CBSE Class 10 Mathematics Real Numbers Assignment Set 13

Previous Year Board Questions

Question 1. Which of the following numbers has terminating decimal expansion? [1 Mark, CBSE 2013]
(A) \( \frac{37}{45} \)   (B) \( \frac{21}{2^3 5^6} \)   (C) \( \frac{17}{49} \)   (D) \( \frac{89}{2^2 3^2} \)
Answer:
Ans. (B) \( \frac{21}{2^3 5^6} \)
A fraction in lowest terms terminates only if its denominator has no prime factor other than 2 and 5. 45 = 3² × 5, 49 = 7², \( 2^2 3^2 \) all contain other primes; only \( 2^3 5^6 \) does not.
In simple words: Reduce to lowest terms first. | Only 2 and 5 are allowed.

Teacher's Note:
1. Check the denominator's prime factors.

Question 2. If a rational number x = p/q, where p, q are integers, q ≠ 0 and p, q have no common factor (except 1), then the decimal expansion of x is terminating if and only if q has a prime factorisation of the form: [1 Mark, CBSE 2013]
(A) \( 2^m \cdot 5^n \)   (B) \( 2^m \cdot 3^n \)   (C) \( 2^m \cdot 7^n \)   (D) \( 5^m \cdot 3^n \)
Answer:
Ans. (A) \( 2^m \cdot 5^n \)
By the theorem on terminating decimals, q must be of the form \( 2^m \times 5^n \) with m, n non-negative integers.
In simple words: If q has any other prime factor, the expansion repeats forever. | m and n may be zero.

Teacher's Note:
1. This is the standard theorem.

Question 3. Which of the following numbers has non-terminating repeating decimal expansion? [1 Mark, CBSE 2013]
(A) \( \frac{7}{80} \)   (B) \( \frac{17}{320} \)   (C) \( \frac{20}{100} \)   (D) \( \frac{93}{420} \)
Answer:
Ans. (D) \( \frac{93}{420} \)
80 = 2⁴ × 5, 320 = 2⁶ × 5 and 20/100 = 1/5 terminate. \( \frac{93}{420} = \frac{31}{140} \) and 140 = 2² × 5 × 7 has the factor 7, so it is non-terminating repeating.
In simple words: The factor 7 in the denominator decides it. | 31/140 = 0.2214285714285...

Teacher's Note:
1. Reduce 93/420 to 31/140.

Question 4. Use Euclid's division algorithm to find the H.C.F. of 870 and 225. [2 Marks, CBSE 2013]
Answer:
\( 870 = 225 \times 3 + 195 \)
\( 225 = 195 \times 1 + 30 \)
\( 195 = 30 \times 6 + 15 \)
\( 30 = 15 \times 2 + 0 \)
H.C.F. = 15.
In simple words: The last non-zero remainder is the H.C.F.

Teacher's Note:
1. Keep dividing divisor by remainder.
2. Check: 870 = 15 × 58 and 225 = 15 × 15.

Question 5. Explain why 5 × 4 × 3 × 2 × 1 + 3 is a composite number. [2 Marks, CBSE 2013]
Answer:
\( 5 \times 4 \times 3 \times 2 \times 1 + 3 = 120 + 3 = 123 = 3 \times 41 \).
123 has the factors 3 and 41 besides 1 and itself, so it is composite.
In simple words: Take 3 as a common factor.

Teacher's Note:
1. Both 120 and 3 are multiples of 3.
2. So the sum is also a multiple of 3.

Question 6. Prove that \( 3 + \sqrt{2} \) is an irrational number. OR Prove that \( 5\sqrt{2} \) is an irrational number. [3 Marks, CBSE 2013]
Answer:
(i) Suppose \( 3 + \sqrt{2} = \dfrac{a}{b} \) is rational (a, b integers, b ≠ 0). Then \( \sqrt{2} = \dfrac{a - 3b}{b} \) is rational, which contradicts the irrationality of \( \sqrt{2} \). Hence \( 3 + \sqrt{2} \) is irrational.
(ii) Suppose \( 5\sqrt{2} = \dfrac{a}{b} \). Then \( \sqrt{2} = \dfrac{a}{5b} \) is rational, a contradiction. Hence \( 5\sqrt{2} \) is irrational.
In simple words: Assume rational, isolate the surd, contradict.

