# NCERT Exemplar Solution for CBSE Class 10 Mathematics Chapter: Real Numbers (Part-I)

In this article you will get CBSE Class 10 Mathematics chapter 1, Real Numbers: NCERT Exemplar Problems and Solutions (Part-I). Every question has been provided with a detailed solution. All the questions given in this article are very important to prepare for CBSE Class 10 Board Exam 2017-2018.

Created On: Jun 16, 2017 10:37 IST
Modified On: Jun 16, 2017 17:33 IST Here you get the CBSE Class 10 Mathematics chapter 1, Real Numbers: NCERT Exemplar Problems and Solutions (Part-I). This part of the chapter includes solutions for Exercise 1.1 of NCERT Exemplar Problems for Class 10 Mathematics Chapter: Real Numbers. This exercise comprises of only the Multiple Choice Questions (MCQs) framed from various important topics in the chapter. Each question is provided with a detailed solution.

CBSE Class 10 Mathematics Syllabus 2017-2018

NCERT Exemplar problems are a very good resource for preparing the critical questions like Higher Order Thinking Skill (HOTS) questions. All these questions are very important to prepare for CBSE Class 10 Mathematics Board Examination 2017-2018 as well as other competitive exams.

Find below the NCERT Exemplar problems and their solutions for Class 10 Mathematics Chapter, Real Numbers:

Exercise 1.1

Choose the correct answer from the given four options in the following questions:

Question1.For some integer m, every even integer is of the form

(a) m

(b) m + 1

(c) 2m

(d) 2m +1

We know that, an integer divisible by 2 is an even integer.

Suppose m is any integer then,

m is… ‒3, ‒2, ‒1, 0, 1, 2, 3…

⇒ 2 m is …‒3×(2), ‒2×(2), ‒1×(2), 0×(2), 1×(2), 2×(2), 3×(2) which is clearly divisible by 2.

or2 m is …‒6, ‒4, ‒2, 0, 2, 4, 6… is clearly divisible by 2.

So, for some integer m, every even integer is of the form2m.

Question2. For some integer q, every odd integer is of the form

(a) q

(b) q+ 1

(c) 2q

(d) 2q + 1

Let q be an integer.

Then, q is …‒3, ‒2, ‒1, 0, 1, 2, 3…

⇒ 2q will be …‒3×2, ‒2×2, ‒1×2, 0×2, 1×2, 2×2, 3×2…

or 2q will be …‒6, ‒4, ‒2, 0, 2, 4, 6…

Now, 2q + 1 will …‒6 + 1, ‒4 + 1, ‒2 + 1, 0 + 1, 2 + 1, 4 + 1, 6+ 1…

Or 2q + 1 will …‒5, ‒3, ‒1, 1, 3, 5, 7…which is clearly not divisible by 2.

So, for some integer q, every odd integer is of the form 2q + 1.

Question3. n2 ‒1 is divisible by 8, if n is

(a) an integer

(b) a natural number

(c) an odd integer

(d) an even integer

In n2 ‒ 1, n can be either even or odd,

Let x = n2 ‒ 1

CaseI: When n is even i.e., n = 2k, where k is an integer

x =(2k)2 ‒ 1 or x = 4k2 ‒ 1

For k = ‒1,

x = 4 (‒1)2  ‒ 1 = 4‒ 1 = 3, which is not divisible by 8.

For k = 0,

x = 4 (0)2 ‒1 = 0 ‒1 = ‒1, which is also not divisible by 8.

Case II: When n is odd i.e., n = 2k +1, where k is an odd integer.

⟹       x = n2 ‒1

⟹       x = (2k + 1)2 − 1

⟹       x = 4 k2 + 4k + 1‒1

⟹       x = 4k2 + 4 k

⟹       x = 4k (k + 1)

At k = ‒1, x = 4 (‒1) (‒1 + 1) = 0 which is divisible by 8.

At k = 0, x= 4 (0) (0+ 1) = 4 which is divisible by 8.

At k =1, x= 4 (1) (1 + 1) = 8 which is divisible by 8.

Therefore, from above two cases, we can say that, n2‒1 is divisible by 8 is n is an odd integer.

Question4. If the HCF of 65 and 117 is expressible in the form 65m ‒ 117, then the value of m is

(a) 4

(b) 2

(c) 1

(d) 3

From the Euclid’s division algorithm,

We have, dividend (b) = divisor (a) × quotient (q) + remainder (r)

Or        b = aq + r, 0 r <a

⟹       117 = 65 ×1+ 52

⟹       65 = 52 × 1 + 13

⟹       52 = 13 × 4 + 0

HCF (65, 117) = 13 ... (i)

It is given that, HCF (65, 117) = 65 m ‒ 117  ... (ii)

From equations (i) and (ii), we get:

65 m − 117 = 13

⟹       65 m = 130

m = 2.

Question 5. The largest number which divides 70 and 125, leaving remainders 5 and 8 respectively, is

(a) 13

(b) 65

(c) 875

(d) 1750

5 and 8 are the remainders of 70 and 125 respectively when divided by largest number.

We will subtract 5 and 8 from 70 and 125 so that both numbers will be divisible by required number.

⟹  65 (70 – 5) and 117 (125 – 8) will be divisible by required number.

Now, the largest number dividing 65 and 117 = HCF(65, 117)

Applying Euclid’s division algorithm, we have

117 = 65 1 + 52       [as, dividend = divisor quotient + remainder]

65 = 52  1 + 13

52 = 13  4 + 0

HFC = 13

So, 13 is the largest number which divides 70 and 125, leaving remainders 5 and 8 respectively.

Question 6. If two positive integers a and b are written as a = x3y2 and b = xy3 , where x, y are prime numbers, then HCF ( a, b) is

(a) x y

(b) xy2

(c) x3y3

(d) x2y2

Explanation:

Here, a = x3y2 = x × x × x × y × y and b = xy3 = x × y × y × y

As, HCF is the product of the smallest power of each common prime factor involved in the numbers, therefore

HCF of a and b = HCF (x3y2 , xy3) = x × y × y =  xy2

Question7. If two positive integers p and q can be expressed as p = ab2 and q = a3b; where a, b being prime number, then LCM (p, q) is equal to

(a) ab

(b)

(c)

(d)

Given,  p = ab2 = a × b × b and q = a3b =  a × a × a × b

LCM is the product of the greatest power of each prime factor involved in the number

LCM of p and q = LCM(ab2, a3b) = a3b2

Question8.The product of a non-zero rational and an irrational number is

(a) always irrational

(b) always rational

(c) rational or irrational

(d) one

The product of a non-zero rational number and an irrational number is irrational

Question9.The least number that is divisible by all the numbers from 1 to 10 (both inclusive)

(a) 10

(b) 100

(C) 504

(d) 2520

The least number that is divisible by all the numbers from 1 to 10 will be LCM of the numbers 1 to 10.

LCM of number 1 to 10 = LCM (1, 2, 3, 4, 5, 6, 7, 8, 9, 10)

Question10.The decimal expansion of the rational number  will terminate after

(a) one decimal place

(b) two decimal place

(c) three decimal places

(d) four decimal places

You may also like to read:

CBSE Class 10 NCERT Textbooks & NCERT Solutions

NCERT Solutions for CBSE Class 10 Maths

NCERT Exemplar Problems and Solutions Class 10 Science: All Chapters

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