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

this article you will get CBSE Class 10 Mathematics chapter 1, Real Numbers: NCERT Exemplar Problems and Solutions (Part-IV). 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. Here you get the CBSE Class 10 Mathematics chapter 1, Real Numbers: NCERT Exemplar Problems and Solutions (Part-IV). This part of the chapter includes solution of Exercise 1.4 of NCERT Exemplar Problems for Class 10 Mathematics Chapter: Real Numbers. This exercise comprises of the Long Answer Type Questions framed from various important topics in the chapter. Each question is provided with a detailed solution.

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

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.

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

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

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

Exercise 1.4

Long Answer Type Questions

Question1. Show that the cube of a positive integer of the form 6q + r, q is an integer and r = 0, 1, 2, 3, 4, 5 is also of the form 6m + r.

Solution:

Let p be an arbitrary positive integer.

By using Euclid’s division algorithm we can say that, corresponding to the positive integers 'p’ and 6, there exist non-negative integers q and r such that

p = 6 q + r, where, 0 ≤ r < 6

p3 = (6q + r)3 =216q3 + r3 + 3 × 6q × r(6q + r)   [Using (a + b)3= a3 + b3 + 3ab (a + b)]

p3 = (216 q3 + 108q2r + 18qr2) + r3           where, 0 ≤ r < 6

When r = 0, then

p3 =216q3 =6(36q3) = 6m, where, m = 36q3 is an integer.

When r = 1, then

p3 = (216q3 + 108q3 + 18q)+ 1=6(36q3 + 18q3 + 3q)+1

p3 = 6m + 1, where m = (36q3 + 1 8q3 + 3q) is an integer.

When r = 2, then

p3 = (216q3+216q2 + 72q)+ 8

p3 =(216q3 +216q2 +72q + 6)+2

p3 = 6(36q3 + 36q2 + 12q + 1)+ 2 =6m+ 2 where, m = (36q2 + 36q2 + 12q + 1) is an integer.

When r = 3, then

p3 =(216q3 + 324q2 + 162q)+27=(216q3 + 324q2 + 162q +24)+ 3

p3= 6 (36q3 + 54q2 + 27q + 4) + 3= 6m + 3 where, m = (36q2 + 54q2 +27q + 4) is an integer.

When r = 4, then

p3 = (216q2 + 432q2 + 288q) + 64

p3 = 6(36q3 + 72q2 + 48q)+ 60+ 4

p3= 6(36q3 +72q2 + 48q +10) + 4=6m + 4 where, m = (36q3 +72q2 + 48q +10) is an integer.

When r = 5, then

p3 =(216q3 + 540q2 + 450q) + 125

p3 = (216q3 + 540q2 + 450q) + 120 +5

p3 = 6 (36q3 + 90q2 + 75q + 20) + 5

p3 = 6m + 5where, m = (36q3 + 90q2 + 75q + 20) is an integer.

Therefore, the cube of a positive integer of the form 6q + r, q is an integer and r = 0, 1, 2, 3, 4, 5 is also of the form 6m + r.

CBSE Class 10 Mathematics Syllabus 2017-2018

Question2. Prove that one and only one out of n, (n + 2) and (n + 4) is divisible by 3, where n is any positive integer.

Solution:

Let q be the quotient and r be the remainder we get after dividing ‘n’ by 3

n = 3q + r, where, 0 ≤ r < 3

Or n = 3q + r, where, r = 0 , 1, 2

So, n = 3q or n = 3q + 1 or n = 3q + 2

When n = 3q, then

n is divisible by 3,

n + 2 (= 3q + 2),

n + 4 (= 3q + 4) are not divisible by 3.

Whenn = 3q + 1, then

n is not divisible by 3,

n + 2 (= 3q + 1 + 2 = 3q + 3), which is divisible by 3,

n + 4 (= 3q + 1 + 4 = 3q + 5), which is not divisible by 3.

