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)
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)
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)
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.
√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".
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…
∴ 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.
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.
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: Let's assume p + q = r, where r is a rational number (to reach a contradiction).
Since p is rational, it can be represented as p = a/b, where "a" and "b" are integers and b ≠ 0.
Then, q = r - p
q = r - (a/b)
Now, as q is irrational and r is rational, let's assume r = c/d, where "c" and "d" are integers and d ≠ 0.
So, q = (c/d) - (a/b)
q = (bc - ad)/(bd)
Since both bc and ad are integers (the product of two integers is an integer), let's assume (bc - ad) = x, where "x" is an integer.
q = x/(bd)
Now, q can be expressed as a fraction of two integers "x" and "bd," making it rational. However, this contradicts our assumption that q is irrational.
Therefore, our assumption that r is rational is incorrect. Hence, the sum (p + q) must be 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
q^2 = 144k² / 12
q^2 = 12k²
Now, we see that q² is 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.