CBSE Class 10 Mathematics Real Numbers Assignment Set 15

Exercise 1 Objective: Introduction and Euclid's Division Lemma

Question 1. Which one of the following is true?
(A) \( \pi \) is a rational number.   (B) All rational numbers are irrational numbers.   (C) All real numbers can be represented on a number line.   (D) \( \dfrac{\sqrt{7}}{8} \) is a rational number.
Answer:
Ans. (C) All real numbers can be represented on a number line.
Every real number corresponds to a point on the number line. \( \pi \) is irrational, rational numbers are not irrational, and \( \sqrt{7}/8 \) is irrational.
In simple words: Real = rational + irrational. | Each real number has exactly one point.

Teacher's Note:
1. Know the real number line.

Question 2. Which is not an irrational number?
(A) \( 5 - \sqrt{3} \)   (B) \( \sqrt{2} + \sqrt{5} \)   (C) \( 4 + \sqrt{2} \)   (D) \( 6 + \sqrt{9} \)
Answer:
Ans. (D) \( 6 + \sqrt{9} \)
\( 6 + \sqrt{9} = 6 + 3 = 9 \), a rational number. The other three are irrational.
In simple words: √9 = 3. | The rest contain √2, √3 or √5.

Teacher's Note:
1. Simplify perfect-square roots.

Question 3. If the least prime factor of a is 3 and the least prime factor of b is 7, then the least prime factor of (a + b) is:
(A) 2   (B) 3   (C) 5   (D) 11
Answer:
Ans. (A) 2
a and b are both odd (their least prime factors are 3 and 7, not 2), so a + b is even. The least prime factor of an even number is 2.
In simple words: Least prime factor 3 or 7 means the number is odd. | An even number has least prime factor 2.

Teacher's Note:
1. Odd + odd = even.

Question 4. Euclid's division lemma states that for any positive integers a and b, there exist unique integers q and r such that a = bq + r, where r must satisfy:
(A) \( 1 < r < b \)   (B) \( 0 < r \le b \)   (C) \( 0 \le r < b \)   (D) \( 0 < r < b \)
Answer:
Ans. (C) \( 0 \le r < b \)
The remainder is non-negative and less than the divisor: \( 0 \le r < b \).
In simple words: If r = 0, b divides a. | r is always less than b.

Teacher's Note:
1. r can be 0.

Question 5. \( (n^2 - 1) \) is divisible by 8, if n is:
(A) any natural number   (B) any integer   (C) any odd positive integer   (D) any even positive integer
Answer:
Ans. (C) any odd positive integer
For odd n, \( n^2 - 1 = (n-1)(n+1) \) is the product of two consecutive even numbers, one a multiple of 4, so it is divisible by 8. For n = 2, \( n^2 - 1 = 3 \), so even n fails.
In simple words: n = 1, 3, 5 give 0, 8, 24. | Even n gives an odd result.

Teacher's Note:
1. Test n = 2.

Question 6. A positive integer n when divided by 9 gives 7 as remainder. What will be the remainder when \( (3n - 1) \) is divided by 9?
(A) 1   (B) 2   (C) 3   (D) 4
Answer:
Ans. (B) 2
n = 9q + 7, so \( 3n - 1 = 27q + 20 = 9(3q + 2) + 2 \). Remainder = 2.
In simple words: Test n = 7: 3n − 1 = 20 = 2 × 9 + 2. | The remainder is 2.

Teacher's Note:
1. Substitute n = 9q + 7.

Exercise 1 Objective: Euclid's Division Algorithm

Question 7. If the H.C.F. of 85 and 153 is expressible in the form 85n − 153, then the value of n is:
(A) 3   (B) 2   (C) 4   (D) 1
Answer:
Ans. (B) 2
H.C.F.(85, 153) = 17 (153 = 85 + 68, 85 = 68 + 17, 68 = 17 × 4). So \( 85n - 153 = 17 \), \( 85n = 170 \), \( n = 2 \).
In simple words: 17 divides both 85 and 153. | Then solve 85n = 170.

Teacher's Note:
1. Find the H.C.F. first.

