NCERT Exemplar Solutions Class 10 Maths Real Numbers — Exercise 1.2 Q1–10

NCERT Exemplar Problems, Class 10 Mathematics — Real Numbers, Exercise 1.2. All 10 short-answer questions solved.

Every question in Exercise 1.2 asks for a one-word verdict — yes or no, true or false — and then the reason. The verdict is worth almost nothing on its own; the justification is the answer. That is the whole character of this exercise, and it is why it is set immediately after the multiple-choice run: the same facts, but now you have to say why.

Two techniques do all the work:

• Exhaust the remainder classes. To settle a claim about every integer, write the integer as $bq + r$ and check each of the $b$ possible values of $r$ in turn. Questions 1, 4 and 5 are this.
• Argue from the prime factorisation. Uniqueness of the factorisation is what makes a divisibility claim either certain or impossible. Questions 6, 7, 8, 9 and 10 are this.

A note on the current syllabus. The division form $a = bq + r$ is no longer set out as Euclid’s division lemma in the rationalised textbook, and the terminating-decimal test that Questions 9 and 10 use has no section or exercise of its own any more. Both arguments are elementary and are used constantly in Exercises 1.3 and 1.4, so they are worth keeping; but Questions 9 and 10 in this exact form are exemplar-only material now.

Key insight. “Is it true for every positive integer?” is never settled by examples. It is settled by splitting the integers into finitely many remainder classes and checking them all — which is a complete proof, and short.

Question 1

Write whether every positive integer can be of the form $4q + 2$, where $q$ is an integer. Justify your answer.

Solution. No.

Divide any positive integer by $4$. The remainder is $0$, $1$, $2$ or $3$, so every positive integer is of exactly one of the forms

$$4q, \qquad 4q+1, \qquad 4q+2, \qquad 4q+3$$

The form $4q+2$ is only one of these four classes, so it cannot describe them all. Concretely, $1$, $3$, $4$, $5$, $7$, $8$ are none of them of the form $4q+2$.

There is an even quicker reason worth noticing: $4q + 2 = 2(2q+1)$ is always even, so no odd number at all can be written that way.

Question 2

“The product of two consecutive positive integers is divisible by $2$.” Is this statement true or false? Give reasons.

Solution. True.

Let the two consecutive integers be $n$ and $n+1$. Exactly one of them is even: if $n$ is even we are done, and if $n$ is odd then $n+1$ is even. Either way the product $n(n+1)$ contains a factor of $2$, so it is divisible by $2$.

Stated in remainder form, $n$ is either $2q$ or $2q+1$, giving

$$n(n+1) = 2q(2q+1) \qquad \text{or} \qquad (2q+1)(2q+2) = 2(2q+1)(q+1)$$

and both are visibly multiples of $2$.

Question 3

“The product of three consecutive positive integers is divisible by $6$.” Is this statement true or false? Justify your answer.

Solution. True.

Take the three integers as $n$, $n+1$, $n+2$. Since $6 = 2 \times 3$ and $2$ and $3$ are coprime, it is enough to show the product is divisible by $2$ and by $3$ separately.

• By $2$: among any two consecutive integers one is even, and three consecutive integers certainly contain two consecutive ones. So a factor of $2$ is present.
• By $3$: among any three consecutive integers the remainders on division by $3$ are $0$, $1$ and $2$ in some order, so exactly one of them is a multiple of $3$.

Having both a factor of $2$ and a factor of $3$, and those being coprime, the product is divisible by $6$.

Question 4

Write whether the square of any positive integer can be of the form $3m + 2$, where $m$ is a natural number. Justify your answer.

Solution. No.

Every positive integer is $3q$, $3q+1$ or $3q+2$. Square each:

$$(3q)^2 = 9q^2 = 3\left(3q^2\right) = 3m$$

$$(3q+1)^2 = 9q^2 + 6q + 1 = 3\left(3q^2 + 2q\right) + 1 = 3m + 1$$

$$(3q+2)^2 = 9q^2 + 12q + 4 = 3\left(3q^2 + 4q + 1\right) + 1 = 3m + 1$$

So a perfect square leaves a remainder of $0$ or $1$ on division by $3$, never $2$. The three cases above are exhaustive, so this is a complete proof and not merely evidence.