Teacher's Note:
1. The sum, difference or product (non-zero rational) of a rational and an irrational is irrational.
2. \( \sqrt{2} \) is proved irrational in the textbook.

Question 7. Show that \( 5^n \) cannot end with the digit 2 for any natural number n. [3 Marks, CBSE 2013]
Answer:
\( 5^n \) is a multiple of 5, so its last digit is 0 or 5. Also \( 5^n \) is odd, so it cannot end in 0 (it is not even), hence the last digit is always 5.
So \( 5^n \) can never end with 2. (A number ending in 2 is even, but \( 5^n \) is odd.)
In simple words: Powers of 5 are odd multiples of 5.

Teacher's Note:
1. Every power of 5 ends in 5: 5, 25, 125, ...
2. A number ending in 2 is even.

Question 8. The H.C.F. of two numbers is 145 and their L.C.M. is 2175. If one number is 725, then the other number is: [1 Mark, CBSE 2014]
(A) 415   (B) 425   (C) 435   (D) 445
Answer:
Ans. (C) 435
Other number = \( \dfrac{145 \times 2175}{725} = 435 \).
In simple words: 145 × 2175 = 315375, and 315375 ÷ 725 = 435. | Check: H.C.F.(725, 435) = 145.

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

Question 9. If H.C.F.(96, 404) = 4, then L.C.M.(96, 404) is: [1 Mark, CBSE 2014]
(A) 9626   (B) 9696   (C) 9656   (D) 9676
Answer:
Ans. (B) 9696
L.C.M. = \( \dfrac{96 \times 404}{4} = 9696 \).
In simple words: 96 × 404 = 38784, divided by 4 gives 9696. | Always check the answer is a multiple of both numbers.

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

Question 10. Check whether \( 6^n \) can end with the digit 0 for any natural number n. [2 Marks, CBSE 2014]
Answer:
\( 6^n = 2^n \times 3^n \), which has no factor 5. A number ending in 0 is divisible by 5, so \( 6^n \) never ends with 0.
In simple words: A number ending in 0 must have 2 and 5 as factors.

Teacher's Note:
1. Check the prime factors.
2. 5 is absent.

Question 11. Prove that \( \sqrt{3} + \sqrt{5} \) is an irrational number. OR Prove that \( 5 + \sqrt{2} \) is an irrational number. [3 Marks, CBSE 2014]
Answer:
(i) Suppose \( \sqrt{3} + \sqrt{5} = r \) is rational. Squaring, \( 8 + 2\sqrt{15} = r^2 \), so \( \sqrt{15} = \dfrac{r^2 - 8}{2} \) is rational, which is false because 15 is not a perfect square. Hence \( \sqrt{3} + \sqrt{5} \) is irrational.
(ii) If \( 5 + \sqrt{2} = \dfrac{a}{b} \), then \( \sqrt{2} = \dfrac{a - 5b}{b} \) is rational, a contradiction. Hence \( 5 + \sqrt{2} \) is irrational.
In simple words: Square to remove one surd level.

Teacher's Note:
1. Square roots of non-square integers are irrational.
2. Always end with the contradiction.

Question 12. Use Euclid's division algorithm to find the H.C.F. of 10224 and 9648. [3 Marks, CBSE 2015]
Answer:
\( 10224 = 9648 \times 1 + 576 \)
\( 9648 = 576 \times 16 + 432 \)
\( 576 = 432 \times 1 + 144 \)
\( 432 = 144 \times 3 + 0 \)
H.C.F. = 144.
In simple words: Last non-zero remainder.

