CBSE Class 10 · Maths
CBSE Class 10 Maths Real Numbers Previous Year Questions
Help your child master CBSE Class 10 Maths Real Numbers with this comprehensive collection of previous year questions and step-by-step solutions. These questions are carefully sourced from real CBSE board exams to give your child the most authentic exam practice. Each question comes with a detailed explanation to build deep understanding and scoring confidence.
CBSE Class 10 Maths Real Numbers — Previous Year Questions with Solutions
The exponent of 5 in the prime factorisation of 3750 is?
Solution
To find the exponent of 5 in the prime factorization of 3750, follow the steps:
Step 1: Prime factorization of 3750
Start dividing 3750 by the smallest prime numbers:
• 3750 ÷ 2 = 1875 (since 3750 is even, divide by 2).
• 1875 ÷ 3 = 625 (digit sum of 1875 is 21, divisible by 3).
• 625 ÷ 5 = 125, 125 ÷ 5 = 25, 25 ÷ 5 = 5, 5 ÷ 5 = 1.
Thus: \[ 3750 = 2 \times 3 \times 5^4 \] Step 2: Exponent of 5
From the factorization $3750 = 2 \times 3 \times 5^4$, the exponent of 5 is 4.
Step 1: Prime factorization of 3750
Start dividing 3750 by the smallest prime numbers:
• 3750 ÷ 2 = 1875 (since 3750 is even, divide by 2).
• 1875 ÷ 3 = 625 (digit sum of 1875 is 21, divisible by 3).
• 625 ÷ 5 = 125, 125 ÷ 5 = 25, 25 ÷ 5 = 5, 5 ÷ 5 = 1.
Thus: \[ 3750 = 2 \times 3 \times 5^4 \] Step 2: Exponent of 5
From the factorization $3750 = 2 \times 3 \times 5^4$, the exponent of 5 is 4.
The HCF and the LCM of 12, 21, 15 respectively are?
Solution
Step 1: Prime Factorization
$12 = 2^2 \times 3$, $21 = 3 \times 7$, $15 = 3 \times 5$
Step 2: HCF
The only common prime factor across all three numbers is $3$.
So, HCF = 3.
Step 3: LCM
Take the highest powers of all prime factors involved:
$\text{LCM} = 2^2 \times 3 \times 5 \times 7 = 4 \times 3 \times 5 \times 7 = 420$
Final Answer: HCF = 3, LCM = 420.
$12 = 2^2 \times 3$, $21 = 3 \times 7$, $15 = 3 \times 5$
Step 2: HCF
The only common prime factor across all three numbers is $3$.
So, HCF = 3.
Step 3: LCM
Take the highest powers of all prime factors involved:
$\text{LCM} = 2^2 \times 3 \times 5 \times 7 = 4 \times 3 \times 5 \times 7 = 420$
Final Answer: HCF = 3, LCM = 420.
The LCM of the smallest two-digit composite number and the smallest composite number is:
Solution
Step 1: Smallest two-digit composite number = 10 (divisible by 1, 2, 5, 10).
Step 2: Smallest composite number = 4 (divisible by 1, 2, 4).
Step 3: LCM of 10 and 4
$10 = 2 \times 5$, $4 = 2^2$
$\text{LCM} = 2^2 \times 5 = 4 \times 5 = \mathbf{20}$
Step 2: Smallest composite number = 4 (divisible by 1, 2, 4).
Step 3: LCM of 10 and 4
$10 = 2 \times 5$, $4 = 2^2$
$\text{LCM} = 2^2 \times 5 = 4 \times 5 = \mathbf{20}$
The LCM of two numbers is 2400. Which of the following cannot be their HCF?
Solution
The HCF must always be a divisor of the LCM.
Factorize: $2400 = 2^6 \times 3 \times 5^2$
Check each option:
• $300 = 2^2 \times 3 \times 5^2$ → divides 2400 ✓
• $400 = 2^4 \times 5^2$ → divides 2400 ✓
• $500 = 2^2 \times 5^3$ → has $5^3$ but 2400 only has $5^2$ → does NOT divide 2400 ✗
• $600 = 2^3 \times 3 \times 5^2$ → divides 2400 ✓
Final Answer: HCF cannot be 500.
