CBSE Class 10 Maths Advanced Real Numbers Very Short Answer Questions Set 02

Very Short Answer Questions

Question 1. Can we have any natural number n, where 7ⁿ ends with the digit zero.
Answer:
A number ending in 0 is divisible by 10, so its prime factors include both 2 and 5.
The only prime factor of 7ⁿ is 7, so it has neither 2 nor 5.
Answer: No, 7ⁿ never ends with 0.
In simple words: A number ending in zero must have 2 and 5 among its prime factors, and 7ⁿ has only 7.

Teacher's Note:
1. Use the uniqueness of prime factorisation.
2. The last digits of powers of 7 cycle through 7, 9, 3, 1.

Question 2. Find the [HCF × LCM] for the numbers 105 and 120.
Answer:
HCF × LCM equals the product of the two numbers.
105 × 120 = 12600.
In simple words: No need to find the HCF and LCM separately; multiply the numbers.

Teacher's Note:
1. Check: HCF = 15 and LCM = 840, and 15 × 840 = 12600.
2. This relation holds for any two positive integers.

Question 3. Without actually performing the long division, state whether the following rational number will have a terminating decimal expansion or non-terminating decimal expansion : (i) \( \frac{77}{210} \)   (ii) \( \frac{15}{1600} \)
Answer:
(i) Dividing by 7: 77/210 = 11/30. Since 30 = 2 × 3 × 5 contains the prime 3, the decimal expansion is non-terminating (repeating).
(ii) 15/1600 = 3/320 and 320 = 2⁶ × 5, which has only 2 and 5, so the decimal expansion is terminating.
In simple words: Reduce the fraction and look for primes other than 2 and 5 in the denominator.

Teacher's Note:
1. Cancel common factors before checking the denominator.
2. Part (i) fails because of the factor 3.

Question 4. What can you say about the prime factorization of the denominators of the following rationals : (i) 25.234567   (ii) \( 25.\overline{345678} \)
Answer:
(i) The decimal terminates, so the denominator (in lowest terms) is of the form 2^m × 5ⁿ, with m, n non-negative integers.
(ii) The decimal is non-terminating and repeating, so the denominator has at least one prime factor other than 2 and 5.
In simple words: A terminating decimal means only the primes 2 and 5 in the denominator; a repeating one means some other prime appears.

Teacher's Note:
1. Link the type of decimal to the denominator's primes.
2. Remember the fraction must be in lowest terms.

Question 5. Let ‘a’ and ‘b’ be any two positive integers. Then, there exists unique integers ‘q’ and ‘r’ such that a = bq + r. If b = 5, then find the possible values of r.
Answer:
The remainder satisfies 0 ≤ r < b, so 0 ≤ r < 5.
The possible values are 0, 1, 2, 3, 4.
In simple words: On division by 5 the remainder can only be one of the five numbers from 0 to 4.

Teacher's Note:
1. r is always smaller than the divisor.
2. r = 0 means 5 divides a exactly.

Question 6. Check whether the number 21q + 18 is of the form 7m + 4, for some integer q.
Answer:
21q + 18 = 21q + 14 + 4 = 7(3q + 2) + 4.
With m = 3q + 2 (an integer), this is of the form 7m + 4.
Answer: Yes.
In simple words: Split 18 as 14 + 4 so that 7 can be taken out.

Teacher's Note:
1. Always try to pull out the multiple of the divisor.
2. Test q = 1: 39 = 7 × 5 + 4.

Question 7. If n = 3q + 2, then check whether n + 7 is divisible by 3, for some integer q.
Answer:
n + 7 = 3q + 2 + 7 = 3q + 9 = 3(q + 3).
This is a multiple of 3.
Answer: Yes, n + 7 is divisible by 3.
In simple words: Substituting gives 3(q + 3), a multiple of 3.

Teacher's Note:
1. Just substitute the given form of n.
2. Since 2 + 7 = 9, the remainder part is a multiple of 3.

Question 8. If n = 5q + 4, then check whether n² − 1 is divisible by 5, for some integer q.
Answer:
n² − 1 = (5q + 4)² − 1 = 25q² + 40q + 16 − 1 = 25q² + 40q + 15.
This equals 5(5q² + 8q + 3), so it is a multiple of 5.
Answer: Yes.
In simple words: Expanding shows every term is a multiple of 5.

