MPBSE · High School · 2021 – 2026
MP Board Class 10 Maths
Most Repeated Previous Year Questions with Solutions — Chapter-wise
Built by reading all six MPBSE Class 10 Mathematics papers (Set A) of 2021, 2022, 2023, 2024, 2025 and 2026 — every objective part and every OR question.
216 solved questions · 14 chapters · at least 15 in each · figures from the papers and a diagram in every geometry solution
Chapter 1Real Numbers
15 solved questions · most repeated first
Find the H.C.F. of 6 and 20 by the prime factorisation method.
Also asked as a 1-mark MCQ in 2023 (options 6 / 20 / 2 / 120) and 2025 (options 2 / 6 / 20 / 1).
Step 1 — write each number as a product of primes:
6 = 2 × 3 20 = 2 × 2 × 5 = 22 × 5Step 2 — HCF = product of the smallest power of each common prime factor.
The only prime common to both is 2, and its smallest power is 21.
HCF(6, 20) = 21 = 2Check: LCM = 22 × 3 × 5 = 60, and HCF × LCM = 2 × 60 = 120 = 6 × 20 ✓
In the MCQ form, the correct option is the one that reads 2. (6 is a factor of 6 but not of 20; 120 is the product, not the HCF.)
Express each number as a product of its prime factors:
- (i) 140 [2021, 2023, 2024]
- (ii) 148 [2025]
- (iii) 156 [2026]
Keep dividing by the smallest prime that divides the number. Stop when the quotient is itself a prime.
(i) 140 ÷ 2 = 70, 70 ÷ 2 = 35, 35 ÷ 5 = 7 (prime)
(ii) 148 ÷ 2 = 74, 74 ÷ 2 = 37 (prime)
148 = 2 × 2 × 37 = 22 × 37(iii) 156 ÷ 2 = 78, 78 ÷ 2 = 39, 39 ÷ 3 = 13 (prime)
156 = 2 × 2 × 3 × 13 = 22 × 3 × 13Check by multiplying back: 4 × 5 × 7 = 140, 4 × 37 = 148, 4 × 3 × 13 = 156. Do not stop at a composite factor such as 35 or 39 — every factor must be prime.
- (i) √2 is a ______ number. (rational / irrational) [2023, 2025]
- (ii) √3 is a ______ number. [2024]
- (iii) True or False: √5 is an irrational number. [2021]
(i) irrational (ii) irrational (iii) True
2, 3 and 5 are prime numbers, and √p is irrational for every prime p. So √2, √3 and √5 cannot be written in the form pq with integers p, q and q ≠ 0.
Do not over-apply the rule: √4 = 2 and √9 = 3 are rational, because 4 and 9 are perfect squares. The proof for √3 is Q.15.
- (i) Show that 3√2 is irrational. [2023]
- (ii) Prove that 7√5 is an irrational number. [2024]
(i) Given fact: √2 is irrational.
Suppose, to the contrary, that 3√2 is rational. Then we can find co-prime integers a and b (b ≠ 0) such that
3√2 = ab ⇒ √2 = a3b (dividing both sides by 3)
Since a, 3 and b are integers, a3b is rational. So √2 would be rational.
But this contradicts the fact that √2 is irrational. So our assumption was wrong.
Hence 3√2 is irrational.
(ii) Given fact: √5 is irrational.
Suppose, to the contrary, that 7√5 is rational. Then 7√5 = ab, where a, b are co-prime integers and b ≠ 0.
Dividing both sides by 7: √5 = a7b
a7b is rational (a ratio of integers), so √5 would be rational. This contradicts the fact that √5 is irrational.
Hence 7√5 is irrational.
The key line for marks: a non-zero rational number × an irrational number is always irrational. Always write the given fact (“√2 is irrational”) first.
Given that HCF (306, 657) = 9, find LCM (306, 657).
Formula: for any two positive integers a and b,
HCF(a, b) × LCM(a, b) = a × b ⇒ LCM(306, 657) = 306 × 6579 = 34 × 657 (306 ÷ 9 = 34) = 22338Check by prime factorisation: 306 = 2 × 32 × 17 and 657 = 32 × 73.
LCM = 2 × 32 × 17 × 73 = 22338 ✓ and HCF = 32 = 9 ✓
Divide first (306 ÷ 9 = 34), then multiply — the numbers stay small and you make fewer mistakes. This formula is true only for two numbers.
Find the L.C.M. and H.C.F. of 8, 9 and 25 by applying the prime factorisation method. [2021 asked the L.C.M. only]
HCF: there is no prime factor common to all three numbers, so the only common factor is 1.
HCF(8, 9, 25) = 1LCM = product of the greatest power of each prime that appears:
LCM(8, 9, 25) = 23 × 32 × 52 = 8 × 9 × 25 = 1800For three numbers, do not find the LCM as “product ÷ HCF” — that rule is only for two numbers. Here the LCM equals 8 × 9 × 25 only because the three numbers have no common prime factor.
Find the L.C.M. of 6, 72 and 120 using the prime factorisation method.
| Prime | in 6 | in 72 | in 120 | Greatest power (for LCM) |
|---|---|---|---|---|
| 2 | 21 | 23 | 23 | 23 |
| 3 | 31 | 32 | 31 | 32 |
| 5 | — | — | 51 | 51 |
(For practice: HCF = smallest power of the common primes = 2 × 3 = 6.)
LCM: take every prime that appears, with its highest power. HCF: take only the common primes, with their lowest power.
Write the statement of the Fundamental Theorem of Arithmetic.
Fundamental Theorem of Arithmetic: Every composite number can be expressed (factorised) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur.
Example: 140 = 2 × 2 × 5 × 7 = 22 × 5 × 7.
Whether we start with 140 = 10 × 14 or 140 = 4 × 35, we always end with the same primes: two 2s, one 5 and one 7. Only their order can change.
Write both parts of the statement: (1) it can be written as a product of primes, and (2) this is unique apart from the order. Missing the word “unique” costs marks.
Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
A composite number has at least one factor other than 1 and itself. Take the common factor out.
First number:
7 × 11 × 13 + 13 = 13 × (7 × 11 + 1) = 13 × (77 + 1) = 13 × 78 = 10141014 has the factors 13 and 78 (besides 1 and 1014). So it is composite.
Second number:
7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 5 × (7 × 6 × 4 × 3 × 2 × 1 + 1) = 5 × (1008 + 1) = 5 × 1009 = 50455045 has the factor 5 (besides 1 and 5045). So it is also composite.
You do not need to multiply everything first. Showing a common factor (13 or 5) is enough to prove that the number is composite.
Find the H.C.F. of 96 and 404 by the prime factorisation method.
96 ÷ 2 = 48, ÷ 2 = 24, ÷ 2 = 12, ÷ 2 = 6, ÷ 2 = 3 (prime)
96 = 2 × 2 × 2 × 2 × 2 × 3 = 25 × 3404 ÷ 2 = 202, ÷ 2 = 101…
