Board Previous Year Questions (Board PYQs)

How to Approach Board PYQs

This section collects the types of questions on Real Numbers that have repeatedly appeared in CBSE and State Board examinations, with full step-by-step solutions. Working through these 26 problems is the single best way to predict what your exam will ask.

What examiners love in this chapter:

1. Prove $\sqrt{p}$ is irrational (2–3 marks) — almost every year.
2. HCF & LCM by prime factorisation, plus the relation $\text{HCF}\times\text{LCM} = a\times b$ (1–3 marks).
3. Terminating / non-terminating decision from the denominator (1 mark MCQ).
4. Word problems on HCF (largest size) and LCM (next time together).

Tag note: Questions below are tagged by exam (e.g. [CBSE Board]). Where the precise year of a specific question could not be confirmed, only the exam name is given — this keeps the content trustworthy.

Exam Tip: Always show the prime factorisation or each division step. Board markers award method marks even if the final number has a small slip.

Solved Previous Year Questions

PYQ 1: Prove irrationality (2 marks)

Prove that $\sqrt{5}$ is an irrational number. [CBSE Board]

Solution:

1. Assume $\sqrt{5} = \dfrac{a}{b}$, where $a, b$ are co-prime integers, $b \neq 0$.
2. Squaring: $5b^2 = a^2$, so $5 \mid a^2 \Rightarrow 5 \mid a$. Let $a = 5c$.
3. Then $5b^2 = 25c^2 \Rightarrow b^2 = 5c^2 \Rightarrow 5 \mid b$.
4. So 5 divides both $a$ and $b$, contradicting co-primeness.

Final Answer: $\sqrt{5}$ is irrational. $\blacksquare$

Takeaway: State 'co-prime' at the start — the contradiction depends on it.

PYQ 2: HCF and LCM (3 marks)

Find the HCF and LCM of 404 and 96 and verify that HCF × LCM = product of the two numbers. [CBSE Board]

Solution:

1. $404 = 2^2 \times 101$; $96 = 2^5 \times 3$.
2. HCF $= 2^2 = 4$ (lowest power of common prime 2).
3. LCM $= 2^5 \times 3 \times 101 = 9696$.
4. Verify: HCF × LCM $= 4 \times 9696 = 38784$, and $404 \times 96 = 38784$. ✓

Final Answer: HCF = 4, LCM = 9696; relation verified.

Takeaway: Always do the verification step when the question asks for it — it carries marks.

PYQ 3: Terminating decimal (1 mark)

Without actual division, state whether $\dfrac{13}{3125}$ has a terminating decimal expansion. [CBSE Board]

Solution:

1. $3125 = 5^5 = 2^0 \times 5^5$.
2. The denominator has only the prime 5, i.e. it is of the form $2^n \times 5^m$.
3. Hence the decimal terminates.

Final Answer: Terminating.

Takeaway: Only 2's and 5's in the (reduced) denominator ⇒ terminating.

PYQ 4: HCF word problem (3 marks)

Find the largest number that divides 2053 and 967 and leaves a remainder of 5 and 7 respectively. [CBSE Board]

Solution:

1. Subtract remainders: $2053 - 5 = 2048$, $967 - 7 = 960$.
2. Required = HCF(2048, 960). $2048 = 2^{11}$, $960 = 2^6 \times 3 \times 5$.
3. Common at lowest power: $2^6 = 64$.

Final Answer: 64.

Takeaway: Reduce by the remainders, then take the HCF.

PYQ 5: Prove a combination irrational (3 marks)

Prove that $3 + 2\sqrt{5}$ is irrational, given that $\sqrt{5}$ is irrational. [CBSE Board]

Solution:

1. Assume $3 + 2\sqrt{5}$ is rational, say $= r$.
2. Then $2\sqrt{5} = r - 3$, so $\sqrt{5} = \dfrac{r-3}{2}$.
3. The right side is rational (operations on rationals), so $\sqrt{5}$ would be rational.
4. This contradicts the given fact that $\sqrt{5}$ is irrational.

Final Answer: $3 + 2\sqrt{5}$ is irrational. $\blacksquare$

Takeaway: Use the given irrational result explicitly — the proof rests on it.

