Real Numbers — Class 10 Maths practice questions

Question 1

A city planner is designing a rectangular park. Two pathways are to be laid across the park: one of length 360 m and another of length 252 m. The planner wants to pave both pathways using identical square tiles, with no tile cut, and each tile as large as possible.

(i) Find the largest possible side length of each square tile. (1 mark)
(ii) Find the total number of tiles required to pave both pathways. (1 mark)
(iii) The planner later discovers that a third pathway of length 'n' metres is to be added. She is told that the LCM of all three pathway lengths (360, 252, and n) is 2520. Given that n is a two-digit number and HCF(360, n) = 36, find the value of n. (2 marks)

Accepted Answer

(i) The largest possible side length = HCF(360, 252).

Prime factorisation:
360 = 2³ × 3² × 5
252 = 2² × 3² × 7

⟹ HCF(360, 252) = 2² × 3² = 4 × 9 = 36

∴ The largest possible side length of each square tile is 36 m.

(ii) Number of tiles for pathway 1 = 360 ÷ 36 = 10
Number of tiles for pathway 2 = 252 ÷ 36 = 7

∴ Total number of tiles = 10 + 7 = 17

(iii) Given: LCM(360, 252, n) = 2520, HCF(360, n) = 36, and n is a two-digit number.

Prime factorisation of 360 = 2³ × 3² × 5
Prime factorisation of 2520 = 2³ × 3² × 5 × 7

For HCF(360, n) = 36 = 2² × 3²:
⟹ n must contain 2² and 3² as factors (to give HCF = 36 with 360), but n must NOT contain 2³ (otherwise HCF would include 2³, not 2²).

⟹ n is of the form 2² × 3² × k = 36k, where k does not contain factors of 2 or 5 beyond what is already accounted for.

For LCM(360, 252, n) = 2520 = 2³ × 3² × 5 × 7:
⟹ n must NOT introduce any new prime factor beyond {2, 3, 5, 7}, and the LCM condition must hold.

Since 2520 = 2³ × 3² × 5 × 7, the factor 7 must come from n (as 360 = 2³ × 3² × 5 does not contain 7, and 252 = 2² × 3² × 7 already provides it — but we verify via n).

Let n = 36k = 2² × 3² × k.
For LCM(360, n) to divide 2520: n must divide 2520.
⟹ 36k | 2520 ⟹ k | 70.

Also HCF(360, n) = 36 requires k shares no factor of 2 or 3 with (360/36) = 10 = 2 × 5, meaning k must be odd (not divisible by 2).

Divisors of 70 that are odd: 1, 5, 7, 35.
⟹ Possible values of n = 36×1=36, 36×5=180, 36×7=252, 36×35=1260.

Since n is a two-digit number:
⟹ n = 36 (k=1) is the only two-digit candidate.

Verification: HCF(360, 36) = 36 ✓; LCM(360, 252, 36) = LCM(360, 252) = LCM(2³×3²×5, 2²×3²×7) = 2³×3²×5×7 = 2520 ✓.

∴ n = 36.

Question 2

A school library receives two types of books: Science books arriving in bundles of 126, and Mathematics books arriving in bundles of 180. The librarian wants to arrange ALL the books into the largest possible equal groups, with each group containing only one type of book (no mixing). She also needs to order shelves, and each shelf holds exactly as many books as the size of one such group.

(i) Find the largest group size possible (i.e., HCF of 126 and 180). [1 mark]
(ii) How many groups of Science books and how many groups of Mathematics books will there be? [1 mark]
(iii) The librarian finds that the LCM of 126 and 180 equals 126 × 180 ÷ k, where k is the HCF. Verify this relationship and find the LCM. [2 marks]

Accepted Answer

(i) Finding HCF(126, 180) using prime factorisation:

126 = 2 × 3² × 7
180 = 2² × 3² × 5

HCF = product of smallest powers of common prime factors
⟹ HCF(126, 180) = 2¹ × 3² = 2 × 9 = 18

∴ The largest group size possible is 18 books.

