Welcome to HSLC Guru! In this article we provide complete ASSEB Class 10 Mathematics Chapter 1 — Real Numbers question answer in English medium. This chapter explores the structure of real numbers through Euclid’s Division Lemma, the Fundamental Theorem of Arithmetic, irrationality proofs, and the rules that decide whether the decimal expansion of a rational number terminates or recurs. Every textbook exercise (1.1, 1.2, 1.3, 1.4) is solved step by step, followed by additional MCQs, fill in the blanks, true/false questions, and a glossary, so that ASSEB students of the Assam Board can revise the chapter quickly and confidently.

Board: ASSEB (Assam State School Education Board)
Class: 10 (HSLC)
Subject: Mathematics
Chapter: 1 — Real Numbers
Medium: English

Chapter Summary

The set of real numbers $\mathbb{R}$ is the union of rational and irrational numbers. In Class 9 we accepted the existence of irrationals; in Class 10 we begin to prove their existence and study the divisibility structure of integers more carefully. The two pillars of Chapter 1 are:

1. Euclid’s Division Lemma and Algorithm — used to compute the HCF (Highest Common Factor) of two positive integers and to establish the divisibility properties of integers.
2. The Fundamental Theorem of Arithmetic — every composite number can be expressed (factorised) as a product of primes, and this factorisation is unique apart from the order of the prime factors.

Using these tools we prove that numbers such as $\sqrt{2}$, $\sqrt{3}$, $\sqrt{5}$ are irrational, and we determine the conditions under which a rational number $\dfrac{p}{q}$ has a terminating decimal expansion. The chapter also revisits the fundamental relation $\text{HCF}(a,b)\times \text{LCM}(a,b)=a\times b$ for any two positive integers $a$ and $b$.

Key Concepts and Formulas

1. Euclid’s Division Lemma: Given positive integers $a$ and $b$, there exist unique integers $q$ and $r$ such that

$$a = bq + r,\qquad 0 \le r < b.$$

Here $a$ is the dividend, $b$ the divisor, $q$ the quotient and $r$ the remainder.

2. Euclid’s Division Algorithm (to find HCF of two positive integers $c$ and $d$ with $c>d$):

• Apply the lemma: $c = d q + r$, $0 \le r < d$.
• If $r = 0$ then $\text{HCF}(c,d)=d$.
• If $r \ne 0$ then apply the lemma to $d$ and $r$.
• Continue until the remainder is zero. The divisor at that stage is the required HCF.

3. Fundamental Theorem of Arithmetic: Every composite number can be expressed as a product of primes, and this factorisation is unique except for the order of the prime factors.

4. HCF and LCM relation: For any two positive integers $a$ and $b$,

$$\text{HCF}(a,b)\times \text{LCM}(a,b) = a \times b.$$

5. Irrational numbers: A number that cannot be written in the form $\dfrac{p}{q}$ where $p,q$ are integers and $q\ne 0$ is called irrational. If $p$ is a prime, then $\sqrt{p}$ is irrational.

6. Terminating decimal condition: Let $x=\dfrac{p}{q}$ be a rational number in lowest terms. Then $x$ has a terminating decimal expansion if and only if

$$q = 2^m \cdot 5^n,$$

where $m,n$ are non-negative integers. Otherwise, the expansion is non-terminating but recurring.

Detailed Theory Notes

1. Why Euclid’s Division Lemma matters. The lemma is the cornerstone of all elementary number theory you will meet in Class 10. It packages the everyday idea of “long division” into a precise statement: dividing $a$ by $b$ produces a unique quotient $q$ and a unique remainder $r$ with $0 \le r < b$. Existence is established by considering the largest multiple of $b$ that does not exceed $a$, and uniqueness comes from the fact that two different remainders would differ by a non-zero multiple of $b$ smaller than $b$ in absolute value, which is impossible.

2. From the Lemma to the Algorithm. If $d$ divides both $a$ and $b$, then $d$ also divides $r = a – bq$. Conversely, any common divisor of $b$ and $r$ also divides $a$. Therefore the set of common divisors of $(a,b)$ is the same as the set of common divisors of $(b,r)$. Replacing the pair $(a,b)$ by $(b,r)$ keeps the HCF unchanged but gives strictly smaller numbers; hence the procedure must end after finitely many steps. The last non-zero remainder is the HCF.

3. Fundamental Theorem of Arithmetic — sketch of proof. Let $n>1$ be a natural number. If $n$ is prime, the result is immediate. If not, $n$ has a divisor strictly between $1$ and $n$; choose the smallest such divisor and call it $p_1$. Then $p_1$ is necessarily prime (any proper divisor of $p_1$ would also divide $n$, contradicting the minimality of $p_1$). Repeating this argument with $n/p_1$ gives a sequence of primes whose product is $n$. Uniqueness follows from Euclid’s lemma: if a prime $p$ divides a product, it must divide at least one factor.

4. The role of irrational numbers. The Pythagoreans famously discovered that not every length is the ratio of two whole numbers — the diagonal of a unit square has length $\sqrt{2}$, which is irrational. The proof technique used in Chapter 1 (proof by contradiction together with the Fundamental Theorem of Arithmetic) is one of the most powerful in mathematics. It allows you to show, with very little machinery, that there are infinitely many irrational numbers nestled between any two rationals.

5. Decimal expansions. Any rational $\dfrac{p}{q}$ either terminates or eventually repeats. If $q$ has only $2$ and $5$ as prime factors, multiplying numerator and denominator by suitable powers of $2$ or $5$ converts the denominator to a power of $10$, giving a terminating decimal. If $q$ has any other prime factor, no such conversion is possible, and the long-division process produces a recurring block whose length is at most $q-1$ digits.

6. HCF $\times$ LCM relation. For two positive integers $a,b$ written via prime factorisation, the HCF takes the smaller exponent of each prime, while the LCM takes the larger exponent. The sum of the smaller and larger exponents equals the sum of the two exponents themselves. This is exactly why $\text{HCF}(a,b)\times \text{LCM}(a,b) = a\times b$. Note the relation does not generalise to three or more numbers in this simple form.

Exercise 1.1 — Solutions

Q1. Use Euclid’s division algorithm to find the HCF of:

(i) 135 and 225

Answer: Since $225>135$, apply the lemma to $225$ and $135$.

$225 = 135\times 1 + 90$

$135 = 90\times 1 + 45$

$90 = 45\times 2 + 0$

The remainder is now $0$, so $\text{HCF}(135,225)=45$.

