Here you get the CBSE Class 10 Mathematics chapter 1, Real Numbers: NCERT Exemplar Problems and Solutions (Part-II). This part of the chapter includes solution of Exercise 1.3 of NCERT Exemplar Problems for Class 10 Mathematics Chapter: Real Numbers. This exercise comprises of the Short 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)
Find below the NCERT Exemplar problems and their solutions for Class 10 Mathematics Chapter, Real Numbers:
Exercise 1.3
Short Answer Type Questions
Question1. Show that the square of any positive integer is either of the form 4q or 4q + 1 for some integer q.
Solution:
Suppose a is any positive integer. Using Euclid’s division algorithm, for positive integers a and 4, there exist non-negative integers k and r, such that
a = 4k + r, where 0 ≤ r <4
⟹ a^{2} = (4k + r)^{2} = 16k^{2} + r^{2} + 8kr …..(i)
When r = 0, then Eq…(i) becomes,
a^{2} = 16k^{2} = 4(4k^{2}) = 4q [Where, q = 4k^{2} is an integer]
When r = 1, then Eq…(i) becomes,
a^{2} = 16k^{2} + 1 + 8k = 4(4k^{2} + 2k) + 1
= 4q + 1 [Where, q = (4k^{2} + 2k)^{ } is an integer]
When r = 2, then Eq…(i) becomes,
a^{2} = 16k^{2} + 4 + 16k = 4(4k^{2} + 2k + 1)
= 4q [Where, q = (4k^{2} + 2k +1) is an integer]
When r = 3, then Eq…(i) becomes,
a^{2} = 16k^{2} + 9 + 24k = 16k^{2} + 24k + 8 + 1
= 4(4k^{2} + 6k + 2) + 1
= 4q + 1 [Where, q = (4k^{2} + 6k + 2)^{ } is an integer]
Therefore, the square of any positive integer is either of the form 4 q or 4q + 1 for some integer q.
Question2. Show that cube or any positive integer is of the form 4m, 4m + 1 or 4m + 3, for some integer m.
Solution:
Suppose a is any positive integer. Using Euclid’s division algorithm, for some positive integers a and 4, there exist non-negative integers k and r, such that
a = 4k + r, where 0 ≤ r <4
⟹ a^{3} = (4k + r)^{3} = 64k^{3} + r^{3} + 12kr^{2} + 48k^{2}r [Using (a + b)^{3} = a^{3} + b^{3} + 3a^{2}b + 3ab^{2}]
⟹ a^{3} = (64k^{3} + 48k^{2}r + 12kr^{2}) + r^{3} ……(i)
When r = 0, then Eq…(i) becomes,
a^{3} = 64k^{3} = 4(16k^{3}) or a^{3} = 4m [Where, m = 16k^{3} is an integer]
When r = 1, then Eq…(i) becomes,
a^{3} = 64k^{3 }+ 48k^{2 }+ 12k + 1 = 4(16k^{3 }+ 12k^{2} + 3k) +1
= 4m + 1 [Where, m = (16k^{3 }+ 12k^{2} + 3k) is an integer]
When r = 2, then Eq…(i) becomes,
a^{3} = 64k^{3 }+ 196k^{2 }+ 36k + 4 = 4 (16k^{3 }+ 49k^{2 }+ 9k + 1)
= 4m [Where, m = (16k^{3 }+ 49k^{2 }+ 9k + 1) is an integer]
When r = 3, then Eq…(i) becomes,
a^{3} = 64k^{3 }+ 144k^{2 }+ 108k + 27 = 64k^{3 }+ 144k^{2 }+ 108k + 24 + 3
= 4 (16k^{3 }+ 36k^{2 }+ 27k + 6) + 3
= 4m + 3 [Where, m = (16k^{3 }+ 36k^{2 }+ 27k + 6) is an integer]
Therefore, the cube of any positive integer is of the form 4m, 4m + 1 or 4m + 3 for some integer m.
CBSE Class 10 Mathematics Syllabus 2017-2018
Question3. Show that the square of any positive integer cannot be of the form 5q + 2 or 5q + 3 for any integer q.
