NCERT Exemplar Solutions Class 10 Maths Chapter 1 Real Numbers

Read the word ``even''. This question is just the definition of an even number written as an algebra rule.

• Definition: a number is even when 2 divides it, which is the same as saying it equals 2 times an integer.
• Form needed: the rule must carry the factor 2 and no extra +1, so it has to look like 2m.
• Reject the rest: that one observation throws out m, m+1 and 2m+1 at once, leaving 2m. Nothing here needs computing; you read the answer straight off the meaning of the word.

Option (C), 2m.

Companion to Q1. The previous question fixed the even form, and this one just claims the leftover.

• Two slots: dividing by 2 puts every integer into exactly one of two boxes, 2q or 2q+1, and odd numbers sit in the second slot.
• Leftover one: the even form already took 2q, so the only thing left for odd numbers is the remainder 1, giving 2q+1.
• Quick check: a single numerical test, q=3 giving 7 which is odd, confirms the form that carries the +1 is the answer.

Option (D), 2q+1.

Counter-example first, then prove the survivor. The smart route is to kill the broad options with one bad case, then settle the rest.

• Odd works: for odd n, the neighbours n-1 and n+1 are consecutive even numbers, so they are 2t and 2(t+1); their product is 4t(t+1), and t(t+1) is even, which supplies the missing factor and gives the full 8.
• Even fails: for even n the value n²-1 is odd, so it is not even divisible by 2, let alone by 8.
• Why not ``integer'': that sharp contrast is exactly why the answer must be the narrow odd-integer option and not the broader ``integer'' or ``natural number'' choice.

Option (C).

Algorithm, then one line of algebra. Two small jobs sit inside this question, and you must do them in order.

• Run Euclid: the chain 117→ 65→ 52→ 13→ 0 delivers the HCF as 13 in just three short divisions.
• Plug in: the phrase ``expressible as 65m-117'' is just a disguised identity, so substituting the HCF gives 65m=130 and hence m=2.
• Sanity check: notice 13 divides both numbers exactly, as 65=13× 5 and 117=13× 9, confirming the HCF before you ever compute m.

Option (B), m=2.

Translate the words into divisibility. The trick is to read each remainder phrase as an exact-division statement.

• Adjust the numbers: ``leaves remainder 5'' means the divisor goes evenly into 70-5=65, and ``leaves remainder 8'' means it goes evenly into 125-8=117.
• Take the HCF: the biggest number that fits both is the highest common factor of 65 and 117, which is 13 from the same Euclid chain as Q4.
• Match the remainders: dividing, 13 into 70 is 5 remainder 5, and 13 into 125 is 9 remainder 8, exactly the remainders the question stated.

Option (A), 13.

Compare exponents prime by prime. With both numbers already in prime-power form, the HCF is a quick exponent comparison.

• List the powers: the prime x appears as 3 and 1, and the prime y appears as 2 and 3.
• Take the smaller: the HCF must divide both numbers, so for each prime it can use at most the smaller exponent, namely x¹ and y²; multiplying gives xy².
• Reject the big ones: the larger choices x³y³ and x²y² would not divide b at all, so they are impossible.

Option (B), xy².

Mirror image of the HCF question. The LCM rule is the opposite of the HCF rule, which makes this one fast.

• Take the larger: where the HCF kept the smaller exponents, the LCM keeps the larger ones, so for a the bigger power is 3 and for b it is 2, giving a³b².
• Verify the product: the neat check is that highest common factor times lowest common multiple equals ab× a³b²=a⁴b³, the same as p× q, so the product identity holds and the answer is confirmed.

Option (C), a³b².

Proof by contradiction in one breath. The fastest way here is to assume the opposite and watch it collapse.

• Assume rational: suppose the product were rational; then dividing it by the non-zero rational factor would turn the irrational number into a rational number, which is impossible.
• Forced answer: that contradiction forces the product to be irrational, so the result is never rational once the rational factor is non-zero.
• Concrete witness: the single example 12 √5=52 shows the same thing, and option (C) fails because no such product can ever land on a rational value.