Factorize: $2400 = 2^6 \times 3 \times 5^2$
Check each option:
• $300 = 2^2 \times 3 \times 5^2$ → divides 2400 ✓
• $400 = 2^4 \times 5^2$ → divides 2400 ✓
• $500 = 2^2 \times 5^3$ → has $5^3$ but 2400 only has $5^2$ → does NOT divide 2400 ✗
• $600 = 2^3 \times 3 \times 5^2$ → divides 2400 ✓
Final Answer: HCF cannot be 500.
The greatest number which divides 1251, 9377, and 15628 leaving remainders 1, 2, and 3 respectively is:
Solution
Subtract each remainder from the respective number:
$1251 – 1 = 1250$, $9377 – 2 = 9375$, $15628 – 3 = 15625$
The required number must divide all three results, so find GCD(1250, 9375, 15625).
Prime factorizations:
$1250 = 2 \times 5^4$, $9375 = 3 \times 5^5$, $15625 = 5^6$
Common factor: $5^4 = 625$
Checking options: 625 divides all three numbers evenly.
Final Answer: 625
$1251 – 1 = 1250$, $9377 – 2 = 9375$, $15628 – 3 = 15625$
The required number must divide all three results, so find GCD(1250, 9375, 15625).
Prime factorizations:
$1250 = 2 \times 5^4$, $9375 = 3 \times 5^5$, $15625 = 5^6$
Common factor: $5^4 = 625$
Checking options: 625 divides all three numbers evenly.
Final Answer: 625
Three alarm clocks ring their alarms at regular intervals of 20 min, 25 min, and 30 min respectively. If they first beep together at 12 noon, at what time will they beep again for the first time?
Solution
Find LCM(20, 25, 30):
$20 = 2^2 \times 5$, $25 = 5^2$, $30 = 2 \times 3 \times 5$
$\text{LCM} = 2^2 \times 3 \times 5^2 = 4 \times 3 \times 25 = 300$ minutes
$300 \text{ min} = 5 \text{ hours}$
$12:00 \text{ noon} + 5 \text{ hours} = \mathbf{5:00 \text{ PM}}$
$20 = 2^2 \times 5$, $25 = 5^2$, $30 = 2 \times 3 \times 5$
$\text{LCM} = 2^2 \times 3 \times 5^2 = 4 \times 3 \times 25 = 300$ minutes
$300 \text{ min} = 5 \text{ hours}$
$12:00 \text{ noon} + 5 \text{ hours} = \mathbf{5:00 \text{ PM}}$
The ratio of LCM and HCF of the least composite and the least prime numbers is:
Solution
Least prime number = 2; Least composite number = 4.
$4 = 2^2$, $2 = 2^1$
$\text{HCF}(4, 2) = 2^1 = 2$
$\text{LCM}(4, 2) = 2^2 = 4$
$\dfrac{\text{LCM}}{\text{HCF}} = \dfrac{4}{2} = 2$
Ratio = 2:1
$4 = 2^2$, $2 = 2^1$
$\text{HCF}(4, 2) = 2^1 = 2$
$\text{LCM}(4, 2) = 2^2 = 4$
$\dfrac{\text{LCM}}{\text{HCF}} = \dfrac{4}{2} = 2$
Ratio = 2:1
The exponent of 5 in the prime factorisation of 3750 is:
Solution
Divide 3750 by 5 repeatedly:
$3750 \div 5 = 750$, $750 \div 5 = 150$, $150 \div 5 = 30$, $30 \div 5 = 6$
Now $6 = 2 \times 3$, so the complete factorization is: $$3750 = 5^4 \times 2 \times 3$$ From the factorization $3750 = 2 \times 3 \times 5^4$, the exponent of 5 is 4.
$3750 \div 5 = 750$, $750 \div 5 = 150$, $150 \div 5 = 30$, $30 \div 5 = 6$
Now $6 = 2 \times 3$, so the complete factorization is: $$3750 = 5^4 \times 2 \times 3$$ From the factorization $3750 = 2 \times 3 \times 5^4$, the exponent of 5 is 4.