(ii) 196 and 38220

Answer: $38220 = 196\times 195 + 0$. Hence $\text{HCF}(196,38220)=196$.

(iii) 867 and 255

Answer:

$867 = 255\times 3 + 102$

$255 = 102\times 2 + 51$

$102 = 51\times 2 + 0$

$\therefore\ \text{HCF}(867,255)=51$.

(iv) 272 and 1032

Answer:

$1032 = 272\times 3 + 216$

$272 = 216\times 1 + 56$

$216 = 56\times 3 + 48$

$56 = 48\times 1 + 8$

$48 = 8\times 6 + 0$

$\therefore\ \text{HCF}(272,1032)=8$.

(v) 405 and 2520

Answer:

$2520 = 405\times 6 + 90$

$405 = 90\times 4 + 45$

$90 = 45\times 2 + 0$

$\therefore\ \text{HCF}(405,2520)=45$.

(vi) 155 and 1385

Answer:

$1385 = 155\times 8 + 145$

$155 = 145\times 1 + 10$

$145 = 10\times 14 + 5$

$10 = 5\times 2 + 0$

$\therefore\ \text{HCF}(155,1385)=5$.

(vii) 384 and 1296

Answer:

$1296 = 384\times 3 + 144$

$384 = 144\times 2 + 96$

$144 = 96\times 1 + 48$

$96 = 48\times 2 + 0$

$\therefore\ \text{HCF}(384,1296)=48$.

(viii) 1848 and 3058

Answer:

$3058 = 1848\times 1 + 1210$

$1848 = 1210\times 1 + 638$

$1210 = 638\times 1 + 572$

$638 = 572\times 1 + 66$

$572 = 66\times 8 + 44$

$66 = 44\times 1 + 22$

$44 = 22\times 2 + 0$

$\therefore\ \text{HCF}(1848,3058)=22$.

Q6. Find the largest number that will divide $625$ and $325$ exactly without leaving a remainder.

Answer: Required number $= \text{HCF}(625, 325)$.

$625 = 325\times 1 + 300$

$325 = 300\times 1 + 25$

$300 = 25\times 12 + 0$

$\therefore\ \text{HCF}(625,325) = 25$. The required largest number is $\mathbf{25}$.

Q7. Find the longest length of rope (in cm) that can measure exactly two ropes of lengths $64$ cm and $80$ cm.

Answer: Required length $= \text{HCF}(64,80)$.

$80 = 64\times 1 + 16$

$64 = 16\times 4 + 0$

$\therefore\ \text{HCF}(64,80) = 16$. The longest such rope is $\mathbf{16}$ cm.

Q2. Show that any positive odd integer is of the form $6q+1$, $6q+3$ or $6q+5$, where $q$ is some integer.

Answer: Let $a$ be any positive integer and $b=6$. By Euclid’s division lemma, $a = 6q + r$ where $0 \le r < 6$, so $r$ is one of $0,1,2,3,4,5$. Therefore $a$ is of the form $6q,\ 6q+1,\ 6q+2,\ 6q+3,\ 6q+4$ or $6q+5$. The numbers $6q$, $6q+2$ and $6q+4$ are even (each is divisible by $2$). Hence any positive odd integer must be of the form $6q+1$, $6q+3$ or $6q+5$.

Q3. An army contingent of $616$ members is to march behind an army band of $32$ members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?

Answer: The required number of columns is the HCF of $616$ and $32$.

$616 = 32\times 19 + 8$

$32 = 8\times 4 + 0$

$\therefore\ \text{HCF}(616,32)=8$. They can march in a maximum of $\mathbf{8}$ columns.

Q4. Use Euclid’s division lemma to show that the square of any positive integer is either of the form $3m$ or $3m+1$ for some integer $m$.

Answer: Let $a$ be any positive integer. By the lemma with $b=3$, $a = 3q + r$ where $r=0,1$ or $2$.

Case 1: $a=3q$. Then $a^2 = 9q^2 = 3(3q^2) = 3m$, where $m=3q^2$.

Case 2: $a=3q+1$. Then $a^2 = 9q^2 + 6q + 1 = 3(3q^2 + 2q) + 1 = 3m+1$.

Case 3: $a=3q+2$. Then $a^2 = 9q^2 + 12q + 4 = 3(3q^2 + 4q + 1) + 1 = 3m+1$.

Hence $a^2$ is of the form $3m$ or $3m+1$.

Q5. Use Euclid’s division lemma to show that the cube of any positive integer is of the form $9m$, $9m+1$ or $9m+8$.

Answer: Let $a$ be any positive integer. With $b=3$, $a=3q+r$, where $r=0,1,2$.

If $a=3q$, then $a^3 = 27q^3 = 9(3q^3) = 9m$.

If $a=3q+1$, then $a^3 = 27q^3 + 27q^2 + 9q + 1 = 9(3q^3+3q^2+q) + 1 = 9m+1$.

If $a=3q+2$, then $a^3 = 27q^3 + 54q^2 + 36q + 8 = 9(3q^3+6q^2+4q) + 8 = 9m+8$.

Hence the cube of any positive integer is of the form $9m,\ 9m+1$ or $9m+8$.

Exercise 1.2 — Solutions

Q1. Express each number as a product of its prime factors.

(i) $140$

Answer: $140 = 2\times 2\times 5\times 7 = 2^2\times 5\times 7$.

(ii) $156$

Answer: $156 = 2\times 2\times 3\times 13 = 2^2\times 3\times 13$.

(iii) $3825$

Answer: $3825 = 3\times 3\times 5\times 5\times 17 = 3^2\times 5^2\times 17$.

(iv) $5005$

Answer: $5005 = 5\times 7\times 11\times 13$.

(v) $7429$

Answer: $7429 = 17\times 19\times 23$.

Q2. Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers.

(i) $26$ and $91$

Answer: $26 = 2\times 13$, $91 = 7\times 13$. So $\text{HCF}=13$ and $\text{LCM}=2\times 7\times 13 = 182$.

Verification: $\text{HCF}\times \text{LCM} = 13\times 182 = 2366 = 26\times 91$. ✓

(ii) $510$ and $92$

Answer: $510 = 2\times 3\times 5\times 17$, $92 = 2^2\times 23$. $\text{HCF}=2$ and $\text{LCM}=2^2\times 3\times 5\times 17\times 23 = 23460$.