Option (A).

Build the LCM, don't multiply everything. The least common multiple is assembled prime by prime, not by brute multiplication.

• Avoid the trap: multiplying all of 1 through 10 together would over-count the shared factors and give a number far too big.
• Strongest demand: instead take only the strongest demand from each prime, so 8 needs 2³, 9 needs 3², and 5 and 7 contribute one each, leaving 2³3²57=2520.
• Screen the options: a quick look also helps, since 2520 is the only choice divisible by both 7 and 9, which any valid answer must be.

Option (D), 2520.

Read it off the denominator. The number of decimal places is decided by the denominator alone, not by any division.

• Factor first: since 1250=2¹5⁴, the fraction terminates, and the deciding number is the bigger exponent, which is 4.
• Confirm by division: the actual division gives 14587/1250=11.6696, which has four digits after the point, just as predicted.
• General rule: no long division is really needed, because once the denominator is 2^m5ⁿ the place count is simply the larger of m and n, here equal to 4.

Option (D), four decimal places.

One counter-example settles it. A ``for every'' claim dies the moment a single number escapes it.

• Find an escapee: take the number 1, which is 4(0)+1 and never 4q+2 for any integer q, so the universal claim fails at once.
• Four classes: dividing by 4 produces four residue classes, and 4q+2 is only the remainder-2 class among them.
• Honest verdict: the answer is No, with the three other classes 4q, 4q+1 and 4q+3 standing as the cases the claim leaves out.

No. 4q+2 is only one of the four residue classes mod 4.

Parity does the work. The whole result rests on one fact about two numbers in a row.

• One is even: among n and n+1, one is even and one is odd, no matter what n is, so a factor of 2 is always present.
• Write it out: call the even one 2k; then the product is 2k times the other number, which is plainly divisible by 2.
• Always true: because this argument never depends on the value of n, the statement holds for every positive integer.

True; one factor is always even.

Two guarantees stacked together. Divisibility by 6 is really two separate promises that you prove one at a time and then combine.

• A multiple of two: three numbers in a row always hide an even number, because parity alternates as you step from one integer to the next, so the factor 2 is guaranteed.
• A multiple of three: across any three steps the residues 0, 1 and 2 each appear exactly once, so one of the three numbers is certainly a multiple of 3.
• Combine the primes: because 2 and 3 are distinct primes, having both factors forces divisibility by their product 6. A glance at the example 456=120=620 shows the rule working on real numbers.

True; the product is always a multiple of 2 and of 3.

Squares dodge remainder 2 mod 3. Working with remainders turns this into a tiny three-line check.

• Square the residues: reduce each integer mod 3 and square, giving 0²0, 1²1 and 2²41.
• Only two outcomes: the only squared residues are 0 and 1, so a perfect square can be 3m or 3m+1 but never 3m+2.
• No escape: because the three-case check is exhaustive, no integer can dodge it, so the answer is a firm No.

No; square residues mod 3 are only 0 and 1.

Direct substitution, no cases needed. With the form already fixed, there is nothing to split into cases.

• Just square it: the number is pinned as 3q+1, so squaring gives 9q²+6q+1 straight away.
• Collect the threes: the first two terms share a factor of 3, leaving remainder 1, so the square is locked into 3m+1 and the forms 3m or 3m+2 are impossible.
• Link to Q14: this is the single-case version of the previous question, where the input residue is fixed at 1 and its square stays at residue 1.

No; the square is always 3m+1.

Pick the top of the common list, then verify. The name itself tells you what to do, and a prime check confirms it.

• Read the name: highest common factor literally means the biggest shared divisor, so from the list of common divisors 3,5,15,25,75 the answer is the largest entry, 75.
• Prime check: factorising shows 525=35²7 shares only 35² with 3000=2³35³, and 325=75.
• Both agree: the listed divisors and the prime method give the same number, which is a reassuring cross-check.

HCF=75.