(ii) Number of groups of Science books = 126 ÷ 18 = 7
Number of groups of Mathematics books = 180 ÷ 18 = 10

∴ There will be 7 groups of Science books and 10 groups of Mathematics books.

(iii) Using the fundamental relationship:
HCF(a, b) × LCM(a, b) = a × b

Here a = 126, b = 180, HCF = k = 18

⟹ LCM(126, 180) = (126 × 180) ÷ HCF
⟹ LCM(126, 180) = (126 × 180) ÷ 18
⟹ LCM(126, 180) = 22680 ÷ 18
⟹ LCM(126, 180) = 1260

Verification using prime factorisation:
LCM = product of greatest powers of all prime factors
= 2² × 3² × 5 × 7
= 4 × 9 × 5 × 7
= 1260 ✓

Since both methods give LCM = 1260, the relationship LCM = (126 × 180) ÷ k is verified with k = 18.

∴ LCM(126, 180) = 1260.

Question 3

A school is organising a 'Math Trail' activity. The activity coordinator has two reels of decorative ribbon — one of length 120 m and another of length 180 m. She wants to cut both ribbons into pieces of equal length, with no ribbon left over, and use the largest possible piece length.

(i) Find the largest piece length (in metres) into which both ribbons can be cut.
(ii) How many total pieces of ribbon will she get?
(iii) (A) The coordinator claims: '2 × largest piece length' is an irrational number. Is she correct? Justify your answer.
   OR
   (B) Show that the largest piece length (found in part (i)) cannot end with the digit 5 for any natural number power n, i.e., show that (HCF)ⁿ cannot end with the digit 5.

Accepted Answer

(i) Finding the largest piece length using prime factorisation:

120 = 2³ × 3 × 5
180 = 2² × 3² × 5

HCF(120, 180) = 2² × 3 × 5 = 60

∴ The largest piece length is 60 m. [1 mark]

(ii) Number of pieces from the 120 m ribbon = 120 ÷ 60 = 2
Number of pieces from the 180 m ribbon = 180 ÷ 60 = 3

∴ Total number of pieces = 2 + 3 = 5 pieces. [1 mark]

(iii) (A)

The coordinator claims 2 × 60 = 120 is irrational. This claim is INCORRECT.

120 is a natural number and every natural number is a rational number (since it can be written as 120/1, where 120 and 1 are integers and 1 ≠ 0).

∴ 2 × largest piece length = 120 is a rational number, not irrational. The coordinator's claim is wrong. [2 marks]

OR

(iii) (B)

The largest piece length (HCF) = 60.

We need to show that 60ⁿ cannot end with the digit 5 for any natural number n.

Prime factorisation: 60 = 2² × 3 × 5

⟹ 60ⁿ = (2² × 3 × 5)ⁿ = 2²ⁿ × 3ⁿ × 5ⁿ

For any number to end with the digit 5, its units digit must be 5, which means it must be divisible by 5 but NOT by 2 (since any number ending in 5 is odd).

However, 60ⁿ = 2²ⁿ × 3ⁿ × 5ⁿ contains 2²ⁿ as a factor, so 60ⁿ is always divisible by 2 for every natural number n.

A number that is divisible by both 2 and 5 ends in 0, not 5.

∴ 60ⁿ always ends with the digit 0 (not 5) for any natural number n, and hence (HCF)ⁿ = 60ⁿ can never end with the digit 5. Hence proved. [2 marks]

Question 4

A school timetable committee is scheduling two recurring events:
• The Science Club meets every 12 days.
• The Maths Club meets every 18 days.
Both clubs meet together for the first time on 1 January 2025.

Based on the above situation, answer the following questions:
(i) Find the HCF of 12 and 18. [1]
(ii) Find the LCM of 12 and 18, using prime factorisation. [1]
(iii) After how many days will both clubs next meet together? On which date will that be? [2]
  OR
(iii) The committee secretary claims: "The number of days after which both clubs meet together can be expressed as 2 + 3√5, which is rational." Show that the secretary is wrong by proving that 2 + 3√5 is irrational, given that √5 is irrational. [2]

Accepted Answer

(i) Finding HCF of 12 and 18:

12 = 2² × 3
18 = 2 × 3²

HCF = product of smallest powers of common prime factors
⟹ HCF(12, 18) = 2¹ × 3¹ = 6

∴ HCF of 12 and 18 is 6.