PYQ 6: LCM word problem (3 marks)

Three bells ring at intervals of 4, 7 and 14 minutes. All three rang together at 6 a.m. When will they ring together again? [CBSE Board]

Solution:

1. Next time together = LCM(4, 7, 14) minutes.
2. $4 = 2^2$, $7 = 7$, $14 = 2 \times 7$. LCM $= 2^2 \times 7 = 28$ minutes.
3. $6{:}00$ a.m. + 28 min = $6{:}28$ a.m.

Final Answer: 6:28 a.m.

Takeaway: 'Together again' uses LCM of the intervals.

PYQ 7: Validity of HCF and LCM (2 marks)

The HCF of two numbers is 18 and their LCM is 760. Is this possible? Justify. [CBSE Board]

Solution:

1. For any two numbers, the HCF must always divide the LCM.
2. Here, does 18 divide 760? $760 \div 18 = 42.2\dots$ — not an integer.
3. So 18 does not divide 760, which is impossible.

Final Answer: No, it is not possible (HCF must divide LCM).

Takeaway: A quick validity check: HCF always divides LCM.

PYQ 8: Composite number reasoning (2 marks)

Explain why $7 \times 11 \times 13 + 13$ and $7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5$ are composite numbers. [CBSE Board]

Solution:

1. $7 \times 11 \times 13 + 13 = 13(7 \times 11 + 1) = 13 \times 78$ — has a factor 13, so composite.
2. $7! + 5 = 5040 + 5 = 5045$. Take 5 common: $5(7\times6\times4\times3\times2\times1 + 1) = 5 \times 1009$ — has a factor 5, so composite.

Final Answer: Both are composite (each has a factor other than 1 and itself).

Takeaway: Spot the common factor to show compositeness.

PYQ 9: Maximum stack size (3 marks)

Three sets of English, Hindi and Mathematics books have to be stacked: 96 English, 240 Hindi and 336 Mathematics. Find the maximum number of books in each stack so that each stack has the same number and books of one subject only. [CBSE Board]

Solution:

1. Maximum per stack = HCF(96, 240, 336).
2. $96 = 2^5 \times 3$; $240 = 2^4 \times 3 \times 5$; $336 = 2^4 \times 3 \times 7$.
3. Common at lowest powers: $2^4 \times 3 = 48$.

Final Answer: 48 books per stack.

Takeaway: 'Same number, maximum per group' → HCF.

PYQ 10: Decimal expansion classification (2 marks)

The decimal expansion of $\dfrac{43}{2^4 \times 5^3}$ will terminate after how many places of decimal? [CBSE Board]

Solution:

1. The denominator is $2^4 \times 5^3$, so $n = 4$, $m = 3$.
2. Number of decimal places $= \max(n, m) = \max(4, 3) = 4$.

Final Answer: After 4 decimal places.

Takeaway: Decimal places $= \max(n, m)$ for denominator $2^n 5^m$.

PYQ 11: HCF and LCM by factor tree (3 marks)

Find the HCF and LCM of 26 and 91 and verify HCF × LCM = product of the numbers. [CBSE Board]

Solution:

1. $26 = 2 \times 13$; $91 = 7 \times 13$.
2. HCF $= 13$ (common prime).
3. LCM $= 2 \times 7 \times 13 = 182$.
4. Verify: $13 \times 182 = 2366$ and $26 \times 91 = 2366$. ✓

Final Answer: HCF = 13, LCM = 182; verified.

Takeaway: Two numbers sharing a single prime have that prime as HCF.

PYQ 12: Prove $\sqrt{2}$ irrational (3 marks)

Prove that $\sqrt{2}$ is irrational. [CBSE Board]

Solution:

1. Assume $\sqrt{2} = \dfrac{a}{b}$ with $a, b$ co-prime, $b \neq 0$.
2. $2b^2 = a^2 \Rightarrow 2 \mid a^2 \Rightarrow 2 \mid a$. Let $a = 2c$.
3. $2b^2 = 4c^2 \Rightarrow b^2 = 2c^2 \Rightarrow 2 \mid b$.
4. So 2 divides both $a$ and $b$ — contradiction.

Final Answer: $\sqrt{2}$ is irrational. $\blacksquare$

Takeaway: The most-asked proof in the chapter — memorise the six steps.

