Solved Examples

How to Use This Section

This is your practice powerhouse for Real Numbers. Below are 30+ fully worked problems spanning the whole chapter — number classification, Euclid's algorithm, prime factorisation, HCF & LCM, irrationality proofs, and decimal expansions, arranged roughly from easy to hard.

How to study: First try each problem yourself with the solution covered. Then check the steps. The goal is not just the answer, but the method — in the board exam, marks are awarded step by step.

Keep these formulae handy as you work:

• Euclid's lemma: $a = bq + r,\ 0 \le r < b$
• $\text{HCF}(a,b) \times \text{LCM}(a,b) = a \times b$ (two numbers only)
• HCF = product of lowest powers of common primes; LCM = product of highest powers of all primes
• Terminating decimal ⇔ denominator (lowest terms) $= 2^n \times 5^m$

Solved Examples

Example 1: Classify the number

Is $\dfrac{15}{1600}$ rational, and does it terminate?

Solution:

1. Simplify: $\dfrac{15}{1600} = \dfrac{3}{320}$.
2. $320 = 2^6 \times 5$. Only primes 2 and 5.
3. So it is rational with a terminating decimal.

Final Answer: Rational, terminating.

Takeaway: Simplify, then check the denominator's primes.

Example 2: HCF by Euclid's algorithm

Find HCF(4052, 12576).

Solution:

1. $12576 = 4052 \times 3 + 420$.
2. $4052 = 420 \times 9 + 272$.
3. $420 = 272 \times 1 + 148$.
4. $272 = 148 \times 1 + 124$; $148 = 124 \times 1 + 24$; $124 = 24 \times 5 + 4$; $24 = 4 \times 6 + 0$.
5. Last divisor = 4.

Final Answer: HCF $= 4$.

Takeaway: Keep applying the lemma; the last non-zero remainder is the HCF.

Example 3: HCF and LCM by factorisation

Find HCF and LCM of 510 and 92.

Solution:

1. $510 = 2 \times 3 \times 5 \times 17$; $92 = 2^2 \times 23$.
2. HCF = common primes at lowest power = $2^1 = 2$.
3. LCM = all primes at highest power = $2^2 \times 3 \times 5 \times 17 \times 23 = 23460$.

Final Answer: HCF $= 2$, LCM $= 23460$.

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

Example 4: Find the missing number

The LCM and HCF of two numbers are 180 and 6. If one number is 36, find the other.

Solution:

1. $\text{HCF}\times\text{LCM} = a \times b \Rightarrow 6 \times 180 = 36 \times b$.
2. $b = \dfrac{1080}{36} = 30$.

Final Answer: The other number is 30.

Takeaway: The product relation instantly gives the fourth quantity.

Example 5: Largest number dividing exactly

Find the largest number that divides 70 and 125 leaving remainders 5 and 8 respectively.

Solution:

1. Subtract remainders: $70 - 5 = 65$, $125 - 8 = 117$.
2. Required = HCF(65, 117). $117 = 65\times1 + 52$; $65 = 52\times1 + 13$; $52 = 13\times4 + 0$.
3. HCF = 13.

Final Answer: 13.

Takeaway: Subtract the remainders first, then take the HCF.

Example 6: Bells ringing together

Three bells toll at intervals of 6, 12 and 18 minutes. If they toll together at 6 a.m., when will they next toll together?

Solution:

1. Next time = LCM(6, 12, 18) minutes.
2. $6 = 2\times3$, $12 = 2^2\times3$, $18 = 2\times3^2$. LCM $= 2^2\times3^2 = 36$ minutes.
3. $6{:}00$ a.m. + 36 min = $6{:}36$ a.m.

Final Answer: 6:36 a.m.

Takeaway: 'Together again' → LCM.

