NCERT Exemplar Solutions Class 10 Maths Real Numbers — Exercise 1.1 Q1–10

NCERT Exemplar Problems, Class 10 Mathematics — Real Numbers, Exercise 1.1. All 10 multiple-choice questions solved.

Exercise 1.1 is the exemplar’s opening run of ten multiple-choice questions on Real Numbers, and it is narrower than it looks. Only three ideas are being tested:

• The division form. Every integer can be written as $a = bq + r$ with $0 \le r < b$, which sorts all the integers into exactly $b$ remainder classes. Questions 1, 2 and 3 are this and nothing else.
• The Fundamental Theorem of Arithmetic, which makes the prime factorisation of a number unique — so HCF is the product of the smallest power of each shared prime, and LCM the product of the largest power of every prime that appears. That settles Questions 4 to 7 and 9.
• The denominator test: a rational number $\frac{p}{q}$ in lowest terms has a terminating decimal expansion exactly when $q = 2^m 5^n$. Question 10.

A note on the current syllabus. Two of those ideas have been trimmed from the textbook. The rationalised Class 10 chapter runs Introduction → The Fundamental Theorem of Arithmetic → Revisiting Irrational Numbers, and sets just two exercises; neither Euclid’s division algorithm nor the terminating-decimal test has a section or a question of its own any more. The chapter’s own introduction still announces both — a paragraph the rationalisation left behind. They are worth doing anyway, because the division form is the cleanest tool for every divisibility argument in Exercises 1.2 to 1.4, but do not expect Question 10’s exact wording in a board paper.

Key insight. Each of these ten questions is decided by writing the number in the one form that exposes the property being asked about — either $a = bq + r$, or the prime factorisation. There is almost no arithmetic here; there is a choice of representation, and once it is made the answer is visible.

Choose the correct answer from the given four options in the following questions:

Question 1

For some integer $m$, every even integer is of the form

Solution. “Even” means divisible by $2$, so an even integer is exactly a multiple of $2$ — and as $m$ runs over all the integers, $2m$ runs over all the even integers and nothing else.

The point of the question is that it asks for a form that captures every even integer and only even integers. (A) and (B) both fail the second half: as $m$ runs over the integers, so do $m$ and $m+1$, so those forms describe every integer, odd ones included. (D) is $2m+1$, which is always odd.

Question 2

For some integer $q$, every odd integer is of the form

Solution. Divide any integer by $2$. The remainder is $0$ or $1$ and nothing else, so every integer is either $2q$ or $2q+1$. The first case is the even one, so the odd integers are precisely those of the form $2q+1$.

Note that this covers the negative odd integers too, which is why the question says integer and not natural number: $-5 = 2(-3) + 1$.

Question 3

$n^2 – 1$ is divisible by $8$, if $n$ is

Solution. Factorise first: $n^2 – 1 = (n-1)(n+1)$. Now the answer depends entirely on the parity of $n$.

If $n$ is odd, then $n-1$ and $n+1$ are two consecutive even numbers. Of any two consecutive even numbers one is a multiple of $4$, and the other is at least a multiple of $2$, so their product carries a factor of $4 \times 2 = 8$. Writing $n = 2q+1$ makes this explicit:

$$n^2 – 1 = (2q+1)^2 – 1 = 4q^2 + 4q = 4q(q+1)$$

and $q(q+1)$ is a product of two consecutive integers, hence even. So $n^2-1 = 4 \times (\text{even}) = 8m$.

If $n$ is even, $n^2-1$ is odd and cannot be divisible by $8$ at all. That kills (D), and it also kills (A) and (B), since “an integer” and “a natural number” both include the even ones — a single counterexample such as $n=2$, giving $n^2-1 = 3$, is enough to discard them.

