Real Numbers – Class 10 Mathematics
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Define: Natural Numbers, Whole Numbers, Integers, Rational Numbers, Irrational Numbers, Real Numbers.
- Natural Numbers (N): 1, 2, 3, 4, 5, …
- Whole Numbers (W): 0, 1, 2, 3, 4, …
- Integers (Z or I): …, -3, -2, -1, 0, 1, 2, 3, …
- Rational Numbers (Q): Numbers expressible as \(\frac{p}{q}\) where p and q are integers, q ≠ 0.
- Irrational Numbers: Numbers that cannot be expressed as \(\frac{p}{q}\). Examples: \(\sqrt{2}, \sqrt{3}, \pi\).
- Real Numbers (R): The union of Rational and Irrational Numbers.
What is a prime number? Give examples. Is 1 a prime number?
A prime number is a positive integer greater than 1 that has exactly two distinct positive divisors: 1 and itself.
Examples: 2 (divisors: 1, 2), 3 (1, 3), 5 (1, 5), 7 (1, 7), 11 (1, 11), 13 (1, 13).
1 is NOT a prime number because it has only one divisor (1).
What is a composite number? Give the smallest composite number with example.
A composite number is a positive integer that has more than two distinct positive divisors.
Smallest composite number is 4 (divisors: 1, 2, 4).
Other examples: 6 (1, 2, 3, 6), 8 (1, 2, 4, 8).
State the Fundamental Theorem of Arithmetic.
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 in which the prime factors occur.
Example: \(60 = 2^2 \times 3 \times 5\) (unique prime factorisation).
If \(7560 = 2^3 \times 3^p \times q \times 7\), find \(p\) and \(q\).
Given: \(7560 = 2^3 \times 3^p \times q \times 7\).
First, prime factorise 7560:
\[7560 = 2^3 \times 3^3 \times 5 \times 7\]
Comparing with given form: \(p = 3\), \(q = 5\).
Find HCF and LCM of 404 and 96. Verify HCF × LCM = Product of the two numbers.
Prime Factorisation:
\(404 = 2^2 \times 101\)
\(96 = 2^5 \times 3\)
HCF = Product of smallest powers of common primes = \(2^2 = 4\)
LCM = Product of greatest powers of all primes = \(2^5 \times 3 \times 101 = 9696\)
Verification: \(HCF \times LCM = 4 \times 9696 = 38784\)
Product of numbers = \(404 \times 96 = 38784\). Verified.
On a morning walk, three persons step off together. Their steps measure 40 cm, 42 cm, and 45 cm respectively. What is the minimum distance each should walk so that each can cover the same distance in complete steps?
We need the LCM of 40, 42, and 45 to find the minimum common distance.
\(40 = 2^3 \times 5\)
\(42 = 2 \times 3 \times 7\)
\(45 = 3^2 \times 5\)
\(LCM = 2^3 \times 3^2 \times 5 \times 7 = 8 \times 9 \times 5 \times 7 = 2520 \text{ cm}\)
Each should walk 2520 cm (or 25.2 m).
During a sale, colour pencils are sold in packs of 24 each and crayons in packs of 32 each. If you want full packs of both and the same number of pencils and crayons, how many packets of each would you need to buy?
We need the LCM of 24 and 32 to find the smallest common number of items.
\(24 = 2^3 \times 3\)
\(32 = 2^5\)
\(LCM = 2^5 \times 3 = 32 \times 3 = 96\)
Number of pencil packets = \(\frac{96}{24} = 4\)
Number of crayon packets = \(\frac{96}{32} = 3\)
Buy 4 packets of pencils and 3 packets of crayons.
Explain how to find HCF using the prime factorisation method.
HCF (Highest Common Factor) using prime factorisation:
- Express each number as a product of prime factors.
- Identify all common prime factors.
- Take the smallest power of each common prime factor.
- Multiply them together to get the HCF.
Example: For 12 (\(2^2 \times 3\)) and 18 (\(2 \times 3^2\)), common primes: 2 and 3. Smallest powers: \(2^1\) and \(3^1\). HCF = \(2 \times 3 = 6\).
Explain how to find LCM using the prime factorisation method.
LCM (Least Common Multiple) using prime factorisation:
- Express each number as a product of prime factors.
- Identify all prime factors present in any of the numbers.
- Take the greatest power of each prime factor.
- Multiply them together to get the LCM.
Example: For 12 (\(2^2 \times 3\)) and 18 (\(2 \times 3^2\)), primes: 2 and 3. Greatest powers: \(2^2\) and \(3^2\). LCM = \(4 \times 9 = 36\).
Two tankers contain 850 litres and 680 litres of petrol respectively. Find the maximum capacity of a container which can measure the petrol of either tanker in exact number of times.
Solution:
We need to find HCF(850, 680) to get the maximum capacity of container that can measure both quantities exactly.
Prime factorisation:
\(850 = 2 \times 5^2 \times 17\)
\(680 = 2^3 \times 5 \times 17\)
HCF = \(2 \times 5 \times 17 = 170\)
Verification:
850 ÷ 170 = 5 (exact)
680 ÷ 170 = 4 (exact)
Answer: Maximum capacity = 170 litres
If HCF of 144 and 180 is expressed in the form \((13m – 2)\), find the value of \(m\).
Solution:
First, find HCF(144, 180):
Prime factorisation:
\(144 = 2^4 \times 3^2\)
\(180 = 2^2 \times 3^2 \times 5\)
HCF = \(2^2 \times 3^2 = 4 \times 9 = 36\)
Given: \(13m – 2 = 36\)
\(13m = 36 + 2 = 38\)
\(m = \frac{38}{13}\)
This gives \(m = \frac{38}{13}\) which is not an integer, suggesting there might be an error in the problem statement or the expression.
Alternative approach: If the expression was meant to be linear, let’s re-examine:
If HCF = 36, and \(13m – 2 = 36\), then:
\(13m = 38\) ⇒ \(m = \frac{38}{13} \approx 2.92\)
This is unusual for such problems. Possibly the expression should be different, or the numbers 144 and 180 should yield HCF that fits an integer \(m\).
Checking actual HCF:
144 = 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144
180 = 1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 30, 36, 45, 60, 90, 180
Common factors: 1, 2, 3, 4, 6, 9, 12, 18, 36
HCF = 36 ✓
Given the context, if the problem statement is correct:
\(m = \frac{38}{13}\)