Question 8. If the H.C.F. of 79 and 97 is expressible in the form 97n − 79m, then the value of m − n is:
(A) 3   (B) 2   (C) 5   (D) 1
Answer:
Ans. (C) 5
H.C.F. = 1. By Euclid's steps: \( 97 = 79 + 18 \), \( 79 = 18 \times 4 + 7 \), \( 18 = 7 \times 2 + 4 \), \( 7 = 4 + 3 \), \( 4 = 3 + 1 \). Back-substituting, \( 1 = 22 \times 97 - 27 \times 79 \), so n = 22, m = 27 and m − n = 5 (this is the pair that comes from the algorithm).
In simple words: The pair (n, m) is not unique, but the algorithm gives (22, 27). | 22 × 97 − 27 × 79 = 2134 − 2133 = 1.

Teacher's Note:
1. Back-substitute the algorithm.

Question 9. If the H.C.F. of 144 and 90 is expressible in the form 144x + 90y, then the value of x − y is:
(A) 3   (B) 2   (C) 5   (D) 1
Answer:
Ans. (C) 5
H.C.F. = 18. \( 18 = 54 - 36 = 54 - (90 - 54) = 2 \times 54 - 90 = 2(144 - 90) - 90 = 2 \times 144 - 3 \times 90 \). So x = 2, y = −3 and x − y = 5.
In simple words: 144 = 90 + 54, 90 = 54 + 36, 54 = 36 + 18. | Check: 288 − 270 = 18.

Teacher's Note:
1. Back-substitute the algorithm.

Question 10. If the H.C.F. of 420 and 130 is expressible in the form 420p + 130q, then the value of p + q is:
(A) 17   (B) 6   (C) 5   (D) 9
Answer:
Ans. (D) 9
H.C.F. = 10: \( 420 = 130 \times 3 + 30 \), \( 130 = 30 \times 4 + 10 \). \( 10 = 130 - 4 \times 30 = 130 - 4(420 - 3 \times 130) = 13 \times 130 - 4 \times 420 \). So p = −4, q = 13 and p + q = 9.
In simple words: Check: −1680 + 1690 = 10. | The algorithm gives this particular pair.

Teacher's Note:
1. Back-substitute the algorithm.

Question 11. For positive integers a and b, if a = bq + r, then:
(A) every common divisor of 'a' and 'q' is a common divisor of 'b' and 'r', and vice-versa   (B) every common divisor of 'a' and 'b' is a common divisor of 'q' and 'r', and vice-versa   (C) every common divisor of 'a' and 'b' is a common divisor of 'b' and 'r', and vice-versa   (D) None of these
Answer:
Ans. (C) every common divisor of 'a' and 'b' is a common divisor of 'b' and 'r', and vice-versa
This is the result on which Euclid's algorithm is based: H.C.F.(a, b) = H.C.F.(b, r).
In simple words: Common divisors of a and b divide r = a − bq. | Common divisors of b and r divide a.

Teacher's Note:
1. Recall the basis of the algorithm.

Exercise 1 Objective: Fundamental Theorem of Arithmetic

Question 12. The [H.C.F. × L.C.M.] for the numbers 125 and 80 is:
(A) 100   (B) 1000   (C) 10000   (D) 500
Answer:
Ans. (C) 10000
H.C.F. × L.C.M. = 125 × 80 = 10000.
In simple words: H.C.F. = 5, L.C.M. = 2000. | 5 × 2000 = 10000.

Teacher's Note:
1. Product formula.

Question 13. If \( x = 2^3 \times 3 \times 5^2 \) and \( y = 2^2 \times 3^3 \), then H.C.F.(x, y) is:
(A) 12   (B) 108   (C) 6   (D) 36
Answer:
Ans. (A) 12
H.C.F. = \( 2^2 \times 3 = 12 \).
In simple words: 5 is not common. | 2² and 3¹ are the smallest.

Teacher's Note:
1. Smallest power of each common prime.

Question 14. Given that H.C.F.(253, 440) = 11 and L.C.M.(253, 440) = 253 × R. The value of R is:
(A) 400   (B) 40   (C) 440   (D) 253
Answer:
Ans. (B) 40
L.C.M. = \( \dfrac{253 \times 440}{11} = 253 \times 40 \), so R = 40.
In simple words: 440 ÷ 11 = 40. | R = 40.

Teacher's Note:
1. L.C.M. = product ÷ H.C.F.

