From a rope 15 1/2 m long, three pieces of lengths 1 2/5 m, 2 2/3 m and 4 5/9 m are cut off. what is the length of the remaining rope.
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From a rope 15 1/2 m long, three pieces of lengths 1 2/5 m, 2 2/3 m and 4 5/9 m are cut off. what is the length of the remaining rope.
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Rope is 15 1/2 m long = 31/2
Three pieces of its ropes are as
• 1 2/5 m = 7/5 m
• 2 2/3 m = 8/3 m
• 4 5/9 m = 41/9 m
Remaining lengths of rope is
31/2 - ( 7/5+8/3+41/9)
31/2-(63+120+205/45)
31/2-388/45
1395-776/45
619/45 = 13 34/45 m is remaining rope
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Verified answer
[tex]\mathfrak{\huge{\purple{\underline{Given:-}}}}[/tex]
Total length of the rope = [tex]\sf 15\dfrac{1}{2} \: m[/tex]
Length cut of from the rope = [tex]\sf 1\dfrac{2}{5} \: m, \: 2\dfrac{2}{3}\: m, \: 4\dfrac{5}{9} \: m[/tex]
[tex]\mathfrak{\huge{\purple{\underline{To \: Find:-}}}}[/tex]
The length of the remaining rope.
[tex]\mathfrak{\huge{\purple{\underline{Analysis:-}}}}[/tex]
Add the length of the rope cut off and subtract it from the total length of the rope to find the remaining length.
[tex]\mathfrak{\huge{\purple{\underline{Solution:-}}}}[/tex]
Total length of the rope cut off = [tex]\sf 1\dfrac{2}{5} + 2\dfrac{2}{3}+ 4\dfrac{5}{9}[/tex]
Convert to improper fractions,
[tex]\implies \sf \dfrac{7}{5} + \dfrac{8}{3} +\dfrac{41}{9}[/tex]
Now, find the LCM of the denominators,
LCM = 45
Next we have to make the denominators equal in order to add them up
[tex]\sf \dfrac{7}{5} \times 9=\boxed{\dfrac{63}{45} }[/tex]
[tex]\sf \dfrac{8}{3} \times 15 = \boxed{\dfrac{120}{45}}[/tex]
[tex]\sf \dfrac{41}{9} \times 5=\boxed{\dfrac{205}{45}}[/tex]
Now, add all the fractions,
[tex]\sf \dfrac{63}{45}+\dfrac{120}{45} + \dfrac{205}{45} =\dfrac{388}{45} \: m[/tex]
[tex]\implies \sf 8 \dfrac{28}{45}\: m[/tex]
Total length of rope = [tex]\sf 15\dfrac{1}{2} \: m[/tex]
Remaining length of rope = [tex]\sf 15\dfrac{1}{2} -\dfrac{388}{45}[/tex]
That is, [tex]\dfrac{31}{2}- \dfrac{388}{45}[/tex]
LCM = 90
[tex]\sf \dfrac{31}{2} \times 45 = \boxed{\dfrac{1395}{90}}[/tex]
[tex]\sf \dfrac{388}{45} \times 2 =\boxed{ \dfrac{776}{90}}[/tex]
Finding the remaining length,
[tex]\sf \dfrac{1395}{90}-\dfrac{776}{90}[/tex]
[tex]\sf =\dfrac{619}{90}[/tex]
[tex]\sf = \underline{\underline{6 \dfrac{79}{90} \: m}}[/tex]
Therefore, [tex]\sf 6 \dfrac{79}{90} \: m[/tex] of rope is remaining.