The prime factorisation of 3825 is:
Solution
Divide step by step:
$3825 \div 3 = 1275$ (digit sum = 18, divisible by 3)
$1275 \div 3 = 425$ (digit sum = 15, divisible by 3)
$425 \div 5 = 85$ (ends in 5)
$85 \div 5 = 17$ (ends in 5)
17 is prime — stop here.
$$3825 = 3^2 \times 5^2 \times 17$$
$3825 \div 3 = 1275$ (digit sum = 18, divisible by 3)
$1275 \div 3 = 425$ (digit sum = 15, divisible by 3)
$425 \div 5 = 85$ (ends in 5)
$85 \div 5 = 17$ (ends in 5)
17 is prime — stop here.
$$3825 = 3^2 \times 5^2 \times 17$$
The sum of exponents of prime factors in the prime factorisation of 196 is?
Solution
$196 \div 2 = 98$, $98 \div 2 = 49$, $49 \div 7 = 7$, $7 \div 7 = 1$
$$196 = 2^2 \times 7^2$$ Exponents: 2 (for prime 2) and 2 (for prime 7).
Sum of exponents $= 2 + 2 = \mathbf{4}$
$$196 = 2^2 \times 7^2$$ Exponents: 2 (for prime 2) and 2 (for prime 7).
Sum of exponents $= 2 + 2 = \mathbf{4}$
HCF (132, 77) is:
Solution
Prime factorization of 132:
$132 \div 2 = 66$, $66 \div 2 = 33$, $33 \div 3 = 11$
$132 = 2^2 \times 3 \times 11$
Prime factorization of 77:
$77 \div 7 = 11$ → $77 = 7 \times 11$
Common prime factor: 11
$$\text{HCF}(132, 77) = 11$$
$132 \div 2 = 66$, $66 \div 2 = 33$, $33 \div 3 = 11$
$132 = 2^2 \times 3 \times 11$
Prime factorization of 77:
$77 \div 7 = 11$ → $77 = 7 \times 11$
Common prime factor: 11
$$\text{HCF}(132, 77) = 11$$
The HCF of smallest 2-digit number and the smallest composite number is:
Solution
Smallest 2-digit number = 10
Smallest composite number = 4 (since $4 = 2 \times 2$)
Factors of 10: 1, 2, 5, 10
Factors of 4: 1, 2, 4
Common factors: 1, 2 → HCF = 2
Smallest composite number = 4 (since $4 = 2 \times 2$)
Factors of 10: 1, 2, 5, 10
Factors of 4: 1, 2, 4
Common factors: 1, 2 → HCF = 2
HCF × LCM for the numbers 40 and 30 is:
Solution
Using the fundamental property: $\text{HCF} \times \text{LCM} = \text{Product of the two numbers}$
$40 \times 30 = 1200$
Verification: HCF(40, 30) = 10; LCM(40, 30) = 120
$10 \times 120 = 1200$ ✓
Answer: 1200
$40 \times 30 = 1200$
Verification: HCF(40, 30) = 10; LCM(40, 30) = 120
$10 \times 120 = 1200$ ✓
Answer: 1200
If $$ p^2 = \frac{32}{50} $$, then $ p $ is a/an:
Solution
Step 1: Simplify the fraction
$$p^2 = \frac{32}{50} = \frac{16}{25}$$
Step 2: Take the square root
$$p = \pm\sqrt{\frac{16}{25}} = \pm\frac{4}{5}$$
Since $\frac{4}{5}$ and $-\frac{4}{5}$ can both be expressed as a fraction of two integers, $ p $ is a rational number.
$$p^2 = \frac{32}{50} = \frac{16}{25}$$
Step 2: Take the square root
$$p = \pm\sqrt{\frac{16}{25}} = \pm\frac{4}{5}$$
Since $\frac{4}{5}$ and $-\frac{4}{5}$ can both be expressed as a fraction of two integers, $ p $ is a rational number.
The exponent of 5 in the prime factorisation of 3750 is?
