NCERT Solutions for Class 10 Maths Unit 1

Real Numbers Class 10

Unit 1 Real Numbers Exercise 1.1, 1.2, 1.3, 1.4 Solutions

In mathematics, a real number is a value that represents a quantity along a continuous line. The real numbers include all the rational numbers, such as the integer -5 and the fraction 4/3, and all the irrational numbers such as √2 (1.41421356, the square root of two, an irrational algebraic number) and π (3.14159265, a transcendental number). Real numbers can be thought of as points on an infinitely long line called the number line or real line, where the points corresponding to integers are equally spaced. Any real number can be determined by a possibly infinite decimal representation such as that of 8.632, where each consecutive digit is measured in units one tenth the size of the previous one. The real line can be thought of as a part of the complex plane, and complex numbers include real numbers.

A real number may be either rational or irrational; either algebraic or transcendental; and either positive, negative, or zero. Real numbers are used to measure continuous quantities. They may be expressed by decimal representations that have an infinite sequence of digits to the right of the decimal point; these are often represented in the same form as 324.823122147 The ellipsis (three dots) indicates that there would still be more digits to come.
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Exercise 1.1 : Solutions of Questions on Page Number : 7

Q1 :  

Use Euclid's division algorithm to find the HCF of:


Answer :

(i) 135 and 225

Since 225 > 135, we apply the division lemma to 225 and 135 to obtain

225 = 135 x 1 + 90

Since remainder 90 ≠ 0, we apply the division lemma to 135 and 90 to obtain

135 = 90 x 1 + 45

We consider the new divisor 90 and new remainder 45, and apply the division lemma to obtain

90 = 2 x 45 + 0

Since the remainder is zero, the process stops.

Since the divisor at this stage is 45,

Therefore, the HCF of 135 and 225 is 45.

(ii)196 and 38220

Since 38220 > 196, we apply the division lemma to 38220 and 196 to obtain

38220 = 196 x 195 + 0

Since the remainder is zero, the process stops.

Since the divisor at this stage is 196,

Therefore, HCF of 196 and 38220 is 196.

(iii)867 and 255

Since 867 > 255, we apply the division lemma to 867 and 255 to obtain

867 = 255 x 3 + 102

Since remainder 102 ≠ 0, we apply the division lemma to 255 and 102 to obtain

255 = 102 x 2 + 51

We consider the new divisor 102 and new remainder 51, and apply the division lemma to obtain

102 = 51 x 2 + 0

Since the remainder is zero, the process stops.

Since the divisor at this stage is 51,

Therefore, HCF of 867 and 255 is 51.

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Q2 :  

Show that any positive odd integer is of the form , or , or , where q is some integer.


Answer :

Let a be any positive integer and b = 6. Then, by Euclid's algorithm,

a = 6q + rfor some integer q ≥ 0, and r = 0, 1, 2, 3, 4, 5 because 0 r < 6.

Therefore, a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5

Also, 6q + 1 = 2 x 3q + 1 = 2k1 + 1, wherek1 is a positive integer

6q + 3 = (6q + 2) + 1 = 2 (3q + 1) + 1 = 2k2 + 1, where k2 is an integer

6q + 5 = (6q + 4) + 1 = 2 (3q + 2) + 1 = 2k3 + 1, where k3 is an integer

Clearly, 6q + 1, 6q + 3, 6q + 5 are of the form 2k + 1, where k is an integer.

Therefore, 6q + 1, 6q + 3, 6q + 5 are not exactly divisible by 2. Hence, these expressions of numbers are odd numbers.

And therefore, any odd integer can be expressed in the form 6q + 1, or 6q + 3,

or 6q + 5

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Q3 :  

An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?


Answer :

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Q4 :  

Use Euclid's division lemma to show that the square of any positive integer is either of form 3m or 3m + 1 for some integer m.

[Hint: Let x be any positive integer then it is of the form 3q, 3q + 1 or 3q + 2. Now square each of these and show that they can be rewritten in the form 3m or 3m + 1.]


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Q5 :  

Use Euclid's division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.


Answer :

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Exercise 1.2 : Solutions of Questions on Page Number : 11

Q1 :  

Express each number as product of its prime factors:


Answer :

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Q2 :  

Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers.


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Q3 :  

Find the LCM and HCF of the following integers by applying the prime factorisation method.


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Q4 :  

Given that HCF (306, 657) = 9, find LCM (306, 657).


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Q5 :  

Check whether 6n can end with the digit 0 for any natural number n.


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Q6 :  

Explain why 7 x 11 x 13 + 13 and 7 x 6 x 5 x 4 x 3 x 2 x 1 + 5 are composite numbers.


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Q7 :  

There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?


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Exercise 1.3 : Solutions of Questions on Page Number : 14

Q1 :  

Prove that is irrational.


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Q2 :  

Prove that is irrational.


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Q3 :  

Prove that the following are irrationals:


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Exercise 1.4 : Solutions of Questions on Page Number : 17

Q1 :  

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:


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Q2 :  

Write down the decimal expansions of those rational numbers in Question 1 above which have terminating decimal expansions.


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Q3 :  

The following real numbers have decimal expansions as given below. In each case, decide whether they are rational or not. If they are rational, and of the form , what can you say about the prime factor of q?

(i) 43.123456789 (ii) 0.120120012000120000… (iii)


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Maths : CBSE NCERT Exercise Solutions for Class 10th for Real Numbers ( Exercise 1.1, 1.2, 1.3, 1.4 ) will be available online in PDF book form soon. The solutions are absolutely Free. Soon you will be able to download the solutions.

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