"Mathematics — Number System" — Practice

Question 1

Evaluate: (−12) + 9 − (−5).
(a) 2  (b) −8  (c) −16  (d) 8

Accepted Answer

First, (−12)+9: unlike signs, so subtract magnitudes (12−9=3) and keep the sign of the larger magnitude (negative) → −3. Then −3−(−5) = −3+5 = 2.

Answer: (a) 2. Trap (b) −8 comes from treating '−(−5)' as simply '−5' (a double-negative sign error): −3−5=−8. Trap (c) −16 comes from mishandling the first step — adding the magnitudes instead of subtracting them when the signs differ (−12+9 wrongly treated as −21), then correctly adding 5: −21+5=−16. Trap (d) 8 comes from correctly computing −3−(−5)=2 in structure, but dropping the negative sign off the intermediate −3 and treating it as +3: 3+5=8.

Question 2

Three bells toll at intervals of 6, 8 and 12 minutes respectively. If all three toll together at 9:00 a.m., at what time will they next toll together?
(a) 9:24 a.m.  (b) 9:02 a.m.  (c) 9:48 a.m.  (d) 9:20 a.m.

Accepted Answer

The bells toll together again after LCM(6,8,12) minutes. Prime factorising: 6=2×3, 8=2³, 12=2²×3. LCM takes each prime at its highest power: 2³×3=24. They next toll together after 24 minutes, i.e. at 9:24 a.m.

Answer: (a) 9:24 a.m. Trap (b) 9:02 a.m. comes from computing the HCF of the three intervals (HCF=2) instead of the LCM — exactly the HCF/LCM word-problem confusion this chapter warns about, mistaking 'toll together again' for a 'largest common measure' question. Trap (c) 9:48 a.m. comes from doubling the correct LCM, perhaps from wrongly assuming the bells need to complete two full repeats of the largest interval before recombining. Trap (d) 9:20 a.m. comes from an arithmetic slip while combining the prime factors — loosely combining only two of the three intervals instead of correctly taking the LCM of all three.

Question 3

Given (2/3 + 1/4) + 3/5 = 2/3 + (1/4 + 3/5), which property of rational numbers does this equation illustrate?
(a) Commutative property of addition  (b) Associative property of addition  (c) Distributive property  (d) Closure property

Accepted Answer

The terms and their order are unchanged on both sides — only the grouping (which pair is added first) has changed. Regrouping without reordering is exactly the associative property.

Answer: (b) Associative property of addition. Trap (a) confuses regrouping with reordering — commutativity would require the terms themselves to swap positions (e.g. 2/3+1/4 becoming 1/4+2/3), which isn't what's shown here. Trap (c) misapplies distributivity, a property that links multiplication to addition (a(b+c)=ab+ac); no multiplication appears in this equation at all. Trap (d) misapplies closure, which is about the result staying within the rational numbers, not about how terms are grouped.

Question 4

For the number 3 4 x 7 2 to be divisible by 9, the digit x must be:
(a) 2  (b) 4  (c) 7  (d) 9

Accepted Answer

Digit sum of the known digits: 3+4+7+2=16. For the whole number to be divisible by 9, the total digit sum (16+x) must be a multiple of 9. The next multiple of 9 above 16 is 18, so x=2 (since 16+2=18).

Answer: (a) 2. Trap (b) 4 comes from miscomputing the known digit sum as 14 instead of 16 (an arithmetic slip, e.g. missing one digit), which would need x=4 to reach 18. Trap (c) 7 comes from mixing up the digit-sum rule for 9 with the alternating odd/even-place rule for 11, landing on a different digit through the wrong test. Trap (d) 9 comes from a conceptual confusion — assuming the missing digit itself must be a multiple of 9, rather than understanding that it's the total digit SUM across the whole number that must be divisible by 9.

Question 5

Which of the following numbers is certainly NOT a perfect square?
(a) 1296  (b) 1024  (c) 1458  (d) 2025

Accepted Answer

Perfect squares never end in 2, 3, 7 or 8. 1458 ends in 8, so it is certainly not a perfect square, without needing to check it further. (For reference: 1296=36², 1024=32², 2025=45² — all genuine perfect squares.)

