This page gives you complete exercise 1.1 class 10 maths ncert solutions — all seven questions from the Real Numbers chapter, solved step by step. Every answer opens with the reasoning behind the method, not just the final numbers, so you can repeat the same steps on any fresh set of numbers in your exam.
Exercise 1.1 runs on one central idea: the Fundamental Theorem of Arithmetic. Once you can break a composite number into its unique product of primes, both HCF and LCM come out of the same factorisation.
The questions build in difficulty — pure prime factorisation (Q1), HCF and LCM of pairs (Q2), HCF and LCM of triples (Q3), and then reasoning questions that test whether you understand why the method works (Q4–Q7).
Each question below is reproduced word for word from the NCERT Class 10 Mathematics textbook, pages 5–6, so you can follow along with the book open or closed.
If you ever want to check the original wording, the official Rationalised Class 10 Mathematics textbook is on the NCERT portal at ncert.nic.in — open the PDF, turn to Chapter 1 (Real Numbers), and compare page by page before you trust any other source.
Exercise 1.1 Class 10 Maths NCERT Solutions: All Seven Questions Solved
Exercise 1.1 is the first task of the Real Numbers chapter and it tests one skill repeatedly: writing a composite number as a product of primes. Question 1 is pure practice of that skill. Questions 2 and 3 apply it to find HCF and LCM of pairs and triples.
Question 4 uses the identity \( \text{HCF}(a,b) \times \text{LCM}(a,b) = a \times b \) to avoid factorising altogether. Question 5 tests divisibility by 2 and 5, Question 6 tests recognising a common factor, and Question 7 is a real-life LCM story problem. Together they cover everything the board can ask from this section.
Theorem 1.1 (Fundamental Theorem of Arithmetic, NCERT, p. 3): every composite number can be expressed as a product of primes, and this factorisation is unique apart from the order of the factors. This uniqueness is the reason the prime factorisation method never misses a prime and never counts one twice.
- HCF rule (NCERT, p. 4): product of the smallest power of each common prime factor.
- LCM rule (NCERT, p. 4): product of the greatest power of each prime factor involved.
- Identity (NCERT, p. 4): for any two positive integers \(a\) and \(b\), \( \text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b \).
- Zero-ending rule (NCERT, p. 4, Example 1): a number ending in the digit 0 must be divisible by both 2 and 5, so its prime factorisation must contain the prime 5.
- Three-number warning (NCERT, p. 5): the product of three numbers is not equal to HCF × LCM of the three; the correct three-number formulas sit in the A NOTE TO THE READER box on p. 9.
Question 1: Express each number as a product of its prime factors: (i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429
- (i) 140
- (ii) 156
- (iii) 3825
- (iv) 5005
- (v) 7429
Concept: The Fundamental Theorem of Arithmetic (NCERT, p. 3) guarantees that every composite number has exactly one prime factorisation, apart from the order of the primes. So you can divide by the smallest prime that works, repeat, and be sure no different factorisation can appear from another route.
Part (i): 140 is even, so divide by 2 first. \( 140 = 2 \times 70 = 2 \times 2 \times 35 = 2 \times 2 \times 5 \times 7 \). Collecting the repeated 2 gives the answer: \( 140 = 2^2 \times 5 \times 7 \).
Part (ii): 156 is even too. \( 156 = 2 \times 78 = 2 \times 2 \times 39 \), and \( 39 = 3 \times 13 \). So \( 156 = 2^2 \times 3 \times 13 \).
Part (iii): 3825 ends in 5, so 5 divides it. \( 3825 = 3 \times 1275 = 3 \times 3 \times 425 = 3 \times 3 \times 5 \times 85 = 3 \times 3 \times 5 \times 5 \times 17 \). Writing the repeats as powers: \( 3825 = 3^2 \times 5^2 \times 17 \).
Part (iv): 5005 ends in 5. \( 5005 = 5 \times 1001 \), and \( 1001 = 7 \times 143 = 7 \times 11 \times 13 \). So \( 5005 = 5 \times 7 \times 11 \times 13 \). All four primes are distinct, so no powers appear.
