CBSE Class 10 Mathematics Real Numbers Assignment Set 14

Exercise 1: Introduction and Euclid's Division Lemma

Question 1. Write a rational number between \( \sqrt{2} \) and \( \sqrt{3} \).
Answer:
\( \sqrt{2} \approx 1.414 \) and \( \sqrt{3} \approx 1.732 \). The rational number \( \dfrac{3}{2} = 1.5 \) lies between them (other answers such as 1.5 or 1.6 are also correct).
In simple words: Use decimal values of the roots.

Teacher's Note:
1. Any terminating decimal between 1.414 and 1.732 works.
2. There are infinitely many.

Question 2. Write whether \( \dfrac{2\sqrt{45} + 3\sqrt{20}}{2\sqrt{5}} \) on simplification gives a rational or an irrational number.
Answer:
\( 2\sqrt{45} = 6\sqrt{5} \) and \( 3\sqrt{20} = 6\sqrt{5} \). So the expression is \( \dfrac{12\sqrt{5}}{2\sqrt{5}} = 6 \), a rational number.
In simple words: Simplify the surds first.

Teacher's Note:
1. Write √45 = 3√5 and √20 = 2√5.
2. The √5 cancels.

Question 3. Use Euclid's Division Lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.
Answer:
Let a = 3q + r, r = 0, 1, 2. \( (3q)^2 = 3(3q^2) \); \( (3q+1)^2 = 3(3q^2+2q) + 1 \); \( (3q+2)^2 = 3(3q^2+4q+1) + 1 \).
Hence every square is 3m or 3m + 1.
In simple words: Check the remainders 0, 1, 2.

Teacher's Note:
1. Squares leave remainder 0 or 1 mod 3.
2. Never 2.

Question 4. Show that of the numbers n, n + 2 and n + 4, only one of them is divisible by 3.
Answer:
Let n = 3q + r. If r = 0, n is divisible by 3 and n + 2, n + 4 leave remainders 2 and 1. If r = 1, \( n + 2 = 3(q+1) \) is divisible by 3 and n, n + 4 leave remainders 1 and 2. If r = 2, \( n + 4 = 3(q+2) \) is divisible by 3 and n, n + 2 leave remainders 2 and 1.
In each case exactly one of the three is divisible by 3.
In simple words: Take n = 3q, 3q + 1, 3q + 2.

Teacher's Note:
1. The three numbers leave different remainders.
2. Exactly one remainder is 0.

Question 5. Let n = 640640640643. Without actually computing \( n^2 \), prove that \( n^2 \) leaves remainder 1 when divided by 8.
Answer:
n ends in 3, so n is odd. For an odd n = 2k + 1, \( n^2 - 1 = 4k(k+1) \), and k(k + 1) is even, so \( n^2 - 1 \) is divisible by 8.
Hence \( n^2 = 8m + 1 \): the remainder is 1.
In simple words: Odd squares are 8m + 1.

Teacher's Note:
1. Use n² − 1 = (n − 1)(n + 1).
2. Two consecutive even numbers: one is a multiple of 4.

Question 6. A number leaves remainder 3 when divided by 6. What will be the remainder if the square of the same number is divided by 6?
Answer:
Let the number be \( n = 6q + 3 \). Then \( n^2 = 36q^2 + 36q + 9 = 6(6q^2 + 6q + 1) + 3 \).
The remainder is 3.
In simple words: Square 6q + 3 and take 6 common.

Teacher's Note:
1. 9 = 6 + 3, so the remainder stays 3.
2. Check: 3² = 9 leaves 3.

Exercise 1: Euclid's Division Algorithm

Question 7. Using Euclid's division algorithm, find the H.C.F. of 210 and 55.
Answer:
\( 210 = 55 \times 3 + 45 \)
\( 55 = 45 \times 1 + 10 \)
\( 45 = 10 \times 4 + 5 \)
\( 10 = 5 \times 2 + 0 \)
H.C.F. = 5.
In simple words: Last non-zero remainder.

Teacher's Note:
1. 210 = 5 × 42 and 55 = 5 × 11.
2. 42 and 11 are co-prime.

Question 8. Using Euclid's division algorithm, find the H.C.F. of 101 and 1277.
Answer:
\( 1277 = 101 \times 12 + 65 \)
\( 101 = 65 \times 1 + 36 \)
\( 65 = 36 \times 1 + 29 \)
\( 36 = 29 \times 1 + 7 \)
\( 29 = 7 \times 4 + 1 \)
\( 7 = 1 \times 7 + 0 \)
H.C.F. = 1.
In simple words: 101 is prime and does not divide 1277.

