CBSE Class 10 Maths Advanced Real Numbers Very Short Answer Questions Set 01

Very Short Answer Questions

Question 1. Find the largest number which divides 70 and 125, leaving remainders 5 and 8, respectively.
Answer:
Subtract the remainders: 70 − 5 = 65 and 125 − 8 = 117. The required number is the H.C.F. of 65 and 117.
117 = 65 × 1 + 52
65 = 52 × 1 + 13
52 = 13 × 4 + 0
H.C.F. = 13, and 13 is larger than both remainders (5 and 8), as it must be.
Answer: 13.
In simple words: The number must divide 65 and 117 exactly, so it is their H.C.F.

Teacher's Note:
1. Always remove the remainders first.
2. The divisor has to exceed every remainder; 13 does.

Question 2. If two positive integers a and b are written as a = x³y² and b = xy³ ; x, y are prime numbers, then find the HCF (a, b)
Answer:
Take each common prime with its smaller power: x appears as x³ and x¹, so x¹; y appears as y² and y³, so y².
H.C.F.(a, b) = \( xy^2 \).
In simple words: Smaller power of each shared prime gives the HCF.

Teacher's Note:
1. Both numbers contain only the primes x and y.
2. Larger powers would give the LCM instead.

Question 3. If two positive integers p and q can be expressed as p = ab² and q = a³b ; a, b being prime numbers, then find the LCM (p, q)
Answer:
Take each prime with its larger power: a appears as a¹ and a³, so a³; b appears as b² and b¹, so b².
L.C.M.(p, q) = \( a^3b^2 \).
In simple words: Larger power of each prime gives the LCM.

Teacher's Note:
1. Use H.C.F. × L.C.M. = pq as a check: HCF is ab, and ab × a³b² = a⁴b³ = pq.
2. Do not add the powers; pick the bigger one.

Question 4. Find the least number that is divisible by all the numbers from 1 to 10 (both inclusive).
Answer:
The required number is the L.C.M. of 1, 2, ..., 10.
Highest powers of primes needed: 2³ (from 8), 3² (from 9), 5 and 7.
L.C.M. = 8 × 9 × 5 × 7 = 2520.
In simple words: Take the biggest power of every prime up to 10 and multiply.

Teacher's Note:
1. 8 already covers 2, 4 and 6's factor 2; 9 covers 3 and 6.
2. The same number 2520 appears in the step-length examples of this chapter.

Question 5. The decimal expansion of the rational number \( \frac{14587}{1250} \) will terminate after how many decimal places ?
Answer:
1250 = 2 × 5⁴, and 14587 has no common factor with it, so the fraction is in lowest terms.
The larger exponent is 4 (of the prime 5), so multiply top and bottom by 2³: \( \frac{14587\times8}{10^4} = \frac{116696}{10000} = 11.6696 \).
The decimal ends after four places.
In simple words: The decimal has as many places as the larger power of 2 or 5 in the denominator.

Teacher's Note:
1. Write the denominator as 2^m × 5ⁿ and take max(m, n).
2. Here the 5⁴ forces four places.

Question 6. Can two numbers have 18 as their HCF and 380 as their LCM ? Give reasons.
Answer:
The H.C.F. of two numbers always divides their L.C.M. (every common divisor is a divisor of each number, and each number divides the L.C.M.).
Here 380 ÷ 18 is not a whole number (18 × 21 = 378), so 18 does not divide 380.
Answer: No.
In simple words: An HCF must be a factor of the LCM, and 18 is not a factor of 380.

Teacher's Note:
1. Test whether the LCM is divisible by the HCF.
2. 380 = 2² × 5 × 19 has no factor 3, so it cannot be a multiple of 18.

Question 7. “The product of two consecutive positive integers is divisible by 2”. Is this statement true or false? Give reasons.
Answer:
Two consecutive positive integers are n and n + 1. One of them is even.
So their product n(n + 1) contains a factor 2.
Answer: True.
In simple words: Of any two consecutive numbers, one is even, so the product is even.

Teacher's Note:
1. Try 3 × 4 = 12 and 7 × 8 = 56.
2. The same idea shows n(n + 1)(n + 2) is divisible by 6.

Question 8. Use Euclid's division algorithm to find HCF of 870 and 225. [CBSE 10th Board 2013, 1 Mark]
Answer:
870 = 225 × 3 + 195
225 = 195 × 1 + 30
195 = 30 × 6 + 15
30 = 15 × 2 + 0
The last divisor is 15, so the H.C.F. is 15.
In simple words: Keep dividing the divisor by the remainder until the remainder is 0.

