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Unit Test (Solutions): Real Numbers | Mathematics (Maths) Class 10 PDF Download

Time: 1 hour

M.M. 30

Attempt all questions.

  • Question numbers 1 to 5 carry 1 mark each.
  • Question numbers 6 to 8 carry 2 marks each.
  • Question numbers  9 to 11 carry 3 marks each.
  • Question number 12 & 13 carry 5 marks each.

Q1: Which of the following numbers is irrational?  (1 Mark)  
(a) √25
(b) 3.14
(c) 0.333...
(d) -7

Ans: (c)
Sol: An irrational number is a number that cannot be expressed as a fraction of two integers. Option c) 0.333... is an example of an irrational number as it represents a non-repeating and non-terminating decimal (1/3), making it irrational.

Q2: What is the value of (5² + 12²)?  (1 Mark)  
(a) 169
(b) 144
(c) 25
(d) 169√2

Ans: (a)
Sol:The given expression is (5² + 12²) = (25 + 144) = 169.

Q3: Which one is not a prime number?  (1 Mark)  
(a) 1
(b) 2
(c) 3
(d) 5

Ans: (a)
Sol: A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. However, 1 does not meet this criteria, as it only has one positive divisor.

Q4: State whether "√16" is a rational number or not.  (1 Mark)  

Ans:  "√16" is a rational number.
Sol: √16 = 4, which is a rational number since it can be expressed as the fraction 4/1.

Q5: Without performing the actual division, state whether "1089" is divisible by "9" or not.  (1 Mark)

Ans: "1089" is divisible by "9".
Sol: A number is divisible by 9 if the sum of its digits is divisible by 9. Here, 1 + 0 + 8 + 9 = 18, and 18 is divisible by 9.

Q6: Write a rational number between √5 and √6.

Ans: √5 = 2.236… and √6 = 2.449…
Sol: ∴ A rational no between 2.24 and 2.44 (approximately) is 2.3 or 2.31 or 2.32 etc.
Note: Take the lower limit slightly greater than √5 and upper limit slightly lesser than √6 .
⇒ One number between √5 and √6 = 2.3

Q7: Express 0.37 as a fraction in its simplest form.  (2 Marks)

Ans: 0.37 can be expressed as the fraction 37/100.
Sol: To convert a decimal to a fraction, we remove the decimal point and place the digits after the decimal over the appropriate place value (in this case, 37/100). We simplify the fraction, but in this case, it is already in its simplest form.

Q8: Find the LCM (Least Common Multiple) of 15 and 20.  (2 Marks)

Ans: The LCM of 15 and 20 is 60.
Sol: To find the LCM, we can use the prime factorization method. 
The prime factorization of 15 is 3 x 5, and the prime factorization of 20 is 2 x 2 x 5. 
The LCM is the product of the highest powers of all the prime factors involved, which is 2 x 2 x 3 x 5 = 60.

Q9: Prove that the square of any positive integer of the form (5k + 1) is one more than a multiple of 8, where "k" is an integer.  (3 Marks) 

Ans: Let's assume the positive integer be "n" in the form of (5k + 1).
Step 1: Square of "n"
n² = (5k + 1)² = 25k² + 10k + 1 = 5(5k² + 2k) + 1
Step 2: Express (5k² + 2k) as an integer "m"

Let (5k² + 2k) = m (where m is an integer)
Step 3: Express n² in terms of "m"
n² = 5m + 1
Step 4: Prove that n² is one more than a multiple of 8

n² = 5m + 1 = 8k + (5m - 8k + 1)
Since (5m - 8k + 1) is an integer, let's say it equals "p"
n² = 8k + p
Thus, n² is one more than a multiple of 8.

Q10: Find the HCF (Highest Common Factor) of 72 and 96 using the prime factorization method.  (3 Marks) 

Ans: The HCF of 72 and 96 is 24.
To find the HCF, we can use the prime factorization method.
Prime factorization of 72: 72 = 2³ x 3²
Prime factorization of 96: 96 = 2⁵ x 3¹
HCF = Product of the common prime factors with their lowest powers
HCF = 2³ x 3¹ = 8 x 3 = 24

Q11: Check whether 6n can end with the digit 0 for any natural number n. (3 Marks) 

Ans:  If the number 6n ends with the digit zero (0), then it should be divisible by 5, as we know any number with a unit place as 0 or 5 is divisible by 5.

Prime factorization of 6n = (2 × 3)n

Therefore, the prime factorization of 6n doesn’t contain the prime number 5.