Teacher's Note:
1. So the numbers are co-prime.
2. Continue until the remainder is 0.

Question 9. If d is the H.C.F. of 56 and 72, find x and y satisfying d = 56x + 72y. Also show that x and y are not unique.
Answer:
\( 72 = 56 \times 1 + 16 \), \( 56 = 16 \times 3 + 8 \), \( 16 = 8 \times 2 + 0 \). So d = 8.
\( 8 = 56 - 16 \times 3 = 56 - 3(72 - 56) = 4 \times 56 - 3 \times 72 \). So x = 4 and y = -3.
Also \( 8 = 56(4 + 72t) + 72(-3 - 56t) \) for every integer t. For t = 1: x = 76 and y = -59 (check: 56 × 76 − 72 × 59 = 4256 − 4248 = 8). Hence x and y are not unique.
In simple words: Back-substitute the Euclid steps.

Teacher's Note:
1. Add and subtract 56 × 72 to get new pairs.
2. General solution: x = 4 + 72t, y = −3 − 56t.

Question 10. Using Euclid's division algorithm, find the greatest number that divides 445, 572 and 699 leaving remainders 4, 5 and 6 respectively.
Answer:
The number divides \( 445 - 4 = 441 \), \( 572 - 5 = 567 \) and \( 699 - 6 = 693 \).
\( 567 = 441 \times 1 + 126 \), \( 441 = 126 \times 3 + 63 \), \( 126 = 63 \times 2 + 0 \): H.C.F.(441, 567) = 63.
\( 693 = 63 \times 11 + 0 \). So the required number is 63.
In simple words: Subtract remainders, then H.C.F.

Teacher's Note:
1. 63 is greater than 6.
2. So the remainders are valid.

Question 11. Find the largest number that divides 245 and 1029 leaving a remainder 5 in each case.
Answer:
The number divides \( 245 - 5 = 240 \) and \( 1029 - 5 = 1024 \). \( 1024 = 240 \times 4 + 64 \), \( 240 = 64 \times 3 + 48 \), \( 64 = 48 \times 1 + 16 \), \( 48 = 16 \times 3 + 0 \).
H.C.F. = 16.
In simple words: Subtract the common remainder.

Teacher's Note:
1. 16 > 5.
2. 240 = 16 × 15 and 1024 = 16 × 64.

Exercise 1: Fundamental Theorem of Arithmetic

Question 12. Can we have any natural number n, where \( 7^n \) ends with the digit zero?
Answer:
\( 7^n \) has only the prime factor 7. A number ending in 0 is divisible by 5 (and 2), which are not factors of \( 7^n \). So no.
In simple words: Ending in zero needs the factors 2 and 5.

Teacher's Note:
1. The prime factorisation of 7ⁿ is just 7ⁿ.
2. Uniqueness of factorisation settles it.

Question 13. Find the [H.C.F. × L.C.M.] for the numbers 105 and 120.
Answer:
H.C.F. × L.C.M. = product of the numbers = 105 × 120 = 12600.
In simple words: Use the product formula.

Teacher's Note:
1. H.C.F. = 15 and L.C.M. = 840.
2. 15 × 840 = 12600.

Question 14. Find the H.C.F. and L.C.M. of the following using the Fundamental Theorem of Arithmetic method: (i) 426 and 576 (ii) 625, 1125 and 2125.
Answer:
(i) \( 426 = 2 \times 3 \times 71 \), \( 576 = 2^6 \times 3^2 \). H.C.F. = \( 2 \times 3 = 6 \). L.C.M. = \( 2^6 \times 3^2 \times 71 = 40896 \).
(ii) \( 625 = 5^4 \), \( 1125 = 3^2 \times 5^3 \), \( 2125 = 5^3 \times 17 \). H.C.F. = \( 5^3 = 125 \). L.C.M. = \( 5^4 \times 3^2 \times 17 = 95625 \).
In simple words: Lowest powers for H.C.F., highest for L.C.M.

Teacher's Note:
1. Check (i): 6 × 40896 = 426 × 576.
2. Check (ii): 125 × 95625 is not required; verify L.C.M. is a multiple of each number.

