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Three men are gambling at a bar. They start with money in the ratio 7:6:5 and finish with sums of money in the ratio 6:5:4 (in the same order as before). One of them won $12. How many dollars did he start with?

 

a) 420                      b) 1080                         c) 432                               d) 120                              e) 90

 

Please show your working and explain( if you can)

 Mar 12, 2018
 #1
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I will take a crack at it !!

 

Let the total amount they started with = m

m / [7+6+5] + 12 = m / [6+5+4]

m =$1,080 - the amount of money they started with as follows:

1,080 / 18 = 60 - average share of each, but we have a ratio of 7:6:5, so:

$420:$360:$300

 

1,080 / 15 = 72 - the average share of each AFTER the game!!, but we have 6:5:4, and:

72 / 6 = $12 - this is winning of the gambler who started with:

$420 and ended up winning $12, since the ratios are in the same order, so they ended up with:

$12, $10, $8 =$30 in total winning !!!!

Note: Somebody should check these crazy numbers!! But then, I don't gamble!!!!.

 Mar 12, 2018
 #2
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I think the above answer is correct, because it just happens that:

The LCM of {7, 6, 5, 4} =420 and the ratio being 7:6:5 or:

$420:$360:$300 =$1,080.

 

Since $1,080 / [7+6+5] =$60 average per man when they started gambling, and:

Since $1080 /  [6+5+4] =$72 average per man if they hadn't lost any money, or:

$432:$360:$288, but one of them ended up with:

$432 - $420 =$12 for the first man, for the ratio of:

$12:$10:$8 =$30 - total wininings for the 3 men, or is it:

$300 - $288 =$12 - Left by the last man, in which case the ratio will be:

$18:$15:$12 =$45 - total winnings for 3 men ????. This last one seems to make more sense, but can't be right because that would mean that he started with $300, which is NOT one of the answers!!!.

 Mar 12, 2018
edited by Guest  Mar 12, 2018
edited by Guest  Mar 12, 2018
 #3
avatar+98123 
+2

Note that the total amount of money, A, doesn't change.....just the distribution of it does

So....this is a "zero-sum" game....the total amts won + total amts  lost   = 0

 

7 : 6 : 5

Means that  the first person  started with (7/18)A

The second (6/18)A  =  (1/3)A

The third (5/18)A

 

In the end   ⇒   6 : 5 : 4

The first person ends up  with (6/15)A  = ( 2/5)A

The second ends  up  with (5/15)A  = (1/3)A

The last ends  up wth (4/15)A

 

Note that the second person breaks even...he starts with (1/3)A  and ends up with the same amount

The first person must have won the $12  because  (2/5)A > (7/18)A

And the difference between (2/5A - 7/18A)  = 1/90A  =  $12

So.... $12  is 1/90 of the total amount  = $1080

 

So....the first person must have started with (7/18) * 1080  = $420

And he ends up with (2/5) * 1080  =  $432

So.....he makes $12

 

{Which means that since the second person breaks even....the third person must have lost the $12 }

 

cool cool cool

 Mar 12, 2018

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