Example 7: Prove $\sqrt{7}$ is irrational

Solution:

1. Assume $\sqrt{7} = \dfrac{a}{b}$, $a,b$ co-prime.
2. $7b^2 = a^2 \Rightarrow 7 \mid a^2 \Rightarrow 7 \mid a$. Write $a = 7c$.
3. $7b^2 = 49c^2 \Rightarrow b^2 = 7c^2 \Rightarrow 7 \mid b$.
4. So 7 divides both $a$ and $b$ — contradiction.

Final Answer: $\sqrt{7}$ is irrational.

Takeaway: Same template; the prime here is 7.

Example 8: Terminating check

Without dividing, decide whether $\dfrac{23}{8}$ terminates and find it.

Solution:

1. $8 = 2^3$ → only 2's → terminates.
2. Multiply by $5^3$: $\dfrac{23 \times 125}{1000} = \dfrac{2875}{1000} = 2.875$.

Final Answer: Terminating, $2.875$.

Takeaway: Power-of-10 trick gives the decimal directly.

Example 9: Composite number check

Show that $5 \times 11 \times 13 + 11$ is composite.

Solution:

1. Take 11 common: $5\times11\times13 + 11 = 11(5\times13 + 1) = 11 \times 66$.
2. It factors as $11 \times 66$, both greater than 1.

Final Answer: Composite (equals 726).

Takeaway: A common factor reveals compositeness immediately.

Example 10: Maximum students per row

44 boys and 32 girls are to be seated in rows with only one gender per row and equal numbers per row. Find the minimum number of rows.

Solution:

1. Maximum students per row = HCF(44, 32). $44 = 2^2\times11$, $32 = 2^5$. HCF $= 2^2 = 4$.
2. Rows of boys $= 44/4 = 11$; rows of girls $= 32/4 = 8$.
3. Minimum rows $= 11 + 8 = 19$.

Final Answer: 19 rows.

Takeaway: Largest equal group → HCF; then add the counts.

Example 11: Prove $2 - \sqrt{5}$ irrational

Solution:

1. Assume $2 - \sqrt{5} = r$ (rational).
2. Then $\sqrt{5} = 2 - r$, a difference of rationals → rational.
3. Contradicts $\sqrt{5}$ irrational.

Final Answer: $2 - \sqrt{5}$ is irrational.

Takeaway: Isolate the surd; the rest must be rational.

Example 12: HCF expressed as a linear combination

Find HCF(196, 38220) and verify by factorisation.

Solution:

1. Euclid: $38220 = 196 \times 195 + 0$, so HCF = 196.
2. Factorise: $196 = 2^2 \times 7^2$; $38220 = 2^2 \times 3 \times 5 \times 7^2 \times 13$.
3. Common at lowest powers: $2^2 \times 7^2 = 196$. ✓

Final Answer: HCF $= 196$.

Takeaway: Two methods, same answer — a good way to self-check.

Example 13: Smallest number divisible by 8, 9, 10

Find the smallest number divisible by 8, 9 and 10.

Solution:

1. LCM(8, 9, 10). $8 = 2^3$, $9 = 3^2$, $10 = 2\times5$.
2. LCM $= 2^3 \times 3^2 \times 5 = 8 \times 9 \times 5 = 360$.

Final Answer: 360.

Takeaway: Smallest common multiple → LCM.

Example 14: Decimal places

How many decimal places does $\dfrac{14587}{2^4 \times 5^3}$ have?

Solution:

1. Denominator $= 2^4 \times 5^3$, so $n=4$, $m=3$.
2. Places $= \max(4,3) = 4$.

Final Answer: 4 decimal places.

Takeaway: Places $= \max(n, m)$.

Example 15: $6^n$ ending digit

Can $6^n$ end in the digit 5 for any natural number $n$?

Solution:

1. $6^n = 2^n \times 3^n$ — always even, so it ends in an even digit.
2. The digit 5 is odd, so $6^n$ can never end in 5.

Final Answer: No.

Takeaway: An even number can only end in 0, 2, 4, 6, or 8.