Spot the common factor. The point of the question is the structure, not the final number.

• Factor it out: both terms share the factor 7, so the sum becomes 7(15+1)=716, a product of two whole numbers.
• Apply the definition: any number that can be written as a product of two integers each bigger than 1 is composite by definition, which this clearly is.
• Structure, not value: the arithmetic value 112 is even and far from prime, but the real reason is the visible factorisation 716 rather than the size of the answer.

Composite, since it equals 716.

Use the divisibility rule, not trial pairs. You can settle this without ever hunting for the two numbers.

• The chain of divisors: the highest common factor divides both numbers, and both numbers divide the lowest common multiple, so by transitivity the HCF must also divide the LCM.
• One division: dividing 380 by 18 leaves remainder 2, so 18 does not divide 380 and the pair simply cannot exist.
• No search needed: there is no point looking for actual numbers, because that single failed division is already a complete proof that the pair is impossible.

No; 18380.

Cancel, then inspect the primes. The order of operations matters here: reduce first, judge second.

• Cancel first: the raw denominator 10500 carries factors 3 and 7, but those cancel against the numerator and leave the reduced fraction 47500.
• Inspect the primes: since 500=2²5³ contains only the primes 2 and 5, the decimal terminates.
• Why reduce first: skipping the cancellation would wrongly suggest the 3 and 7 cause non-termination, which is exactly why reducing to lowest terms before judging is essential.

Terminating; reduced denominator is 2²5³.

Terminating fixes the denominator's primes. A decimal that stops puts a hard limit on what primes the denominator may hold.

• Write as a power of ten: a four-place decimal is an integer over 10⁴, and 10⁴=2⁴5⁴, so only 2s and 5s appear underneath.
• Cancel safely: after cancelling common factors the denominator q can keep only 2s and 5s, never a 3 or a 7 or any other prime.
• Conclusion: the prime factors of q are therefore exactly some twos and fives, that is, q=2^m5ⁿ, which is the standard terminating-decimal form.

Only the primes 2 and 5 (q=2^m5ⁿ).

Parity split, then read the remainder. Splitting by even and odd covers every integer in just two cases.

• Even case: squaring the even number 2m gives 4m², a clean multiple of 4, so the square is of the form 4q.
• Odd case: squaring the odd number gives 4(m²+m)+1, which is of the form 4q+1, and there is no third option.
• Spot check: the examples 6²=36=49 and 7²=49=412+1 match the two cases exactly, confirming every perfect square is 4q or 4q+1.

n²=4q or 4q+1 for every integer n.

Cube the four residues mod 4. Working with remainders avoids the heavy binomial expansions entirely.

• Cube each residue: modulo 4 you get 0³0, 1³1, 2³=80 and 3³=273.
• Three outcomes: a cube can therefore only be 0, 1 or 3 mod 4, which means 4m, 4m+1 or 4m+3, and the remainder 2 is impossible.
• Why it works: this residue shortcut replaces the long algebraic expansions yet reaches the very same three-form conclusion.

n³∈4m, 4m+1, 4m+3.

Square the residues mod 5. A short residue table answers the whole question.

• Build the table: the residues 0,1,2,3,4 square to 0,1,4,9,16, which reduce to 0,1,4,4,1 modulo 5.
• Allowed values: a perfect square therefore lands only on 0, 1 or 4 modulo 5, while the values 2 and 3 stay permanently out of reach.
• Finish: that immediately rules out the forms 5q+2 and 5q+3, exactly as the question required, with no further work needed.

No square is 5q+2 or 5q+3.

Residue table mod 6. The same table trick handles divisor 6, just with six rows instead of five.

• Square them all: squaring 0,1,2,3,4,5 gives 0,1,4,9,16,25, which reduce to 0,1,4,3,4,1 modulo 6.
• Reachable set: the only remainders a square can reach are 0,1,3,4, so the values 2 and 5 are excluded.
• Conclusion: hence no perfect square can be written as 6m+2 or 6m+5, which proves the statement without any long algebraic expansion.