Verification: $2\times 23460 = 46920 = 510\times 92$. ✓

(iii) $336$ and $54$

Answer: $336 = 2^4\times 3\times 7$, $54 = 2\times 3^3$. $\text{HCF}=2\times 3 = 6$ and $\text{LCM}=2^4\times 3^3\times 7 = 3024$.

Verification: $6\times 3024 = 18144 = 336\times 54$. ✓

Q3. Find the LCM and HCF of the following integers by applying the prime factorisation method.

(i) $12$, $15$ and $21$

Answer: $12 = 2^2\times 3$, $15 = 3\times 5$, $21 = 3\times 7$. $\text{HCF}=3$. $\text{LCM} = 2^2\times 3\times 5\times 7 = 420$.

(ii) $17$, $23$ and $29$

Answer: All three are primes. $\text{HCF}=1$. $\text{LCM} = 17\times 23\times 29 = 11339$.

(iii) $8$, $9$ and $25$

Answer: $8 = 2^3$, $9 = 3^2$, $25 = 5^2$. They share no common prime, so $\text{HCF}=1$. $\text{LCM} = 2^3\times 3^2\times 5^2 = 1800$.

Q4. Given that $\text{HCF}(306,\ 657) = 9$, find $\text{LCM}(306,\ 657)$.

Answer: Using $\text{HCF}\times \text{LCM} = a\times b$,

$$\text{LCM} = \dfrac{306\times 657}{9} = \dfrac{201042}{9} = 22338.$$

Q5. Check whether $6^n$ can end with the digit $0$ for any natural number $n$.

Answer: If a number ends in $0$, it must be divisible by $10$, i.e., its prime factorisation must contain $2$ and $5$. Now $6^n = (2\times 3)^n = 2^n \times 3^n$. The factor $5$ never appears, so by the uniqueness of prime factorisation, $6^n$ can never end with the digit $0$.

Q6. Explain why $7\times 11\times 13 + 13$ and $7\times 6\times 5\times 4\times 3\times 2\times 1 + 5$ are composite numbers.

Answer:

$7\times 11\times 13 + 13 = 13(7\times 11 + 1) = 13\times 78$. Since the number has factors other than $1$ and itself, it is composite.

$7! + 5 = 5(7\times 6\times 4\times 3\times 2\times 1 + 1) = 5\times 1009$. Hence it is composite.

Q7. There is a circular path around a sports field. Sonia takes $18$ minutes to drive one round of the field, while Ravi takes $12$ minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?

Answer: They meet again at the starting point after $\text{LCM}(18,12)$ minutes.

$18 = 2\times 3^2$, $12 = 2^2\times 3$. $\text{LCM}=2^2\times 3^2 = 36$. They meet after $\mathbf{36}$ minutes.

Q8. A regiment has $300$ soldiers who can be made to stand in rows of $15$, $20$ or $25$ with no soldier left behind. Verify that $300$ is indeed the smallest such number.

Answer: The smallest number of soldiers is $\text{LCM}(15,20,25)$. $15 = 3\times 5$, $20 = 2^2\times 5$, $25 = 5^2$. $\text{LCM} = 2^2\times 3\times 5^2 = 300$. So $300$ is correct.

Exercise 1.3 — Solutions

Q1. Prove that $\sqrt{5}$ is irrational.

Answer: Suppose, on the contrary, that $\sqrt{5}$ is rational. Then we can write

$$\sqrt{5} = \dfrac{a}{b},$$

where $a,b$ are integers with $b\ne 0$ and $\gcd(a,b)=1$. Squaring both sides,

$$5 = \dfrac{a^2}{b^2}\ \Rightarrow\ a^2 = 5b^2.$$

Therefore $5$ divides $a^2$, and hence $5$ divides $a$ (as $5$ is prime). Write $a = 5c$. Then $a^2 = 25c^2$, so $25c^2 = 5b^2$, i.e., $b^2 = 5c^2$. Hence $5$ divides $b^2$, and so $5$ divides $b$. Thus $5$ is a common factor of $a$ and $b$, contradicting $\gcd(a,b)=1$. Our assumption is wrong; therefore $\sqrt{5}$ is irrational.

Q2. Prove that $3 + 2\sqrt{5}$ is irrational.

Answer: Assume that $3 + 2\sqrt{5}$ is rational. Then $3 + 2\sqrt{5} = \dfrac{a}{b}$ for some integers $a,b$ with $b\ne 0$. Hence

$$2\sqrt{5} = \dfrac{a}{b} – 3 = \dfrac{a-3b}{b},\quad \sqrt{5} = \dfrac{a-3b}{2b}.$$

The right side is rational, so $\sqrt{5}$ would be rational. This contradicts the result of Q1. Hence $3+2\sqrt{5}$ is irrational.

Q3. Prove that the following are irrational.

(i) $\dfrac{1}{\sqrt{2}}$

Answer: Assume $\dfrac{1}{\sqrt{2}}$ is rational. Then $\dfrac{1}{\sqrt{2}} = \dfrac{a}{b}$, giving $\sqrt{2} = \dfrac{b}{a}$, which would make $\sqrt{2}$ rational. Since $\sqrt{2}$ is irrational, our assumption is wrong, so $\dfrac{1}{\sqrt{2}}$ is irrational.

(ii) $7\sqrt{5}$

Answer: Assume $7\sqrt{5}$ is rational. Then $7\sqrt{5} = \dfrac{a}{b}$, so $\sqrt{5} = \dfrac{a}{7b}$, which is rational. This contradicts the irrationality of $\sqrt{5}$. Hence $7\sqrt{5}$ is irrational.

(iii) $6 + \sqrt{2}$

Answer: Assume $6+\sqrt{2}$ is rational. Then $6+\sqrt{2} = \dfrac{a}{b}$, so $\sqrt{2} = \dfrac{a-6b}{b}$, which is rational. This contradicts the irrationality of $\sqrt{2}$. Hence $6+\sqrt{2}$ is irrational.

Exercise 1.4 — Solutions

Q1. Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion.

(Recall: a rational $\dfrac{p}{q}$ in lowest form has a terminating decimal expansion if and only if $q = 2^m\cdot 5^n$.)

(i) $\dfrac{13}{3125}$

Answer: $3125 = 5^5 = 2^0\times 5^5$. Denominator is of the form $2^m 5^n$, so the expansion is terminating.

(ii) $\dfrac{17}{8}$

Answer: $8 = 2^3$. Hence terminating.