Example 16: Word problem with HCF

Two ropes of length 64 m and 80 m are to be cut into pieces of equal maximum length. Find that length and the total number of pieces.

Solution:

1. Maximum length = HCF(64, 80). $64 = 2^6$, $80 = 2^4\times5$. HCF $= 2^4 = 16$ m.
2. Pieces $= 64/16 + 80/16 = 4 + 5 = 9$.

Final Answer: 16 m each; 9 pieces.

Takeaway: Max equal piece length → HCF; count pieces by division.

Example 17: Prove $\sqrt{2} + 5$ irrational

Solution:

1. Assume $\sqrt{2} + 5 = r$ (rational).
2. Then $\sqrt{2} = r - 5$, rational.
3. Contradiction, since $\sqrt{2}$ is irrational.

Final Answer: $\sqrt{2} + 5$ is irrational.

Takeaway: Rational + irrational = irrational.

Example 18: Largest number with given remainders (two numbers)

Find the greatest number that divides 1251, 9377 and 15628 leaving remainders 1, 2 and 3 respectively.

Solution:

1. Subtract remainders: $1251-1 = 1250$, $9377-2 = 9375$, $15628-3 = 15625$.
2. Required = HCF(1250, 9375, 15625). $1250 = 2\times5^4$, $9375 = 3\times5^5$, $15625 = 5^6$.
3. Common at lowest power: $5^4 = 625$.

Final Answer: 625.

Takeaway: Subtract each remainder, then take the HCF of all reduced numbers.

Example 19: LCM word problem (circular track)

Two runners complete one round of a track in 18 and 12 minutes. If they start together, after how long will they meet again at the start?

Solution:

1. They meet at the start after LCM(18, 12) minutes.
2. $18 = 2\times3^2$, $12 = 2^2\times3$. LCM $= 2^2\times3^2 = 36$ minutes.

Final Answer: After 36 minutes.

Takeaway: Meeting again at the start → LCM of lap times.

Example 20: Form of an integer

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

Solution:

1. Any integer is $3q$, $3q+1$, or $3q+2$.
2. $(3q)^3 = 27q^3 = 9(3q^3) = 9m$.
3. $(3q+1)^3 = 27q^3 + 27q^2 + 9q + 1 = 9m + 1$.
4. $(3q+2)^3 = 27q^3 + 54q^2 + 36q + 8 = 9m + 8$.

Final Answer: The cube is $9m$, $9m+1$, or $9m+8$.

Takeaway: Test the three residue cases and expand.

Example 21: Co-prime check

Are 847 and 2160 co-prime?

Solution:

1. $847 = 7 \times 11^2$; $2160 = 2^4 \times 3^3 \times 5$.
2. They share no common prime factor.
3. So HCF = 1.

Final Answer: Yes, they are co-prime.

Takeaway: No shared prime ⇒ HCF 1 ⇒ co-prime.

Example 22: Prove $3 + 2\sqrt{5}$ irrational

Solution:

1. Assume $3 + 2\sqrt{5} = r$ (rational).
2. $2\sqrt{5} = r - 3 \Rightarrow \sqrt{5} = \dfrac{r-3}{2}$, rational.
3. Contradicts $\sqrt{5}$ irrational.

Final Answer: $3 + 2\sqrt{5}$ is irrational.

Takeaway: Subtract the rational, divide the coefficient, isolate the surd.

Example 23: Number of decimal places & value

Express $\dfrac{7}{80}$ as a decimal.

Solution:

1. $80 = 2^4 \times 5$. Only 2's and 5's → terminates.
2. Balance: multiply top and bottom by $5^3$: $\dfrac{7 \times 125}{10000} = \dfrac{875}{10000} = 0.0875$.

Final Answer: $0.0875$.

Takeaway: $\max(4,1)=4$ places, matching 0.0875.

Example 24: Stacking books

There are 96 maths, 240 physics and 336 chemistry books. They are stacked so each stack has books of one subject and all stacks have the same height (number of books). Find the maximum books per stack and the number of stacks.