Hence, it is clear that for any natural number n, 6n is not divisible by 5 and thus it proves that 6n cannot end with the digit 0 for any natural number n.

Q12: Given that p is a rational number and q is an irrational number, prove that their sum (p + q) is an irrational number.  (5 Marks) 

Ans:

  1. Definitions:
  • A rational number p can be expressed as a/b, where a and b are integers and b is not equal to 0.
  • An irrational number q cannot be expressed as a fraction of integers.
  1. Assumption: Assume for the sake of contradiction that p + q is rational. This means we can express p + q as c/d, where c and d are integers and d is not equal to 0.

  2. Rearranging the Equation: From our assumption, we have:
    p + q = c/d
    Rearranging this gives:
    q = c/d - p

  3. Substituting the Rational Number: Since p is rational, we can express it as a/b:
    q = c/d - a/b
    To combine these fractions, we find a common denominator:
    q = (cb - ad) / bd

  4. Conclusion: Since cb - ad and bd are both integers (as they are products and sums of integers), q can be expressed as a fraction of two integers, which contradicts our original statement that q is irrational.

  5. Final Statement: Thus, our assumption that p + q is rational must be false. Therefore, p + q is irrational.

Q13: Prove that 5√3 - 3√75 is an irrational number.  (5 Marks)  

Ans: To prove that 5√3 - 3√75 is an irrational number, we assume the contrary, i.e., let's assume 5√3 - 3√75 is a rational number. 
So, it can be expressed as 5√3 - 3√75 = p/q, where p and q are co-prime integers (i.e., they have no common factors other than 1) and q ≠ 0.
Now, let's work on simplifying the expression:
5√3 - 3√75
Step 1: Factorize the numbers inside the radicals.
√3 cannot be simplified further as it is a prime number.
√75 = √(5 * 5 * 3) = 5√3
Step 2: Substitute the factorized value back into the expression.
5√3 - 3√75 = 5√3 - 3 * 5√3
Step 3: Combine like terms.
5√3 - 3√75 = (5 - 3)√3 = 2√3
Now, let's express 2√3 as a rational number:
2√3 = p/q
Squaring both sides, we get:
4 * 3 = (p/q)2
12 = p2 / q2
From the above equation, we can see that p2 is a multiple of 12, which means p must also be a multiple of 12 (since 12 is not a prime number). Let's write p as p = 12k, where k is an integer.
Substituting the value of p back into our equation, we get:
12 = (12k)2 / q2
12 = 144k2  / q2
q2 = 144k2 / 12
q2 = 12k2
Now, we see that qis also a multiple of 12, which implies that q must also be a multiple of 12.
But this contradicts our initial assumption that p and q are co-prime (i.e., they have no common factors other than 1) because both p and q are divisible by 12. Hence, our initial assumption that 5√3 - 3√75 is rational is incorrect. Therefore, 5√3 - 3√75 must be an irrational number.

The document Unit Test (Solutions): Real Numbers | Mathematics (Maths) Class 10 is a part of the Class 10 Course Mathematics (Maths) Class 10.
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FAQs on Unit Test (Solutions): Real Numbers - Mathematics (Maths) Class 10

1. What are real numbers and how are they classified?
Ans. Real numbers are all the numbers that can be found on the number line. They include both rational numbers (like integers and fractions) and irrational numbers (like the square root of 2 or pi). Real numbers can be classified into several categories: natural numbers, whole numbers, integers, rational numbers, and irrational numbers.
2. How do you perform operations with real numbers?
Ans. Operations with real numbers, such as addition, subtraction, multiplication, and division, follow the same rules as with other types of numbers. When performing these operations, it’s important to pay attention to the properties of real numbers, such as the commutative and associative properties, to ensure accurate results.
3. What is the importance of irrational numbers in the set of real numbers?
Ans. Irrational numbers are crucial because they fill in the gaps between rational numbers on the number line. They represent quantities that cannot be expressed as simple fractions, allowing for a more complete understanding of measurements, such as lengths and areas, especially in geometry.
4. Can real numbers be negative, and what does that imply?
Ans. Yes, real numbers can be negative. Negative real numbers represent values less than zero and are important in various mathematical contexts, such as debt in finance or temperature below freezing. They are part of the number line extending to the left of zero.
5. How do you determine if a number is rational or irrational?
Ans. To determine if a number is rational or irrational, check if it can be expressed as a fraction of two integers (where the denominator is not zero). If it can, it is rational. If it cannot be written in such a form and has a non-repeating, non-terminating decimal expansion, it is irrational.
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