SOLUTION: A man is carrying rice on a journey. He passes through three customs stations. At the first, he gives up a third of his rice, at the second, a fifth of what was left, and at the th

Algebra ->  Linear-equations -> SOLUTION: A man is carrying rice on a journey. He passes through three customs stations. At the first, he gives up a third of his rice, at the second, a fifth of what was left, and at the th      Log On


   

Question 670955: A man is carrying rice on a journey. He passes through three customs stations. At the first, he gives up a third of his rice, at the second, a fifth of what was left, and at the third, a seventh of what was left. After he passes through all three stations, he has five pounds of rice left. How much rice did he begin with?
Found 2 solutions by mananth, josmiceli:
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
Let the man have x pounds of rice
I
he gave 1/3 of x =x/3
so he had balance x-(x/3) = 2x/3
II
He gave (1/5) (2x/3) =2x/15
Balance = (2x/3)-(2x/15)= 8x/15
III
He gives 1/7
(1/7) (8x/15) = 8x/105
Balance = (8x/15)-(8x/105) = 48x/105
48x/105 = 5
48x= 525
x=525/48

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +x+ = pounds of rice he started with
After 1st station, he has left
+x+-+%281%2F3%29%2Ax+=+%282%2F3%29%2Ax+
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After the 2nd station, he has left
+%282%2F3%29%2Ax+-+%281%2F5%29%2A%282%2F3%29%2Ax+
+%284%2F5%29%2A%282%2F3%29%2Ax+=+%288%2F15%29%2Ax+
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After the 3rd station, he has left
+%288%2F15%29%2Ax+-+%281%2F7%29%2A%288%2F15%29%2Ax+
+%286%2F7%29%2A%288%2F15%29%2Ax+
+%2848%2F105%29%2Ax+
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Also given:
+%2848%2F105%29%2Ax+=+5+
+x+=+%28105%2F48%29%2A5+
+x+=+10.9375+
He started with 10.9375 pounds
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check:
After 1st station
7.29203 pounds
After 2nd station
5.83362
After 3rd station:
5.00025
close enough