Solution
Divide 3750 by 5 repeatedly:
$$3750 \div 5 = 750$$ $$750 \div 5 = 150$$ $$150 \div 5 = 30$$ $$30 \div 5 = 6$$ Now $6 = 2 \times 3$, so: $$3750 = 5^4 \times 2 \times 3$$ The exponent of 5 in the prime factorization of 3750 is 4.
$$3750 \div 5 = 750$$ $$750 \div 5 = 150$$ $$150 \div 5 = 30$$ $$30 \div 5 = 6$$ Now $6 = 2 \times 3$, so: $$3750 = 5^4 \times 2 \times 3$$ The exponent of 5 in the prime factorization of 3750 is 4.
If xy = 180 and HCF (x, y) = 3, then find the LCM (x, y)?
Answer
We know the fundamental theorem:
\[\text{HCF}(x, y) \times \text{LCM}(x, y) = x \times y\] Given: $xy = 180$ and $\text{HCF}(x, y) = 3$
Substituting:
\[3 \times \text{LCM}(x, y) = 180\] \[\text{LCM}(x, y) = \frac{180}{3} = \boxed{60}\]
\[\text{HCF}(x, y) \times \text{LCM}(x, y) = x \times y\] Given: $xy = 180$ and $\text{HCF}(x, y) = 3$
Substituting:
\[3 \times \text{LCM}(x, y) = 180\] \[\text{LCM}(x, y) = \frac{180}{3} = \boxed{60}\]
The LCM of two numbers is 182 and their HCF is 13. If one of the numbers is 26, find the other?
Answer
Product of two numbers = LCM × HCF
Let the other number be $x$:
$$x \times 26 = 182 \times 13$$ $$x = \frac{182 \times 13}{26} = \frac{2366}{26} = \boxed{91}$$ The other number is 91.
Let the other number be $x$:
$$x \times 26 = 182 \times 13$$ $$x = \frac{182 \times 13}{26} = \frac{2366}{26} = \boxed{91}$$ The other number is 91.
Find the sum of exponents of prime factors in the prime factorisation of 196.
Answer
Prime factorisation of 196:
$$196 = 14 \times 14 = (2 \times 7) \times (2 \times 7) = 2^2 \times 7^2$$
Exponents: 2 (for prime 2) and 2 (for prime 7)
$$\text{Sum of exponents} = 2 + 2 = \boxed{4}$$
$$\text{Sum of exponents} = 2 + 2 = \boxed{4}$$
Find the LCM of smallest two-digit composite number and smallest composite number?
Answer
Smallest composite number = 4, Smallest two-digit composite number = 10
Prime factorisation: $$4 = 2^2, \quad 10 = 2 \times 5$$ LCM includes highest powers of all primes:
Highest power of 2 → $2^2$; Highest power of 5 → $5^1$ $$\text{LCM} = 2^2 \times 5 = 4 \times 5 = \boxed{20}$$
Prime factorisation: $$4 = 2^2, \quad 10 = 2 \times 5$$ LCM includes highest powers of all primes:
Highest power of 2 → $2^2$; Highest power of 5 → $5^1$ $$\text{LCM} = 2^2 \times 5 = 4 \times 5 = \boxed{20}$$
What is the HCF of smallest prime number and the smallest composite number?
Answer
Smallest prime number = 2; Smallest composite number = 4
Factors of 2: 1, 2
Factors of 4: 1, 2, 4
Common factors = 1, 2 → Greatest common factor = 2 $$\boxed{2}$$
Factors of 2: 1, 2
Factors of 4: 1, 2, 4
Common factors = 1, 2 → Greatest common factor = 2 $$\boxed{2}$$
A forester wants to plant 66 apple trees, 88 banana trees, and 110 mango trees in equal rows (in terms of number of trees). Also, he wants to make distinct rows of the trees (only one type of tree in one row). Find the minimum number of rows required?
Answer
To minimise rows, each row must have the maximum equal number of trees → find HCF(66, 88, 110).