Teacher's Note:
1. Divide repeatedly.
2. Check: 10224 = 144 × 71 and 9648 = 144 × 67.

Question 13. Which of the following is not a rational number? [1 Mark, CBSE 2015]
(A) \( \sqrt{3} \)   (B) \( \sqrt{9} \)   (C) \( \sqrt{16} \)   (D) \( \sqrt{25} \)
Answer:
Ans. (A) \( \sqrt{3} \)
\( \sqrt{9} = 3 \), \( \sqrt{16} = 4 \), \( \sqrt{25} = 5 \) are rational; \( \sqrt{3} \) is irrational.
In simple words: 3 is not a perfect square. | So √3 is irrational.

Teacher's Note:
1. Perfect squares give rational roots.

Question 14. Prove that \( \sqrt{7} \) is an irrational number. OR Prove that \( 3 - \sqrt{5} \) is an irrational number. [3 Marks, CBSE 2015]
Answer:
(i) Suppose \( \sqrt{7} = \dfrac{a}{b} \) with a, b co-prime. Then \( a^2 = 7b^2 \), so 7 divides a; put a = 7c to get \( b^2 = 7c^2 \), so 7 divides b. Contradiction with co-primality. Hence \( \sqrt{7} \) is irrational.
(ii) If \( 3 - \sqrt{5} = \dfrac{a}{b} \), then \( \sqrt{5} = \dfrac{3b - a}{b} \) is rational, contradicting the irrationality of \( \sqrt{5} \). Hence \( 3 - \sqrt{5} \) is irrational.
In simple words: Use the prime-divides-square result.

Teacher's Note:
1. Recall: p
2. a² ⇒ p
3. a.
4. A non-zero rational minus an irrational is irrational.

Question 15. 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. [3 Marks, CBSE 2015]
Answer:
Let a = 3q + r with r = 0, 1, 2.
r = 0: \( a^2 = 9q^2 = 3(3q^2) = 3m \). r = 1: \( a^2 = 9q^2 + 6q + 1 = 3(3q^2 + 2q) + 1 = 3m + 1 \). r = 2: \( a^2 = 9q^2 + 12q + 4 = 3(3q^2 + 4q + 1) + 1 = 3m + 1 \).
So the square is of the form 3m or 3m + 1.
In simple words: Squares leave remainder 0 or 1 mod 3.

Teacher's Note:
1. Never 3m + 2.
2. Same idea as the mod 4 and mod 5 results.

Question 16. The product of the H.C.F. and L.C.M. for the numbers 50 and 20 is: [1 Mark, CBSE 2016]
(A) 10   (B) 1000   (C) 100   (D) 110
Answer:
Ans. (B) 1000
H.C.F. × L.C.M. = product of the numbers = 50 × 20 = 1000.
In simple words: H.C.F. = 10, L.C.M. = 100. | 10 × 100 = 1000.

Teacher's Note:
1. H.C.F. × L.C.M. = a × b.

Question 17. If the H.C.F. of 210 and 55 is expressible in the form \( 210 \times 5 + 55p \), then the value of p is: [1 Mark, CBSE 2016]
(A) - 17   (B) - 18   (C) - 20   (D) - 19
Answer:
Ans. (D) - 19
H.C.F.(210, 55) = 5. So \( 1050 + 55p = 5 \), giving \( 55p = -1045 \) and \( p = -19 \).
In simple words: 210 = 55 × 3 + 45, 55 = 45 + 10, 45 = 10 × 4 + 5, 10 = 5 × 2. | So the H.C.F. is 5.

Teacher's Note:
1. Find the H.C.F. first, then solve for p.

