# NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.4

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## Class 10 Maths Chapter 1 Real Numbers Ex 1.4

Exercise 1.4 class 10th Question 1

1. Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion:

(i) \frac{\mathbf{13}}{\mathbf{3125}} (ii) \frac{\mathbf{17}}{\mathbf{8}} (iii) \frac{\mathbf{64}}{\mathbf{455}} (iv) \frac{\mathbf{15}}{\mathbf{1600}} (v)  \frac{\mathbf{29}}{\mathbf{343}}

(iv) 23/23 5(vii) 129/22 57 75  (viii) \frac{\mathbf{6}}{\mathbf{15}} (ix)  \frac{\mathbf{35}}{\mathbf{50}} (x)  \frac{\mathbf{77}}{\mathbf{210}}

Solution:

According to the theory we will take x = \frac{p}{q} as a rational number, and the prime factorization will be in the form 2n 5m. Note that here, n and m are non- negative integers. Then x will have a decimal expression which terminates.
(i) \frac{\mathbf{13}}{\mathbf{3125}}
By factorizing the denominator we will get,
3125 = 5x5x5x5x5 = 55
Here, the denominator is in the form of 5m so, \frac{\mathbf{13}}{\mathbf{3125}} is terminating.
(ii)  \frac{\mathbf{17}}{\mathbf{8}}
By factorizing the denominator we will get,
8 = 2x2x2= 23
Here, the denominator is in the form of 2n so, \frac{\mathbf{17}}{\mathbf{8}} is terminating
(iii)  \frac{\mathbf{64}}{\mathbf{455}}
By factorizing the denominator we will get,
455 = 5x7x13
Here, the denominator is in the form of 2n 5m so, \frac{\mathbf{64}}{\mathbf{455}}  is not terminating
(iv) \frac{\mathbf{15}}{\mathbf{1600}}
By factorizing the denominator we will get,
1600 = 2x2x2x2x2x2x5x5 = 26 52
Here, the denominator is in the form of 2n 5m so, \frac{\mathbf{15}}{\mathbf{1600}} is terminating
(v)  \frac{\mathbf{29}}{\mathbf{343}}
By factorizing the denominator we will get,
343 = 7x7x7 = 73
Here, the denominator is in the form of 2n 5m so, \frac{\mathbf{29}}{\mathbf{343}} is terminating
(iv) 23/23 52
Here, the denominator is in the form of 2n 5m so, 23/23 52 23/23 52 is terminating
(vii) 129/22 57 75
Here, the denominator is in the form of 2n 5m so, 129/22 57 75 is not terminating
(viii)  \frac{\mathbf{6}}{\mathbf{15}}
Let’s divide both the numerator and denominator by 3 and we will get \frac{\mathbf{3}}{\mathbf{15}}
that is why the denominator is in the form of 5m so, \frac{\mathbf{6}}{\mathbf{15}} is terminating.
(ix)  \frac{\mathbf{35}}{\mathbf{50}}
Let’s divide both the numerator and denominator by 5 and we will get \frac{\mathbf{7}}{\mathbf{10}}
now factorize the denominator and we will get
10 = 2×5
Here, the denominator is in the form of 2n 5m so, \frac{\mathbf{35}}{\mathbf{50}} is terminating
(x) \frac{\mathbf{77}}{\mathbf{210}}
Let’s divide both the numerator and denominator by 7 and we will get \frac{\mathbf{11}}{\mathbf{30}}
now factorize the denominator and we will get
30 = 2x3x5
Here, the denominator is not in the form of 2n 5m so, \frac{\mathbf{77}}{\mathbf{210}} is non-terminating repeating

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