Answer: (c) 1458. Trap (a) 1296 would only seem suspect to someone wrongly recalling that numbers ending in 6 are excluded — 6 is in fact an allowed ending digit for squares (e.g. 4²=16, 6²=36). Trap (b) 1024 similarly relies on a wrong belief that numbers ending in 4 are excluded — 4 is allowed (e.g. 2²=4, 8²=64). Trap (d) 2025 relies on wrongly treating numbers ending in 5 as excluded — 5 is allowed, and any number ending in 5 has a square ending in 25 (e.g. 15²=225).

Question 6

Find the smallest number by which 72 must be multiplied so that the product becomes a perfect cube.
(a) 3  (b) 2  (c) 6  (d) 9

Accepted Answer

72 = 2³ × 3². For a perfect cube, every prime's exponent must be a multiple of 3. The factor 2³ already has exponent 3 — no adjustment needed. The factor 3² needs exactly one more factor of 3 to reach 3³. So the smallest multiplier is 3.

Answer: (a) 3. Trap (b) 2 comes from misapplying the fix to the wrong prime — 2 already satisfies the exponent-multiple-of-3 condition and needs no adjustment at all. Trap (c) 6 comes from multiplying the 'needed' factors of both primes together (2×3=6), even though only the factor 3 was actually deficient. Trap (d) 9 comes from over-correcting — multiplying by 3×3=9, as if two more factors of 3 were needed, when 3² only needs exactly one more factor of 3 to reach the next multiple of 3 (exponent 3).

Question 7

Which of the following fractions has a terminating decimal expansion?
(a) 7/12  (b) 5/16  (c) 4/15  (d) 11/21

Accepted Answer

A fraction in lowest terms terminates only if its denominator's sole prime factors are 2 and/or 5. Checking each denominator: 12=2²×3 (has a 3 — recurs); 16=2⁴ (only 2s — terminates); 15=3×5 (has a 3 — recurs); 21=3×7 (has 3 and 7 — recurs).

Answer: (b) 5/16. Trap (a) 7/12 comes from noticing 12 contains a factor of 2 and stopping there, ignoring that 12 also contains a factor of 3 — the rule requires the denominator's ONLY prime factors to be 2 and/or 5, not merely that 2 appears among them. Trap (c) 4/15 comes from a superficial, digit-based judgement — 15 ends in 5, so it may look 'terminating-friendly', but the correct test is prime factorisation of the denominator, and 15 also contains a factor of 3. Trap (d) 11/21 comes from not checking 21's prime factorisation at all (21=3×7, neither of which is 2 or 5).

Question 8

A three-digit number has hundreds digit a, tens digit b and units digit c. Which of the following correctly expresses the number in generalised (place-value) form?
(a) a+b+c  (b) 100a+10b+c  (c) abc  (d) 100c+10b+a

Accepted Answer

Place value assigns 100 to the hundreds digit, 10 to the tens digit, and 1 to the units digit, so the number equals 100a+10b+c.

Answer: (b) 100a+10b+c. Trap (a) a+b+c wrongly computes the DIGIT SUM of the number, not the number's own value — a mistake that transplants the digit-sum idea used for divisibility rules into a context where it doesn't belong. Trap (c) 'abc' misreads the juxtaposed digits as a product or bare concatenation rather than a weighted place-value expansion. Trap (d) 100c+10b+a correctly uses place-value structure but assigns the weights in reverse order — it actually generalises the digits-reversed number, not the original one.

Question 9

A teacher wants students to discover the divisibility rule for 9 inductively rather than memorising it as a stated fact. Which classroom activity best achieves this?
(a) Write the rule on the board and ask students to copy it into their notebooks for memorisation.
(b) Give students a range of numbers, have them check divisibility by 9 through actual division, compute each number's digit sum alongside it, sort the numbers accordingly, and ask them to look for a pattern connecting the two.
(c) Assign twenty problems applying the rule for homework, without first explaining why the rule works.
(d) Present the modular-arithmetic (mod 9) proof of the rule on the board before showing any worked examples.

Accepted Answer

Option (b) has students generate and sort their own data (actual divisibility results alongside digit sums) and notice the connecting pattern themselves before the rule is ever named — the defining structure of inductive, discovery-based learning that NCF 2005 asks CTET-certified teachers to default to.

Answer: (b). Trap (a) is straightforward rule transmission and rote memorisation — the exact opposite of inductive discovery, even though it 'covers' the same rule. Trap (c) assumes the rule already and drills its application; repeated practice of a rule is not the same activity as discovering the underlying pattern in the first place. Trap (d) offers a mathematically rigorous but purely deductive justification (the mod-9 congruence argument) presented rule-first and proof-first, rather than pattern-first — correct mathematics, but not an inductive activity, and pitched well beyond what a Class VI-VIII learner is expected to follow at this stage.