Prime factorisation:
$66 = 2 \times 3 \times 11$
$88 = 2^3 \times 11$
$110 = 2 \times 5 \times 11$
Common factors: $2^1 \times 11^1 = 22$
$$\text{HCF} = 22 \text{ trees per row}$$ Number of rows:
Apple: $66 \div 22 = 3$ rows
Banana: $88 \div 22 = 4$ rows
Mango: $110 \div 22 = 5$ rows
Total: $3 + 4 + 5 = \boxed{12 \text{ rows}}$
Prime factorisation:
$66 = 2 \times 3 \times 11$
$88 = 2^3 \times 11$
$110 = 2 \times 5 \times 11$
Common factors: $2^1 \times 11^1 = 22$
$$\text{HCF} = 22 \text{ trees per row}$$ Number of rows:
Apple: $66 \div 22 = 3$ rows
Banana: $88 \div 22 = 4$ rows
Mango: $110 \div 22 = 5$ rows
Total: $3 + 4 + 5 = \boxed{12 \text{ rows}}$
3 bells ring at an interval of 4, 7, and 14 minutes. All three bells rang at 6 am. When will the three bells ring together next?
Answer
The three bells will next ring together at the LCM of their individual intervals.
$4 = 2 \times 2$, $7 = 7 \times 1$, $14 = 2 \times 7$
$$\text{LCM} = 2 \times 2 \times 7 = 28 \text{ minutes}$$ The three bells will ring together again at $6:28 \text{ am}$.
$4 = 2 \times 2$, $7 = 7 \times 1$, $14 = 2 \times 7$
$$\text{LCM} = 2 \times 2 \times 7 = 28 \text{ minutes}$$ The three bells will ring together again at $6:28 \text{ am}$.
Write the smallest number which is divisible by both 306 and 657.
Answer
The smallest number divisible by both = LCM(306, 657)
Prime factorisation:
$306 = 2 \times 3^2 \times 17$
$657 = 3^2 \times 73$
LCM includes highest powers of all primes:
$\text{LCM} = 2^1 \times 3^2 \times 17 \times 73$
Calculating step by step:
$3^2 = 9$, $9 \times 2 = 18$, $18 \times 17 = 306$, $306 \times 73 = 22338$ $$\boxed{22338}$$
Prime factorisation:
$306 = 2 \times 3^2 \times 17$
$657 = 3^2 \times 73$
LCM includes highest powers of all primes:
$\text{LCM} = 2^1 \times 3^2 \times 17 \times 73$
Calculating step by step:
$3^2 = 9$, $9 \times 2 = 18$, $18 \times 17 = 306$, $306 \times 73 = 22338$ $$\boxed{22338}$$
Find the HCF and LCM of 90 and 144 by the method of prime factorisation.
Answer
Prime factorisation of 90:
$90 \div 2 = 45$, $45 \div 3 = 15$, $15 \div 3 = 5$, 5 is prime.
$$90 = 2 \times 3^2 \times 5$$ Prime factorisation of 144:
$144 \div 2 = 72$, $72 \div 2 = 36$, $36 \div 2 = 18$, $18 \div 2 = 9$, $9 \div 3 = 3$, $3 \div 3 = 1$
$$144 = 2^4 \times 3^2$$ HCF (lowest powers of common primes):
Common primes: 2 and 3
$$\text{HCF} = 2^1 \times 3^2 = 2 \times 9 = \mathbf{18}$$ LCM (highest powers of all primes):
$$\text{LCM} = 2^4 \times 3^2 \times 5 = 16 \times 9 \times 5 = \mathbf{720}$$
$90 \div 2 = 45$, $45 \div 3 = 15$, $15 \div 3 = 5$, 5 is prime.
$$90 = 2 \times 3^2 \times 5$$ Prime factorisation of 144:
$144 \div 2 = 72$, $72 \div 2 = 36$, $36 \div 2 = 18$, $18 \div 2 = 9$, $9 \div 3 = 3$, $3 \div 3 = 1$
$$144 = 2^4 \times 3^2$$ HCF (lowest powers of common primes):
Common primes: 2 and 3
$$\text{HCF} = 2^1 \times 3^2 = 2 \times 9 = \mathbf{18}$$ LCM (highest powers of all primes):
$$\text{LCM} = 2^4 \times 3^2 \times 5 = 16 \times 9 \times 5 = \mathbf{720}$$
Prove that $ 2 – \sqrt{3} $ is irrational, given that $\sqrt{3}$ is an irrational number.
