Question

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.​

Answer 1

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

Hope it helps u

Mark me as Brainiliest answer

Answer 2

$\mathfrak{\huge{\purple{\underline{Given:-}}}}$

Total length of the rope = $\sf 15\dfrac{1}{2} \: m$

Length cut of from the rope = $\sf 1\dfrac{2}{5} \: m, \: 2\dfrac{2}{3}\: m, \: 4\dfrac{5}{9} \: m$

$\mathfrak{\huge{\purple{\underline{To \: Find:-}}}}$

The length of the remaining rope.

$\mathfrak{\huge{\purple{\underline{Analysis:-}}}}$

Add the length of the rope cut off and subtract it from the total length of the rope to find the remaining length.

$\mathfrak{\huge{\purple{\underline{Solution:-}}}}$

Total length of the rope cut off = $\sf 1\dfrac{2}{5} + 2\dfrac{2}{3}+ 4\dfrac{5}{9}$

Convert to improper fractions,

$\implies \sf \dfrac{7}{5} + \dfrac{8}{3} +\dfrac{41}{9}$

Now, find the LCM of the denominators,

LCM = 45

Next we have to make the denominators equal in order to add them up

$\sf \dfrac{7}{5} \times 9=\boxed{\dfrac{63}{45} }$

$\sf \dfrac{8}{3} \times 15 = \boxed{\dfrac{120}{45}}$

$\sf \dfrac{41}{9} \times 5=\boxed{\dfrac{205}{45}}$

Now, add all the fractions,

$\sf \dfrac{63}{45}+\dfrac{120}{45} + \dfrac{205}{45} =\dfrac{388}{45} \: m$

$\implies \sf 8 \dfrac{28}{45}\: m$

Total length of rope = $\sf 15\dfrac{1}{2} \: m$

Remaining length of rope = $\sf 15\dfrac{1}{2} -\dfrac{388}{45}$

That is, $\dfrac{31}{2}- \dfrac{388}{45}$

LCM = 90

$\sf \dfrac{31}{2} \times 45 = \boxed{\dfrac{1395}{90}}$

$\sf \dfrac{388}{45} \times 2 =\boxed{ \dfrac{776}{90}}$

Finding the remaining length,

$\sf \dfrac{1395}{90}-\dfrac{776}{90}$

$\sf =\dfrac{619}{90}$

$\sf = \underline{\underline{6 \dfrac{79}{90} \: m}}$

Therefore, $\sf 6 \dfrac{79}{90} \: m$ of rope is remaining.