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Mathematics

Express the following numbers in the form pq\dfrac{p}{q}, where p and q are both integers and q ≠ 0.

(i)0.3(ii)5.2(iii)0.404040…(iv)0.47(v)0.134(vi)0.001\begin{matrix} \text{(i)} & 0.\overline{3} \\[1.5em] \text{(ii)} & 5.\overline{2} \\[1.5em] \text{(iii)} & 0.404040… \\[1.5em] \text{(iv)} & 0.4\overline{7} \\[1.5em] \text{(v)} & 0.1\overline{34} \\[1.5em] \text{(vi)} & 0.\overline{001} \\[1.5em] \end{matrix}

Rational Irrational Nos

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Answer

(i) Let x = 0.30.\overline{3} = 0.333333… ….(i)\qquad \text{….(i)}

As there is one repeating digit after decimal point,

So multiplying both sides of (i) by 10

we get,

10x = 3.3333…….(ii)\qquad \text{….(ii)}

Subtracting (i) from (ii), we get

9x = 3

x = 39\dfrac{3}{9} = 13\bold{\dfrac{1}{3}},

which is in the form of pq\dfrac{p}{q}, q ≠ 0.

(ii) Let x = 5.25.\overline{2} = 5.2222… ….(i)\qquad \text{….(i)}

As there is one repeating digit after decimal point,

So multiplying both sides of (i) by 10

we get,

10x = 52.2222…….(ii)\qquad \text{….(ii)}

Subtracting (i) from (ii), we get

9x = 47

x = 479\bold{\dfrac{47}{9}},

Which is in the form of pq\dfrac{p}{q}, q ≠ 0.

(iii) Let x = 0.400.\overline{40} = 0.4040… ….(i)\qquad \text{….(i)}

As there are two repeating digit after decimal point,

So multiplying both sides of (i) by 100

we get,

100x = 40.4040…….(ii)\qquad \text{….(ii)}

Subtracting (i) from (ii), we get

99x = 40

x = 4099\bold{\dfrac{40}{99}},

Which is in the form of pq\dfrac{p}{q}, q ≠ 0

(iv) Let x = 0.470.4\overline{7} = 0.477777… ….(i)\qquad \text{….(i)}

As there is one repeating digit after decimal point ,

So multiplying both sides of (i) by 10

we get,

10x=4.7777…….(ii)\qquad \text{….(ii)}

Multiply by 100 on both sides

100x=47.77777…..….(iii)\qquad \text{….(iii)}

subtracting (ii) from (iii), we get

100x-10x=47.7777.. -4.777…

90x= 43

x = 4390\bold{\dfrac{43}{90}}

Which is in the form of pq\dfrac{p}{q}, q ≠ 0

(v) Let x = 0.1340.1\overline{34} = 0.13434 … ….(i)\qquad \text{….(i)}

So multiplying both sides of (i) by 10

we get,

10x=1.343434…….(ii)\qquad \text{….(ii)}

Again multiply by 100 on both sides ,

1000x =134.3434…..….(iii)\qquad \text{….(iii)}

Subtracting (ii) from (iii), we get

1000x - 10x = 134.3434… - 1.3434…

990x = 133

x = 133990\bold{\dfrac{133}{990}}

which is in the form of pq\dfrac{p}{q}, q ≠ 0

(vi) Let x = 0.0010.\overline{001} = 0.001001001… ….(i)\qquad \text{….(i)}

So multiplying both sides of (i) by 1000,

we get,

1000x = 1.001001…….(ii)\qquad \text{….(ii)}

Subtracting (i) from (ii), we get

1000x - x = 1.001001… - 0.001001…

999x = 1

x = 1999\bold{\dfrac{1}{999}}

which is in the form of pq\dfrac{p}{q}, q ≠ 0.

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