NCERT Solutions: Real Numbers (Exercise 1.1)

# NCERT Solutions for Class 10 Maths Chapter 1 - Real Numbers (Exercise 1.1)

Q1. Express each number as a product of its prime factors:
(i) 140
(ii) 156
(iii) 3825
(iv) 5005
(v) 7429
Solutions:
(i) 140
Prime factors of 140 = 2, 2, 5, 7
= 22 x 5 x 7

(ii) 156
Prime factors of 156 = 2 x 2 x 3 x 13
= 22 x 3 x 13

(iii) 3825
Prime factors of 3825 = 3 x 3 x 5 x 5 x 17
= 3x 52 x 17

(iv) 5005
Prime factors of 5005 = 5 x 7 x 11 x 13

(v) 7429
Prime factors of 7429 = 17 x 19 x 23

Q2. Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers.
(i) 26 and 91
(ii) 510 and 92
(iii) 336 and 54
Solutions:
(i) 26 and 91
Prime factors of 26 = 2 x 13
Prime factors of 91 = 7 x 13
HCF of 26 and 91 = 13
LCM of 26 and 91 = 2 x 7 x 13
= 14 x 17
= 182
Product of two numbers = 26 x 91
= 2366
LCM x HCF = 182 x13 = 2366
So, product of two numbers = LCM x HCF

(ii) 510 and 92
Prime factors of 510 = 2 x 3 x 5 x17
Prime factors of 92 = 2x 2 x 23
HCF of two numbers = 2
LCM of two numbers = 2 x 2 x 3 x 5 x 17 x 23
= 23460
Product of two numbers = 510x92
= 46920
LCM x HCF = 2 x 23460
=46920
Product of two numbers = LCM x HCF

(iii) 336 and 54
Prime factors of 336 = 2 x 2 x 2 x 2 x 3 x 7
Prime factors of 54 = 2 x 3 x 3 x 3
HCF of two numbers = 6
LCM of two numbers = 2 x 2 x 2 x 2 x 3 x 3 x 3 x 7
= 24 x 33 x 7 = 3024
Product of two numbers = 336 x 54
=18144
LCM x HCF = 3024 x 6 =18144
Product duct of two numbers = LCM x HCF

Q3. Find the LCM and HCF of the following integers by applying the prime factorisation method.
(i) 12, 15 and 21
(ii) 17, 23 and 29
(iii) 8, 9 and 25
Solutions:
(i) 12, 15 and 21
Prime factors of 12 = 2 x 2 x 3
= 22 x 3
Prime factors of 15 = 3 x 5
Prime factors of 21 = 2 x 2 x 3
HCF of 12,15 and 21 = 3
LCM of 12,15 and 21 = 22 x 3 x 5 x 7
= 420

(ii) 17, 23 and 29
Prime factors of 17 = 17 x 1
Prime factors of 23 = 23 x 1
Prime factors of 29 = 29 x 1
HCF of 17, 23 and 29 = 1
LCM of 17, 23 and 29 = 17 x 23 x 29
= 11339

(iii) 8, 9 and 25
Prime factors of 8 = 2 x 2 x 2 x1
= 23 x 1
Prime factors of 9 = 3 x 3 x1
= 32 x 1
Prime factors of 25 = 5 x 5 x 1
= 52 x 1
HCF of 8, 9 and 25 = 1
LCM of 8,9 and 25 = 2 x 2 x 2 x 3 x 3 x 5 x 5
= 1800

Q4. Given that HCF (306, 657) = 9, find LCM (306, 657).
Solution:
Given, HCF (306, 657) = 9.
We have to find,
LCM (306, 657) = ?
We know that
LCM x HCF = Product of two numbers
LCM x 9 = 306x 657

LCM = 34 x 657
LCM = 22338

Q5. Check whether 6n can end with the digit 0 for any natural number n.
Solution:
Prime factors of 6n = (2 x 3)n = (2)n (3)n
You can observe clearly, 5 is not in the prime factors of 6n.
That means 6n will not be divisible by 5.
Therefore, 6n cannot end with the digit 0 for any natural number n.