Part (v): 7429 is not even and does not end in 5. Testing primes: \( 7429 \div 17 = 437 \), and \( 437 = 19 \times 23 \). Hence \( 7429 = 17 \times 19 \times 23 \).
Check: multiply each factorisation back. For example \( 3^2 \times 5^2 \times 17 = 9 \times 25 \times 17 = 3825 \). If the product does not return the original number, one of your divisions is wrong.
Question 2: 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
Concept: After factorising both numbers, the HCF keeps only the primes the two share, at their smallest power; the LCM takes every prime that appears in either number, at its greatest power.
The identity \( \text{HCF} \times \text{LCM} = a \times b \) (NCERT, p. 4) is exactly the verification the question asks for — if the two sides do not match, a factorisation is wrong.
Part (i): \( 26 = 2 \times 13 \) and \( 91 = 7 \times 13 \). The only common prime is 13, so \( \text{HCF}(26, 91) = 13 \). The LCM takes 2, 7 and 13 once each: \( \text{LCM} = 2 \times 7 \times 13 = 182 \). Verify: \( 13 \times 182 = 2366 \) and \( 26 \times 91 = 2366 \). The sides match.
Part (ii): \( 510 = 2 \times 3 \times 5 \times 17 \) and \( 92 = 2^2 \times 23 \). The only common prime is 2, at its smallest power \(2^1\), so \( \text{HCF}(510, 92) = 2 \). The LCM takes every prime involved at its greatest power: \( \text{LCM} = 2^2 \times 3 \times 5 \times 17 \times 23 = 23460 \). Verify: \( 2 \times 23460 = 46920 \) and \( 510 \times 92 = 46920 \).
Part (iii): \( 336 = 2^4 \times 3 \times 7 \) and \( 54 = 2 \times 3^3 \). Common primes are 2 and 3; the smallest powers are \(2^1\) and \(3^1\), so \( \text{HCF}(336, 54) = 2 \times 3 = 6 \). The LCM takes the greatest powers: \( \text{LCM} = 2^4 \times 3^3 \times 7 = 3024 \). Verify: \( 6 \times 3024 = 18144 \) and \( 336 \times 54 = 18144 \).
| Pair | Prime factorisation | HCF | LCM | Verification |
|---|---|---|---|---|
| 26, 91 | \(26 = 2 \times 13;\); \(91 = 7 \times 13\) | 13 | 182 | \(13 \times 182 = 26 \times 91\) |
| 510, 92 | \(510 = 2 \times 3 \times 5 \times 17;\); \(92 = 2^2 \times 23\) | 2 | 23460 | \(2 \times 23460 = 510 \times 92\) |
| 336, 54 | \(336 = 2^4 \times 3 \times 7;\); \(54 = 2 \times 3^3\) | 6 | 3024 | \(6 \times 3024 = 336 \times 54\) |
Check: the verification line is not optional — it is the whole point of the question. If \( \text{HCF} \times \text{LCM} \) ever differs from the product, re-read your factorisation before touching the answer.
Question 3: 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
Concept: The same two rules extend to three numbers. The HCF is still the product of the smallest power of every prime that appears in all three; the LCM is the product of the greatest power of every prime that appears in any of them.
Warning: for three numbers the relation \( \text{HCF} \times \text{LCM} = \) product does not hold (NCERT, p. 5), so never verify a triple that way.
Part (i): \( 12 = 2^2 \times 3 \), \( 15 = 3 \times 5 \), \( 21 = 3 \times 7 \). The only prime in all three is 3, so \( \text{HCF}(12, 15, 21) = 3 \). Every prime present goes into the LCM at its greatest power: \( \text{LCM} = 2^2 \times 3 \times 5 \times 7 = 420 \).
Part (ii): 17, 23 and 29 are all primes. No prime appears in more than one number, so \( \text{HCF}(17, 23, 29) = 1 \) and \( \text{LCM} = 17 \times 23 \times 29 = 11339 \). Here and only here the LCM equals the plain product of the three.
