+0  
 
+1
79
2
avatar

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
avatar+391 
+1

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

 Oct 15, 2021
 #2
avatar+103 
+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

9 Online Users