Question 15. If the least prime factor of a is 5 and the least prime factor of b is 11, then the least prime factor of (a + b) is:
(A) 2   (B) 3   (C) 5   (D) 11
Answer:
Ans. (A) 2
Both a and b are odd, so a + b is even and its least prime factor is 2.
In simple words: Neither a nor b is even. | Hence a + b is even.

Teacher's Note:
1. Odd + odd = even.

Question 16. How many prime factors are there in the prime factorisation of 5005?
(A) 2   (B) 4   (C) 6   (D) 7
Answer:
Ans. (B) 4
\( 5005 = 5 \times 7 \times 11 \times 13 \): four prime factors.
In simple words: 5005 = 5 × 1001 and 1001 = 7 × 11 × 13. | Four distinct primes.

Teacher's Note:
1. Divide by 5 first.

Question 17. The product of the H.C.F. and L.C.M. of the smallest prime number and the smallest composite number is:
(A) 2   (B) 4   (C) 6   (D) 8
Answer:
Ans. (D) 8
The numbers are 2 and 4, so H.C.F. × L.C.M. = 2 × 4 = 8.
In simple words: Smallest prime: 2. Smallest composite: 4. | H.C.F. = 2, L.C.M. = 4.

Teacher's Note:
1. Product formula.

Exercise 1 Objective: Proof of Irrationality and Decimal Representation

Question 18. The decimal expansion of the rational number \( \dfrac{31}{2^2 \times 5} \) will terminate after:
(A) one decimal place   (B) two decimal places   (C) three decimal places   (D) more than three decimal places
Answer:
Ans. (B) two decimal places
\( \dfrac{31}{20} = 1.55 \): two decimal places (the larger of the powers of 2 and 5, here 2).
In simple words: m = 2, n = 1. | 31/20 = 1.55.

Teacher's Note:
1. Number of places = max(m, n).

Question 19. Which of the following is a non-terminating repeating decimal?
(A) \( \frac{35}{14} \)   (B) \( \frac{14}{35} \)   (C) \( \frac{1}{7} \)   (D) \( \frac{7}{8} \)
Answer:
Ans. (C) \( \frac{1}{7} \)
\( \frac{35}{14} = \frac{5}{2} \), \( \frac{14}{35} = \frac{2}{5} \) and \( \frac{7}{8} \) terminate. \( \frac{1}{7} = 0.\overline{142857} \) does not.
In simple words: Only 1/7 has a prime other than 2, 5 in the denominator. | 7 is the culprit.

Teacher's Note:
1. Reduce each fraction.

Question 20. The decimal representation of \( \dfrac{27}{400} \) is:
(A) Terminating   (B) Non terminating recurring   (C) Non terminating non recurring   (D) None of these
Answer:
Ans. (A) Terminating
\( 400 = 2^4 \times 5^2 \), so the decimal terminates (0.0675).
In simple words: Only 2 and 5. | 27/400 = 0.0675.

Teacher's Note:
1. Factorise the denominator.

Question 21. How many rational numbers exist between any two distinct rational numbers?
(A) 2   (B) 3   (C) 11   (D) Infinite
Answer:
Ans. (D) Infinite
Between any two distinct rational numbers there are infinitely many rational numbers (for instance repeated averaging).
In simple words: The average of two rationals is rational. | Keep taking averages.

Teacher's Note:
1. Density property.

Question 22. 3.24636363... is:
(A) an integer   (B) a rational number   (C) an irrational number   (D) None of these
Answer:
Ans. (B) a rational number
The decimal is non-terminating repeating (the block 63 repeats), so it is a rational number.
In simple words: Any repeating decimal can be written as p/q. | Irrationals do not repeat.

Teacher's Note:
1. Repeating means rational.

Question 23. A rational number can be expressed as a terminating decimal if the denominator has the factor:
(A) 2, 3 or 5   (B) 3 or 5   (C) 2 or 3   (D) 2 or 5
Answer:
Ans. (D) 2 or 5
The denominator (in lowest terms) must have no prime factors other than 2 and 5, that is, only factors 2 or 5.
In simple words: Other primes cause repeating. | m and n can be zero.

Teacher's Note:
1. Only 2 and 5 allowed.