Part (iii): \( 8 = 2^3 \), \( 9 = 3^2 \), \( 25 = 5^2 \). The three share no common prime, so \( \text{HCF}(8, 9, 25) = 1 \). The LCM takes every prime at its greatest power: \( \text{LCM} = 2^3 \times 3^2 \times 5^2 = 1800 \).
| Triple | Prime factorisation | HCF | LCM | Why |
|---|---|---|---|---|
| 12, 15, 21 | \(2^2 \times 3;\); \(3 \times 5;\); \(3 \times 7\) | 3 | 420 | all three share prime 3 |
| 17, 23, 29 | all prime | 1 | 11339 | pairwise coprime |
| 8, 9, 25 | \(2^3;\); \(3^2;\); \(5^2\) | 1 | 1800 | no common prime |
Check: when the HCF of three numbers is 1, the numbers are called pairwise coprime and the LCM is simply their product. Do not try to verify a triple with \( \text{HCF} \times \text{LCM} \) — that identity is for pairs only.
Question 4: Given that HCF (306, 657) = 9, find LCM (306, 657).
Concept: This question is built to reward you for knowing the identity \( \text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b \) (NCERT, p. 4). You do not need to factorise 306 or 657 at all — re-arranging the identity gives the LCM directly from the HCF you are handed.
Substitute \(a = 306\), \(b = 657\), and \( \text{HCF} = 9 \) into the rearranged form:
\[ \text{LCM}(306, 657) = \frac{306 \times 657}{\text{HCF}(306, 657)} = \frac{306 \times 657}{9} \]
Since \( 657 \div 9 = 73 \), the division works out cleanly: \( \text{LCM}(306, 657) = 306 \times 73 = 22338 \).
Check: confirm the identity backwards: \( 9 \times 22338 = 201042 \) and \( 306 \times 657 = 201042 \). Matching products mean the answer is right, no factorisation needed.
Question 5: Check whether 6^n can end with the digit 0 for any natural number n.
Concept: A number that ends in zero is divisible by 10, and \( 10 = 2 \times 5 \). So any zero-ending number must have both the prime 2 and the prime 5 in its prime factorisation (NCERT, p. 4, Example 1). Testing a handful of values of \(n\) can never settle the question — you need the factorisation argument.
Write \( 6^n \) in prime factors: \( 6^n = (2 \times 3)^n = 2^n \times 3^n \). The only primes in this factorisation are 2 and 3. There is no prime 5 anywhere.
By the uniqueness part of the Fundamental Theorem of Arithmetic, no other prime can appear in the factorisation of \( 6^n \) no matter how large \(n\) grows. Since the prime 5 is absent, \( 6^n \) can never be divisible by 5, and therefore it can never end with the digit 0 for any natural number \(n\).
Check: see the pattern with small powers — \(6^1 = 6\), \(6^2 = 36\), \(6^3 = 216\) always end in 6, never 0. The argument above is the proof that the pattern continues forever.
Question 6: Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
Concept: A number is composite when it has a factor other than 1 and itself. The fastest way to prove that is to pull out a shared factor — look for a prime that appears in every term of the sum.
First expression: the prime 13 appears in both terms. Factor it out: \( 7 \times 11 \times 13 + 13 = 13(7 \times 11 + 1) = 13(77 + 1) = 13 \times 78 \). Since \( 78 = 2 \times 3 \times 13 \), the full factorisation is \( 13 \times 78 = 2 \times 3 \times 13^2 \). The expression is a product of integers greater than 1, so it is composite.
Second expression: the prime 5 appears in every term. Factor it out: \( 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5 = 5(7 \times 6 \times 4 \times 3 \times 2 \times 1 + 1) \). The bracket works out to \( 1008 + 1 = 1009 \), so the expression equals \( 5 \times 1009 \), which is divisible by 5. Since it has a factor other than 1 and itself, it is composite regardless of whether 1009 happens to be prime.
