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Question 101946: apples are collected in a basket for six people. One-third, one-forth, one-eighth and one-fifth are given to four people, respectively. The fifth person gets ten apples with one apple remaining for the sixth person. Find the original number of apples in the basket.
Answer by ptaylor(2198)   (Show Source):
You can put this solution on YOUR website! Let x=original number of apples in the basket
1st person gets (1/3)x
2nd person gets(1/4)x
3rd person gets(1/8)x
4th person gets (1/5)x
5th person gets ---10 apples
6th person gets-------1 apple
After the first four persons get their apples, we'll find out what fraction of the apples are left for the 5th and 6th persons and we are told that whatever that fraction is, will equal 11:
(1/3)x+(1/4)x+(1/8)x+(1/5)x= Multiply each term by 120/120, the LCM:
(40/120)x+(30/120)x+(15/120)x+(24/120)x=(109/120)x
So, the apples that are left for the 5th & 6th persons equals:
(120/120)x-(109/120)x=(11/120)x ---and this equals 11.
(11/120)x=11 multiply both sides by 120
11x=1320 divide both sides by 11
x=120--------------------------number of apples
CK
(40/120)*120+(30/120)*120+(15/120)*120+(24/120)*120+11=120 or all the apples
40+30+15+24+11=120
109+11=120
120=120
Another way to work this problem is as follows:
(1/3)x+(1/4)x+(1/8)x+(1/5)x+11=x
(109/120)x-x=-11
-(11/120)x=-11
-11x=-1320
x=120
Hope this helps---ptaylor
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