Question

Class 9
Mathematics
Chapter 1

Write in p/q form

Note : Refer the attachment for the " Question " .​

Answer 1

Answer:

5/2

take out the point from the question and write in p/q form

hope it will help you..

Answer 2

Suppose a number with repeating decimal part but integer part is 0, where an n - digit number is repeating in decimal part. Then its p/q form is the repeating n - digit number, divided by 10ⁿ - 1.

The number 10ⁿ - 1 is actually the number with 'n' number of 9's (10² - 1 = 99).

E.g.: In the number $\displaystyle0.\overline{237},$ the number 237 is repeating and it's a 3 digit number, so $\displaystyle n=3.$

Therefore,

$\displaystyle\longrightarrow 0.\overline{237}=\dfrac{237}{999}$

1.

$\displaystyle\longrightarrow5.\overline{2}=5+0.\overline{2}$

$\displaystyle\longrightarrow5.\overline{2}=5+\dfrac{2}{9}$

$\displaystyle\longrightarrow\underline{\underline{5.\overline{2}=\dfrac{47}{9}}}$

2.

$\displaystyle\longrightarrow12.\overline{48}=12+0.\overline{48}$

$\displaystyle\longrightarrow12.\overline{48}=12+\dfrac{48}{99}$

$\displaystyle\longrightarrow\underline{\underline{12.\overline{48}=\dfrac{1236}{99}=\dfrac {412}{33}}}$

3.

$\displaystyle\longrightarrow\underline{\underline{0.\overline{531}=\dfrac{531}{999}=\dfrac {59}{111}}}$

4.

$\displaystyle\longrightarrow\underline{\underline{0.\overline{134}=\dfrac{134}{999}}}$

5.

$\displaystyle\longrightarrow2.\overline{2612}=2+0.\overline{2612}$

$\displaystyle\longrightarrow2.\overline{2612}=2+\dfrac{2612}{9999}$

$\displaystyle\longrightarrow\underline{\underline{2.\overline{2612}=\dfrac{22610}{9999}}}$

A decimal number with n - digit decimal part, can be written in p/q form, by dividing the whole number (ignoring the decimal point) by 10ⁿ.

The number 10ⁿ is 1 followed by 'n' number of 0's (10² = 100).

E.g.: In the number 3.0237, the decimal part 0237 is a 4 digit number (never ignore 0).

Therefore,

$\displaystyle\longrightarrow 3.0237=\dfrac{30237}{10000}$

6.

$\displaystyle\longrightarrow\underline {\underline {0.2532=\dfrac{2532}{10000}=\dfrac{633}{2500}}}$

7.

$\displaystyle\longrightarrow\underline {\underline {0.24=\dfrac{24}{100}=\dfrac{6}{25}}}$

8.

$\displaystyle\longrightarrow\underline {\underline {8.03=\dfrac{803}{100}}}$

9.

$\displaystyle\longrightarrow\underline {\underline {6.025=\dfrac{6025}{1000}=\dfrac{241}{40}}}$

10.

$\displaystyle\longrightarrow\underline {\underline {0.4640=\dfrac{4640}{10000}=\dfrac{58}{125}}}$