PYQ 13: Smallest number problem (3 marks)

Find the smallest number which when increased by 17 is exactly divisible by both 520 and 468. [CBSE Board]

Solution:

1. The number + 17 must be a common multiple of 520 and 468, so take LCM.
2. $520 = 2^3 \times 5 \times 13$; $468 = 2^2 \times 3^2 \times 13$. LCM $= 2^3 \times 3^2 \times 5 \times 13 = 4680$.
3. So number + 17 = 4680 ⇒ number $= 4680 - 17 = 4663$.

Final Answer: 4663.

Takeaway: 'Increased by k, divisible by all' ⇒ LCM − k.

PYQ 14: HCF and LCM in symbolic form (2 marks)

If two positive integers $p$ and $q$ are written as $p = a^2 b^3$ and $q = a^3 b$, where $a, b$ are prime numbers, then find HCF$(p, q)$ and LCM$(p, q)$. [CBSE Board]

Solution:

1. HCF = product of lowest powers: $a^{\min(2,3)} b^{\min(3,1)} = a^2 b$.
2. LCM = product of highest powers: $a^{\max(2,3)} b^{\max(3,1)} = a^3 b^3$.

Final Answer: HCF $= a^2 b$, LCM $= a^3 b^3$.

Takeaway: With symbolic prime powers, take min for HCF and max for LCM of each exponent.

PYQ 15: Prove $\sqrt{3}$ irrational (3 marks)

Prove that $\sqrt{3}$ is an irrational number. [CBSE Board]

Solution:

1. Assume $\sqrt{3} = \dfrac{a}{b}$ with $a, b$ co-prime, $b \neq 0$.
2. $3b^2 = a^2 \Rightarrow 3 \mid a^2 \Rightarrow 3 \mid a$. Let $a = 3c$.
3. $3b^2 = 9c^2 \Rightarrow b^2 = 3c^2 \Rightarrow 3 \mid b$.
4. So 3 divides both $a$ and $b$ — contradiction.

Final Answer: $\sqrt{3}$ is irrational. $\blacksquare$

Takeaway: Same template as $\sqrt2$, with the prime 3.

PYQ 16: HCF by Euclid's algorithm (2 marks)

Using Euclid's division algorithm, find the HCF of 96 and 404. [State Board]

Solution:

1. $404 = 96 \times 4 + 20$.
2. $96 = 20 \times 4 + 16$.
3. $20 = 16 \times 1 + 4$.
4. $16 = 4 \times 4 + 0$. Last divisor = 4.

Final Answer: HCF $= 4$.

Takeaway: Apply the lemma repeatedly; the last non-zero remainder is the HCF.

PYQ 17: Prove $5 - \sqrt{3}$ irrational (3 marks)

Prove that $5 - \sqrt{3}$ is irrational, given that $\sqrt{3}$ is irrational. [CBSE Board]

Solution:

1. Assume $5 - \sqrt{3} = r$ (rational).
2. Then $\sqrt{3} = 5 - r$, a difference of rationals → rational.
3. This contradicts the given fact that $\sqrt{3}$ is irrational.

Final Answer: $5 - \sqrt{3}$ is irrational. $\blacksquare$

Takeaway: Isolate the surd; the rest must be rational, giving the contradiction.

PYQ 18: Greatest number with given remainder (3 marks)

Find the greatest number that divides 245 and 1029, leaving remainder 5 in each case. [CBSE Board]

Solution:

1. Subtract the remainder: $245 - 5 = 240$, $1029 - 5 = 1024$.
2. Required = HCF(240, 1024). $240 = 2^4 \times 3 \times 5$, $1024 = 2^{10}$.
3. Common at lowest power: $2^4 = 16$.

Final Answer: 16.

Takeaway: 'Leaves the same remainder' → subtract it, then take the HCF.

PYQ 19: Find the other number (2 marks)

The HCF of two numbers is 9 and their LCM is 360. If one number is 45, find the other. [CBSE Board]

Solution:

1. Use $\text{HCF} \times \text{LCM} = a \times b$.
2. $9 \times 360 = 45 \times b \Rightarrow b = \dfrac{3240}{45} = 72$.

Final Answer: The other number is 72.

Takeaway: The product relation gives the fourth quantity at once (valid for two numbers).