Teacher's Note:
1. The last non-zero remainder (15) is the H.C.F.
2. Check: 870 = 15 × 58 and 225 = 15 × 15.

Question 9. Find the LCM and HCF of 510 and 92 and verify that LCM × HCF = product of the two numbers. [CBSE 10th Board 2017, 1 Mark]
Answer:
510 = 2 × 3 × 5 × 17 and 92 = 2² × 23.
H.C.F. = 2 (the only common prime, smallest power). L.C.M. = 2² × 3 × 5 × 17 × 23 = 23460.
Verification: L.C.M. × H.C.F. = 23460 × 2 = 46920 and 510 × 92 = 46920, which are equal.
So HCF = 2, LCM = 23460.
In simple words: Factorise both numbers, take smallest common powers for the HCF and biggest powers for the LCM.

Teacher's Note:
1. After finding both, always multiply them to match the product of the numbers.
2. 92 = 4 × 23 and 23 is a prime.

Question 10. Write a rational number between \( \sqrt{2} \) and \( \sqrt{3} \).
Answer:
√2 ≈ 1.414 and √3 ≈ 1.732.
The rational number 1.5 = \( \frac{3}{2} \) lies between them.
Answer: \( \frac{3}{2} \) (any other fraction such as 1.6 or 1.7 also works).
In simple words: Choose a terminating decimal between 1.414 and 1.732.

Teacher's Note:
1. Use decimal approximations of the roots to see the interval.
2. Check by squaring: 1.5² = 2.25, which lies between 2 and 3.

Question 11. Write whether \( \frac{2\sqrt{45}+3\sqrt{20}}{2\sqrt5} \) on simplification gives a rational or an irrational number.
Answer:
√45 = 3√5 and √20 = 2√5, so the numerator is 2 × 3√5 + 3 × 2√5 = 12√5.
The expression is \( \frac{12\sqrt5}{2\sqrt5} = 6 \).
Answer: it is a rational number (equal to 6).
In simple words: The roots cancel completely, leaving the whole number 6.

Teacher's Note:
1. Simplify each root to a multiple of √5 first.
2. Do not decide before simplifying.

Question 12. There is remainder of 3 when a number is 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.
n² = 36q² + 36q + 9 = 6(6q² + 6q + 1) + 3.
The remainder is 3.
In simple words: Square the form 6q + 3 and read off the remainder.

Teacher's Note:
1. Example: 9 leaves 3 on division by 6, and 81 = 6 × 13 + 3.
2. Since 9 = 6 + 3, it contributes the remainder 3.

Question 13. Using Euclid's Division algorithm, find the HCF of 210 and 55.
Answer:
210 = 55 × 3 + 45
55 = 45 × 1 + 10
45 = 10 × 4 + 5
10 = 5 × 2 + 0
The last divisor is 5, so the H.C.F. is 5.
In simple words: Repeat the division lemma till the remainder becomes 0.

Teacher's Note:
1. The divisor in the final step is the HCF.
2. 210 = 5 × 42 and 55 = 5 × 11.

Question 14. Using Euclid's Division algorithm, find the HCF of 101 and 1277.
Answer:
1277 = 101 × 12 + 65
101 = 65 × 1 + 36
65 = 36 × 1 + 29
36 = 29 × 1 + 7
29 = 7 × 4 + 1
7 = 1 × 7 + 0
The last divisor is 1, so the H.C.F. is 1 (the numbers are co-prime).
In simple words: The chain ends with divisor 1, so the two numbers have no common factor other than 1.

Teacher's Note:
1. 101 is a prime number that does not divide 1277.
2. Start with the larger number as the dividend.

Question 15. Find the largest number that divides 245 and 1029 leaving a remainder of 5 in each case.
Answer:
Subtract the remainder: 245 − 5 = 240 and 1029 − 5 = 1024. The number is H.C.F.(240, 1024).
1024 = 240 × 4 + 64
240 = 64 × 3 + 48
64 = 48 × 1 + 16
48 = 16 × 3 + 0
H.C.F. = 16, which is greater than the remainder 5.
Answer: 16.
In simple words: Remove the common remainder and find the HCF of what is left.

Teacher's Note:
1. The answer must exceed the remainder 5.
2. 240 = 16 × 15 and 1024 = 16 × 64.