Answer
Let $ 2 – \sqrt{3} $ be a rational number.
Then we can find co-prime integers $ a $ and $ b $ (where $ b \neq 0 $) such that: $$2 – \sqrt{3} = \frac{a}{b}$$ Rearranging: $$2 – \frac{a}{b} = \sqrt{3} \quad \Rightarrow \quad \frac{2b – a}{b} = \sqrt{3}$$ Since $ a $ and $ b $ are integers, $\frac{2b – a}{b}$ is a rational number.
But $\sqrt{3}$ is an irrational number. A rational number cannot equal an irrational number — this is a contradiction.
Therefore, our assumption is false, and $ 2 – \sqrt{3} $ is irrational. Hence Proved.
Then we can find co-prime integers $ a $ and $ b $ (where $ b \neq 0 $) such that: $$2 – \sqrt{3} = \frac{a}{b}$$ Rearranging: $$2 – \frac{a}{b} = \sqrt{3} \quad \Rightarrow \quad \frac{2b – a}{b} = \sqrt{3}$$ Since $ a $ and $ b $ are integers, $\frac{2b – a}{b}$ is a rational number.
But $\sqrt{3}$ is an irrational number. A rational number cannot equal an irrational number — this is a contradiction.
Therefore, our assumption is false, and $ 2 – \sqrt{3} $ is irrational. Hence Proved.
Prove that $ \dfrac{2 + \sqrt{3}}{5} $ is an irrational number, given that $ \sqrt{3} $ is an irrational number.
Answer
Let $\dfrac{2 + \sqrt{3}}{5}$ be a rational number.
Therefore, we can write it in the form $\dfrac{p}{q}$, where $p$ and $q$ are integers: $$\frac{2 + \sqrt{3}}{5} = \frac{p}{q} \quad \Rightarrow \quad 2 + \sqrt{3} = \frac{5p}{q}$$ Rearranging: $$\sqrt{3} = \frac{5p}{q} – 2 = \frac{5p – 2q}{q}$$ Since $p$ and $q$ are co-prime integers, $\dfrac{5p – 2q}{q}$ is a rational number.
But this contradicts the fact that $\sqrt{3}$ is an irrational number.
So our assumption is wrong. Therefore, $\dfrac{2 + \sqrt{3}}{5}$ is an irrational number. Hence Proved.
Therefore, we can write it in the form $\dfrac{p}{q}$, where $p$ and $q$ are integers: $$\frac{2 + \sqrt{3}}{5} = \frac{p}{q} \quad \Rightarrow \quad 2 + \sqrt{3} = \frac{5p}{q}$$ Rearranging: $$\sqrt{3} = \frac{5p}{q} – 2 = \frac{5p – 2q}{q}$$ Since $p$ and $q$ are co-prime integers, $\dfrac{5p – 2q}{q}$ is a rational number.
But this contradicts the fact that $\sqrt{3}$ is an irrational number.
So our assumption is wrong. Therefore, $\dfrac{2 + \sqrt{3}}{5}$ is an irrational number. Hence Proved.
Prove that $\sqrt{5}$ is an irrational number.
Answer
Let $\sqrt{5}$ be a rational number.
Then $\sqrt{5} = \dfrac{p}{q}$, where $p$ and $q$ are co-prime integers and $q \neq 0$.
Squaring both sides: $$5 = \frac{p^2}{q^2} \quad \Rightarrow \quad p^2 = 5q^2$$ Thus $p^2$ is divisible by 5, and therefore $p$ is divisible by 5.
Let $p = 5r$ for some positive integer $r$. Then: $$p^2 = 25r^2 \quad \Rightarrow \quad 5q^2 = 25r^2 \quad \Rightarrow \quad q^2 = 5r^2$$ Thus $q^2$ is divisible by 5, and therefore $q$ is divisible by 5.
