Question 1187617
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A gold bar is first cut into 2 pieces in the ratio of 2:5 . 
The heavier piece is then cut into 2 pieces in the ratio of 5:6. 
The difference between the mass of the heaviest piece and lightest piece was 24 grams. 
What was the mass of the gold bar before it was cut.
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<pre>
Let x be the mass of the gold bar before it was cut.


After the first cut we have the pieces of  {{{(2/7)x}}}  and  {{{(5/7)x}}}  masses.



After the second cut, the heavier piece of  {{{(5/7)x}}}  is disintegrated into the parts of

{{{(5/11)*(5/7)x}}} = {{{(25/77)x}}}  and  {{{(6/11)*(5/7)x}}} = {{{(30/77)x}}}  of mass.



Of three pieces,  {{{(2/7)x}}} = {{{(22/77)x}}},  {{{(25/77)x}}}  and  {{{(30/77)x}}},  the heaviest part 

is  {{{(30/77)x}}},  while the lightest part is  {{{(22/77)x}}}.



The difference is  {{{(30/77)x}}} - {{{(22/77)x}}} = {{{(8/77)x}}},  so


    {{{(8/77)x}}} = 24.


It gives  x = {{{(24*77)/8}}} = 3*77 = 231 grams.



<U>ANSWER</U>.  The mass of the gold bar was initially 231 grams.
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Solved.