Question

Without actually performing the long division, state whether state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion.
(i) $\frac{23}{8}$
(ii) $\frac{125}{441}$
(iii) $\frac{35}{50} [NCERT]$
(iv) $\frac{77}{210} [NCERT]$
(v) $\frac{129}{2^{2} * 5^{7} * 7^{17}}$
(vi) $\frac{987}{10500}$

Answer 1

If the factors of denominator of the given rational number is of form 2ⁿ 5^m ,where n and m are non negative integers, then the decimal expansion of the rational number is terminating otherwise non terminating recurring.

SOLUTION:

(i) Given : 23/8

23/ 2³

Here, the factors of the denominator 8  are 2³ , which is in the form 2ⁿ 5^m .

So , 23/8 has terminating  decimal expansion.

23/8 = 23 × 5³ /2³ × 5³ = 23 × 125 / (2×5)³

= 2875/10³ = 2875/1000 = 2.875  

Hence, the decimal expansion of 23/8 is 2.875.

(ii) Given : 125/441

125 / 3² x 7²

Here, the factors of the denominator 441 are 3² × 7², which is not in the form 2ⁿ 5^m .

So , 125/441 has a non-terminating repeating decimal expansion.

(iii) Given : 35/50

It is not in the simplest form. We have to make it in the simplest form by taking the HCF of both the numbers and divide both the numbers by their HCF.  HCF(35,50) = 5

35÷5 /50÷5 = 7/10

Now,7/10 is in the simplest form.

Here, the factors of the denominator 10 are 2× 5, which is in the form 2ⁿ 5^m .

So , 35/50 has  a terminating decimal expansion.

35/50 = 7/10 = 0.7  

Hence, the decimal expansion of 7/10 is 0.7  

(iv) Given : 77/210

It is not in the simplest form. We have to make it in the simplest form by taking the HCF of both the numbers and divide both the numbers by their HCF.  HCF(77,210) = 7

77÷7 /210÷7 = 11/30

Now,11/30 is in the simplest form.

Here, the factors of the denominator 30 are 2×3 × 5, which is not in the form 2ⁿ 5^m .

So , 77/210 has a non-terminating repeating decimal expansion.

(v) Given : 129/2² × 5^7 × 7^17

Here, the factors of the denominator  are 2² × 5^7 × 7^17, which is not in the form 2ⁿ 5^m .

So , 129/2² × 5^7 × 7^17 has a non-terminating repeating decimal expansion.

(vi) Given : 987/10500

It is not in the simplest form. We have to make it in the simplest form by taking the HCF of both the numbers and divide both the numbers by their HCF.  HCF(987,10500) = 21

987 ÷ 21 /10500 ÷ 21 = 47/500

Here, the factors of the denominator 500  are 5³ × 2², which is in the form 2ⁿ 5^m .

So , 987/10500 has terminating  decimal expansion

987/10500 = 47/500 = 47× 2 / 5³ × 2² × 2¹

= 94/(5×2)³ = 94 /1000 = 0.094  

Hence, the decimal expansion of 987/10500 is 0.094.

HOPE THIS ANSWER WILL HELP YOU…

Answer 2

(i)

Given : 23/8

⇒ 23/2^3

⇒ 23/2³ * 5⁰

Here, denominator is in the form of 2^n * 5^m.

Therefore, 23/8 is a terminating decimal expansion.

(ii)

Given : 125/441

⇒ On Factorizing the denominator, we get

441 = 3 * 3 * 7 * 7

⇒ 125/3^2 * 7^2.

here, Denominator is not of the form 2^n * 5^m.

Therefore, 125/441 is a non-terminating repeating decimal.

(iii)

Given : 35/50

⇒ On Factorizing the denominator, we get

50 = 2 * 25.

⇒ 35/2^0 * 5^2

Here, Denominator is of the form 2^n * 5^m.

Therefore, 35/50 is a terminating decimal expansion.

(iv)

Given : 77/210 = 11/30

⇒ On Factorizing the denominator, we get

30 = 2 * 3 * 5

⇒ 11/2 * 3 * 5

Here, Denominator is not of the form 2^n * 5^m.

Therefore, 77/210 is a non-terminating repeating decimal.

(v)

Given : 129/2^2 * 5^7 * 7^17

Denominator should be in the form of 2^n * 5^m. Here, it also has 7 as its factor.

Therefore, It is a non-terminating repeating decimal.

(vi)

Given : 987/10500 = 47/500

⇒ On Factorizing the denominator, we get

500 = 2 * 2 * 5 * 5 * 5

⇒ 47/2^2 *5^3.

Here, Denominator is of the form 2^n * 5^m.

Therefore, It is a terminating decimal expansion.

Hope this helps!