Question 4

If the HCF of $65$ and $117$ is expressible in the form $65m – 117$, then the value of $m$ is

Solution. Find the HCF first, then solve for $m$ — in that order, because the equation is useless until the left-hand side is a number.

$$117 = 65 \times 1 + 52, \qquad 65 = 52 \times 1 + 13, \qquad 52 = 13 \times 4 + 0$$

The last non-zero remainder is $13$, so $\operatorname{HCF}(65, 117) = 13$. (The prime factorisations $65 = 5 \times 13$ and $117 = 3^2 \times 13$ give the same thing.) Therefore

$$65m – 117 = 13 \implies 65m = 130 \implies m = 2$$

Question 5

The largest number which divides $70$ and $125$, leaving remainders $5$ and $8$, respectively, is

Solution. A remainder is what is left over, so subtract it and the division becomes exact. If $d$ divides $70$ leaving remainder $5$, then $d$ divides $70 – 5 = 65$ exactly; likewise $d$ divides $125 – 8 = 117$ exactly. The largest such $d$ is therefore

$$\operatorname{HCF}(65, 117) = 13$$

which is the same computation as Question 4 — the exemplar has set the identical pair of numbers twice, dressed differently.

One check is worth making every time: a remainder must be smaller than the divisor. Here the remainders are $5$ and $8$, and $13 > 8$, so $13$ is genuinely admissible. Options (C) $875 = 5^3 \times 7$ and (D) $1750$ are far larger than the numbers being divided, and (B) $65$ does not divide $117$.

Question 6

If two positive integers $a$ and $b$ are written as $a = x^3 y^2$ and $b = x y^3$, where $x$, $y$ are prime numbers, then $\operatorname{HCF}(a, b)$ is

Solution. Because $x$ and $y$ are prime, the expressions given are the prime factorisations, and the Fundamental Theorem of Arithmetic says there is no other. A common divisor may therefore use each prime only up to the power available in both numbers, so the HCF takes the smaller exponent of each prime:

$$\operatorname{HCF}(a, b) = x^{\min(3,\,1)} \, y^{\min(2,\,3)} = x^1 y^2 = xy^2$$

Option (C) is the LCM, which takes the larger exponents — the single most common slip on this pair of questions.

The condition that $x$ and $y$ be prime is doing real work here. If $x$ and $y$ were allowed to share a factor the exponent rule would not apply: with $x = 2$ and $y = 4$, $a = 8 \times 16 = 128$ and $b = 2 \times 64 = 128$, whose HCF is $128$, not $xy^2 = 32$.

Question 7

If two positive integers $p$ and $q$ can be expressed as $p = ab^2$ and $q = a^3 b$, where $a$, $b$ are prime numbers, then $\operatorname{LCM}(p, q)$ is

Solution. A common multiple must contain enough of each prime to be divisible by both numbers, so the LCM takes the larger exponent of each prime:

$$\operatorname{LCM}(p, q) = a^{\max(1,\,3)} \, b^{\max(2,\,1)} = a^3 b^2$$

Option (D) $a^3b^3$ is a common multiple, but not the least one — it is $b$ times too big. That is the trap: “is a multiple of both” is not the same as “is the smallest multiple of both”.

The identity $\operatorname{HCF} \times \operatorname{LCM} = pq$ confirms it. Here $\operatorname{HCF}(p,q) = ab$, and $ab \times a^3b^2 = a^4b^3 = (ab^2)(a^3b) = pq$.

Question 8

The product of a non-zero rational and an irrational number is

Solution. Suppose $r$ is rational and non-zero, $s$ is irrational, and their product $rs$ were rational — call it $t$. Since $r \ne 0$ we may divide, and

$$s = \frac{t}{r}$$

is a quotient of two rationals, hence rational. That contradicts $s$ being irrational. So $rs$ can never be rational.

Everything turns on $r \ne 0$, which is why the question says “non-zero”: $0 \times \sqrt{2} = 0$ is rational, and that single case is what makes option (C) look plausible. Exclude it and the result is absolute.