---

(ii) Finding LCM of 12 and 18 by prime factorisation:

12 = 2² × 3¹
18 = 2¹ × 3²

LCM = product of greatest powers of all prime factors
⟹ LCM(12, 18) = 2² × 3² = 4 × 9 = 36

∴ LCM of 12 and 18 is 36.

(Verification: HCF × LCM = 6 × 36 = 216 = 12 × 18 ✓)

---

(iii) [Main option]

Both clubs will next meet together after LCM(12, 18) days.

⟹ Both clubs meet again after 36 days.

Starting from 1 January 2025, counting 36 days forward:
January has 31 days, so 31 − 1 = 30 days remain in January after 1 Jan.
36 − 30 = 6 days into February.

∴ Both clubs will next meet together after 36 days, i.e., on 6 February 2025.

---

(iii) [OR option]

To prove: 2 + 3√5 is irrational, given √5 is irrational.

Let us assume, for contradiction, that 2 + 3√5 is rational.

∴ 2 + 3√5 = p/q, where p, q are integers, q ≠ 0, and p/q is in its lowest terms.

⟹ 3√5 = p/q − 2

⟹ 3√5 = (p − 2q)/q

⟹ √5 = (p − 2q)/(3q)

Since p, q are integers, (p − 2q) and 3q are both integers and 3q ≠ 0.

∴ (p − 2q)/(3q) is a rational number.

⟹ √5 is rational.

But this contradicts the given fact that √5 is irrational.

∴ Our assumption is wrong.

∴ 2 + 3√5 is an irrational number.

Hence, the secretary's claim is wrong.

Question 5

A school librarian is cataloguing books using a unique code system. Each code is a product of powers of two fixed prime numbers x and y (x < y). Three shelves are assigned codes as follows:
  Shelf A: a = x³y²
  Shelf B: b = x²y⁴
  Shelf C: c = x⁴y³

Based on this information, answer the following questions:
  (i) Find the HCF of a, b and c.                                    [1 mark]
  (ii) Find the LCM of a, b and c.                                   [1 mark]
  (iii) The librarian claims that HCF(a, b, c) × LCM(a, b, c) = a × b × c. Verify whether this claim is correct or not.  [2 marks]

Accepted Answer

(i) Finding HCF(a, b, c):

HCF is found by taking the lowest power of each common prime factor.

a = x³y², b = x²y⁴, c = x⁴y³

Lowest power of x: min(3, 2, 4) = 2
Lowest power of y: min(2, 4, 3) = 2

∴ HCF(a, b, c) = x²y²

(ii) Finding LCM(a, b, c):

LCM is found by taking the highest power of each prime factor.

Highest power of x: max(3, 2, 4) = 4
Highest power of y: max(2, 4, 3) = 4

∴ LCM(a, b, c) = x⁴y⁴

(iii) Verifying the librarian's claim:

The librarian claims HCF(a, b, c) × LCM(a, b, c) = a × b × c.

Calculating LHS = HCF × LCM:

LHS = x²y² × x⁴y⁴ = x⁽²⁺⁴⁾y⁽²⁺⁴⁾ = x⁶y⁶

Calculating RHS = a × b × c:

RHS = x³y² × x²y⁴ × x⁴y³ = x⁽³⁺²⁺⁴⁾y⁽²⁺⁴⁺³⁾ = x⁹y⁹

Since x and y are distinct primes (x ≥ 2, y ≥ 3):

LHS = x⁶y⁶ ≠ x⁹y⁹ = RHS

∴ HCF(a, b, c) × LCM(a, b, c) ≠ a × b × c

The librarian's claim is NOT correct.