(iii) $\dfrac{64}{455}$

Answer: $455 = 5\times 7\times 13$. The factor $7$ (and $13$) appears, so non-terminating repeating.

(iv) $\dfrac{15}{1600}$

Answer: $1600 = 2^6\times 5^2$. Hence terminating.

(v) $\dfrac{29}{343}$

Answer: $343 = 7^3$. Hence non-terminating repeating.

(vi) $\dfrac{23}{2^3\,5^2}$

Answer: Denominator is $2^3\times 5^2$, of the required form. Hence terminating.

(vii) $\dfrac{129}{2^2\,5^7\,7^5}$

Answer: The factor $7^5$ is present in the denominator, so non-terminating repeating.

(viii) $\dfrac{6}{15}$

Answer: $\dfrac{6}{15} = \dfrac{2}{5}$. Denominator $5 = 5^1$. Hence terminating.

(ix) $\dfrac{35}{50}$

Answer: $\dfrac{35}{50} = \dfrac{7}{10}$. Denominator $10 = 2\times 5$. Hence terminating.

(x) $\dfrac{77}{210}$

Answer: $\dfrac{77}{210} = \dfrac{11}{30}$ since $\gcd(77,210)=7$. Denominator $30 = 2\times 3\times 5$ contains the prime $3$. Hence non-terminating repeating.

Q2. Write down the decimal expansions of those rational numbers in Question 1 above which have terminating decimal expansions.

(i) $\dfrac{13}{3125}$

Answer: $\dfrac{13}{3125} = \dfrac{13\times 2^5}{3125\times 2^5} = \dfrac{416}{100000} = 0.00416$.

(ii) $\dfrac{17}{8}$

Answer: $\dfrac{17}{8} = 2.125$.

(iv) $\dfrac{15}{1600}$

Answer: $\dfrac{15}{1600} = \dfrac{15\times 5^4}{2^6\times 5^6} = \dfrac{15\times 625}{10^6} = \dfrac{9375}{1000000} = 0.009375$.

(vi) $\dfrac{23}{2^3\,5^2}$

Answer: Multiply numerator and denominator by $5$: $\dfrac{23\times 5}{2^3\times 5^3} = \dfrac{115}{1000} = 0.115$.

(viii) $\dfrac{6}{15} = \dfrac{2}{5}$

Answer: $\dfrac{2}{5} = 0.4$.

(ix) $\dfrac{35}{50} = \dfrac{7}{10}$

Answer: $\dfrac{7}{10} = 0.7$.

Q3. The following real numbers have decimal expansions as given below. In each case decide whether they are rational or not. If they are rational, and of the form $\dfrac{p}{q}$, what can you say about the prime factors of $q$?

(i) $43.123456789$

Answer: The decimal terminates, so the number is rational. Writing $43.123456789 = \dfrac{43123456789}{10^9} = \dfrac{43123456789}{2^9\times 5^9}$, the prime factors of the denominator $q$ are only $2$ and $5$.

(ii) $0.120120012000120000\ldots$

Answer: The expansion is non-terminating and non-repeating (the gaps grow longer), so the number is irrational.

(iii) $43.\overline{123456789}$ (the block $123456789$ repeats)

Answer: The expansion is non-terminating but recurring, so the number is rational. However the prime factorisation of its denominator $q$ must contain a prime other than $2$ or $5$ (otherwise it would terminate).

Worked Examples (Important Textbook Examples)

Example A. Use Euclid’s algorithm to find the HCF of $4052$ and $12576$.

Answer: Apply the lemma successively.

$12576 = 4052\times 3 + 420$

$4052 = 420\times 9 + 272$

$420 = 272\times 1 + 148$

$272 = 148\times 1 + 124$

$148 = 124\times 1 + 24$

$124 = 24\times 5 + 4$

$24 = 4\times 6 + 0$

The remainder is now $0$, so $\text{HCF}(4052,12576) = 4$.

Example B. Show that every positive even integer is of the form $2q$ and every positive odd integer is of the form $2q+1$.

Answer: Let $a$ be any positive integer. By Euclid’s lemma with $b=2$, $a = 2q + r$ where $r=0$ or $r=1$. If $r=0$ then $a = 2q$ (even). If $r=1$ then $a = 2q + 1$ (odd). Hence the result.

Example C. Find the LCM and HCF of $6$ and $20$ by the prime factorisation method.

Answer: $6 = 2\times 3$ and $20 = 2^2\times 5$.

$\text{HCF}(6,20) = 2^1 = 2$ (the lowest power of each common prime factor).

$\text{LCM}(6,20) = 2^2\times 3\times 5 = 60$ (the greatest power of each prime factor of either number).

Verification: $\text{HCF}\times \text{LCM} = 2\times 60 = 120 = 6\times 20$.

Example D. Prove that $\sqrt{3}$ is irrational.

Answer: Suppose, on the contrary, that $\sqrt{3}$ is rational. Then we can write $\sqrt{3} = \dfrac{a}{b}$ where $a,b$ are integers, $b\ne 0$ and $\gcd(a,b)=1$. Squaring,

$$3 = \dfrac{a^2}{b^2}\ \Longrightarrow\ a^2 = 3 b^2.$$

Hence $3$ divides $a^2$, and so $3$ divides $a$ (since $3$ is prime). Let $a = 3c$. Then $a^2 = 9c^2$, so $9c^2 = 3b^2$, giving $b^2 = 3c^2$. Therefore $3$ divides $b^2$ and hence $b$. So $3$ divides both $a$ and $b$, contradicting $\gcd(a,b)=1$. Thus $\sqrt{3}$ is irrational.

Example E. Show that $5 – \sqrt{3}$ is irrational.

Answer: Assume $5 – \sqrt{3}$ is rational. Then $5 – \sqrt{3} = \dfrac{a}{b}$ for some integers $a,b$ with $b\ne 0$. Rearranging,

$$\sqrt{3} = 5 – \dfrac{a}{b} = \dfrac{5b – a}{b}.$$

The right-hand side is rational, which would make $\sqrt{3}$ rational. This contradicts the result of Example D. Hence $5 – \sqrt{3}$ is irrational.

Example F. Without performing long division, decide whether $\dfrac{13}{3125}$ has a terminating decimal expansion. If so, write the decimal.