Question 5

A positive integer is of the form $3q + 1$, $q$ being a natural number. Can you write its square in any form other than $3m + 1$, i.e. $3m$ or $3m + 2$ for some integer $m$? Justify your answer.

Solution. No.

$$(3q+1)^2 = 9q^2 + 6q + 1 = 3\left(3q^2 + 2q\right) + 1$$

Writing $m = 3q^2 + 2q$, which is an integer, the square is exactly $3m + 1$. There is no choice in the matter — the algebra produces a remainder of $1$ and nothing else, so the square can be neither $3m$ nor $3m+2$.

This is the middle case of Question 4, isolated.

A misprint in the answer key. The printed key gives this as $(3q+1)^2 = 9q^2 + 6q + 1 = 3(3q^2 + 2q) = 3m + 1$, with the middle expression missing its $+\,1$. As printed the chain says $3(3q^2+2q) = 3m+1$, which is false — that quantity is $3m$. The intended line is $9q^2 + 6q + 1 = 3(3q^2+2q) + 1$, as above. The final answer in the key is right; only the working line is mis-set.

Question 6

The numbers $525$ and $3000$ are both divisible only by $3$, $5$, $15$, $25$ and $75$. What is $\operatorname{HCF}(525, 3000)$? Justify your answer.

Solution. The question is telling you the common divisors of the pair, not all the divisors of each ($3000$ is divisible by $2$, after all, and both are divisible by $1$). Read that way it is exactly right: the common divisors of $525$ and $3000$ are $1, 3, 5, 15, 25$ and $75$.

The HCF is by definition the greatest of the common divisors, so it is the largest number in that list:

$$\operatorname{HCF}(525, 3000) = 75$$

The factorisations confirm it, taking the smaller power of each shared prime:

$$525 = 3 \times 5^2 \times 7, \qquad 3000 = 2^3 \times 3 \times 5^3$$

$$\operatorname{HCF} = 3^1 \times 5^2 = 75$$

Notice too that the listed divisors are precisely the divisors of $75$. That is not a coincidence: every common divisor of two numbers divides their HCF.

Question 7

Explain why $3 \times 5 \times 7 + 7$ is a composite number.

Solution. Do not multiply it out first — factorise it. Every term carries a $7$:

$$3 \times 5 \times 7 + 7 = 7\left(3 \times 5 + 1\right) = 7 \times 16 = 112$$

A number is composite when it has a factor other than $1$ and itself. Here $112 = 7 \times 16$ exhibits such a factorisation directly, so $112$ is composite. (Its full prime factorisation is $2^4 \times 7$, giving it ten divisors in all.)

The general lesson is what the question is really after: any expression of the form $a \times b \times c + c$ equals $c(ab + 1)$ and is therefore composite whenever both $c > 1$ and $ab + 1 > 1$ — no arithmetic required.

Question 8

Can two numbers have $18$ as their HCF and $380$ as their LCM? Give reasons.

Solution. No.

The HCF of two numbers divides each of them, and each of them divides the LCM. So the HCF must divide the LCM. Here

$$380 = 18 \times 21 + 2$$

leaves a remainder of $2$, so $18 \nmid 380$, and no such pair of numbers exists.

The factorisations say the same thing more sharply: $18 = 2 \times 3^2$ contains a factor of $9$, while $380 = 2^2 \times 5 \times 19$ contains no factor of $3$ at all. A number divisible by $9$ cannot divide one that is not divisible by $3$.

Question 9

Without actually performing the long division, find if $\dfrac{987}{10500}$ will have a terminating or non-terminating (repeating) decimal expansion. Give reasons for your answer.

