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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?

Oct 15, 2021

#1
+1

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

Oct 15, 2021
#2
+1

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

Nov 6, 2021