Question 18. Prove that \( \sqrt{5} \) is an irrational number. OR Prove that \( 5 + 3\sqrt{2} \) is an irrational number. [3 Marks, CBSE 2016]
Answer:
(i) Suppose \( \sqrt{5} = \dfrac{a}{b} \) with a, b co-prime. Then \( a^2 = 5b^2 \), so 5 divides a; with a = 5c, \( b^2 = 5c^2 \), so 5 divides b: contradiction. Hence \( \sqrt{5} \) is irrational.
(ii) If \( 5 + 3\sqrt{2} = \dfrac{a}{b} \), then \( \sqrt{2} = \dfrac{a - 5b}{3b} \) is rational, contradicting the irrationality of \( \sqrt{2} \). Hence \( 5 + 3\sqrt{2} \) is irrational.
In simple words: Prime divides square, then contradiction.

Teacher's Note:
1. The same pattern proves any square root of a prime irrational.
2. A rational plus a non-zero multiple of an irrational is irrational.

Question 19. Prove that for any positive integer n, \( n^3 - n \) is divisible by 6. [3 Marks, CBSE 2016]
Answer:
\( n^3 - n = n(n-1)(n+1) \), the product of three consecutive integers \( (n-1), n, (n+1) \).
One of three consecutive integers is divisible by 3 and at least one is even, so the product is divisible by 2 and by 3, hence by 6.
In simple words: Factorise into three consecutive integers.

Teacher's Note:
1. Among three consecutive integers there is a multiple of 3 and a multiple of 2.
2. 2 and 3 are co-prime, so 6 divides the product.

Question 20. Find the L.C.M. and H.C.F. of 510 and 92 and verify that L.C.M. × H.C.F. = product of the two numbers. [3 Marks, CBSE 2017]
Answer:
\( 510 = 2 \times 3 \times 5 \times 17 \) and \( 92 = 2^2 \times 23 \).
H.C.F. = 2 and L.C.M. = \( 2^2 \times 3 \times 5 \times 17 \times 23 = 23460 \).
Check: H.C.F. × L.C.M. = 2 × 23460 = 46920 = 510 × 92. Verified.
In simple words: Prime factorise both numbers.

Teacher's Note:
1. Highest powers for L.C.M., lowest common powers for H.C.F.
2. 510 × 92 = 46920.

Question 21. The largest number that will divide 398, 436 and 542 leaving remainders 7, 11 and 15 respectively is: [1 Mark, CBSE 2017]
(A) 11   (B) 17   (C) 34   (D) 51
Answer:
Ans. (B) 17
The number divides 391, 425 and 527. \( 391 = 17 \times 23 \), \( 425 = 17 \times 25 \), \( 527 = 17 \times 31 \), so H.C.F. = 17 (which is larger than every remainder).
In simple words: 17 > 15, so valid. | 34 and 51 do not divide 391.

Teacher's Note:
1. Subtract the remainders and find the H.C.F.

Question 22. Prove that \( \dfrac{7}{3}\sqrt{5} \) is an irrational number. OR Prove that \( 5 - 2\sqrt{3} \) is an irrational number. [3 Marks, CBSE 2017]
Answer:
(i) Suppose \( \dfrac{7}{3}\sqrt{5} = \dfrac{a}{b} \) is rational. Then \( \sqrt{5} = \dfrac{3a}{7b} \) is rational, contradicting the irrationality of \( \sqrt{5} \).
(ii) Suppose \( 5 - 2\sqrt{3} = \dfrac{a}{b} \). Then \( \sqrt{3} = \dfrac{5b - a}{2b} \) is rational, a contradiction. Hence both numbers are irrational.
In simple words: Isolate the surd and contradict.

Teacher's Note:
1. Use √5 and √3 irrational.
2. Nonzero rational multiples keep irrationality.

Question 23. Prove that \( n^2 - n \) is divisible by 2 for any positive integer n. [3 Marks, CBSE 2017]
Answer:
\( n^2 - n = n(n - 1) \), a product of two consecutive integers. One of them is even, so the product is divisible by 2.
In simple words: Two consecutive integers include an even number.

Teacher's Note:
1. n(n − 1) is always even.
2. Parallel to n³ − n divisible by 6.