Exercise 1.1 — Real Numbers

1. 13 marksNCERT Cl-10 Maths, Ex 1.1, Q1

   Write each of these numbers as a product of prime factors: (i) 140, (ii) 156, (iii) 3825, (iv) 5005, (v) 7429.

   Hint. Divide by the smallest prime that goes in, and keep dividing the quotient. Stop when the quotient is itself prime.

   The method is the same every time: keep pulling out the smallest prime that divides the number.

   (i) 140. 140 ÷ 2 = 70, 70 ÷ 2 = 35, and 35 = 5 × 7.

   ✦ 140 = 2² × 5 × 7

   (ii) 156. 156 ÷ 2 = 78, 78 ÷ 2 = 39, and 39 = 3 × 13.

   ✦ 156 = 2² × 3 × 13

   (iii) 3825. It is odd, so 2 is out. The digits add to 3+8+2+5 = 18, which is divisible by 9, so 3 goes in twice: 3825 ÷ 3 = 1275, 1275 ÷ 3 = 425. Then 425 ÷ 5 = 85 and 85 ÷ 5 = 17, which is prime.

   ✦ 3825 = 3² × 5² × 17

   (iv) 5005. Ends in 5, so start there: 5005 ÷ 5 = 1001. And 1001 is worth memorising — it is 7 × 11 × 13.

   ✦ 5005 = 5 × 7 × 11 × 13

   (v) 7429. This one has no small factors, which is what makes it awkward. Work upwards through the primes: 7, 11, 13 all fail. 17 works — 7429 ÷ 17 = 437. Now factor 437: 19 × 23 = 437.

   ✦ 7429 = 17 × 19 × 23

   Where students slip. Stopping at 3825 = 9 × 425. Nine and 425 are not prime, so the factorisation is not finished. Every factor in your final answer must be a prime.

   Another way. A factor tree gives the same answer and is easier to check — the primes are whatever ends up at the tips of the branches.

2. 24 marksNCERT Cl-10 Maths, Ex 1.1, Q2

   For each pair below, find the HCF and the LCM, then confirm that HCF × LCM equals the product of the two numbers: (i) 26 and 91, (ii) 510 and 92, (iii) 336 and 54.

   Hint. Factorise both numbers first. HCF takes the common primes to the lowest power present; LCM takes every prime to the highest power present.

   (i) 26 and 91.

   Step 1 — Factorise: 26 = 2 × 13 and 91 = 7 × 13.

   Step 2 — The only shared prime is 13, so HCF = 13. For the LCM take every prime at its highest power: 2 × 7 × 13 = 182.

   Step 3 — Check: 13 × 182 = 2366, and 26 × 91 = 2366. ✓

   ✦ HCF = 13, LCM = 182

   (ii) 510 and 92.

   Step 1 — Factorise: 510 = 2 × 3 × 5 × 17 and 92 = 2² × 23.

   Step 2 — The only shared prime is 2, and the lower power is 2¹, so HCF = 2. For the LCM: 2² × 3 × 5 × 17 × 23 = 23460.

   Step 3 — Check: 2 × 23460 = 46920, and 510 × 92 = 46920. ✓

   ✦ HCF = 2, LCM = 23460

   (iii) 336 and 54.

   Step 1 — Factorise: 336 = 2⁴ × 3 × 7 and 54 = 2 × 3³.

   Step 2 — Shared primes are 2 and 3. Take the lower power of each: 2¹ × 3¹ = 6, so HCF = 6. For the LCM take the higher power of each: 2⁴ × 3³ × 7 = 3024.

   Step 3 — Check: 6 × 3024 = 18144, and 336 × 54 = 18144. ✓

   ✦ HCF = 6, LCM = 3024

   Where students slip. Swapping the rules — taking the highest power for the HCF. Remember it by meaning: the HCF must divide *both* numbers, so it can only use what both of them actually have.

   Another way. Once you have the HCF, you can skip factorising for the LCM: LCM = (product of the two numbers) ÷ HCF. For (iii) that is 18144 ÷ 6 = 3024.

3. 33 marksNCERT Cl-10 Maths, Ex 1.1, Q3

   Use prime factorisation to find the HCF and LCM of each set of three numbers: (i) 12, 15 and 21, (ii) 17, 23 and 29, (iii) 8, 9 and 25.

   Hint. The same two rules work for three numbers. For the HCF, a prime must appear in all three factorisations to count.

   (i) 12, 15 and 21.

   Step 1 — Factorise: 12 = 2² × 3, 15 = 3 × 5, 21 = 3 × 7.

   Step 2 — Only 3 appears in all three, at power 1 each time, so HCF = 3.

   Step 3 — LCM takes every prime at its highest power: 2² × 3 × 5 × 7 = 420.

   ✦ HCF = 3, LCM = 420

   (ii) 17, 23 and 29.

   Step 1 — All three are already prime, and they are different primes.

   Step 2 — Nothing is shared, so HCF = 1. Numbers like these are co-prime.

   Step 3 — With no overlap, the LCM is just the product: 17 × 23 × 29 = 11339.

   ✦ HCF = 1, LCM = 11339

   (iii) 8, 9 and 25.

   Step 1 — Factorise: 8 = 2³, 9 = 3², 25 = 5².

   Step 2 — Three different primes, nothing shared, so HCF = 1 again.

   Step 3 — LCM = 2³ × 3² × 5² = 8 × 9 × 25 = 1800.

   ✦ HCF = 1, LCM = 1800

   Where students slip. Writing HCF = 0 when the numbers share no factor. Every number is divisible by 1, so the HCF can never drop below 1.