Now both $p$ and $q$ are divisible by 5, which contradicts the fact that $p$ and $q$ are co-primes.
Hence our assumption is false. Therefore, $\sqrt{5}$ is an irrational number. Hence Proved.
Then $\sqrt{5} = \dfrac{p}{q}$, where $p$ and $q$ are co-prime integers and $q \neq 0$.
Squaring both sides: $$5 = \frac{p^2}{q^2} \quad \Rightarrow \quad p^2 = 5q^2$$ Thus $p^2$ is divisible by 5, and therefore $p$ is divisible by 5.
Let $p = 5r$ for some positive integer $r$. Then: $$p^2 = 25r^2 \quad \Rightarrow \quad 5q^2 = 25r^2 \quad \Rightarrow \quad q^2 = 5r^2$$ Thus $q^2$ is divisible by 5, and therefore $q$ is divisible by 5.
Now both $p$ and $q$ are divisible by 5, which contradicts the fact that $p$ and $q$ are co-primes.
Hence our assumption is false. Therefore, $\sqrt{5}$ is an irrational number. Hence Proved.
If HCF of 144 and 180 is expressed in the form 13m − 16, find the value of m.
Answer
Prime factors of $180 = 2^2 \times 3^2 \times 5$
Prime factors of $144 = 2^4 \times 3^2$
$$\text{HCF}(180, 144) = 2^2 \times 3^2 = 4 \times 9 = 36$$ Given: $36 = 13m – 16$
Solving: $$36 + 16 = 13m \quad \Rightarrow \quad 52 = 13m \quad \Rightarrow \quad m = \frac{52}{13} = \boxed{4}$$
Prime factors of $144 = 2^4 \times 3^2$
$$\text{HCF}(180, 144) = 2^2 \times 3^2 = 4 \times 9 = 36$$ Given: $36 = 13m – 16$
Solving: $$36 + 16 = 13m \quad \Rightarrow \quad 52 = 13m \quad \Rightarrow \quad m = \frac{52}{13} = \boxed{4}$$
Case Study: Seminar Seating Arrangement
Read the following and answer any four questions from Q1 to Q5.
A seminar is being conducted by an Educational Organisation, where the participants will be educators of different subjects. The number of participants in Hindi, English and Mathematics are 60, 84 and 108 respectively.
Q1. In each room, the same number of participants are to be seated and all of them being in the same subject, hence maximum number of participants that can be accommodated in each room is:
(A) 14 (B) 12 (C) 16 (D) 18
Q2. What is the minimum number of rooms required during the event?
(A) 11 (B) 31 (C) 41 (D) 21
Q3. The LCM of 60, 84 and 108 is:
(A) 3780 (B) 3680 (C) 4780 (D) 4680
Q4. The product of HCF and LCM of 60, 84 and 108 is:
(A) 55360 (B) 35360 (C) 45500 (D) 45360
Q5. 108 can be expressed as a product of its primes as:
(A) $2^3 \times 3^2$ (B) $2^3 \times 3^3$ (C) $2^2 \times 3^2$ (D) $2^2 \times 3^3$
A seminar is being conducted by an Educational Organisation, where the participants will be educators of different subjects. The number of participants in Hindi, English and Mathematics are 60, 84 and 108 respectively.
Q1. In each room, the same number of participants are to be seated and all of them being in the same subject, hence maximum number of participants that can be accommodated in each room is:
(A) 14 (B) 12 (C) 16 (D) 18
Q2. What is the minimum number of rooms required during the event?
(A) 11 (B) 31 (C) 41 (D) 21
Q3. The LCM of 60, 84 and 108 is:
(A) 3780 (B) 3680 (C) 4780 (D) 4680
Q4. The product of HCF and LCM of 60, 84 and 108 is:
(A) 55360 (B) 35360 (C) 45500 (D) 45360
Q5. 108 can be expressed as a product of its primes as:
(A) $2^3 \times 3^2$ (B) $2^3 \times 3^3$ (C) $2^2 \times 3^2$ (D) $2^2 \times 3^3$
Answer
Ans. Q1 → Option (B) 12
The maximum number of participants per room = HCF(60, 84, 108).
