A cable of length 72m is cut into four pieces, such that the length of the the shortest piece is 20% of the length of the longest piece. if the sum of the lengths of the remaining two pieces is equal to the sum of the shortest and the longest pieces , then find the length (in m) of the longest piece.
A. 30
B. 10
C. 20
D. 15
Answer: Option A
Solution (By Examveda Team)
Let length of the longest piece = XShortest Piece = 20% of X = 0.2 X
Now,
Sum Total = 72
X + 0.2 + Two other pieces = 72
1.2X + 1.2X = 72 (Given, Long + short = Other two pieces)
2.4X = 72
X = 30 m.
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