate boxes, A and B. When 2/5 of the balls
were transferred from Box A to Box B, Box A still had 10 balls more than Box B.
How many balls were there in each of the boxes at first?
+677 I will answer your question...
In box A there were (at first) 250 balls and
in box B were (at first) 40 balls (\(250+40\)).
I suggested you to get trys and then get an answer,
but if you're trying please note that the balls on
each side must be divided by 5.
Use for example:
| in-box A | in-box B |
| 270 balls | 20 balls |
| 162 balls | 128 balls |
(\(162-128=34\))
| in-box A | in-box B |
| 260 balls | 30 balls |
| 156 balls | 134 balls |
(\(156-134=22\))
You see the differences beetween the first and the second chart they're only (\(34-22=\)) (\(12\)).
And the differences are ALWAYS 12, so (\(22-12=10\)) and 10 is the difference we should get.
I subtract now 260 balls to 250 balls (like I did) and add these 10 to the 30 balls (\(260-10=250, 30+10=40\))
I know that we are getting the difference of 10, but still I make a chart for a proof:
| in-box A | in-box B |
| 250 balls | 40 balls |
| 150 balls | 140 balls |
The difference is now (\(150-140=\)) 10 and that is what we should get.
Straight
+129 Zachary kept 290 balls in two boxes.
A as B
Let A ---> x balls => x+y=290
B ---> y
Now, 2/5A ---> B B has total
=y+2x/5
A has x-2x/5=3x/5
Still A has 10 balls more than box B
3x/5 = y+2x/5+10
=> 3x/5 - 2x/5 = y+10
=> x/5 = y+10
=> x - 5y = 50
x+y=290
x-5y=50
- + -
--------------- |
6y=240 | x = 290 - 40
y=240/6=40 | = 250
|
There were 250 balls in A
40 balls in B at first
