Divide 184 int two such parts that one- third of one part exceeds one-seventh of another part by 8
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let x be one part and (184-x) be another part.
By the problem,
x /3= (184-x)/7+8
or, x/3- (184-x)/7=8
or, 7x-3(184-x)/21=8
or, 7x-552+3x/21=8
or, 10x-552/21=8
or,10x-552=8×21
or,10x=168+552
or, x= 720/10
=72
Therefore , one part = 72
another part = 184-72= 112 (Ans)
By the problem,
x /3= (184-x)/7+8
or, x/3- (184-x)/7=8
or, 7x-3(184-x)/21=8
or, 7x-552+3x/21=8
or, 10x-552/21=8
or,10x-552=8×21
or,10x=168+552
or, x= 720/10
=72
Therefore , one part = 72
another part = 184-72= 112 (Ans)
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