Q6. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
Solution:
It can be observed that,
7 x 11 x 13 + 13 = 13 (7 x11 +1)
= 13( 77 + 1)
= 13 x 78
= 13 x13 x 6 x1
= 13 x13 x 2 x 3 x1
The given number has 2,3,13 and 1 as its factors. Therefore, it is a composite number.
7 x 6 x 5 x 4 x 3 x 2 x1 + 5 = 5 x( 7 x 6 x 4 x 3 x 2 x 1 + 1)
= 5 x(1008 + 1)
=5x1009x1
1009 cannot be factorised further. Therefore, the given expression has 5,1009 and 1 as its factors. Hence, it is a composite number.

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?
Solution:
LCM of 18 and 12,
18 = 2 x 3 x 3
12 = 2 x 2 x 3
LCM of 12 and 18 = 2 x 2 x 3 x 3 = 36
Therefore, Ravi and Sonia will meet together at starting point after 36 minutes.

## Deleted Exercise of NCERT

Q1. Use Euclid’s division algorithm to find the HCF of
(i) 135 and 225
(ii) 196 and 38220
(iii) 867 and 255
Ans:

(i) HCF of 135 and 225
Step 1: Since 225 > 135, we apply the division lemma to 225 and 135 to obtain
⇒ 225 = 135 × 1 + 90
Step 2: Since remainder 90 ≠ 0, we apply the division lemma to 135 and 90 to obtain
135 = 90 × 1 + 45
Step 3: We consider the new divisor 90 and new remainder 45 and apply the division lemma to obtain: 90 = 2 × 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) HCF of 196 and 38220
Step 1: Since 38220 > 196, we apply the division lemma to 38220 and 196 to obtain
⇒ 38220 = 196 × 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) HCF of 867 and 255
Step 1: Since 867 > 255, we apply the division lemma to 867 and 255 to obtain
⇒ 867 = 255 × 3 + 102
Step 2: Since remainder 102 ≠ 0, we apply the division lemma to 255 and 102 to obtain
⇒ 255 = 102 × 2 + 51
Step 3: We consider the new divisor 102 and new remainder 51 and apply the division lemma to obtain
⇒ 102 = 51 × 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.

Q2. Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.
Ans: Let us consider a positive odd integer as ‘a’.
On dividing ‘a’ by 6, Let ‘a’ be the quotient and ‘r’ be the remainder.
∴ Using Euclid’s lemma, we have:
a = 6q + r where 0 ≤ r < 6 i.e., r = 0, 1, 2, 3, 4, or 5

Now substituting the value of r, we get:
If r = 0, then a = 6q
Similarly, for r= 1, 2, 3, 4 and 5, the value of a is 6q+1, 6q+2, 6q+3, 6q+4 and 6q+5, respectively.
If a = 6q, 6q+2, 6q+4, then a is an even number and divisible by 2.
A positive integer can be either even or odd.
Therefore, any positive odd integer is of the form of 6q+1, 6q+3 and 6q+5, where q is some integer.

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?
Ans: Given,
Number of army contingent members = 616
Number of army band members = 32
If the two groups have to march in the same column, we have to find out the highest common factor between the two groups.
HCF(616, 32) gives the maximum number of columns in which they can march.

By Using Euclid’s algorithm to find their HCF, we get,
Since, 616 > 32
∴ 616 = 32 × 19 + 8
Since, 8 ≠ 0, therefore, taking 32 as the new divisor, we have,
32 = 8 × 4 + 0
Now we have got remainder as 0, therefore, HCF (616, 32) = 8.
Hence, the maximum number of columns in which they can march is 8.

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 form 3m or 3m + 1.]

Ans: Let us consider an arbitrary positive integer as ‘x’ such that it is of the form 3q, (3q + 1) or (3q + 2).