Answer: Here $3125 = 5^5$, so the denominator (in lowest terms) is of the form $2^m \cdot 5^n$. Hence the expansion terminates. Multiplying numerator and denominator by $2^5 = 32$:

$$\dfrac{13}{3125} = \dfrac{13\times 32}{3125\times 32} = \dfrac{416}{100000} = 0.00416.$$

Additional Practice — MCQs

Q1. The HCF of two consecutive natural numbers is: (a) $0$   (b) $1$   (c) $2$   (d) the smaller number Answer: (b) $1$.

Q2. The decimal expansion of $\dfrac{14587}{1250}$ will: (a) terminate after one decimal place   (b) terminate after two decimal places   (c) terminate after four decimal places   (d) not terminate Answer: (c) terminate after four decimal places, since $1250 = 2\times 5^4$.

Q3. If $\text{HCF}(a,b)=12$ and $a\times b = 1800$, then $\text{LCM}(a,b)$ is: (a) $90$   (b) $150$   (c) $900$   (d) $216$ Answer: (b) $150$. Using $\text{HCF}\times \text{LCM} = a\times b$, $\text{LCM} = 1800/12 = 150$.

Q4. The product of a non-zero rational and an irrational number is: (a) always rational   (b) always irrational   (c) rational or irrational   (d) one Answer: (b) always irrational.

Q5. The exponent of $2$ in the prime factorisation of $144$ is: (a) $2$   (b) $3$   (c) $4$   (d) $5$ Answer: (c) $4$, since $144 = 2^4\times 3^2$.

Q6. If two positive integers $a$ and $b$ are written as $a = x^3 y^2$ and $b = x y^3$ where $x,y$ are primes, then $\text{HCF}(a,b)$ is: (a) $xy$   (b) $xy^2$   (c) $x^3 y^3$   (d) $x^2 y^2$ Answer: (b) $xy^2$.

Q7. The largest number which divides $70$ and $125$, leaving remainders $5$ and $8$ respectively, is: (a) $13$   (b) $65$   (c) $875$   (d) $1750$ Answer: (a) $13$. We need $\text{HCF}(70-5,\ 125-8)=\text{HCF}(65,117)=13$.

Q8. $\sqrt{2}+\sqrt{3}$ is: (a) rational   (b) irrational   (c) zero   (d) integer Answer: (b) irrational.

Q9. The number $0.\overline{6}$ in the form $\dfrac{p}{q}$ is: (a) $\dfrac{6}{10}$   (b) $\dfrac{2}{3}$   (c) $\dfrac{3}{5}$   (d) $\dfrac{6}{99}$ Answer: (b) $\dfrac{2}{3}$.

Q10. The smallest composite number is: (a) $1$   (b) $2$   (c) $3$   (d) $4$ Answer: (d) $4$.

Q11. Which of the following is a rational number? (a) $\sqrt{2}$   (b) $\sqrt{3}$   (c) $\sqrt{4}$   (d) $\sqrt{5}$ Answer: (c) $\sqrt{4} = 2$ is rational.

Q12. The decimal expansion of $\dfrac{2}{11}$ is: (a) terminating   (b) non-terminating recurring   (c) non-terminating non-recurring   (d) integer Answer: (b) non-terminating recurring, since $11$ is not of the form $2^m 5^n$.

Q13. The greatest number that divides $437$ and $377$ leaving remainders $5$ and $7$ respectively is: (a) $26$   (b) $32$   (c) $36$   (d) $48$ Answer: Required HCF $=\text{HCF}(437-5,\ 377-7)=\text{HCF}(432,370)$. Since $432 = 370\times 1 + 62$, $370 = 62\times 5 + 60$, $62 = 60\times 1 + 2$, $60 = 2\times 30$, the HCF is $2$. (Hence none of (a)–(d); the typical question gives a meaningful HCF — students should focus on the procedure.)

Q14. If $a = 2^3\times 3$, $b = 2\times 3\times 5$, $c = 3^n\times 5$ and $\text{LCM}(a,b,c)=2^3\times 3^2\times 5$, then $n$ equals: (a) $1$   (b) $2$   (c) $3$   (d) $4$ Answer: (b) $2$. The highest power of $3$ in the LCM is $2$, which must come from $c$.

Q15. The number of zeros at the end of $5! = 120$ is: (a) $0$   (b) $1$   (c) $2$   (d) $3$ Answer: (b) $1$ (because $5! = 2^3\times 3\times 5$).

Q16. Two natural numbers whose sum is $85$ and whose LCM is $102$ are: (a) $30,55$   (b) $17,68$   (c) $35,55$   (d) $51,34$ Answer: (d) $51,34$. Check: $51+34=85$ and $\text{LCM}(51,34)=102$.

Q17. $\dfrac{17}{30}$ has a decimal expansion that is: (a) terminating   (b) non-terminating recurring   (c) irrational   (d) zero Answer: (b), since $30 = 2\times 3\times 5$ contains the prime $3$.

Q18. If $\dfrac{p}{q}$ is a rational number with $q = 2^3\times 5^4$, then the maximum number of digits after the decimal point in its expansion is: (a) $3$   (b) $4$   (c) $7$   (d) $12$ Answer: (b) $4$ (the larger of the powers $3$ and $4$).

Q19. Which of the following numbers is irrational? (a) $\sqrt{16}$   (b) $\sqrt[3]{27}$   (c) $\sqrt{12}$   (d) $\sqrt{0.04}$ Answer: (c) $\sqrt{12} = 2\sqrt{3}$, which is irrational.

Q20. If $\text{HCF}(72, 120) = 24$, then $\text{LCM}(72,120)$ equals: (a) $240$   (b) $360$   (c) $480$   (d) $720$ Answer: (b) $360$, using $\text{LCM}=72\times 120/24 = 360$.

Additional Practice — Fill in the Blanks

1. Euclid’s division lemma states that for any two positive integers $a$ and $b$ there exist unique integers $q$ and $r$ such that $a = bq + r$, where ____. Answer: $0 \le r < b$.

2. Every composite number can be expressed as a product of ____, and this factorisation is unique apart from the order. Answer: primes.

3. If $p$ is a prime number, then $\sqrt{p}$ is ____. Answer: irrational.

4. For any two positive integers $a$ and $b$, $\text{HCF}(a,b)\times \text{LCM}(a,b) = $ ____. Answer: $a\times b$.

5. A rational number $\dfrac{p}{q}$ (in lowest terms) has a terminating decimal expansion if and only if $q$ is of the form ____. Answer: $2^m\cdot 5^n$ where $m,n$ are non-negative integers.