Solution:
Suppose a is any positive integer. Using Euclid’s division algorithm, for some positive integers a and 5, there exist non-negative integers k and r, such that
a = 5k + r, where 0 ≤ r <5
Now, a^{2 }= (5k + r)^{2} = 25k^{2} + r^{2} + 10kr
⟹ a^{2 }= 5(5k^{2} + 2kr) + r^{2} ……(i)
When r = 0, then Eq…(i) becomes,
a^{2 }= 5(5k^{2}) = 5q [Where, q = 5k^{2} is an integer]
When r = 1, then Eq…(i) becomes,
a^{2 }= 5(5k^{2} + 2k) + 1
= 5q +1 [Where, q = (5k^{2} + 2kr) is an integer]
When r = 2, then Eq…(i) becomes,
a^{2} = 5(5k^{2} + 4k) + 4
= 5q + 4 [Where, q = (5k^{2} + 4k) is an integer]
When r = 3, then Eq…(i) becomes,
a^{2} = 5(5k^{2} + 6k) + 9
= 5(5k^{2} + 6k) + 5 + 4
= 5(5k^{2} + 6k + 1) + 4 = 5q + 4 [Where, q = (5k^{2} + 6k + 1)^{ } is an integer]
When r = 4, then Eq…(i) becomes,
a^{2} = 5(5k^{2} + 8k) + 16
= 5(5k^{2} + 8k) + 15 + 1
= 5(5k^{2} + 8k + 3) + 1 = 5q + 1 [Where, q = (5k^{2} + 8k + 3)^{ } is an integer]
Therefore, the square of any positive integer can’t be in the form 5q + 2 or 5q + 3 where q is any integer.
Question4.Show that the square of any positive integer cannot be of the form 6m + 2 or 6m + 5 for any integer m.
Solution:
Suppose a is any positive integer. Using Euclid’s division algorithm, for some positive integers a and 6, there exist non-negative integers k and r, such that
a = 6k + r, where 0 ≤ r <6
Now, a^{2 }= (6k + r)^{2} = 36k^{2} + r^{2} + 12kr
⟹ a^{2 }= 6(6k^{2} + 2kr) + r^{2} ……(i)
When r = 0, then Eq…(i) becomes,
a^{2 }= 6(6k^{2}) = 6m [Where, m = 5k^{2} is an integer]
When r = 1, then Eq…(i) becomes,
a^{2 }= 6(6k^{2} + 2k) + 1
= 6m +1 [Where, m = (6k^{2} + 2k) is an integer]
When r = 2, then Eq…(i) becomes,
a^{2} = 6(6k^{2} + 4k) + 4
= 6m + 4 [Where, m = (6k^{2} + 4k) is an integer]
When r = 3, then Eq…(i) becomes,
a^{2} = 6(6k^{2} + 6k) + 9
= 6(6k^{2} + 6k) + 6 + 3
= 6(6k^{2} + 6k + 1) + 3 = 6m + 3 [Where, m = (6k^{2} + 6k + 1)^{ } is an integer]
When r = 4, then Eq…(i) becomes,
a^{2} = 6(6k^{2} + 8k) + 16
= 6(6k^{2} + 8k) + 12 + 4
= 6(6k^{2} + 8k + 2) + 4 = 6m + 4 [Where, m = (6k^{2} + 8k + 2)^{ } is an integer]
When r = 5, then Eq…(i) becomes,
a^{2} = 6(6k^{2} + 10k) + 25
= 6(6k^{2} + 8k) + 24 + 1
= 6(6k^{2} + 8k + 4) + 1 = 6m + 1 [Where, m = (6k^{2} + 8k + 4)^{ } is an integer]
Therefore, the square of any positive integer can’t be of the form 6m + 2 or 6 m + 5 for any integer m.
Question5. Show that the square of any odd integer is of the form 4m + 1, for some integer m.
Solution:
By Euclid’s division algorithm,
We have a = bq+ r … (i) Where 0 r < 4 and a, b are positive integers
On putting b = 4 in Eq… (i), we have
a = 4q + r, Where 0 r < 4 i.e., r = 0, 1,2,3
When r = 0 then a = 4q which is divisible by is even.
When r = 1 then a = 4q + 1which is not divisible by 2.
When r = 2 then a = 4q + 2 = 2(2q + 1), where (2q + 1) is divisible by is even.
When r = 3 then a = 4q + 3which is not divisible by 2.
Therefore, for any positive integer q, we will get (4q + 1) and (4q + 3) as odd integers.