Square residues mod 6 are 0,1,3,4 only.

One form, one expansion. A single odd form and one square is all this proof needs.

• Expand once: an odd number is 2k+1, and squaring gives 4k²+4k+1, where the first two terms hold a common factor of 4.
• Read the remainder: pulling that 4 out leaves remainder 1, so every odd square is of the form 4q+1.
• Sharper result: going one step further, k(k+1) is always even, so the same square is in fact 8q+1, the result that board-level proofs often use.

n²=4q+1 for odd n.

Two consecutive integers carry an extra 2. The factor of 8 comes from a visible 4 plus one hidden 2.

• Substitute: putting n=2k+1 gives n²-1=4k(k+1), which already shows the explicit factor of 4.
• Find the hidden two: because k(k+1) is a product of consecutive integers it is even, contributing a second factor of 2, so the 4 and this hidden 2 multiply to 8.
• Link to Q3: this is the same identity behind Q3, here stated as a divisibility result, showing 8 divides n²-1.

8n²-1 for every odd integer n.

Add the residues mod 4. Reusing the odd-square fact makes this a one-line addition.

• Each square is one: from Q25 every odd square is 1 modulo 4, so the sum is 1+1=2 modulo 4.
• Read the remainder: a remainder of 2 means the total is divisible by 2 but not by 4, which is precisely the claim.
• Match the expansion: the explicit form 4(m²+m+n²+n)+2 shows the same thing, since the bracket is a clean multiple of 4 and the trailing +2 keeps the sum even while blocking division by 4.

x²+y² is even, with remainder 2 mod 4.

Two Euclid chains, one for each pair. For three numbers you pair them up and carry the result forward.

• First pair: the chain 567441126630 gives 63 as the common factor of the first two numbers.
• Carry forward: then 693=6311 exactly, so the third number contributes no smaller factor and the overall HCF stays at 63.
• Prime cross-check: factorising agrees, since 441=3²7², 567=3⁴7 and 693=3²711 share the common part 3²7=63.

HCF=63.

Clear the remainders, then run Euclid. Strip the remainders first, and the structure of the numbers becomes obvious.

• Subtract first: after removing the remainders, the numbers 1250, 9375 and 15625 are all heavy in powers of five, namely 25⁴, 35⁵ and 5⁶.
• Common part: their shared factor is 5⁴=625, which the Euclid chains confirm directly.
• Final check: dividing back, 625 leaves remainder 1 in 1251, remainder 2 in 9377 and remainder 3 in 15628, matching the question exactly.

625.

Force a surd to equal a fraction. The whole proof hangs on trapping an irrational number into a rational shape.

• Assume rational: suppose the sum were rational; then moving 3 across and squaring expresses 3 as the ratio r²-22r of two rational quantities.
• Hit the wall: that is impossible, because 3 is irrational and can never equal a ratio of rationals, so the assumption collapses and the sum must be irrational.
• Reusable recipe: the exact same isolate-and-square method handles 2+3 or any sum a+b where a and b are non-square, so it is worth memorising.

Irrational.

Uniqueness of prime factors does the proof. The last digit of a number is decided by the primes it is built from.

• Fix the primes: by the Fundamental Theorem of Arithmetic the prime factors of 12ⁿ=2²ⁿ3ⁿ are exactly 2 and 3, and no factor of 5 can ever appear at any power n.
• What a last digit needs: a number ending in 0 needs both 2 and 5, and a number ending in 5 needs 5, so either ending demands a factor of 5 that simply is not present.
• Same idea elsewhere: this is the very argument that shows 6ⁿ never ends in 5, just applied here to the base 12 instead.

12ⁿ never ends in 0 or 5.

Translate the walk into an LCM. The word ``complete steps'' is the signal that this is really an LCM problem.