Solution:

1. Max per stack = HCF(96, 240, 336). $96 = 2^5\times3$, $240 = 2^4\times3\times5$, $336 = 2^4\times3\times7$.
2. Common at lowest powers: $2^4\times3 = 48$.
3. Stacks $= 96/48 + 240/48 + 336/48 = 2 + 5 + 7 = 14$.

Final Answer: 48 books per stack; 14 stacks.

Takeaway: Equal-height stacks → HCF; sum the per-subject stack counts.

Example 25: Prove $\dfrac{\sqrt{3}}{5}$ irrational

Solution:

1. Assume $\dfrac{\sqrt{3}}{5} = r$ (rational).
2. Then $\sqrt{3} = 5r$, rational.
3. Contradicts $\sqrt{3}$ irrational.

Final Answer: $\dfrac{\sqrt{3}}{5}$ is irrational.

Takeaway: Irrational divided by a non-zero rational is still irrational.

Example 26: Smallest number leaving a remainder

Find the least number which when divided by 6, 15 and 18 leaves remainder 5 in each case.

Solution:

1. Required = LCM(6, 15, 18) + 5.
2. $6 = 2\times3$, $15 = 3\times5$, $18 = 2\times3^2$. LCM $= 2\times3^2\times5 = 90$.
3. Required $= 90 + 5 = 95$.

Final Answer: 95.

Takeaway: Same remainder on several divisors → LCM + remainder.

Example 27: Identify terminating among many

Which of $\dfrac{13}{125}$, $\dfrac{17}{210}$, $\dfrac{15}{1600}$ are terminating?

Solution:

1. $125 = 5^3$ → T.
2. $210 = 2\times3\times5\times7$ → contains 3, 7 → NT.
3. $1600 = 2^6\times5^2$ → T.

Final Answer: $\dfrac{13}{125}$ and $\dfrac{15}{1600}$ terminate; $\dfrac{17}{210}$ does not.

Takeaway: Factorise each denominator and scan for stray primes.

Example 28: HCF and LCM relation to find LCM

The HCF of 96 and 404 is 4. Find their LCM.

Solution:

1. $\text{LCM} = \dfrac{a \times b}{\text{HCF}} = \dfrac{96 \times 404}{4}$.
2. $= \dfrac{38784}{4} = 9696$.

Final Answer: LCM $= 9696$.

Takeaway: Rearranging the product relation gives the LCM quickly.

Example 29: Prove $7 - 3\sqrt{2}$ irrational

Solution:

1. Assume $7 - 3\sqrt{2} = r$ (rational).
2. $3\sqrt{2} = 7 - r \Rightarrow \sqrt{2} = \dfrac{7-r}{3}$, rational.
3. Contradicts $\sqrt{2}$ irrational.

Final Answer: $7 - 3\sqrt{2}$ is irrational.

Takeaway: Isolate $\sqrt2$ by subtracting and dividing; rationality of the result gives the contradiction.

Example 30: Mixed multi-step problem

The HCF of two numbers is 16 and their product is 3072. Find their LCM.

Solution:

1. $\text{HCF} \times \text{LCM} =$ product of numbers $= 3072$.
2. $\text{LCM} = \dfrac{3072}{16} = 192$.

Final Answer: LCM $= 192$.

Takeaway: Product of the numbers ÷ HCF = LCM.

Example 31: Greatest 3-digit number divisible by 8, 10, 12

Find the greatest 3-digit number exactly divisible by 8, 10 and 12.

Solution:

1. LCM(8, 10, 12): $8 = 2^3$, $10 = 2\times5$, $12 = 2^2\times3$. LCM $= 2^3\times3\times5 = 120$.
2. Greatest 3-digit multiple of 120: $120 \times 8 = 960$ (since $120\times9 = 1080 > 999$).

Final Answer: 960.

Takeaway: Find the LCM, then the largest multiple within the required range.