小學數學求助

ping, let me help you. :-)

Let n be the number of cards that Joe had at first.

That means, Ariel had (420 - n) cards.

The number of cards in the first transferal is (420 - n)/3, then by that time Ariel had 2(420 - n)/3 and Joe had n + (420 - n)/3 = (420 + 2n)/3.

The number of cards in the second transferal is (2/5)*(420 - n)/3 = 2(420 + 2n)/15, then by that time Ariel had 2(420 - n)/3 + 2(420 + 2n)/15 and Joe had (3/5)*(420 + 2n)/3 = (420 + 2n)/5.

Therefore,
2(420 - n)/3 + 2(420 + 2n)/15 = (420 + 2n)/5

Multiply 15 to both sides.
10(420 - n) + 2(420 + 2n) = 3*(420 + 2n)

10(420 - n) = 420 + 2n

4200 - 10n = 420 + 2n

12n = 3780

n = 315

That means, Joe had 315 cards at first.


Verification:

Ariel　　　Joe
105　　　315
105-35　　315+35
70　　　 350
70+140　 350-140
210　　　210

2013-12-09 19:27:08 補充：
數字更正：　（計算沒錯）

The number of cards in the second transferal is (2/5)*(420 + 2n)/3 = 2(420 + 2n)/15

2013-12-10 13:05:48 補充：
YES!!! Godfrey's method is much easier because he directly used 420/2 = 210.

Thanks!

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Let n = # of cards Joe had; Ariel had = (420 – n) cards
After 1st transfer,
Ariel had 2/3 (420 – n), Joe had n + 1/3 (420 – n),
After 2nd transfer, both had 420/2 = 210
Joe had 3/5 [n + 1/3 (420 – n)] = 210
(5/3)(3/5) [n + 1/3 (420 – n)] = 5/3 (210)
n + 140 – n/3 = 350
(2/3) n = 210
n = 315