Asked by Iron Sparrow

2/7 part of a cable was laid on the first day, 1/5 part on the second day and 1/3 part was laid on the third day. If 443 1/3m was laid on the fourth day to complete the work, what is the total length of the cable laid?
Answered by bobpursley
Length of cable=L
2L/7 +L/5+L/3+(443*3+1)/3=L

Multiply by 35*3

2*15L+21L+35L+35*(443*3+1)=35*3L

L(35*3-30-21-35)=443*3 +1
solve for L
Answered by G. Mariappan
Length of Cable=L
(2L/7)+(L/5)+(L/3)+(143*3+1)=L
Multiply by 35*3
2*15L+21L+35L+(35*3*(1330/3)=35*3L
30L+21L+35L+46550=105L
105L-86L=46550
19L=46550
L=46550/19=2450m
Answered by Manisha
2450
2/7+1/5+1/3+1330/3
LCM 105
30+21+35+46550=86/105
105-86=46550
19=46550
46550/19=2450m
Answered by Manisha
A wire was laid on first day=2/7
" " " " " " " " " "second day=1/5
" " " " " " " " " "third day=1/3
Fourth day complete=443ยน/3=1330/3
2/7+1/5+1/3+1330 =105
30+21+35+46550=105
86+46550=105
105-86=46550
19/46550=2450m
Answered by Manisha
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