Teacher's Note:
1. Another route: n² − 1 = (n − 1)(n + 1) = (5q + 3)(5q + 5), and 5q + 5 is a multiple of 5.
2. Try q = 1: 81 − 1 = 80.

Question 9. Find the HCF of 1650 and 847.
Answer:
1650 = 847 × 1 + 803
847 = 803 × 1 + 44
803 = 44 × 18 + 11
44 = 11 × 4 + 0
The last divisor is 11, so H.C.F. = 11.
In simple words: Apply Euclid's division algorithm till the remainder is zero.

Teacher's Note:
1. The remainder 44 in the third step divides exactly by 11.
2. Check: 1650 = 11 × 150 and 847 = 11 × 77.

Question 10. Find the HCF of 2781 and 1242.
Answer:
2781 = 1242 × 2 + 297
1242 = 297 × 4 + 54
297 = 54 × 5 + 27
54 = 27 × 2 + 0
The last divisor is 27, so H.C.F. = 27.
In simple words: Divide the larger number by the smaller, then continue with divisor and remainder.

Teacher's Note:
1. Check: 2781 = 27 × 103 and 1242 = 27 × 46.
2. Write each step as a = bq + r.

Question 11. Find HCF of 13281 and 15844.
Answer:
15844 = 13281 × 1 + 2563
13281 = 2563 × 5 + 466
2563 = 466 × 5 + 233
466 = 233 × 2 + 0
The last divisor is 233, so H.C.F. = 233.
In simple words: Euclid's algorithm gives the HCF in four divisions.

Teacher's Note:
1. Start by dividing the larger number 15844 by 13281.
2. Check: 13281 = 233 × 57 and 15844 = 233 × 68.

Question 12. Find HCF of 97 and 101.
Answer:
101 = 97 × 1 + 4
97 = 4 × 24 + 1
4 = 1 × 4 + 0
The last divisor is 1, so H.C.F. = 1.
In simple words: Both 97 and 101 are primes, so their HCF is 1.

Teacher's Note:
1. Two different primes are always co-prime.
2. The algorithm confirms it in three steps.

Question 13. If the H.C.F. of 408 and 1032 is expressible in the form 1032m − 408 × 5, find m.
Answer:
1032 = 408 × 2 + 216
408 = 216 × 1 + 192
216 = 192 × 1 + 24
192 = 24 × 8 + 0
So H.C.F.(408, 1032) = 24.
Set 1032m − 408 × 5 = 24, i.e. 1032m = 24 + 2040 = 2064.
m = 2064 ÷ 1032 = 2.
In simple words: Find the HCF first, then solve for m in the given form.

Teacher's Note:
1. Check: 1032 × 2 − 2040 = 2064 − 2040 = 24.
2. The linear equation has just one unknown.

Question 14. Check whether 4ⁿ can end with the digit 0 for any natural number n.
Answer:
4ⁿ = (2²)ⁿ = 2²ⁿ, so its only prime factor is 2.
A number that ends with 0 needs 5 as a factor too, which 4ⁿ does not have.
Answer: No, 4ⁿ cannot end with 0.
In simple words: 4ⁿ has only the prime 2, but a number ending in 0 needs the prime 5 as well.

Teacher's Note:
1. Same method as for 6ⁿ in the worked example.
2. The last digit of 4ⁿ is only ever 4 or 6.

Question 15. Without actually performing the long division, state whether \( \frac{343}{875} \) has terminating decimal expansion or not.
Answer:
343 = 7³ and 875 = 5³ × 7, so \( \frac{343}{875} = \frac{7^3}{5^3\times7} = \frac{49}{125} \).
In lowest terms the denominator is 125 = 5³, of the form 2^m × 5ⁿ.
Answer: yes, it has a terminating decimal expansion (0.392).
In simple words: After cancelling the 7s, only 5s remain in the denominator.

Teacher's Note:
1. Reduce to lowest terms before applying the rule.
2. 49 ÷ 125 = 0.392.