Question

please say me the answer fast question no is 20 and 21 both ok and first do this sum in your notebook and then send to me ok please please ​do fast

Answer 1

Question

1. Arrange the following fractions in ascending and descending orders.

(a) $\dfrac{23}{3}, \dfrac{18}{5},\dfrac{67}{9},\dfrac{13}{2},\dfrac{23}{6}$

(b) $\dfrac{7}{8}, \dfrac{4}{9}, \dfrac{5}{6},\dfrac{3}{5},\dfrac{2}{3}$

2. Add the following fractions.

(a) $\dfrac{5}{6}+\dfrac{7}{8} +\dfrac{8}{9}$

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Answer

1. (a) $\dfrac{23}{3}, \dfrac{18}{5},\dfrac{67}{9},\dfrac{13}{2},\dfrac{23}{6}$

First we will make the denominators of all fractions equal to find the ascending and descending order.

Let's find LCM of all the denominators first.

$\begin{array}{c|c} \tt 2 & \sf{ 3,5,9,2,6} \\ \cline{1-2} \tt 3 & \sf {3,5,9,1,3} \\ \cline{1-2} \tt 3 & \sf{ 1,5,3,1,1} \\ \cline{1-2} \tt 5 & \sf{ 1,5,1,1,1} \\ \cline{1-2} \tt & \sf{1,1,1,1,1 }\\ \end{array}$

LCM → 2 × 3 × 3 × 5 = 90

⇒ $\dfrac{23\times 30 }{3\times 30 } = \dfrac{690}{90}$

⇒ $\dfrac{18\times 18 }{5\times 18 } =\dfrac{324}{90}$

⇒ $\dfrac{67\times 10}{9\times 10 } = \dfrac{670}{90}$

⇒ $\dfrac{13\times 45 }{2\times 45 } = \dfrac{585}{90}$

⇒ $\dfrac{23\times 15 }{6\times 15 }=\dfrac{345}{90}$

Now with the help of the numerators we will arrange the following in ascending and descending order respectively.

Ascending Order

→ $\dfrac{324}{90} <\dfrac{345}{90} < \dfrac{585}{90} < \dfrac{670}{90} < \dfrac{690}{90}$

→ $\dfrac{18}{5} < \dfrac{23}{6} < \dfrac{13}{2} < \dfrac{67}{9}< \dfrac{23}{3}$

Descending Order

→ $\dfrac{690}{90}> \dfrac{670}{90} > \dfrac{585}{90} > \dfrac{345}{90} > \dfrac{324}{90}$

→ $\dfrac{23}{3} > \dfrac{67}{9}> \dfrac{13}{2} > \dfrac{23}{6}> \dfrac{18}{5}$

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(b) $\dfrac{7}{8}, \dfrac{4}{9}, \dfrac{5}{6},\dfrac{3}{5},\dfrac{2}{3}$

First we will make the denominators of all fractions equal to find the ascending and descending order.

Let's find LCM of all the denominators first.

$\begin{array}{c|c} \tt 2 & \sf{ 8,9,6,5,3} \\ \cline{1-2} \tt 2 & \sf {4,9,3,5,3} \\ \cline{1-2} \tt 2 & \sf{ 2,9,3,5,3} \\ \cline{1-2} \tt 3 & \sf{ 1,9,3,5,3} \\ \cline{1-2} \tt 3 & \sf{1,3,1,5,1 }\\ \cline{1-2} \tt 5 & \sf 1,1,1,5,1\\ \cline{1-2} \tt & \sf 1,1,1,1,1 \\ \end{array}$

LCM → 2 × 2 × 2 × 3 × 3 × 5 = 360

⇒ $\dfrac{7\times 45 }{8\times 45 } = \dfrac{315}{360}$

⇒ $\dfrac{4\times 40 }{9\times 40 } = \dfrac{160}{360}$

⇒ $\dfrac{5\times 60 }{6\times 60 }= \dfrac{300}{360}$

⇒ $\dfrac{3\times 72 }{5\times 72 } = \dfrac{216}{360}$

⇒ $\dfrac{2\times 120}{3\times 120} = \dfrac{240}{360}$

Ascending Order

→ $\dfrac{160}{360} < \dfrac{216}{360} < \dfrac{240}{360} < \dfrac{300}{360} < \dfrac{315}{360}$

→ $\dfrac{4}{9} < \dfrac{3}{5} <\dfrac{2}{3}<\dfrac{5}{6} < \dfrac{7}{8}$

Descending Order

→ $\dfrac{315}{360} > \dfrac{300}{360} > \dfrac{240}{360} > \dfrac{216}{360} > \dfrac{160}{360}$

→ $\dfrac{7}{8} > \dfrac{5}{6} > \dfrac{2}{3} > \dfrac{3}{5} > \dfrac{4}{9}$

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2. (a) $\dfrac{5}{6}+\dfrac{7}{8} +\dfrac{8}{9}$

Let's find LCM of all the denominators first.

$\begin{array}{c|c} \tt 2 & \sf{6,8,9} \\ \cline{1-2} \tt 2 & \sf {3,4,9} \\ \cline{1-2} \tt 2 & \sf{ 3,2,9} \\ \cline{1-2} \tt 3 & \sf{ 3,1,9} \\ \cline{1-2} \tt 3& \sf{1,1,3}\\ \cline{1-2} \tt & \sf{1,1,1} \\ \end{array}$

LCM → 2 × 2 × 2 × 3 × 3 = 72

⇒ $\dfrac{5\times 12 }{6\times 12 } = \dfrac{60}{72}$

⇒ $\dfrac{7\times 9 }{8\times 9 } =\dfrac{63}{72}$

⇒ $\dfrac{8\times 8 }{9 \times 8 } =\dfrac{64}{72}$

⇒ $\dfrac{60}{72}+\dfrac{63}{72} +\dfrac{64}{72}$

⇒ $\dfrac{187}{72}$

$\bf \therefore \dfrac{5}{6}+\dfrac{7}{8} +\dfrac{8}{9} = \dfrac{187}{72}$

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