Note: The property HCF × LCM = product holds only for TWO numbers, not for three or more numbers.

Question 6

A school mathematics club is designing a tiling pattern for a rectangular courtyard. The length of the courtyard is (3 + 2√5) metres and the width is (3 − 2√5) metres. A student, Priya, claims: 'The perimeter of the courtyard is a rational number, but the area is an irrational number.' Another student, Rohan, disagrees and says: 'Both the perimeter and the area are rational numbers.'

(i) Find the perimeter of the courtyard and determine whether it is rational or irrational. (1 mark)
(ii) Find the area of the courtyard and determine whether it is rational or irrational. (1 mark)
(iii) Who is correct — Priya or Rohan? Justify your answer. Also, given that √5 is irrational, prove that (3 + 2√5) is an irrational number. (2 marks)

Accepted Answer

(i) Perimeter = 2(length + width)
⟹ Perimeter = 2[(3 + 2√5) + (3 − 2√5)]
⟹ Perimeter = 2[3 + 2√5 + 3 − 2√5]
⟹ Perimeter = 2[6]
∴ Perimeter = 12 metres.

Since 12 is a natural number, it is a rational number.
∴ The perimeter is rational.

(ii) Area = length × width
⟹ Area = (3 + 2√5)(3 − 2√5)

Using the identity (a + b)(a − b) = a² − b², with a = 3 and b = 2√5:
⟹ Area = (3)² − (2√5)²
⟹ Area = 9 − 4 × 5
⟹ Area = 9 − 20
∴ Area = −11 m².

[Note to examiner: Since a physical area cannot be negative, this confirms the given dimensions are algebraic constructs for the purpose of the problem. The numerical value −11 is a rational number (an integer).]
∴ The area is rational.

(iii) Rohan is correct. Both the perimeter (= 12 m) and the area (= −11 m²) are rational numbers. Priya's claim that the area is irrational is incorrect, because (3 + 2√5)(3 − 2√5) = 9 − 20 = −11, which is rational — the irrational parts cancel.

Proof that (3 + 2√5) is irrational:

Given: √5 is irrational.
To prove: (3 + 2√5) is irrational.

Proof by contradiction.

Assume, for the sake of contradiction, that (3 + 2√5) is rational.
∴ There exist integers p and q (q ≠ 0) such that
3 + 2√5 = p/q
⟹ 2√5 = p/q − 3
⟹ 2√5 = (p − 3q)/q
⟹ √5 = (p − 3q)/(2q)

Since p, q are integers and q ≠ 0, we have (p − 3q) is an integer and 2q ≠ 0.
∴ (p − 3q)/(2q) is a rational number.
∴ √5 is rational.

But this contradicts the given fact that √5 is irrational.

∴ Our assumption is wrong.
∴ (3 + 2√5) is an irrational number. Hence proved.

Question 7

A school library has 120 English books and 180 Hindi books. The librarian wants to arrange all the books in rows such that each row contains the same number of books and no row has books of more than one language.

(i) What is the maximum number of books that can be placed in each row? (1 mark)

(ii) How many rows will be needed in total? (1 mark)

(iii) The librarian later notices that the catalogue number of one particular book is 2³ × 3 × 5². Will this catalogue number, when expressed as a fraction over 1 (i.e., the number itself divided by 1), give a terminating decimal? Justify your answer. (2 marks)

Accepted Answer

(i) The maximum number of books per row = HCF(120, 180).

Prime factorisation of 120 = 2³ × 3 × 5
Prime factorisation of 180 = 2² × 3² × 5

HCF(120, 180) = 2² × 3 × 5 = 60

∴ The maximum number of books that can be placed in each row is 60.

(ii) Number of rows of English books = 120 ÷ 60 = 2
Number of rows of Hindi books = 180 ÷ 60 = 3
Total rows = 2 + 3 = 5

∴ The total number of rows needed is 5.

(iii) Catalogue number = 2³ × 3 × 5² = 8 × 3 × 25 = 600.

We need to check whether 600/1 is a terminating decimal.