Check: multiply the factored form back to be sure — \( 13 \times 78 = 1014 \) (the original sum) and \( 5 \times 1009 = 5045 \). The factor you pulled out is the proof of compositeness.
Question 7: 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?
Concept: “Meet again at the starting point” is an LCM problem, not a speed problem. Sonia returns to the start every 18 minutes and Ravi every 12 minutes; the first time both are back together is the smallest number of minutes divisible by both 18 and 12 — that is their LCM.
Factorise: \( 18 = 2 \times 3^2 \) and \( 12 = 2^2 \times 3 \). The LCM takes the greatest power of every prime involved:
\[ \text{LCM}(18, 12) = 2^2 \times 3^2 = 4 \times 9 = 36 \]
So both Sonia and Ravi are at the starting point together after 36 minutes.
Check: in 36 minutes Sonia has driven \( 36 \div 18 = 2 \) complete rounds and Ravi has driven \( 36 \div 12 = 3 \) complete rounds — both exactly back at the start. This sanity check confirms the LCM is correct.
Exercise 1.2 of this chapter continues with more properties of numbers, while the exercises of the next chapter are collected in the Polynomials chapter notes.
Method recap: prime factorisation for HCF and LCM
This is the reusable process behind Questions 1–3. Once you can run it on any pair of numbers, the whole exercise is yours.
- Factorise each number into primes in ascending order, using powers for repeats.
- HCF: take the smallest power of each prime the numbers share, and multiply.
- LCM: take the greatest power of every prime that appears in any number, and multiply.
- Verify with \( \text{HCF} \times \text{LCM} = a \times b \) for a pair; for a triple, re-check the two rules directly.
Worked example with fresh numbers (588 and 630): factorise first — \( 588 = 2^2 \times 3 \times 7^2 \) and \( 630 = 2 \times 3^2 \times 5 \times 7 \). The common primes are 2, 3 and 7; their smallest powers are \(2^1, 3^1, 7^1\). So \( \text{HCF}(588, 630) = 2 \times 3 \times 7 = 42 \). Every prime present goes into the LCM at its greatest power: \( \text{LCM}(588, 630) = 2^2 \times 3^2 \times 5 \times 7^2 = 8820 \). Verify: \( 42 \times 8820 = 370440 \) and \( 588 \times 630 = 370440 \).
| Rule | HCF | LCM |
|---|---|---|
| Which primes? | only the primes common to all numbers | every prime that appears in any number |
| Which power? | the smallest power of each common prime | the greatest power of each prime involved |
| Example: 588, 630 | \(2^1 \times 3^1 \times 7^1 = 42\) | \(2^2 \times 3^2 \times 5 \times 7^2 = 8820\) |
Why this works: the Fundamental Theorem of Arithmetic guarantees the factorisation is unique, so when you list primes you can never miss one that belongs, and you never count the same prime twice. That is what makes the method reliable for every pair.
Three-number formulas (A NOTE TO THE READER, NCERT, p. 9) — needed to handle triples like Question 3, because \( \text{HCF} \times \text{LCM} \) of three numbers is not their product (NCERT, p. 5):
\[ \text{LCM}(p, q, r) = \frac{p \cdot q \cdot r \cdot \text{HCF}(p, q, r)}{\text{HCF}(p, q) \cdot \text{HCF}(q, r) \cdot \text{HCF}(p, r)} \]
\[ \text{HCF}(p, q, r) = \frac{p \cdot q \cdot r \cdot \text{LCM}(p, q, r)}{\text{LCM}(p, q) \cdot \text{LCM}(q, r) \cdot \text{LCM}(p, r)} \]
Pair these worked solutions with the Class 10 Maths notes and the Class 10 notes index for quick revision of the whole chapter.
