Chapter 3
The World of Numbers
CBSE Class 9
Mathematics Solutions
Educart Mathematics class Class 9 NCERT Exemplar cover
Question:

Convert the following decimal numbers into the form of pq\frac{p}{q}.

(A) 12.612.6

(B) 0.01200.0120

(C) 3.052‾3.0\overline{52}

(D) 1.235‾1.2\overline{35}

(E) 0.23‾0.\overline{23}

(F) 2.05‾2.0\overline{5}

(G) 2.125‾2.12\overline{5}

(H) 3.125‾3.12\overline{5}

(I) 2.1625‾2.\overline{1625}

Answer: Verified

(A) 12.6=12610=63512.6 = \frac{126}{10} = \frac{63}{5}

(B) 0.0120=12010000=32500.0120 = \frac{120}{10000} = \frac{3}{250}

(C) 3.052‾3.0\overline{52}: Let x=3.0525252...x = 3.0525252...

10x=30.52525......(i)10x = 30.52525... \quad ...(i)

1000x=3052.52525......(ii)1000x = 3052.52525... \quad ...(ii)

Subtract eq (i) from (ii):

1000x−10x=3052.5252...−30.5252...1000x - 10x = 3052.5252... - 30.5252...

=3022= 3022

990x=3022990x = 3022

⇒x=3022990=1511495\Rightarrow x = \frac{3022}{990} = \frac{1511}{495}

(D) 1.235‾1.2\overline{35}: Let x=1.23535353...x = 1.23535353...

10x=12.35353......(i)10x = 12.35353... \quad ...(i)

1000x=1235.3535......(ii)1000x = 1235.3535... \quad ...(ii)

Subtract eq (i) from (ii):

1000x−10x=1235.3535−12.35351000x - 10x = 1235.3535 - 12.3535

=1223= 1223

990x=1223990x = 1223

⇒x=1223990\Rightarrow x = \frac{1223}{990}

(E) 0.23‾:Letx=0.232323......(i)(E) \ 0.\overline{23}: \text{Let} \quad x = 0.232323... \quad ...(i)

100x=23.2323......(ii)100x = 23.2323... \quad ...(ii)

Subtract eq(i) from (ii):

99x=2399x = 23

⇒x=2399\Rightarrow x = \frac{23}{99}

(F) 2.05‾:Let x=2.0555...(F) \ 2.0\overline{5}: \text{Let } x = 2.0555...

10x=20.555......(i)10x = 20.555... \quad ...(i)

100x=205.555......(ii)100x = 205.555... \quad ...(ii)

Subtract eq(i) from (ii):

100x−10x=185100x - 10x = 185

⇒90x=185\Rightarrow 90x = 185

⇒x=18590=3718\Rightarrow x = \frac{185}{90} = \frac{37}{18}

(G) 2.125‾:Letx=2.12555...(G) \ 2.12\overline{5}: \text{Let} \quad x = 2.12555...

100x=212.555...;100x = 212.555...;

1000x=2125.555...1000x = 2125.555...

1000x−100x=2125.555−212.5551000x - 100x = 2125.555 - 212.555

=1913= 1913

900x=1913900x = 1913

⇒x=1913900\Rightarrow x = \frac{1913}{900}

(H) 3.125‾:Letx=3.12555...(H) \ 3.12\overline{5}: \text{Let} \quad x = 3.12555...

100x=312.555......(i)100x = 312.555... \quad ...(i)

1000x=3125.555......(ii)1000x = 3125.555... \quad ...(ii)

Subtract eq(i) from (ii):

1000x−100x=28131000x - 100x = 2813

900x=2813900x = 2813

⇒x=2813900\Rightarrow x = \frac{2813}{900}

(I) 2.1625‾:(I) \ 2.\overline{1625}:

Letx=2.1625162516251625......(i)\text{Let} \quad x = 2.1625162516251625... \quad ...(i)

10000x=21625.1625‾...(ii)10000x = 21625.\overline{1625} \quad ...(ii)

Subtract eq(i) from (ii):

10000x−x=21625.1625...−2.1625...10000x - x = 21625.1625... - 2.1625...

=21623= 21623

9999x=216239999x = 21623

⇒x=216239999\Rightarrow x = \frac{21623}{9999}

Concept Applied
To convert a repeating decimal to pq\frac{p}{q} form, let xx equal the decimal, multiply by a suitable power of 1010 to align the repeating block, then subtract to eliminate it.

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