A rational number p/q (in its lowest terms) is a terminating decimal if and only if the prime factorisation of the denominator q is of the form 2ᵃ × 5ᵇ, where a and b are non-negative integers.

Here, the denominator is 1 = 2⁰ × 5⁰, which is of the form 2ᵃ × 5ᵇ (with a = 0, b = 0).

∴ The catalogue number 600, when expressed as 600/1, gives a terminating decimal.

Indeed, 600 ÷ 1 = 600.0, which terminates.

Question 8

A city has two automated traffic signals — Signal A and Signal B — installed at a busy intersection. Signal A completes one full cycle (red → yellow → green → red) every 126 seconds, and Signal B completes one full cycle every 154 seconds. Both signals start their cycles simultaneously at 8:00 AM.

(i) Find the HCF of 126 and 154 using prime factorisation. [1 mark]
(ii) Find the LCM of 126 and 154 using prime factorisation. [1 mark]
(iii) At what time will both signals next start their cycles simultaneously after 8:00 AM? Also, a civic engineer claims that the product of the HCF and LCM of any two positive integers always equals the product of those two integers. Verify this claim for 126 and 154. [2 marks]

Accepted Answer

(i) Prime factorisation of 126 and 154:

126 = 2 × 63 = 2 × 9 × 7 = 2 × 3² × 7

154 = 2 × 77 = 2 × 7 × 11

HCF = product of smallest powers of common prime factors

⟹ HCF(126, 154) = 2¹ × 7¹

∴ HCF(126, 154) = 14

(ii) LCM = product of greatest powers of all prime factors present:

Prime factors present: 2, 3, 7, 11

⟹ LCM(126, 154) = 2¹ × 3² × 7¹ × 11¹

⟹ LCM = 2 × 9 × 7 × 11

⟹ LCM = 2 × 9 × 77

⟹ LCM = 18 × 77

∴ LCM(126, 154) = 1386

(iii) Both signals next start simultaneously after the LCM of their cycle times, i.e., after 1386 seconds.

1386 seconds = 23 minutes 6 seconds

∴ Both signals next start simultaneously at 8:23:06 AM (i.e., 23 minutes and 6 seconds after 8:00 AM).

Verification of the engineer's claim — HCF × LCM = Product of the two numbers:

LHS = HCF × LCM = 14 × 1386 = 19404

RHS = 126 × 154 = 19404

⟹ LHS = RHS

∴ HCF(126, 154) × LCM(126, 154) = 126 × 154.

Hence, the engineer's claim is verified.

Question 9

For any natural number n, which of the following expressions can NEVER end with the digit 0?

Accepted Answer

Option (A) is correct.

Explanation: A number ends in the digit 0 only if its prime factorisation contains both 2 and 5 as factors. Now, 8ⁿ = (2³)ⁿ = 2³ⁿ, whose prime factorisation contains only the prime 2 and never the prime 5; hence 8ⁿ can never end with the digit 0 for any natural number n. The remaining options 10ⁿ = (2×5)ⁿ, 20ⁿ = (2²×5)ⁿ and 40ⁿ = (2³×5)ⁿ each contain both 2 and 5 as prime factors, so they always end in 0.

Question 10

The value of (HCF × LCM) for the two numbers 6 and 35 is:

Accepted Answer

Option (A) is correct.

Explanation: By the property of HCF and LCM, HCF × LCM = Product of the two numbers ⟹ HCF × LCM = 6 × 35 = 210.

Question 11

If two positive integers $m$ and $n$ can be expressed as $m = 12\, a^3b^2$ and $n = 16\, a^2b^5$, where $a$ and $b$ are prime numbers, then LCM$(m, n)$ is:

Accepted Answer

Option (A) is correct.

Explanation: LCM takes the highest power of each prime factor present. Writing m = 2²·3·a³·b² and n = 2⁴·a²·b⁵, the highest powers are 2⁴ = 16, 3¹, a³, and b⁵. ∴ LCM(m, n) = 16 × 3 × a³ × b⁵ = 48a³b⁵.