6. The HCF of $96$ and $404$ is ____. Answer: $4$.

7. The LCM of $6$, $72$ and $120$ is ____. Answer: $360$.

8. The decimal expansion of $\dfrac{3}{8}$ is ____. Answer: $0.375$.

9. $\sqrt{2}\times \sqrt{3} = $ ____. Answer: $\sqrt{6}$.

10. The number of prime factors of $1729$ is ____. Answer: three (since $1729 = 7\times 13\times 19$).

Additional Practice — True / False

1. The product of two irrational numbers is always irrational. Answer: False. For example, $\sqrt{2}\times \sqrt{2} = 2$, which is rational.

2. Every integer is a rational number. Answer: True. Any integer $n$ can be written as $\dfrac{n}{1}$.

3. $\dfrac{22}{7}$ is an irrational number because it equals $\pi$. Answer: False. $\dfrac{22}{7}$ is a rational approximation of $\pi$; it is itself rational.

4. The HCF of two prime numbers is always $1$. Answer: True (assuming the two primes are distinct).

5. The number $4^n$ ends with the digit $0$ for some natural number $n$. Answer: False. $4^n = 2^{2n}$ has no factor $5$, so it can never end with $0$.

6. Every terminating decimal is a rational number. Answer: True.

7. If $\text{HCF}(a,b)=1$ then $a$ and $b$ are both prime. Answer: False. For example, $\text{HCF}(8,9)=1$ but neither $8$ nor $9$ is prime.

8. $\sqrt{3}$ is a rational number. Answer: False. $\sqrt{3}$ is irrational.

9. A non-terminating recurring decimal is always rational. Answer: True.

10. The Fundamental Theorem of Arithmetic guarantees uniqueness of prime factorisation up to the order of factors. Answer: True.

Additional Practice — Short Answer Questions

SA1. Find the HCF of $96$ and $404$ by Euclid’s algorithm. Hence find their LCM. Answer: $404 = 96\times 4 + 20$; $96 = 20\times 4 + 16$; $20 = 16\times 1 + 4$; $16 = 4\times 4 + 0$. Hence $\text{HCF}=4$. $\text{LCM} = \dfrac{96\times 404}{4} = \dfrac{38784}{4} = 9696$.

SA2. Find the HCF and LCM of $510$ and $92$ by prime factorisation, and verify that $\text{HCF}\times \text{LCM} = 510\times 92$. Answer: $510 = 2\times 3\times 5\times 17$, $92 = 2^2\times 23$. $\text{HCF}=2$ and $\text{LCM}= 2^2\times 3\times 5\times 17\times 23 = 23460$. Verification: $2\times 23460 = 46920 = 510\times 92$. ✓

SA3. Find the largest number that divides $1251$, $9377$ and $15628$ leaving remainders $1$, $2$ and $3$ respectively. Answer: Required number $= \text{HCF}(1251-1,\ 9377-2,\ 15628-3) = \text{HCF}(1250,\ 9375,\ 15625)$. Since $1250 = 2\times 5^4$, $9375 = 3\times 5^5$, $15625 = 5^6$, the HCF is $5^4 = 625$. So the answer is $\mathbf{625}$.

SA4. Three bells ring at intervals of $9$, $12$ and $15$ minutes respectively. If they all ring together at $8\!:\!00$ a.m., when will they next ring together? Answer: Required time $= \text{LCM}(9,12,15)$ minutes. $9 = 3^2$, $12 = 2^2\times 3$, $15 = 3\times 5$. $\text{LCM} = 2^2\times 3^2\times 5 = 180$ minutes $= 3$ hours. They will next ring together at $\mathbf{11\!:\!00}$ a.m.

SA5. Find the smallest number which when divided by $35$, $56$ and $91$ leaves remainder $7$ in each case. Answer: Required number $= \text{LCM}(35,56,91) + 7$. $35 = 5\times 7$, $56 = 2^3\times 7$, $91 = 7\times 13$. $\text{LCM} = 2^3\times 5\times 7\times 13 = 3640$. Hence the answer is $3640 + 7 = \mathbf{3647}$.

SA6. Show that $12^n$ cannot end with the digit $0$ for any natural number $n$. Answer: $12^n = (2^2\times 3)^n = 2^{2n}\times 3^n$. The prime $5$ does not appear in this factorisation, so by the Fundamental Theorem of Arithmetic, $12^n$ is not divisible by $10$ and hence cannot end in $0$.

SA7. Prove that $\sqrt{2}$ is an irrational number. Answer: Suppose $\sqrt{2}$ is rational. Then $\sqrt{2} = \dfrac{a}{b}$ with $\gcd(a,b)=1$ and $b\ne 0$. Squaring, $a^2 = 2b^2$, so $2\mid a^2$ and hence $2\mid a$. Write $a = 2c$. Then $4c^2 = 2b^2$, i.e., $b^2 = 2c^2$. So $2\mid b^2$, hence $2\mid b$. So $2$ divides both $a$ and $b$, contradicting $\gcd(a,b)=1$. Therefore $\sqrt{2}$ is irrational.

SA8. Prove that $\dfrac{1}{\sqrt{3}}$ is irrational. Answer: Suppose $\dfrac{1}{\sqrt{3}}$ is rational. Then $\dfrac{1}{\sqrt{3}} = \dfrac{a}{b}$, so $\sqrt{3} = \dfrac{b}{a}$, which would be rational. This contradicts the fact that $\sqrt{3}$ is irrational. Hence $\dfrac{1}{\sqrt{3}}$ is irrational.

SA9. If two positive integers $p$ and $q$ are written as $p = a^2 b^3$ and $q = a^3 b$ where $a,b$ are prime numbers, find the LCM and HCF of $p$ and $q$. Answer: $\text{HCF}(p,q) = a^2 b$ (lowest power of each common prime). $\text{LCM}(p,q) = a^3 b^3$ (highest power of each prime).

SA10. Without performing long division, state whether $\dfrac{77}{210}$ has a terminating or non-terminating decimal expansion. Answer: $\dfrac{77}{210} = \dfrac{11}{30}$, with denominator $30 = 2\times 3\times 5$. Since $3$ appears, it is non-terminating recurring.