   Another way. Careful with the shortcut here: HCF × LCM = product only works for **two** numbers. For part (i), 3 × 420 = 1260, but 12 × 15 × 21 = 3780. Use factorisation for three numbers.

4. 42 marksNCERT Cl-10 Maths, Ex 1.1, Q4

   You are told that HCF(306, 657) = 9. Use this to find LCM(306, 657).

   Hint. This is exactly the situation the HCF × LCM identity is built for — you are given three of the four quantities.

   Step 1 — For two numbers, HCF × LCM = the product of the numbers. That is the whole question.

   Step 2 — Substitute what you know: 9 × LCM = 306 × 657.

   Step 3 — Multiply out the right side: 306 × 657 = 201042.

   Step 4 — Divide: LCM = 201042 ÷ 9 = 22338.

   ✦ LCM(306, 657) = 22338

   Where students slip. Factorising both numbers from scratch. The HCF was handed to you precisely so you would not have to — spotting that is what the question is testing.

   Another way. Divide before multiplying to keep the numbers small: 306 ÷ 9 = 34, so LCM = 34 × 657 = 22338.

5. 53 marksNCERT Cl-10 Maths, Ex 1.1, Q5

   Can 6ⁿ end in the digit 0 for any natural number n? Justify your answer.

   Hint. Ask what a number must be divisible by in order to end in 0, then check whether 6ⁿ can ever supply it.

   Step 1 — A number ends in 0 exactly when it is divisible by 10, and 10 = 2 × 5. So 6ⁿ would need both a 2 and a 5 in its prime factorisation.

   Step 2 — Factorise the base: 6 = 2 × 3, so 6ⁿ = (2 × 3)ⁿ = 2ⁿ × 3ⁿ. The only primes present are 2 and 3.

   Step 3 — Here is where the Fundamental Theorem of Arithmetic does the real work. It says the prime factorisation of a number is unique. Since 2ⁿ × 3ⁿ is already a prime factorisation of 6ⁿ, and it contains no 5, there is no other factorisation hiding a 5 somewhere.

   Step 4 — With no factor of 5, 6ⁿ is not divisible by 10, so it cannot end in 0.

   ✦ No — 6ⁿ never ends in the digit 0, for any natural number n.

   Where students slip. Answering by testing 6, 36, 216 and stopping. Checking examples shows it has not happened yet, not that it cannot happen. The uniqueness of prime factorisation is what makes the argument cover every n at once.

   Another way. You can also track the last digit: it cycles 6, 6, 6, … because 6 × 6 = 36 always ends in 6. It never reaches 0.

6. 63 marksNCERT Cl-10 Maths, Ex 1.1, Q6

   Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.

   Hint. Do not multiply everything out first. Look for a factor common to both terms and take it outside a bracket.

   A composite number is one with a factor other than 1 and itself. So it is enough to display such a factor — you never need the full factorisation.

   First number: 7 × 11 × 13 + 13.

   Step 1 — Both terms contain 13, so take it out: 7 × 11 × 13 + 13 = 13 × (7 × 11 + 1).

   Step 2 — Simplify the bracket: 7 × 11 + 1 = 78. So the number is 13 × 78 = 1014.

   Step 3 — It is a product of 13 and 78, and both are bigger than 1. That is exactly what composite means.

   ✦ 1014 is composite — 13 divides it.

   Second number: 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5.

   Step 1 — The product runs down through 5, so 5 divides the first term; it obviously divides the second. Take 5 out: 5 × (7 × 6 × 4 × 3 × 2 × 1 + 1).

   Step 2 — The bracket is 1008 + 1 = 1009, so the number is 5 × 1009 = 5045.

   Step 3 — Again a product of two factors above 1.

   ✦ 5045 is composite — 5 divides it.

   Where students slip. Multiplying out to 1014 and 5045 and then hunting for factors by trial. It works but wastes most of the time available; the common factor is visible before any arithmetic happens.

   Another way. For the second number, the last digit is enough on its own: 5040 + 5 = 5045 ends in 5, so it is divisible by 5.

7. 73 marksNCERT Cl-10 Maths, Ex 1.1, Q7

   Two people jog around a circular field, starting together from the same point and going the same way. One completes a lap in 18 minutes, the other in 12 minutes. After how long will they next be at the starting point together?

   Hint. Each is back at the start at multiples of their own lap time. You want the first moment that is a multiple of both — so this is an LCM, not an HCF.

   Step 1 — Decide which one it is. The first runner is at the start at 18, 36, 54, … minutes; the second at 12, 24, 36, … minutes. You want the earliest time appearing in both lists — that is the LCM.

   Step 2 — Factorise: 18 = 2 × 3² and 12 = 2² × 3.

   Step 3 — LCM takes each prime at its highest power: 2² × 3² = 4 × 9 = 36.

   Step 4 — Sanity-check against the lists: at 36 minutes the first has done 2 laps and the second 3. Both are exactly at the start. ✓

   ✦ They meet again at the starting point after 36 minutes.

   Where students slip. Computing the HCF (which is 6) because the word 'common' appears in the question. Ask what the answer means: after 6 minutes neither runner has finished a lap, so 6 cannot possibly be the answer.

   Another way. For a quick check, list multiples of the larger number and test each against the smaller: 18 (not a multiple of 12), 36 (yes) — done.

Solutions written by the tuition.in editorial team and checked against the rationalised NCERT Class 10 Mathematics textbook and the CBSE 2026-27 syllabus (Euclid's division lemma and decimal expansions are no longer part of this chapter). Questions are referenced from the NCERT textbook for identification.