• Why LCM: for all three walkers to land exactly on a step, the distance must be a multiple of 40, 42 and 45 at once, and the smallest such number is their lowest common multiple.
• Build it: taking the highest prime powers 2³, 3², 5 and 7 gives 2520 cm, which is 25.2 metres.
• Whole-step check: at that distance walker one takes 63 steps, walker two takes 60 and walker three takes 56, all whole numbers, which confirms the answer.

2520 cm.

Pad to a power of ten. The decimal falls out the moment the denominator becomes a clean power of ten.

• See the gap: since 5000=2³5⁴, the fives already reach the fourth power while the twos lag behind at only three.
• Balance it: multiplying top and bottom by a single 2 turns the denominator into 2⁴5⁴=10⁴ and the numerator into 514.
• Read it off: the fraction is then 514/10000=0.0514, four decimal places matching the larger exponent, with no long division needed once the denominator is base-ten ready.

2³5⁴; decimal =0.0514.

General version of Q30. This is the same proof as before, only with the primes kept as letters.

• Copy the method: the steps mirror the 3+5 proof exactly, but hold the primes as symbols instead of fixed numbers.
• Reach the clash: assuming rationality and squaring forces q=a²+q-p2a, a ratio of rationals, which clashes with q being irrational for the prime q.
• General and special: the contradiction proves the sum irrational for any two primes p and q, and putting p=3 and q=5 recovers the earlier special case.

Irrational.

Binomial then a residue check. The proof splits neatly into one expansion and one small table.

• Expand: expanding (6q+r)³ produces three terms each divisible by 6 plus the lone r³, so the cube has the form 6k+r³.
• Check the leftover: the only question left is whether r³ keeps the same remainder r modulo 6, and a direct table for r=0 to 5 returns the remainders 0,1,2,3,4,5, identical to the inputs.
• Closure: so the cube carries the very same form 6m+r as the original number, which is a clean closure property of cubing modulo 6.

(6q+r)³=6m+r for every r=0,,5.

They cover all residues mod 3. The slick argument looks at the three numbers as residues rather than as values.

• Reduce mod 3: modulo 3 the three numbers become n, n+2 and n+1, because 4 is the same as 1, so they form the three distinct residues 0,1,2 in some order.
• Exactly one zero: since each residue appears once, exactly one of them is 0 modulo 3, and that single zero residue marks the one multiple of 3.
• Why only one: there can be no second multiple, which is what proves ``one and only one'', and the case table above is just this same idea spelled out longhand.

Precisely one of the three is a multiple of 3.

A complete residue run. Consecutive numbers automatically sweep through every remainder mod 3.

• Sweep the residues: three numbers in a row step through the remainders r, r+1 and r+2 modulo 3, which is the full set 0,1,2.
• One is zero: the residue 0 shows up exactly once, marking the multiple of 3, so one of n, n+1, n+2 is divisible by 3 no matter where the block starts.
• Why it matters: this single fact is the backbone of the result that a product of three consecutive integers is divisible by 6.

One of three consecutive integers is a multiple of 3.

One factorisation, two guarantees. A single clever factorisation turns a cubic into three numbers in a row.

• Factor it: writing n³-n as (n-1)n(n+1) exposes three consecutive integers sitting side by side.
• Two factors appear: such a triple always contains a multiple of 2 and a multiple of 3, exactly the facts proved in Q13 and Q37, so both small factors are present.
• Combine: because 2 and 3 are coprime, their product 6 must divide the whole expression, so 6 divides n³-n for every positive integer n with no exceptions left to check.

6n³-n for all n.

Five offsets, five residues. The five offsets are chosen so they sweep every remainder mod 5 exactly once.

• Reduce mod 5: reducing the five numbers n, n+4, n+8, n+12, n+16 modulo 5 gives r, r+4, r+3, r+2, r+1, which is the complete residue set 0,1,2,3,4 in some order.
• One hits zero: exactly one of these residues equals 0 modulo 5, so exactly one of the five numbers is divisible by 5 and never two.
• Table agrees: the five-case table above just confirms this offset by offset, with the divisible member shifting as the starting residue r changes.

One and only one of the five is divisible by 5.