CBSE Class 10 Mathematics Question Bank Set 06

Question 1. Euclid's Division lemma states that for any two positive integers a and b, there exist unique integers q and r such that \(a = bq + r\), where r must satisfy.
Answer: \(0 \le r < b\)
In simple words: The remainder is always greater than or equal to zero and strictly less than the divisor.

Exam Tip: Memorize the strict inequality condition \(0 \le r < b\), as it is frequently tested in objective questions.

Question 2. Express 10010 and 140 as prime factors.
Answer:
For 10010:
\(10010 = 2 \times 5 \times 1001\)
\(1001 = 7 \times 143 = 7 \times 11 \times 13\)
Thus, \(10010 = 2 \times 5 \times 7 \times 11 \times 13\)
For 140:
\(140 = 2 \times 70 = 2 \times 2 \times 35 = 2^2 \times 5 \times 7\)
In simple words: Prime factorization means expressing a composite number as a product of prime numbers.

Exam Tip: Always write the prime factors in ascending order with proper exponents.

Question 3. If \(p/q\) is a rational number (\(q \neq 0\)), what is the condition of q so that the decimal representation of \(p/q\) is terminating?
Answer: The denominator q must be of the form \(2^n \times 5^m\), where n and m are non-negative integers.
In simple words: A fraction has a terminating decimal if its denominator only has prime factors of 2 and/or 5.

Exam Tip: Simplify the rational number to its lowest terms before checking the prime factors of the denominator.

Question 4. Write any one rational number between \(\sqrt{2}\) and \(\sqrt{3}\).
Answer:
Given values are approximately \(\sqrt{2} \approx 1.414\) and \(\sqrt{3} \approx 1.732\).
Any terminating or repeating decimal between these two values is a valid rational number, for example, 1.5321.
In simple words: Choose a terminating decimal that lies between 1.414 and 1.732.

Question 5. Find the \([\text{HCF} \times \text{LCM}]\) for 105 and 120.
Answer:
For any two positive integers a and b, the product of their HCF and LCM is equal to the product of the numbers:
\(\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b\)
Substituting \(a = 105\) and \(b = 120\):
\(\text{HCF} \times \text{LCM} = 105 \times 120 = 12600\)
In simple words: The product of the HCF and LCM of two numbers always equals the product of the numbers themselves.

Exam Tip: Save time by not calculating individual HCF and LCM values when applying this standard property.

Question 6. If two numbers are 26 and 91 and their H.C.F is 13 then LCM is.
Answer:
Using the relation \(\text{HCF} \times \text{LCM} = \text{Product of two numbers}\):
\(13 \times \text{LCM} = 26 \times 91\)
\(\text{LCM} = \frac{26 \times 91}{13} = 2 \times 91 = 182\)
In simple words: Divide the product of the two numbers by their HCF to get the LCM.

Exam Tip: Always cross-check your LCM by dividing it by both original numbers.

Question 7. The decimal expansion of the rational number \(\frac{33}{2^2 \cdot 5}\) will terminate after how many decimal places?
Answer:
The denominator is \(2^2 \cdot 5^1\). The maximum of the exponents of 2 and 5 is 2.
Therefore, the decimal expansion will terminate after two decimal places.
In simple words: Count the higher power between 2 and 5 in the denominator to find the number of decimal places.

Exam Tip: The number of decimal places equals \(\max(n, m)\) when the denominator is expressed as \(2^n 5^m\).

Question 8. HCF of two consecutive integers x and x+1 is.
Answer: 1
In simple words: Consecutive integers share no common factors other than 1.

Exam Tip: Any two consecutive integers are always co-prime.