A water pipe is cut into two pieces. The longer piece is 70% of the length of the pipe. By how much percentage is the longer piece longer than the shorter piece? Choose answer:
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Answered by
29
Solutions :-
Given :
A water pipe is cut into two pieces.
The longer piece is 70% of the length of the pipe.
Let the total length of water pipe be 100 m
Therefore,
Length of longer piece = 70% of 100 m
= 70 m
Shorter piece = (100 - 70) m = 30 m
The longer piece longer than the shorter piece = (70 - 30) m = 40 m
Find percent is the longer piece longer than the shorter piece :-
= ( 40 × 100)/30 %
= 4000/30 %
= 133.33 % approx.
Hence,
133.33 % is the longer piece longer than the shorter piece.
Given :
A water pipe is cut into two pieces.
The longer piece is 70% of the length of the pipe.
Let the total length of water pipe be 100 m
Therefore,
Length of longer piece = 70% of 100 m
= 70 m
Shorter piece = (100 - 70) m = 30 m
The longer piece longer than the shorter piece = (70 - 30) m = 40 m
Find percent is the longer piece longer than the shorter piece :-
= ( 40 × 100)/30 %
= 4000/30 %
= 133.33 % approx.
Hence,
133.33 % is the longer piece longer than the shorter piece.
Answered by
22
HEY MATE HERE IS YOUR ANSWER
____________________________
Given :-
♕The water pipe is cut into 2 pieces.
♕The longer piece is 70% of the length of the pipe.
____________________________
Solution :-
Let total length of pipe be 100m.
Therefore,
Length of longer piece = 70% of 100 = 70m
Short piece = 100 - 70 = 30m
The longer piece = 70 - 30 = 40m
To find percent of shorter piece :-
______________________________
Hope it will help you
@thanksforquestion
@bebrainly
____________________________
Given :-
♕The water pipe is cut into 2 pieces.
♕The longer piece is 70% of the length of the pipe.
____________________________
Solution :-
Let total length of pipe be 100m.
Therefore,
Length of longer piece = 70% of 100 = 70m
Short piece = 100 - 70 = 30m
The longer piece = 70 - 30 = 40m
To find percent of shorter piece :-
______________________________
Hope it will help you
@thanksforquestion
@bebrainly
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