Question 15. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
Answer:
\( 7 \times 11 \times 13 + 13 = 13(77 + 1) = 13 \times 78 \), which has factors other than 1 and itself.
\( 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5 = 5040 + 5 = 5(1008 + 1) = 5 \times 1009 \), also composite.
In simple words: Take the common factor out.

Teacher's Note:
1. A product of two numbers greater than 1 is composite.
2. Both numbers are shown as such products.

Question 16. An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?
Answer:
The number of columns must divide both 616 and 32, so it is H.C.F.(616, 32). \( 616 = 32 \times 19 + 8 \), \( 32 = 8 \times 4 + 0 \).
Maximum number of columns = 8.
In simple words: H.C.F. of the two group sizes.

Teacher's Note:
1. Each column then has 77 and 4 members.
2. 616 ÷ 8 = 77 and 32 ÷ 8 = 4.

Question 17. There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?
Answer:
They meet again after L.C.M.(18, 12) = 36 minutes.
In simple words: L.C.M. of the two times.

Teacher's Note:
1. 18 = 2 × 3², 12 = 2² × 3.
2. L.C.M. = 2² × 3² = 36.

Question 18. Aakash, Kushal and Harish go for a morning walk. They step off together and 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?
Answer:
L.C.M.(40, 42, 45) = \( 2^3 \times 3^2 \times 5 \times 7 = \) 2520 cm (25.2 m).
In simple words: L.C.M. of the three step lengths.

Teacher's Note:
1. They take 63, 60, 56 steps.
2. Convert to metres if asked.

Exercise 1: Proof of Irrationality and Decimal Representation

Question 19. Prove that \( \sqrt{3} \) is irrational.
Answer:
Suppose \( \sqrt{3} = \dfrac{a}{b} \), a, b co-prime. Then \( a^2 = 3b^2 \), so 3 divides a; with a = 3c, \( b^2 = 3c^2 \), so 3 divides b. Contradiction. Hence \( \sqrt{3} \) is irrational.
In simple words: Contradiction with co-primality.

Teacher's Note:
1. Needs: prime p divides a² ⇒ p divides a.
2. Same proof for √2, √5, √7.

Question 20. Prove that \( 5 - 2\sqrt{3} \) is an irrational number.
Answer:
If \( 5 - 2\sqrt{3} = \dfrac{a}{b} \) is rational, then \( \sqrt{3} = \dfrac{5b - a}{2b} \) is rational, contradicting the irrationality of \( \sqrt{3} \).
In simple words: Isolate √3.

Teacher's Note:
1. Rational combinations of √3 stay irrational.
2. Assume, then contradict.

Question 21. Prove that \( \sqrt{2} + \sqrt{3} \) is irrational.
Answer:
Suppose \( \sqrt{2} + \sqrt{3} = r \) is rational. Squaring: \( 5 + 2\sqrt{6} = r^2 \), so \( \sqrt{6} = \dfrac{r^2 - 5}{2} \) is rational. But 6 is not a perfect square, so \( \sqrt{6} \) is irrational. Contradiction.
In simple words: Square both sides.

Teacher's Note:
1. The square root of a non-square integer is irrational.
2. Conclude with the contradiction.

Question 22. Without actually performing the long division, state whether the following rational numbers will have a terminating or a non-terminating decimal expansion: (i) \( \dfrac{77}{210} \) (ii) \( \dfrac{15}{1600} \).
Answer:
(i) \( \dfrac{77}{210} = \dfrac{11}{30} \) and \( 30 = 2 \times 3 \times 5 \) has the factor 3, so the expansion is non-terminating (repeating).
(ii) \( \dfrac{15}{1600} = \dfrac{3}{320} \) and \( 320 = 2^6 \times 5 \), so the expansion is terminating (0.009375).
In simple words: Reduce, then inspect the denominator.

Teacher's Note:
1. Cancel common factors first.
2. 11/30 = 0.3666...

Question 23. What can you say about the prime factorisations of the denominators of the following rationals: (i) 25.234567 (ii) \( 25.\overline{345678} \)?
Answer:
(i) 25.234567 terminates, so the denominator (in lowest terms) is of the form \( 2^m \times 5^n \).
(ii) \( 25.\overline{345678} \) is non-terminating repeating, so the denominator has at least one prime factor other than 2 and 5.
In simple words: Link decimal type to the denominator.

Teacher's Note:
1. Terminating means only 2s and 5s.
2. Repeating means another prime is present.