Question 10

Several students in a Class VII class insist that '0.4 × 20 must be more than 20, because multiplication makes numbers bigger.' Which teaching response most directly addresses the root of this misconception?
(a) Tell students the correct answer (8) and ask them to memorise that multiplying by a decimal less than 1 gives a smaller result.
(b) Have students first estimate and compute several products using whole-number multipliers greater than 1, then repeat with multipliers between 0 and 1, using a number line or area model to see directly why the outcome differs.
(c) Give extra practice problems using only multipliers greater than 1 until fluency improves.
(d) Move directly to an algebraic explanation using variables and inequality signs.

Accepted Answer

Option (b) directly confronts the over-generalised whole-number-only intuition by having students work through both cases (multiplier >1 and multiplier between 0 and 1) side by side, using a concrete or visual model to see for themselves why the result differs — exactly NCERT's concrete-before-abstract approach to correcting a number-sense misconception at its source.

Answer: (b). Trap (a) is a direct correction of the answer without addressing why the original belief seemed reasonable in the first place, so the underlying misconception is likely to resurface on a slightly different problem. Trap (c) is actively counterproductive — practising only multipliers greater than 1 reinforces the very over-generalisation the class is already relying on. Trap (d) jumps straight to abstract algebraic justification for a misconception that is fundamentally about number sense and needs a concrete or visual anchor first, exactly the stage NCF-2005-aligned pedagogy would not skip.

Question 11

A Class VI teacher is introducing fraction comparison for the very first time. Which sequence of activities best reflects NCERT's recommended concrete-to-abstract progression?
(a) State the cross-multiplication rule first, then assign several comparison problems for practice.
(b) Have students fold paper strips or shade fraction circles to physically compare pieces, then represent the same fractions on a number line, and only then introduce the symbolic cross-multiplication or common-denominator rule.
(c) Ask students to memorise a table of common fraction-decimal equivalents before any comparison activity.
(d) Introduce the density of rational numbers (infinitely many fractions between any two given fractions) before comparing two simple given fractions.

Accepted Answer

Option (b) moves through concrete manipulation (paper strips/fraction circles) to a pictorial representation (the number line) and only then to the abstract symbolic rule — exactly the concrete→pictorial→abstract sequence NCERT's own textbook activities are built around.

Answer: (b). Trap (a) begins with the abstract symbolic rule itself, the reverse of the recommended sequence, skipping the concrete and pictorial stages entirely. Trap (c) substitutes memorisation of an unrelated equivalence table for genuine comparison reasoning, addressing a different skill rather than building fraction-comparison understanding. Trap (d) introduces a considerably more advanced idea (density) before the foundational skill (comparing two given fractions) that it actually depends on, sequencing the curriculum backwards.

Question 12

During a fraction-comparison lesson, a teacher notices several students believe 1/8 is greater than 1/4 because 8 is greater than 4. Which activity best helps students correct this specific misconception?
(a) State that the rule is the opposite of what they think, and ask them to memorise it going forward.
(b) Physically divide two identical paper circles — one into 4 equal parts and one into 8 equal parts — and let students directly compare the visible size of one shaded piece from each, connecting piece-size to the value of the denominator.
(c) Show the decimal equivalents (0.125 and 0.25) and ask students to compare the decimals instead of the fractions.
(d) Explain that a larger denominator always signals a 'more precise' fraction, without addressing the size comparison directly.

Accepted Answer

Option (b) makes the actual, visible piece-size difference the object of comparison — cutting the same-sized whole into more equal parts visibly shrinks each part — which directly targets the misconception at its root: that a bigger denominator was mistakenly treated like a bigger whole-number digit rather than a count of smaller parts.

Answer: (b). Trap (a) again just corrects the stated answer without building the underlying size-intuition, so the same reasoning error is likely to resurface with a different pair of fractions. Trap (c) sidesteps the fraction misconception rather than resolving it — comparing decimals may give the right numeric answer, but it doesn't build the piece-size intuition needed for fraction reasoning specifically, and a student could still hold the flawed fraction belief afterward. Trap (d) introduces a vague and inaccurate justification ('more precision') that doesn't actually explain WHY a larger denominator means smaller pieces, and risks planting a new, equally unhelpful misconception in its place.

Practice — "Mathematics — Number System"

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