Class 10 Maths Chapter 1 Real Numbers Practice Paper 5
1The exponent of 3 in the prime factorisation of 5832 is
Check answer
5832 = 8 × 729
8 = 2³ and 729 = 3⁶
So 5832 = 2³ × 3⁶, and the exponent of 3 is 6.
2If m = p²q³ and n = p³q², where p and q are prime numbers, then LCM (m, n) is
Check answer
LCM takes the highest power of each prime.
Highest power of p is p³ (from n), highest power of q is q³ (from m).
LCM = p³q³
3The HCF and the LCM of 16, 20, 24 respectively are
Check answer
16 = 2⁴ | 20 = 2² × 5 | 24 = 2³ × 3
HCF = lowest common power = 2² = 4
LCM = highest power of each prime = 2⁴ × 3 × 5 = 240
4If the HCF of 91 and 156 is expressible in the form 91m − 156, then the value of m is
Check answer
Euclid's division algorithm:
156 = 91 × 1 + 65
91 = 65 × 1 + 26
65 = 26 × 2 + 13
26 = 13 × 2 + 0 → HCF = 13
Check: 91 × 2 − 156 = 182 − 156 = 13 ✓
So m = 2.
5Ravi has a 56 cm green ribbon and a 96 cm yellow ribbon. He cuts each into pieces so that all pieces are of equal length. What is the length of each piece?
Check answer
The piece length must divide both ribbons exactly, and we want the longest such length — that is the HCF.
56 = 2³ × 7 | 96 = 2⁵ × 3
HCF = 2³ = 8 cm
Why not LCM: the LCM would be longer than either ribbon, which is impossible here.
6The largest number which divides 98 and 147 leaving remainders 6 and 9 respectively is
Check answer
Subtract the remainders first — that part is not divisible.
98 − 6 = 92 and 147 − 9 = 138
Required number = HCF(92, 138)
92 = 2² × 23 | 138 = 2 × 3 × 23
HCF = 2 × 23 = 46
7If 7560 = 2m × 3n × 5k × 7p, then m + n + k + p is
Check answer
7560 = 2³ × 3³ × 5 × 7
So m = 3, n = 3, k = 1, p = 1
m + n + k + p = 3 + 3 + 1 + 1 = 8
8If a = 3² × 7, b = 3³ × 5 × 11, c = 5n × 11 and LCM (a, b, c) = 3³ × 5² × 7 × 11, then n is equal to
Check answer
Look only at the power of 5 in the LCM: it is 5².
a has no 5, and b has only 5¹.
So the 5² must come from c = 5n × 11.
Therefore n = 2.
9Assertion–Reason
Assertion (A): If the product of two numbers is 7280 and their HCF is 14, then their LCM is 520.
Reason (R): HCF × LCM = product of the two numbers.
Check answer
LCM = product ÷ HCF = 7280 ÷ 14 = 520 ✓
So A is true, and R is exactly the rule used to get it.
10Assertion–Reason
Assertion (A): 12n ends with the digit zero, where n is a natural number.
Reason (R): A number ends with zero if its prime factorisation contains both 2 and 5.
Check answer
12n = (2² × 3)n = 22n × 3n
There is no factor of 5 anywhere, so 12n can never end in 0 — A is false.
R states the correct condition, so R is true.
11Explain why 5 × 7 × 11 + 11 and 11 × 13 × 17 + 13 × 11 are composite numbers.
Check answer
First number: 5 × 7 × 11 + 11 = 11 × (5 × 7 + 1) = 11 × 36 = 396
396 = 2² × 3² × 11
Second number: 11 × 13 × 17 + 13 × 11 = 11 × 13 × (17 + 1) = 143 × 18 = 2574
2574 = 2 × 3² × 11 × 13
Each is a product of factors other than 1 and itself, so both are composite.
12Two numbers are in the ratio 4 : 5 and their LCM is 320. What is their HCF?
Check answer
Let the numbers be 4x and 5x, where x is the HCF (4 and 5 are co-prime).
LCM = 4 × 5 × x = 20x
20x = 320 → x = 16
So HCF = 16, and the numbers are 64 and 80.
13Show that any number of the form 15n, where n ∈ N, can never end with the digit 0. (2017)
Check answer
A number ends with 0 only if it is divisible by 10 = 2 × 5, that is, only if 2 appears in its prime factorisation.
By the Fundamental Theorem of Arithmetic the factorisation of 15n is unique and contains only 3 and 5 — never 2.
Therefore 15n can never end with 0.
14The LCM of two numbers is 7 times their HCF. The sum of the LCM and HCF is 640. Find the HCF.