Solution. The test applies to the fraction in its lowest terms, so reduce first. This is the step the question is testing, because the unreduced denominator $10500 = 2^2 \times 3 \times 5^3 \times 7$ contains a $3$ and a $7$ and would wrongly suggest a non-terminating expansion.

$$987 = 3 \times 7 \times 47, \qquad 10500 = 2^2 \times 3 \times 5^3 \times 7$$

The common factor is $3 \times 7 = 21$, and cancelling it gives

$$\frac{987}{10500} = \frac{47}{500}$$

Now $500 = 2^2 \times 5^3$ is of the form $2^m 5^n$, so the expansion terminates — after $\max(2,3) = 3$ places:

$$\frac{47}{500} = \frac{47 \times 2}{10^3} = \frac{94}{1000} = 0.094$$

Question 10

A rational number in its decimal expansion is $327.7081$. What can you say about the prime factors of $q$, when this number is expressed in the form $\dfrac{p}{q}$? Give reasons.

Solution. The expansion terminates, so by the converse of the test used in Question 9, the denominator in lowest terms must be of the form $2^m 5^n$ with $m$ and $n$ non-negative integers — the only prime factors $q$ can have are $2$ and $5$.

For this particular number the exponents can be pinned down exactly. Four decimal places means

$$327.7081 = \frac{3277081}{10^4} = \frac{3277081}{2^4 \times 5^4}$$

and $3277081$ is odd and does not end in $0$ or $5$, so no further cancelling is possible. In lowest terms $q = 10000 = 2^4 \times 5^4$: here $m = n = 4$.

Two cautions. First, the conclusion is about $q$ in lowest terms — writing the same number as $\frac{3277081 \times 7}{70000}$ gives a denominator with a factor of $7$, which is why the reduced form has to be specified. Second, $m$ and $n$ in the general statement are non-negative integers, not necessarily positive: $0.5 = \frac{1}{2} = \frac{1}{2^1 5^0}$ terminates with $n = 0$. It happens that both exponents are $4$ here.

Common mistakes

• Answering Question 1 with an example instead of a reason. “No, because $5$ is not of that form” is only half a mark. The reason is that division by $4$ produces four remainder classes and $4q+2$ is one of them; the example illustrates it, it does not prove it.
• Testing Question 4 on a few squares and stopping. $1, 4, 9, 16, 25$ all avoid the form $3m+2$, but a list is not a proof. Squaring $3q$, $3q+1$ and $3q+2$ covers every integer there is, in three lines.
• Forgetting to reduce the fraction in Question 9. $10500$ has $3$ and $7$ in it, so the unreduced denominator says “non-terminating” — the opposite of the right answer. The test is only valid in lowest terms, and this question exists to make that point.
• Not saying “in lowest terms” in Question 10. Without it, $q$ can be given any prime factors you like by multiplying top and bottom. It is the one phrase the answer needs.
• Reading Question 6 as listing all divisors of each number. $3000$ is even and $525$ is divisible by $7$, so neither list is complete on its own. The numbers given are the divisors the two have in common, and the HCF is the largest of them.
• Multiplying out Question 7 before looking at it. Reaching $112$ and then hunting for factors works, but the whole point is that the common factor of $7$ is visible in the expression itself. In the board version of this question with larger numbers, multiplying out first is how the marks are lost.
• Arguing Question 8 from size. “$380$ is bigger than $18$, so it could work” is not a reason. The condition is divisibility: HCF divides LCM, always.

Practise next

• Exemplar Exercises 1.3 and 1.4 — the same remainder-class argument used in Questions 1, 4 and 5, but now written out as full proofs for modulus $4$, $5$ and $6$.
• Exemplar Exercise 1.1 if the HCF and LCM exponent rules behind Questions 6 and 8 need a faster drill.
• NCERT textbook Exercise 1.1 — Question 6 there (“Explain why $7 \times 11 \times 13 + 13$ … are composite numbers”) is Question 7 above with different numbers, and is the version a board paper will set.
• NCERT textbook Exercise 1.2 for irrationality proofs, which use the same “assume the opposite and derive a contradiction” shape as the reasoning in Question 8.