15 multiple-choice questions and 2 case-based questions on the current (rationalised) syllabus, in the style of the board paper's Sections A and E. Try each one first; the answer, a one-line reason and a worked solution open on a tap.
Choose one option. Section A of the board paper has 18 MCQs of 1 mark each, spread over all chapters.
The prime factorisation of \(2340\) is
Why: Divide repeatedly by the smallest prime: 2340 = 2·2·3·3·5·13.
\(2340 = 2 \times 1170 = 2^2 \times 585 = 2^2 \times 3^2 \times 65 = 2^2 \times 3^2 \times 5 \times 13\).
The HCF of \(252\) and \(588\) is
Why: HCF takes the lowest power of each common prime.
\(252 = 2^2 \times 3^2 \times 7\), \(588 = 2^2 \times 3 \times 7^2\). HCF \(= 2^2 \times 3 \times 7 = 84\).
The LCM of \(2^3 \times 5^2 \times 7\) and \(2^2 \times 3 \times 5^3\) is
Why: LCM takes the highest power of every prime that appears in either number.
Highest powers: \(2^3\), \(3^1\), \(5^3\), \(7^1\). LCM \(= 2^3 \times 3 \times 5^3 \times 7\).
The HCF and LCM of two numbers are \(12\) and \(504\). If one of the numbers is \(72\), the other is
Why: For two numbers, HCF × LCM = product of the numbers.
Other number \(= \dfrac{12 \times 504}{72} = \dfrac{6048}{72} = 84\). Check: \(\text{HCF}(72, 84) = 12\).
To prove that \(\sqrt7\) is irrational, Asha assumes \(\sqrt7 = \dfrac{p}{q}\), where \(p\) and \(q\) are co-prime integers and \(q \ne 0\). Squaring gives \(p^2 = 7q^2\). Which conclusion gives the contradiction?
Why: 7 | p² gives 7 | p (7 is prime); then 7 | q² gives 7 | q, so p and q share the factor 7.
\(p^2 = 7q^2\) means \(7 \mid p^2\), and as \(7\) is prime, \(7 \mid p\). Write \(p = 7r\): \(49r^2 = 7q^2 \Rightarrow q^2 = 7r^2\), so \(7 \mid q\). Then \(7\) is a common factor of \(p\) and \(q\), contradicting that they are co-prime. Hence \(\sqrt7\) is irrational.
For which natural number \(k\) does \(12^k\) end with the digit \(0\)?
Why: A number ending in 0 needs both 2 and 5 as prime factors; 12k = 22k·3k has no 5.
\(12^k = 2^{2k} \times 3^k\). By the Fundamental Theorem of Arithmetic this factorisation is unique, so \(5\) is never a factor and \(12^k\) never ends in \(0\).
The smallest natural number divisible by each of \(6\), \(8\) and \(15\) is
Why: The smallest common multiple is the LCM.
\(6 = 2 \times 3\), \(8 = 2^3\), \(15 = 3 \times 5\). LCM \(= 2^3 \times 3 \times 5 = 120\).
The largest number that divides \(223\) and \(367\), leaving a remainder of \(7\) in each case, is
Why: Subtract the remainder, then take the HCF of 216 and 360.
\(223 - 7 = 216 = 2^3 \times 3^3\), \(367 - 7 = 360 = 2^3 \times 3^2 \times 5\). HCF \(= 2^3 \times 3^2 = 72\) (and \(72 \gt 7\)).
If \(p\) and \(q\) are co-prime natural numbers, then \(\text{LCM}(p, q)\) is
Why: Co-prime means HCF = 1, so LCM = product ÷ HCF = pq.
\(\text{HCF}(p,q) \times \text{LCM}(p,q) = pq\) and \(\text{HCF}(p,q) = 1\), so \(\text{LCM}(p,q) = pq\).
If \(p\) and \(q\) are distinct primes, then \(\text{HCF}(p^3q^2,\ p^2q^4)\) is
Why: HCF uses the smaller power of each prime: p² and q².
Smaller powers: \(p^{\min(3,2)} = p^2\), \(q^{\min(2,4)} = q^2\). HCF \(= p^2q^2\).
If \(3087 = 3^m \times 7^n\), where \(m\) and \(n\) are natural numbers, then \(m + 2n\) equals
Why: 3087 = 9 × 343 = 3² × 7³, so m = 2, n = 3.
\(3087 = 3 \times 1029 = 3^2 \times 343 = 3^2 \times 7^3\). So \(m = 2\), \(n = 3\) and \(m + 2n = 8\).
Exactly one of these four numbers is rational. Which one?
Why: (a + b)(a − b) = a² − b² removes the surd: 5 − 4 = 1.
\((\sqrt5+2)(\sqrt5-2) = 5 - 4 = 1\). The others are \(9 + 4\sqrt5\), \(\sqrt5 + 2\) and \(1 + \dfrac{2\sqrt5}{5}\), all irrational.
The sum of a rational number and an irrational number is
Why: If r + x = s with r, s rational, then x = s − r would be rational: a contradiction.
Suppose \(r + x = s\) with \(r, s\) rational and \(x\) irrational. Then \(x = s - r\) is rational, a contradiction. So the sum is always irrational.
For the numbers \(26\) and \(91\), \(\text{HCF} \times \text{LCM}\) equals
Why: HCF × LCM of two numbers equals their product, 26 × 91.
\(26 = 2 \times 13\), \(91 = 7 \times 13\): HCF \(= 13\), LCM \(= 2 \times 7 \times 13 = 182\). \(13 \times 182 = 2366 = 26 \times 91\).
The number \(2^4 \times 5^3 \times 7\) ends with how many zeros?
Why: Each zero needs one pair 2 × 5; there are only three 5s.
\(2^4 \times 5^3 \times 7 = (2 \times 5)^3 \times 2 \times 7 = 14 \times 10^3 = 14000\): three zeros.
Section E of the board paper has three case-based questions of 4 marks: a real-life passage, then parts of 1, 1 and 2 marks, with a choice (OR) on the 2-mark part.
Three city bus routes start from the same depot. A Route A bus leaves every \(16\) minutes, a Route B bus every \(20\) minutes and a Route C bus every \(24\) minutes. At 7:00 a.m. one bus of each route leaves the depot together.
A school club has \(312\) pencils, \(234\) erasers and \(195\) sharpeners. It packs them into identical kits so that every kit has the same number of pencils, the same number of erasers and the same number of sharpeners, with nothing left over, and makes as many kits as possible.
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Original questions written by CBSE Math Revision for the CBSE 2026-27 syllabus; not taken from NCERT, NCERT Exemplar or CBSE papers. Every answer was re-checked by computer algebra and by an independent reviewer. CBSE Math Revision is independent and not affiliated with CBSE or NCERT. Spotted a slip? Tell us and it goes in the corrections log.