When n = 3q + 2, then

n is not divisible by 3,

n + 2(= 3q + 2 + 2 = 3q + 4), which is not divisible by 3,

n + 4 (= 3q + 2 + 4 = 3q + 6), which is divisible by 3.

Hence, one and only one out of n, (n + 2) and (n + 4) is divisible by 3.

Question3. Prove that one of any three consecutive positive integers must be divisible by 3.

Solution:

Any three consecutive positive integers must be of the form n, (n + 1) and (n + 2), where n is any natural number.

Let, a = n, b = n +1 and c = n + 2

Therefore, order triplet is (a, b, c)=(n, n + 1, n + 2),where n = 1, 2, 3,...… (i)

At n = 1; (a, b, c) = (1, 1 + 1, 1 + 2) = (1, 2, 3)

At n = 2; (a, b, c) = (1, 2 + 1, 2 + 2) = (2, 3, 4)

At n = 3; (a, b, c) = (3, 3 + 1, 3 + 2) = (3, 4,5)

At n = 4; (a, b, c) = (4, 4 + 1, 4 + 2) = (4, 5, 6)

At n = 5; (a, b, c) = (5, 5 + 1, 5 + 2) = (5, 6, 7)

At n = 6; (a, b, c)=(6, 6 + 1, 6 + 2) = (6, 7, 8)

At n= 7; (a, b, c) = (7, 7+ 1, 7 + 2) = (7, 8, 9)

At n = 8; (a, b, c) = (8, 8 + 1, 8 + 2) = (8, 9, 10)

We observe that each triplet consist of one and only one number which is divisible by 3.

Hence, one of any three consecutive positive integers must be divisible by 3.

Question4. For any positive integer n, prove that n3n is divisible by 6.

Solution:

Let x=n3n

a = n(n2‒1)

x = n (n ‒ 1) ×(n+ 1)  [Using (a2b2)=(ab)(a + b)]

x=(n ‒1) × n × (n+ 1) ... (i)

We know that, if a number is completely divisible by 2 and 3, then it is also divisible by 6.

Divisibility test for 3:

If the sum of digits of any number is divisible by 3, then it is divisible by 3:

Sum of the digits = (n ‒ 1) + (n) + (n + 1) = 3⇒Number is divisible by 3.

Divisibility test for 2:

If n is odd then (n ‒ 1) and (n + 1) will be even so, (n ‒1) × n × (n + 1) will be divisible by 2.

If n is even then, (n ‒1) × n × (n + 1) will be divisible by 2.

Therefore, for any positive integral value of n, n3 n is divisible by 6.

Question5. Show that one and only one out of n, n + 4,n + 8, n + 12 and n + 16 is divisible by 5, where n is any positive integer.

Solution:

Let q be the quotient and r be the remainder when nis divided by 5,

By Euclid’s division algorithm

n = 5q + r, where 0 ≤ r < 5

n = 5q + r, where r = 0,1,2,3, 4

Therefore, n=5q or 5q + 1 or 5q +2 or 5q + 3 or 5q + 4

If n = 5 q, then

n is divisible by 5,

n + 4 (= 5q + 4) is not divisible by 5,

n + 8 (= 5q + 8) is not divisible by 5,

n + 12 (= 5q + 12) is not divisible by 5,

n + 16 (=5q + 16) is not divisible by 5.

If n= 5q + 1, then

n is not divisible by 5,

n + 4 (= 5q + 1 + 4 = 5q + 5) is divisible by 5,

n + 8 (= 5q + 8 + 1 = 5q + 9) is not divisible by 5,

n + 12 (= 5q + 12 + 2 = 5q + 16) is not divisible by 5,

n + 16 (=5q + 12 + 16 = 5q + 28) is not divisible by 5.

Similarly, if we check for n = 5q +2 or 5q + 3 or 5q + 4 then, there will be one and only one out of n, n + 4, n + 8, n + 12 and n + 16 is divisible by 5.

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