Now, (4q + 1)^{2} = 16q^{2} + 1 + 8q
= 4(4q^{2} + 2q) + 1 = 4m +1 [Where, m = (4q^{2} + 2q)^{ }is an integer]
And (4q + 3)^{2} = 16q^{2} + 9 + 24q
= 4(4q^{2} + 6q + 2) + 1 = 4m +1 [Where, m = (4q^{2} + 6q + 2)^{ }is an integer]
Hence, for some integer m, the square of any odd integer is of the form 4m + 1.
Question6. If n is an odd integer, then show that n^{ 2} – 1 is divisible by 8.
Question7. Prove that, if x and y are both odd positive integers, thenis even but not divisible by 4.
Solution:
Question8. Use Euclid’s division algorithm to find the HCF of 441, 567 and 693.
Solution:
Suppose, a = 693, b = 567 and c = 441
By Euclid’s division algorithms,
(dividend) a = (divisor) b × (quotient) q + (remainder) r
⇒ a = bq + r ….. (i)
First we take two larger number, a = 693 and b = 567,
693 = 567 × 1 + 126
567 = 126 × 4 + 63
126 = 63 × 2 + 0
Thus, HCF (693,567) = 63
Again applying Euclid’s division algorithm and putting a = 441 and say b = 63, we have
a = bq + r = 63 × 7 + 0
⟹ 441 = 63 × 7 + 0
Thus, HCF (693, 441) = 63.
Question 9. Using Euclid’s division algorithm, first the largest number that divides 1251, 9377 and 15628 leaving remainders 1, 2 and 3, respectively.
Solution:
Given, 1, 2 and 3 are the remainders on dividing 1251, 9377 and 15628, respectively by the first largest number.
After subtracting these remainders from the numbers the largest number will these three numbers completely.
Now, 1251 – 1 = 1250, 9377 – 2 = 9375, and 15628 – 3 = 15625
Thus, 1250, 9375 and 15625 will be completely divisible by the first largest number (that asked in question).
Now, required largest number = HCF of 1250, 9375 and 15625
By Euclid’s division algorithm,
dividend (a) = divisor (b) × quotient (q) + remainder (r)
⇒ a = bq + r … (i)
For largest number, we will put a = 15625 and b = 9375 in equation … (i)
⇒15625 = 9375 × 1 + 6250
⇒ 9375=6250 × 1+3125
⇒ 6250 = 3125 × 2 +0
Thus, HCF (15625, 9375,) = 3125
Again applying Euclid’s division algorithm, and taking a = 1250 and b = 3125, in eq… (i) we get,
3125 = 1250 × 2 + 625
⇒ 1250 = 625 × 2 + 0
Thus, HCF (1250, 9375, 15625) = 625
Hence, 625 is the largest number which divides 1251, 9377 and 15628 leaving remainder 1, 2 and 3, respectively.
Question11. Show that 12^{n} cannot end with the digit 0 or 5 for any natural number n.
Solution:
Any number which ends with the digit 0 or 5, then it will be divisible by 5.
If 12^{n} ends with the digit zero it must be divisible by 5.
This is possible only if prime factorisation of 12^{n} contains the prime number 5.
Now, 12 = 2 × 2 × 3 = 2^{2} × 3
⟹ 12^{n }= (2^{2} × 3)^{n} = 2^{2n} × 3n
Since, prime factorisation of 12^{n} doesn’t contain 5 hence, there is no value of natural numaber n for which 12^{n} ends with digit zero or five.
Question12. On a morning walk, three persons step off together and their steps measure 40 cm, 42 cm and 45 cm, respectively. What is the minimum distance each should walk, so that each can cover the same distance in complete steps?
Solution:
The minimum distance each should walk will be the LCM of 40 cm, 42 cm and 45 cm
In terms of prime number, 40 = 2 × 2 × 2 × 5,42 = 2 × 3 × 7 and 45 = 3 × 3 × 5
Thus, LCM (40, 42, 45) =2 ×2 × 2 × 3 × 3 × 5 × 7= 2520.
Minimum distance each should walk is 2520 cm.
Question13. Write the denominator of rational number 257/5000 in the form 2m × 5n, where m = 3 and n = 4 are non-negative integers. Hence, write its decimal expansion, without actual division.
Solution:
You may also like to read:
NCERT Solutions for CBSE Class 10 Maths
CBSE Class 10 NCERT Textbooks & NCERT Solutions
NCERT Exemplar Problems and Solutions Class 10 Science: All Chapters
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