PYQ 20: Form of an odd integer (3 marks)

Show that any positive odd integer is of the form $4q + 1$ or $4q + 3$, where $q$ is some integer. [CBSE Board]

Solution:

1. By Euclid's lemma with divisor 4, any integer is $4q$, $4q+1$, $4q+2$, or $4q+3$.
2. $4q$ and $4q+2$ are even (multiples of 2).
3. So an odd integer must be $4q+1$ or $4q+3$.

Final Answer: Every positive odd integer is $4q+1$ or $4q+3$. $\blacksquare$

Takeaway: List remainder cases, then eliminate the even ones.

PYQ 21: HCF and LCM of three numbers (3 marks)

Find the HCF and LCM of 12, 15 and 21 by the prime factorisation method. [CBSE Board]

Solution:

1. $12 = 2^2 \times 3$; $15 = 3 \times 5$; $21 = 3 \times 7$.
2. HCF: only the prime 3 is common to all → HCF $= 3$.
3. LCM: all primes at highest powers $= 2^2 \times 3 \times 5 \times 7 = 420$.

Final Answer: HCF = 3, LCM = 420.

Takeaway: For HCF of three numbers, a prime must appear in all three.

PYQ 22: Prove $\sqrt{2} + \sqrt{3}$ irrational (3 marks)

Prove that $\sqrt{2} + \sqrt{3}$ is irrational. [CBSE Board]

Solution:

1. Assume $\sqrt{2} + \sqrt{3} = r$ (rational).
2. Square: $2 + 2\sqrt{6} + 3 = r^2 \Rightarrow 2\sqrt{6} = r^2 - 5$.
3. So $\sqrt{6} = \dfrac{r^2 - 5}{2}$, which would be rational.
4. But $\sqrt{6}$ is irrational (6 is not a perfect square). Contradiction.

Final Answer: $\sqrt{2} + \sqrt{3}$ is irrational. $\blacksquare$

Takeaway: Squaring turns a sum of surds into a single surd you can show is irrational.

PYQ 23: Classify decimal expansions (2 marks)

Without dividing, state which of $\dfrac{3}{8}$, $\dfrac{13}{125}$ and $\dfrac{7}{12}$ have terminating decimal expansions. [CBSE Board]

Solution:

1. $\dfrac{3}{8}$: $8 = 2^3$ → only 2's → terminating.
2. $\dfrac{13}{125}$: $125 = 5^3$ → only 5's → terminating.
3. $\dfrac{7}{12}$: $12 = 2^2 \times 3$ → contains 3 → non-terminating recurring.

Final Answer: $\dfrac{3}{8}$ and $\dfrac{13}{125}$ terminate; $\dfrac{7}{12}$ does not.

Takeaway: Factorise the denominator; a prime other than 2 or 5 forces recurrence.

PYQ 24: Army parade (HCF) (3 marks)

An army contingent of 616 members marches behind a band of 32 members. Both groups march in the same number of columns. Find the maximum number of columns. [CBSE Board]

Solution:

1. Maximum columns = HCF(616, 32).
2. $616 = 2^3 \times 7 \times 11$; $32 = 2^5$.
3. Common at lowest power: $2^3 = 8$.

Final Answer: 8 columns.

Takeaway: 'Maximum equal groups from two quantities' → HCF.

PYQ 25: Prove $6 + \sqrt{2}$ irrational (2 marks)

Prove that $6 + \sqrt{2}$ is irrational, given $\sqrt{2}$ is irrational. [CBSE Board]

Solution:

1. Assume $6 + \sqrt{2} = r$ (rational).
2. Then $\sqrt{2} = r - 6$, a difference of rationals → rational.
3. This contradicts $\sqrt{2}$ being irrational.

Final Answer: $6 + \sqrt{2}$ is irrational. $\blacksquare$

Takeaway: Rational + irrational is always irrational.

PYQ 26: Greatest 3-digit multiple (3 marks)

Find the greatest 3-digit number which is exactly divisible by 8, 10 and 12. [CBSE Board]

Solution:

1. The number must be a multiple of LCM(8, 10, 12).
2. $8 = 2^3$, $10 = 2 \times 5$, $12 = 2^2 \times 3$. LCM $= 2^3 \times 3 \times 5 = 120$.
3. Greatest 3-digit multiple of 120: $120 \times 8 = 960$ (since $120 \times 9 = 1080 > 999$).

Final Answer: 960.

Takeaway: Find the LCM, then the largest multiple within the required range.