Additional Practice — Long Answer / HOTS Questions

LA1. Use Euclid’s division lemma to show that the cube of any positive integer is of the form $4m$, $4m+1$ or $4m+3$ for some integer $m$. Answer: Let $a$ be any positive integer. By the lemma with $b=4$, $a$ is of the form $4q,\ 4q+1,\ 4q+2$ or $4q+3$. If $a=4q$ then $a^3 = 64 q^3 = 4(16 q^3) = 4m$. If $a=4q+1$ then $a^3 = 64q^3 + 48q^2 + 12q + 1 = 4(16q^3+12q^2+3q) + 1 = 4m+1$. If $a=4q+2$ then $a^3 = 64q^3 + 96q^2 + 48q + 8 = 4(16q^3+24q^2+12q+2) = 4m$. If $a=4q+3$ then $a^3 = 64q^3 + 144q^2 + 108q + 27 = 4(16q^3+36q^2+27q+6) + 3 = 4m+3$. So $a^3$ is of one of the stated forms. (Note that $4m+2$ does not occur.)

LA2. Show that one and only one out of $n,\ n+2,\ n+4$ is divisible by $3$, where $n$ is any positive integer. Answer: By Euclid’s lemma, $n = 3q,\ 3q+1$ or $3q+2$. If $n=3q$, then $n$ itself is divisible by $3$, while $n+2 = 3q+2$ and $n+4 = 3q+4 = 3(q+1)+1$ are not. If $n=3q+1$, then $n+2 = 3q+3 = 3(q+1)$ is divisible by $3$, but the others are not. If $n=3q+2$, then $n+4 = 3q+6 = 3(q+2)$ is divisible by $3$, but the others are not. Hence exactly one of $n,\ n+2,\ n+4$ is divisible by $3$.

LA3. Prove that $\sqrt{2} + \sqrt{3}$ is irrational. Answer: Assume that $\sqrt{2}+\sqrt{3}$ is rational, say $\sqrt{2}+\sqrt{3} = r$, where $r\in \mathbb{Q}$. Squaring, $5 + 2\sqrt{6} = r^2$, so $\sqrt{6} = \dfrac{r^2 – 5}{2}$, which would be rational. But $\sqrt{6}$ is irrational (since $6$ is not a perfect square — apply the same proof technique used for $\sqrt{2}$). This contradiction shows that $\sqrt{2}+\sqrt{3}$ is irrational.

LA4. Find the smallest natural number that, when divided by $12,15,20$ and $54$, leaves remainder $8$ in each case. Answer: Required number $= \text{LCM}(12,15,20,54) + 8$. $12 = 2^2\times 3$, $15 = 3\times 5$, $20 = 2^2\times 5$, $54 = 2\times 3^3$. $\text{LCM} = 2^2\times 3^3\times 5 = 540$. So the answer is $540 + 8 = \mathbf{548}$.

LA5. Three sets of English, Hindi and Mathematics books have to be stacked in such a way that all the books are stored topic-wise and the height of each stack is the same. The number of English books is $96$, the number of Hindi books is $240$ and the number of Mathematics books is $336$. Assuming that the books are of the same thickness, determine the number of stacks of English, Hindi and Mathematics books. Answer: Number of books per stack $= \text{HCF}(96,240,336)$. $96 = 2^5\times 3$, $240 = 2^4\times 3\times 5$, $336 = 2^4\times 3\times 7$. $\text{HCF} = 2^4\times 3 = 48$. Number of stacks $= 96/48,\ 240/48,\ 336/48 = 2,\ 5,\ 7$. Total $= 14$ stacks.

LA6. The length, breadth and height of a room are $8$ m $25$ cm, $6$ m $75$ cm, and $4$ m $50$ cm respectively. Find the length of the longest rod that can measure all the three dimensions exactly. Answer: Convert to cm: $825,\ 675,\ 450$. Required length $= \text{HCF}(825,675,450)$. $825 = 3\times 5^2\times 11$, $675 = 3^3\times 5^2$, $450 = 2\times 3^2\times 5^2$. $\text{HCF} = 3\times 5^2 = 75$ cm. So the longest rod is $\mathbf{75}$ cm.

LA7. Show that the square of any odd integer is of the form $8q+1$ for some integer $q$. Answer: Let $n$ be any odd integer. Then $n=2m+1$ for some integer $m$. So $n^2 = 4m^2 + 4m + 1 = 4m(m+1) + 1$. Since $m(m+1)$ is the product of two consecutive integers, it is even; let $m(m+1) = 2k$. Then $n^2 = 8k + 1$, of the desired form.

Quick Reference Tables

Concept	Statement / Formula
Euclid’s Division Lemma	For positive integers $a,b$: $a = bq + r,\ 0 \le r < b$ (unique).
Euclid’s Algorithm for HCF	Repeatedly apply the lemma; HCF is the divisor when the remainder first becomes zero.
Fundamental Theorem of Arithmetic	Every integer $n>1$ is uniquely a product of primes, up to order.
HCF–LCM relation (two integers)	$\text{HCF}(a,b)\cdot \text{LCM}(a,b) = a\cdot b$.
Irrationality of $\sqrt{p}$	If $p$ is prime, $\sqrt{p}$ is irrational.
Terminating decimal test	$\dfrac{p}{q}$ in lowest terms terminates iff $q = 2^m\,5^n$.
Number of decimal places (terminating case)	If $q = 2^m\,5^n$, the expansion has $\max(m,n)$ decimal places (after suitable conversion to power of $10$).

Number	Prime factorisation
$36$	$2^2\times 3^2$
$72$	$2^3\times 3^2$
$100$	$2^2\times 5^2$
$144$	$2^4\times 3^2$
$210$	$2\times 3\times 5\times 7$
$1000$	$2^3\times 5^3$
$1024$	$2^{10}$
$1729$	$7\times 13\times 19$ (Hardy–Ramanujan number)

Common Mistakes to Avoid

1. Forgetting that Euclid’s division lemma requires $0 \le r < b$ — many students write $r$ as a negative number, which is wrong.
2. Confusing the HCF and the LCM. The HCF is the divisor at the step where the remainder becomes zero in Euclid’s algorithm; the LCM is computed using the relation $\text{HCF}\times \text{LCM} = a\times b$ for two numbers.
3. Trying to apply $\text{HCF}\times \text{LCM} = a\times b$ to three or more numbers — this identity is valid only for two integers.
4. Saying that “every irrational $+$ irrational is irrational.” For example, $(\sqrt{2}) + (-\sqrt{2}) = 0$ is rational. The correct statement is: irrational $+$ rational is always irrational.
5. Reducing the fraction after testing the denominator. Always reduce $\dfrac{p}{q}$ to lowest terms before checking whether $q = 2^m\cdot 5^n$.
6. Writing $4$ as composite but $1$ as prime. By definition, $1$ is neither prime nor composite, and the smallest prime is $2$.
7. Confusing “non-terminating” with “irrational.” Many rationals (e.g., $\dfrac{1}{3} = 0.\overline{3}$) have non-terminating decimal expansions, but they are still rational because the expansion repeats.

