Math, asked by shubhamsahu86061, 2 months ago

Find the rational numbers between 1 upon 4 and 1 upon 5

Answers

Answered by sahutanishka944
0

Answer:

First rational no.

\begin{gathered} \frac{1}{2} ( \frac{1}{4} + \frac{1}{5} ) \\ \end{gathered}

2

1

(

4

1

+

5

1

)

\begin{gathered} \frac{1}{2} \times ( \frac{5 + 4}{20} ) \\ \end{gathered}

2

1

×(

20

5+4

)

[LCM of 4 and 5 is 20]

\begin{gathered} \frac{1}{2} \times \frac{9}{20} \\ \end{gathered}

2

1

×

20

9

\begin{gathered} \frac{9}{40} \\ \end{gathered}

40

9

Second rational no.

\begin{gathered} \frac{1}{2} \times ( \frac{9}{40} + \frac{1}{4}) \\ \end{gathered}

2

1

×(

40

9

+

4

1

)

\begin{gathered} \frac{1}{2} \times( \frac{9 + 10}{40} ) \\ \end{gathered}

2

1

×(

40

9+10

)

[LCM of 4 and 40 is 40]

\begin{gathered} \frac{1}{2} \times \frac{19}{40} \\ \end{gathered}

2

1

×

40

19

\begin{gathered} \frac{19}{80} \\ \end{gathered}

80

19

Third rational no.

\begin{gathered} \frac{1}{2} \times ( \frac{9}{40} + \frac{1}{5}) \\ \end{gathered}

2

1

×(

40

9

+

5

1

)

\begin{gathered} \frac{1}{2} \times( \frac{9 + 8}{40} ) \\ \end{gathered}

2

1

×(

40

9+8

)

[LCM of 5 and 40 is 40]

\begin{gathered} \frac{1}{2} \times \frac{17}{40} \\ \end{gathered}

2

1

×

40

17

\begin{gathered} \frac{17}{80} \\ \end{gathered}

80

17

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