$60 = 2^2 \times 3 \times 5$, $84 = 2^2 \times 3 \times 7$, $108 = 2^2 \times 3^3$
$\text{HCF} = 2^2 \times 3 = \mathbf{12}$
Ans. Q2 → Option (D) 21
Total participants = $60 + 84 + 108 = 252$
Minimum rooms = $\dfrac{252}{12} = \mathbf{21}$
Ans. Q3 → Option (A) 3780
$\text{LCM}(60, 84, 108) = 2^2 \times 3^3 \times 5 \times 7 = 4 \times 27 \times 35 = \mathbf{3780}$
Ans. Q4 → Option (D) 45360
$\text{HCF} = 12$, $\text{LCM} = 3780$
Product $= 12 \times 3780 = \mathbf{45360}$
Ans. Q5 → Option (D) $2^2 \times 3^3$
$108 = 2 \times 2 \times 3 \times 3 \times 3 = 2^2 \times 3^3$
The maximum number of participants per room = HCF(60, 84, 108).
$60 = 2^2 \times 3 \times 5$, $84 = 2^2 \times 3 \times 7$, $108 = 2^2 \times 3^3$
$\text{HCF} = 2^2 \times 3 = \mathbf{12}$
Ans. Q2 → Option (D) 21
Total participants = $60 + 84 + 108 = 252$
Minimum rooms = $\dfrac{252}{12} = \mathbf{21}$
Ans. Q3 → Option (A) 3780
$\text{LCM}(60, 84, 108) = 2^2 \times 3^3 \times 5 \times 7 = 4 \times 27 \times 35 = \mathbf{3780}$
Ans. Q4 → Option (D) 45360
$\text{HCF} = 12$, $\text{LCM} = 3780$
Product $= 12 \times 3780 = \mathbf{45360}$
Ans. Q5 → Option (D) $2^2 \times 3^3$
$108 = 2 \times 2 \times 3 \times 3 \times 3 = 2^2 \times 3^3$
Frequently Asked Questions
What is Real Numbers about in CBSE Class 10 Maths? ⌄
Real Numbers is the first chapter of CBSE Class 10 Maths and covers the Fundamental Theorem of Arithmetic — which states that every composite number can be expressed as a product of primes in a unique way. Your child will also learn about HCF (Highest Common Factor), LCM (Least Common Multiple), and how to prove that numbers like $\sqrt{2}$, $\sqrt{3}$, and $\sqrt{5}$ are irrational.
How many marks does Real Numbers carry in the CBSE Class 10 board exam? ⌄
Real Numbers falls under the Number Systems unit, which carries approximately 6 marks in the CBSE Class 10 board examination. Questions from this chapter typically appear as 1-mark MCQs, 2-mark short answers, and occasionally as part of case study questions, making it important for your child to have a strong conceptual grip on HCF, LCM, and irrationality proofs.
What are the most important topics in Real Numbers for board exams? ⌄
The most exam-critical topics are: (1) Prime factorisation and finding HCF and LCM using the method of prime factorisation, (2) The relationship HCF × LCM = Product of two numbers, (3) Proving that numbers such as $\sqrt{2}$, $\sqrt{3}$, $\sqrt{5}$, and expressions like $2 – \sqrt{3}$ are irrational, and (4) Applications of LCM in real-life problems such as bells ringing or lights blinking together.
What are common mistakes students make in Real Numbers? ⌄
One of the most frequent errors is confusing HCF and LCM — for example, using LCM where HCF is required in a problem and vice versa. Students also lose marks by writing incomplete proofs when showing a number is irrational; every step including the contradiction must be stated clearly. Another common slip is forgetting that the HCF of three numbers must divide all three numbers, not just two of them.
How does Angle Belearn help students master Real Numbers? ⌄
Angle Belearn’s CBSE specialists have curated a rich bank of previous year and competency-based questions specifically for Class 10 Real Numbers, each accompanied by detailed step-by-step solutions. Our structured approach helps students understand the reasoning behind every answer — not just memorise steps — so they can handle any variation the CBSE board throws at them with confidence.