Common mistakes students make in Exercise 1.1
Each error below is tied to a specific question in this exercise. Fix the rule, then use the check column to catch it before submission.
| Mistake | Correct rule | How to check your answer |
|---|---|---|
| Using the greatest power for HCF or the smallest for LCM (Q2, Q3) | HCF = smallest power of each common prime; LCM = greatest power of every prime involved | Verify with \( \text{HCF} \times \text{LCM} = a \times b \); mismatch means a rule was swapped |
| Testing values of \(n\) instead of factoring in Q5 | A zero-ending number must be divisible by 2 and 5, so it must contain prime 5 | Argue that \(6^n = 2^n \times 3^n\) contains no prime 5, so no \(n\) works |
| Factorising 306 and 657 from scratch in Q4 | Use \( \text{LCM} = \dfrac{a \times b}{\text{HCF}} \) directly | Substitute: \( \dfrac{306 \times 657}{9} = 22338 \) |
| Missing the shared factor in Q6 | Pull out the common prime: \(13(77+1)\) and \(5(1008+1)\) | Expand the factored form — it must multiply back to the original sum |
| Assuming product = \( \text{HCF} \times \text{LCM} \) for three numbers in Q3 | Only true for two numbers; for three, use the p. 9 formulas | Test part (i): product = 3780 but \(3 \times 420 = 1260 \neq 3780\) |
| Treating Q7 as a speed or distance problem | “Meet again at the starting point” means the LCM of the two lap times | \( \text{LCM}(18, 12) = 36 \); check Sonia drives 2 laps, Ravi 3 laps in that time |
Frequently asked questions on Exercise 1.1
Why does a number ending with the digit zero have to be divisible by both 2 and 5, and how does that decide Question 5?
A number ending in zero is divisible by 10, and \( 10 = 2 \times 5 \). So both primes 2 and 5 must divide it, which means both must appear in its prime factorisation. For \( 6^n = (2 \times 3)^n = 2^n \times 3^n \), the only primes are 2 and 3 — no prime 5 exists, and the Fundamental Theorem of Arithmetic forbids one from appearing. So \( 6^n \) can never end in zero.
Does HCF × LCM equal the product of the two numbers for every pair in Exercise 1.1, and how do I use it in Question 4?
Yes — for any two positive integers \(a\) and \(b\), \( \text{HCF}(a,b) \times \text{LCM}(a,b) = a \times b \) (NCERT, p. 4). Question 4 hands you \( \text{HCF}(306, 657) = 9 \), so you substitute straight into the identity: \( \text{LCM}(306, 657) = \frac{306 \times 657}{9} = 22338 \). No factorisation needed — the identity does the whole job.
When is the LCM of three numbers equal to their product, as in Question 3 part (ii) and part (iii)?
When the three numbers are pairwise coprime — no two of them share a single prime factor. In part (ii), 17, 23 and 29 are distinct primes, so HCF = 1 and LCM = \( 17 \times 23 \times 29 = 11339 \). In part (iii), \( 8 = 2^3 \), \( 9 = 3^2 \), \( 25 = 5^2 \) use completely different primes, so LCM = \( 2^3 \times 3^2 \times 5^2 = 1800 \). In part (i), all three numbers contain a 3, so the LCM is not the product — it comes out to 420.
In Question 6, why is 7 × 11 × 13 + 13 a composite number instead of a prime?
Because 13 is a common factor of both terms. Factor it out: \( 7 \times 11 \times 13 + 13 = 13(7 \times 11 + 1) = 13 \times 78 = 2 \times 3 \times 13^2 \). A product of integers greater than 1 has factors besides 1 and itself, so the expression is composite by definition. The second expression works the same way: \( 5(1008 + 1) = 5 \times 1009 \), which is divisible by 5.
How do I find the HCF and LCM of a pair quickly by prime factorisation in Exercise 1.1?
Write both numbers as products of primes in ascending order, using powers for repeats. For HCF, take the smallest power of each prime common to both and multiply. For LCM, take the greatest power of every prime in either number and multiply. Then verify with \( \text{HCF} \times \text{LCM} = a \times b \).
If the two sides differ, one of your factorisations contains an error — hunt it down before moving on. Browse more CBSE notes across classes for related topics.
Reference: NCERT Class 10 Mathematics textbook, chapter Real Numbers.
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