Glossary

Real number: Any number that can be represented on the number line; includes both rational and irrational numbers.

Rational number: A number expressible as $\dfrac{p}{q}$ where $p,q$ are integers and $q\ne 0$.

Irrational number: A real number that cannot be written as $\dfrac{p}{q}$ with $p,q$ integers and $q\ne 0$.

Prime number: A natural number greater than $1$ whose only divisors are $1$ and itself.

Composite number: A natural number greater than $1$ which has at least one divisor other than $1$ and itself.

HCF (Highest Common Factor): The greatest positive integer that divides each of the given integers without leaving a remainder. Also called GCD.

LCM (Least Common Multiple): The smallest positive integer that is divisible by each of the given integers.

Euclid’s Division Lemma: For positive integers $a,b$, there exist unique integers $q,r$ with $a=bq+r$ and $0\le r

Euclid’s Division Algorithm: A repeated application of the lemma used to compute the HCF of two positive integers.

Fundamental Theorem of Arithmetic: Every composite number can be uniquely expressed as a product of primes, apart from the order of the factors.

Terminating decimal: A decimal expansion that ends after a finite number of digits.

Non-terminating recurring decimal: A decimal expansion that does not end but repeats a fixed block of digits indefinitely.

Practice Drill — Mixed Problems with Quick Answers

D1. Find the HCF of $144$ and $180$ by Euclid’s algorithm. Answer: $180 = 144\times 1 + 36$; $144 = 36\times 4 + 0$. So $\text{HCF} = 36$.

D2. Express $429$ as a product of primes. Answer: $429 = 3\times 11\times 13$.

D3. If $\text{LCM}(a,18)=36$ and $\text{HCF}(a,18)=2$, find $a$. Answer: Using $a\times 18 = \text{HCF}\times \text{LCM} = 2\times 36 = 72$, we get $a = 4$.

D4. Find the HCF and LCM of $144$, $180$ and $192$. Answer: $144 = 2^4\times 3^2$, $180 = 2^2\times 3^2\times 5$, $192 = 2^6\times 3$. $\text{HCF}= 2^2\times 3 = 12$ and $\text{LCM} = 2^6\times 3^2\times 5 = 2880$.

D5. Determine whether $\dfrac{257}{5000}$ has a terminating decimal expansion. If yes, write it. Answer: $5000 = 2^3\times 5^4$, so it is terminating. Multiply numerator and denominator by $2$: $\dfrac{257\times 2}{2^4\times 5^4} = \dfrac{514}{10000} = 0.0514$.

D6. Decide whether $\dfrac{17}{12}$ has a terminating decimal expansion. Answer: $12 = 2^2\times 3$ contains the prime $3$, so the expansion is non-terminating recurring.

D7. Express $0.\overline{47}$ as a fraction in lowest terms. Answer: Let $x = 0.\overline{47}$. Then $100x = 47.\overline{47}$ and $99x = 47$, so $x = \dfrac{47}{99}$.

D8. If $a$ and $b$ are two positive integers such that $\text{HCF}(a,b) = 1$, prove that $a^2$ and $b^2$ are also coprime. Answer: If a prime $p$ divided both $a^2$ and $b^2$, then $p$ would divide both $a$ and $b$ (Euclid’s lemma), contradicting $\text{HCF}(a,b)=1$.

D9. Two positive integers are in the ratio $3:4$ and their LCM is $180$. Find the numbers. Answer: Let the numbers be $3k$ and $4k$. $\text{HCF}(3k,4k)=k$ since $\gcd(3,4)=1$. So $\text{LCM} = 12k = 180$, i.e., $k=15$. The numbers are $45$ and $60$.

D10. Show that $7+\sqrt{5}$ is irrational, given that $\sqrt{5}$ is irrational. Answer: If $7+\sqrt{5}$ were rational, say $\dfrac{a}{b}$, then $\sqrt{5} = \dfrac{a}{b} – 7 = \dfrac{a – 7b}{b}$ would be rational — contradiction. Hence $7+\sqrt{5}$ is irrational.

Case-Study / Real-Life Application

Case Study 1. Two tankers contain $850$ litres and $680$ litres of petrol respectively. The owner wants to use a single container to fill both tankers exactly an integer number of times. What is the maximum capacity (in litres) of such a container?

Answer: Required capacity $=\text{HCF}(850,680)$.

$850 = 680\times 1 + 170$

$680 = 170\times 4 + 0$

$\therefore\ \text{HCF}(850,680) = 170$. The maximum capacity is $\mathbf{170}$ litres.

Case Study 2. A trader has $420$ litres of orange juice, $130$ litres of apple juice and $250$ litres of mango juice. He wants to fill them into bottles of equal capacity such that no juice is left over and no two juices are mixed in the same bottle. Find the maximum capacity of each bottle and the total number of bottles required.

Answer: Capacity per bottle $= \text{HCF}(420, 130, 250)$.
$420 = 2^2\times 3\times 5\times 7$, $130 = 2\times 5\times 13$, $250 = 2\times 5^3$.
$\text{HCF} = 2\times 5 = 10$ litres.
Number of bottles $= \dfrac{420}{10} + \dfrac{130}{10} + \dfrac{250}{10} = 42+13+25 = \mathbf{80}$ bottles.

Case Study 3. Two LED bulbs blink at intervals of $40$ seconds and $60$ seconds respectively. They blink together at $9\!:\!00$ p.m. After how long will they blink together again?

Answer: Required time $=\text{LCM}(40,60)$. $40 = 2^3\times 5$, $60 = 2^2\times 3\times 5$. $\text{LCM} = 2^3\times 3\times 5 = 120$ seconds $= 2$ minutes. They blink together again at $\mathbf{9\!:\!02}$ p.m.

Keep practising every problem from the textbook and the extra exercises above. Mastery of Chapter 1 builds the foundation for Polynomials, Pair of Linear Equations, and the rest of your ASSEB Class 10 Mathematics course. Best wishes from HSLC Guru!