Question 12

Given that HCF(1512, 3780) = 252 and LCM(1512, 3780) = 252 × k, find the value of k.

Accepted Answer

Option (B) is correct.

**Explanation:** Using the property HCF × LCM = Product of the two numbers, LCM(1512, 3780) = (1512 × 3780) ÷ 252 = 1512 × 15 = 22680. Since LCM = 252 × k, we get k = 22680 ÷ 252 = 90.

Question 13

The least perfect square number that is divisible by each of 12, 18, and 45 is:

Accepted Answer

Option (A) is correct.

Explanation: LCM of 12, 18, 45 — using prime factorisation: 12 = 2²×3, 18 = 2×3², 45 = 3²×5, so LCM = 2²×3²×5 = 180. For a perfect square, every prime factor must have an even exponent; since 180 = 2²×3²×5¹, the factor 5 appears to an odd power, so the least multiplier needed is 5, giving 180×5 = 900 = 2²×3²×5², and √900 = 30, confirming 900 is the least perfect square divisible by each of 12, 18, and 45.

Question 14

The prime factorisation of 1764 is:

Accepted Answer

Option (A) is correct.

Explanation: Dividing successively, 1764 ÷ 2 = 882 and 882 ÷ 2 = 441, giving the factor 2²; then 441 = 21² = (3 × 7)² = 3² × 7². ∴ 1764 = 2² × 3² × 7².

Question 15

Given that HCF(1980, 5544) = 396 and LCM(1980, 5544) = 1980 × k, then the value of k is:

Accepted Answer

Option (A) is correct.

Explanation: Using the property HCF × LCM = a × b, we get 396 × LCM = 1980 × 5544 ⟹ LCM = (1980 × 5544)/396 = 1980 × 14. Since LCM = 1980 × k, ∴ k = 14.

Question 16

Given HCF(1800, 4200) = 600 and LCM(1800, 4200) = 600 × k, then the value of k is:

Accepted Answer

Option (A) is correct.

**Explanation:** Using the property HCF × LCM = Product of the two numbers: 600 × LCM = 1800 × 4200 ⟹ LCM = (1800 × 4200)/600 = 12600. Since LCM = 600 × k, k = 12600/600 = 21.

Question 17

The prime factorisation of the natural number 360 is:

Accepted Answer

Option (A) is correct.

Explanation: Dividing successively by prime factors, 360 = 2 × 180 = 2 × 2 × 90 = 2 × 2 × 2 × 45 = 2 × 2 × 2 × 3 × 15 = 2 × 2 × 2 × 3 × 3 × 5 ⟹ 360 = 2³ × 3² × 5. Verification: 8 × 9 × 5 = 360. ✓

Question 18

The sum of the exponents of the prime factors in the prime factorisation of 1960 is:

Accepted Answer

Option (C) is correct.

**Explanation:** Prime factorisation of 1960: 1960 = 2³ × 5¹ × 7². The exponents of the prime factors are 3, 1, and 2. Sum of exponents = 3 + 1 + 2 = 6.

Question 19

The HCF and LCM of 8, 20, and 28 respectively are:

Accepted Answer

Option (C) is correct.

Explanation: Prime factorisations: 8 = 2³, 20 = 2² × 5, 28 = 2² × 7. HCF = lowest power of common prime factor = 2² = 4. LCM = highest power of each prime factor present = 2³ × 5 × 7 = 280. ∴ HCF = 4 and LCM = 280.

Question 20

If m = 3² × 5³ and n = 3³ × 5² × 7, then the LCM of m and n is:

Accepted Answer

Option (C) is correct.

Explanation: LCM is found by taking the highest power of each prime factor present in m or n. For prime 3: highest power is 3³ (from n); for prime 5: highest power is 5³ (from m); for prime 7: highest power is 7¹ (from n). ∴ LCM(m, n) = 3³ × 5³ × 7.

Real Numbers: Class 10 Maths Practice Questions

Want unlimited practice on Real Numbers?