Question 9

The least number that is divisible by all the numbers from $1$ to $10$ (both inclusive) is

Solution. “Least number divisible by all of them” is the definition of the LCM, and the LCM is built by taking each prime to the highest power that appears anywhere in the list:

• $2$: the highest power among $1,\dots,10$ is $2^3$, from $8$;
• $3$: the highest is $3^2$, from $9$;
• $5$: just $5^1$;
• $7$: just $7^1$.

$$\operatorname{LCM}(1,2,\dots,10) = 2^3 \times 3^2 \times 5 \times 7 = 8 \times 9 \times 5 \times 7 = 2520$$

The wrong options are each short of one prime power, which is the quickest way to eliminate them: (C) $504 = 2^3 \times 3^2 \times 7$ misses the $5$, so it is not divisible by $5$ or $10$; (B) $100 = 2^2 \times 5^2$ is not divisible by $3$; (A) $10$ is not divisible by $3$ either.

Question 10

The decimal expansion of the rational number $\dfrac{14587}{1250}$ will terminate after

Solution. A fraction in lowest terms terminates exactly when its denominator is of the form $2^m 5^n$, and the number of decimal places is then $\max(m, n)$ — because that is how many factors of $10$ you must manufacture.

Here $1250 = 2 \times 5^4$, and $14587$ is odd and not a multiple of $5$, so the fraction is already in lowest terms. With $m = 1$ and $n = 4$, three more factors of $2$ are needed to reach $10^4$:

$$\frac{14587}{1250} = \frac{14587 \times 2^3}{2^4 \times 5^4} = \frac{116696}{10^4} = 11.6696$$

which stops after four decimal places.

Syllabus note: as explained at the top of this page, this test no longer has a section or an exercise in the rationalised textbook. It remains the fastest way to answer any “will it terminate?” question, and it is still worth knowing.

Common mistakes

• Swapping min and max in Questions 6 and 7. HCF takes the smallest power of each shared prime; LCM takes the largest power of every prime present. Option (C) in Question 6 is the LCM and option (B) in Question 7 is the HCF-flavoured answer, so both wrong options are exactly this slip, pre-prepared.
• Forgetting to subtract the remainders in Question 5. Working with $70$ and $125$ themselves gives $\operatorname{HCF} = 5$, which is not among the options — and would fail anyway, since a divisor of $5$ cannot leave a remainder of $8$.
• Not checking that the divisor exceeds the remainder. Any answer to Question 5 that is $8$ or less is impossible before you compute anything. It is a one-second sanity check that catches a whole class of errors.
• Reading Question 3 as “true for all $n$”. $n^2-1$ is only divisible by $8$ for odd $n$; at $n = 2$ it is $3$. Options (A) and (B) are both defeated by that single value, which is faster than any general argument.
• Counting decimal places as $m + n$ in Question 10. With $1250 = 2^1 5^4$ the answer is $\max(1,4) = 4$, not $1 + 4 = 5$. You are matching the larger exponent up to a power of ten, not adding them.
• Dropping the word “non-zero” in Question 8. It is the entire content of the question. With zero allowed, option (C) would be correct.
• Using the exponent rule in Question 6 without noticing $x$, $y$ are prime. The min/max rule is a consequence of unique factorisation, so it needs the bases to actually be primes. The question states it; a similar-looking question that does not state it is a different question.

Practise next

• Exemplar Exercise 1.2 — the same ideas, but now you have to justify the answer in writing rather than pick an option, which is where most marks are lost.
• Exemplar Exercises 1.3 and 1.4 — the full proofs that Questions 1, 2 and 3 are shortcuts to.
• NCERT textbook Exercise 1.1 for prime factorisation, HCF and LCM set as computations, including the $\operatorname{HCF} \times \operatorname{LCM} = $ product check used in Question 7 above.
• NCERT textbook Exercise 1.2 for irrationality proofs, which is where Question 8’s argument is set out in full.