Check answer
Let HCF = h, so LCM = 7h.
h + 7h = 640
8h = 640
h = 80
15Prove that √7 is an irrational number. (2023)
Check answer
Suppose √7 is rational. Then √7 = p/q where p, q are integers, q ≠ 0, and p and q have no common factor except 1.
Squaring: 7 = p²/q², so p² = 7q².
So 7 divides p², and since 7 is prime, 7 divides p.
Let p = 7k. Then 49k² = 7q², so q² = 7k².
So 7 divides q² and therefore 7 divides q.
Now 7 is a common factor of p and q — this contradicts our assumption.
Hence √7 is irrational.
Marking tip: state the assumption, reach the contradiction, and write the concluding line. All three carry marks.
16Four bells toll together at 7:00 am. They toll after 8, 12, 15 and 18 seconds respectively. How many times will they toll together again in the next 3 hours?
Check answer
They meet again after an interval equal to LCM(8, 12, 15, 18).
8 = 2³ | 12 = 2² × 3 | 15 = 3 × 5 | 18 = 2 × 3²
LCM = 2³ × 3² × 5 = 360 seconds = 6 minutes
3 hours = 180 minutes
Number of times = 180 ÷ 6 = 30
17Given that √7 is irrational, prove that 4 + 3√7 is irrational. (CBSE Sample Paper 2022)
Check answer
Suppose 4 + 3√7 is rational. Call it r.
Then 3√7 = r − 4. Since r and 4 are both rational, r − 4 is rational.
So √7 = (r − 4)/3, a quotient of two rationals, which is rational.
But we are given that √7 is irrational — a contradiction.
Hence 4 + 3√7 is irrational.
18(a) Find the largest positive integer that divides 156, 208 and 286 leaving remainders 8, 10 and 12 respectively. (3)
(b) An army contingent of 864 soldiers is to march behind a band of 48 members, both in the same number of columns. What is the maximum number of columns? (2)
Check answer
Subtract the remainders first:
156 − 8 = 148, 208 − 10 = 198, 286 − 12 = 274
Required number = HCF(148, 198, 274)
148 = 2² × 37 | 198 = 2 × 3² × 11 | 274 = 2 × 137
The only common prime is 2, so HCF = 2.
(b) Answer: 48 columns
Both groups split into the same number of columns, so the answer is HCF(864, 48).
864 = 2⁵ × 3³ | 48 = 2⁴ × 3
HCF = 2⁴ × 3 = 48
19A morning walk helps improve mental clarity and focus. In a morning walk, three students step off together. Their steps measure 84 cm, 96 cm and 108 cm.
(ii) Find the sum of the exponents of the prime factors of the total distance. (1)
(iii) What is the minimum distance each should walk to cover it in complete steps? (2)
Check answer
(i) HCF(84, 108) = 2² × 3 = 12
(ii) Total distance = LCM(84, 96, 108) = 2⁵ × 3³ × 7 = 6048
Sum of exponents = 5 + 3 + 1 = 9
(iii) Each must walk a distance divisible by all three step lengths, so the minimum is the LCM = 6048 cm = 60.48 m
20The length, breadth and height of a family room are 10 m 80 cm, 8 m 40 cm and 6 m 60 cm.
(ii) What is the LCM of the three measurements? (1)
(iii) If HCF(1080, 840) = 120, find LCM(1080, 840). (1)
Check answer
1080 = 2³ × 3³ × 5
840 = 2³ × 3 × 5 × 7
660 = 2² × 3 × 5 × 11
(i) The longest rod that measures all three exactly is the HCF.
HCF = 2² × 3 × 5 = 60 cm
(ii) LCM = 2³ × 3³ × 5 × 7 × 11 = 83160 cm
(iii) HCF × LCM = product of the two numbers
LCM(1080, 840) = (1080 × 840) ÷ 120 = 7560
The six rules this whole paper runs on
1. Every composite number breaks into primes in exactly one way — the Fundamental Theorem of Arithmetic.2. HCF takes the lowest power of each common prime. LCM takes the highest power of every prime present.
3. HCF × LCM = product of the two numbers. This works for two numbers only, never three.
4. A number ends in 0 only if both 2 and 5 are in its prime factorisation.
5. "Largest number dividing a and b leaving remainders r₁ and r₂" means HCF(a − r₁, b − r₂). Subtract first.
6. Cutting into equal pieces or forming equal columns asks for HCF. Meeting again or covering a common distance asks for LCM.
Prepared by Sumeet Sahu at Unique Study Point, Indore, as per the latest NCERT syllabus for Session 2026-27. Download the PDF above to attempt the paper under exam conditions, then come back and check your working.
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