Note:
For any positive integer x and b = 3,
x = 3q + r, where q is quotient and r is remainder such that 0 < r < 3.
If r = 0 then x = 3q
If r = 1 then x = 3q + 1
If r = 2 then x = 3q + 2

For x = 3q, we have
x2 = (3q)2
⇒ x2 = 9q2
= 3 (3q2)
= 3m ...(1) (Putting 3q2 = m where m is an integer)

For x = 3q + 1
x2 = (3q + 1)2
= 9q2 + 6q + 1
= 3 (3q2 + 2q) + 1
= 3m + 1 ...(2) (Puting 3q2 + 2q = m, where m is an integer)

For x = 3q + 2
x2 = (3r + 2)2
= 9q2 + 12q + 4

= (9q2 + 12q + 3) + 1
= 3 (3q2 + 4q + 1) + 1
= 3m + 1 ...(3) (Putting 3q2 + 4q + 1 = m, where m is an integer)
From (1), (2) and (3)
x2 = 3m or 3m + 1
Thus, the square of any positive integer is either of form 3m or 3m + 1 for some integer m.

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.
Ans: Let us consider an arbitrary positive integer x such that it is in the form of 3q, (3q + 1) or (3q + 2).

Note:
For any positive integer x and b = 3,
x = 3q + r, where q is quotient and r is remainder such that 0 < r < 3.
If r = 0 then x = 3q
If r = 1 then x = 3q + 1
If r = 2 then x = 3q + 2

∴ x is of the form 3q, (3q + 1) or (3q + 2).

For x = 3q
x3 = (3q)3
= 27q3
= 9 (3q3)
= 9m ...(1) (Putting 3q3 = m, where m is an integer)

For x = 3q + 1
x3 = (3q + 1)3
= 27q3 + 27q2 + 9q + 1
= 9 (3q3 + 3q2 + q) + 1
= 9m + 1 ...(2) (Putting (3q3 + 3q2 + 1) = m, where m is an integer)

For x = 3q + 2
x3 = (3q + 2)3
= 27q3 + 54q2 + 36q + 8
= 9 (3q3 + 6q2 + 4q) + 8
= 9m + 8 ...(3) (Putting (3q3 + 6q2 + 4q) = m, where m is an integer)
From (1), (2), (3) we have:
x3 = 9m, (9m + 1) or (9m + 8)
Thus, x3 of any positive integer can be in the form 9m, (9m + 1) or (9m + 8).

Check out the NCERT Solutions of all the exercises of Real Numbers:

Exercise 1.2 NCERT Solutions: Real Numbers

Exercise 1.3 NCERT Solutions: Real Numbers

Exercise 1.4 NCERT Solutions: Real Numbers

The document NCERT Solutions for Class 10 Maths Chapter 1 - Real Numbers (Exercise 1.1) is a part of the Class 10 Course Mathematics (Maths) Class 10.
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## Mathematics (Maths) Class 10

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## FAQs on NCERT Solutions for Class 10 Maths Chapter 1 - Real Numbers (Exercise 1.1)

 1. What is Real Numbers in Class 10 Maths?
Ans. Real Numbers are a set of all rational and irrational numbers, which can be expressed in decimal form. It includes all positive and negative integers, fractions, and decimals. In Class 10 Maths, Real Numbers are an important topic that students need to learn thoroughly.
 2. What is the importance of studying Real Numbers?
Ans. Real Numbers are a fundamental concept in Mathematics. They are used in various mathematical operations and formulas. Studying Real Numbers helps students to understand the number system, properties of numbers, and their operations. It also helps in solving complex mathematical problems.
 3. How can I prepare for the Real Numbers exam in Class 10?
Ans. To prepare for the Real Numbers exam in Class 10, students should start by understanding the basic concepts thoroughly. They should practice solving various types of problems from textbooks and reference books. Students can also take the help of online tutorials, sample papers, and previous year question papers for better preparation.
 4. What are the properties of Real Numbers?
Ans. The properties of Real Numbers are commutative, associative, distributive, additive identity, multiplicative identity, additive inverse, and multiplicative inverse. These properties help in solving mathematical problems and simplify complex expressions.
 5. How can I improve my understanding of Real Numbers?
Ans. To improve your understanding of Real Numbers, you should practice solving different types of problems regularly. You can also take the help of online tutorials, video lectures, and reference books. It is also important to clear your doubts and seek help from your teachers whenever required.

## Mathematics (Maths) Class 10

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