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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

Jeffreymars16  Mar 12, 2018
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#1
0

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!!!!.

Guest Mar 12, 2018
#2
+1

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!!!.

Guest Mar 12, 2018
edited by Guest  Mar 12, 2018
edited by Guest  Mar 12, 2018
#3
+84168
+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 }

CPhill  Mar 12, 2018

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