CBSE Class 10 Maths question paper, with every solution worked

Correct option: (a)

(a) 0, 1, 2, 3, 4. Explanation: According to Euclid's division lemma, for any two positive integers $a$ and $b$, there exist unique integers $q$ and $r$ such that $a = bq + r$, where $0 \le r < b$. Here, the divisor $b$ is 5. Therefore, the remainder $r$ must be less than 5 and non-negative. The possible integer values for $r$ are 0, 1, 2, 3, and 4.

Correct option: (c)

(c) 75 minutes. To find when both buses meet at the start, we need the Least Common Multiple (LCM) of their trip durations. $15 = 3 \times 5$ and $25 = 5^2$. $LCM(15, 25) = 3 \times 5^2 = 75$ minutes. Thus, they meet after 75 minutes.

Correct option: (b)

(b) $2^m \times 5^n$. A rational number $\frac{p}{q}$ has a terminating decimal expansion if the prime factorization of $q$ is of the form $2^m \times 5^n$, where $m, n \geq 0$. Any other prime factor in $q$ results in a non-terminating repeating decimal.

Correct option: (b)

(b) 5. Explanation: To find the maximum number of vials per box that divides both batch sizes evenly, we calculate the HCF of 140 and 3825. Prime factorization of 140 is 2² × 5 × 7. Prime factorization of 3825 is 3² × 5² × 17. The only common prime factor is 5. Therefore, HCF(140, 3825) = 5.

Correct option: (c)

(c) The product of a non-zero rational number and an irrational number is always irrational. Explanation: The area is (5/2)√3. Since 5/2 is a non-zero rational number and √3 is an irrational number, their product must be irrational. Approximation by decimals does not make a number rational; rational numbers have terminating or repeating decimal expansions.

Correct option: (a)

(a) 6. Explanation: The HCF is the product of the smallest powers of all common prime factors. The common prime factors are 2 and 3. The smallest power of 2 present in all is $2^1$, and the smallest power of 3 present in all is $3^1$. Therefore, HCF $= 2^1 \times 3^1 = 6$.

Correct option: (c)

(c) A is true but R is false. Explanation: The number $12^n$ ends with 0 only if it is divisible by 5, which requires 5 to be a prime factor. Since $12 = 2^2 imes 3$, the prime factors of $12^n$ are only 2 and 3. Thus, A is true because 5 is not a factor, but R is false because the prime factorisation does not contain 5.

Correct option: (a)

(a) Both A and R are true and R is the correct explanation of A. Explanation: Assertion (A) is true because $\frac{14}{5} = 2.8$. Reason (R) is true and provides the condition for terminating decimals. Since the denominator 5 is of the form $2^0 \times 5^1$, the condition is satisfied, making R the correct explanation for A.

Yes, it has a terminating decimal expansion. We simplify $\frac{35}{50}$ by dividing numerator and denominator by 5, getting $\frac{7}{10}$. The denominator is 10, which factors into $2 \times 5$. Since the prime factorization of the denominator is of the form $2^n \times 5^m$, the decimal expansion terminates. Specifically, $\frac{7}{10} = 0.7$.

We can factor out 11 from the expression: $11(7 \times 13 + 1)$. Calculating the term inside the bracket: $7 \times 13 = 91$, so $91 + 1 = 92$. The number becomes $11 \times 92$. Since 92 can be further factorized into $2^2 \times 23$, the original number is $11 \times 2^2 \times 23$. It has factors other than 1 and itself (such as 2, 11, 23), making it a composite number.

1305 = 3² × 5 × 29. Explanation: Dividing 1305 by 5 gives 261. Dividing 261 by 9 (3×3) gives 29, which is prime.

The decimal expansion of 17/32 terminates after 5 decimal places, as the prime factorisation of the denominator 32 is 2^5.

The relation is a = bq + r, where q is the quotient and r is the remainder such that 0 ≤ r < b.

Let the number be N = 7 × 11 × 13 + 13. We can factor out the common term 13: N = 13 × (7 × 11 + 1). Calculate the value inside the parenthesis: 7 × 11 = 77. So, N = 13 × (77 + 1) = 13 × 78. Now, factorize 78 further into primes: 78 = 2 × 39 = 2 × 3 × 13. Substituting this back, N = 13 × 2 × 3 × 13 = 2 × 3 × 13². Since N can be expressed as a product of prime factors (2, 3, and 13) and has more than two factors, it is a composite number.

The prime factorisation of the denominator 8 is 2^3, which is of the form 2^x * 5^y. Hence, 17/8 has a terminating decimal expansion.

The odd numbers between 20 and 30 are 21, 23, 25, 27, 29. Check each: 21 is divisible by 3 ($3 \times 7$). 23 is prime (not divisible by 2, 3, 5). 25 is divisible by 5 ($5 \times 5$). 27 is divisible by 3 ($3 \times 9$). 29 is prime (not divisible by 2, 3, 5). Thus, there ARE primes (23 and 29) in this range. The statement in the prompt is false; the question asks to show existence or non-existence. Correction: Show that 23 and 29 are prime. 23 has no factors other than 1 and 23. 29 has no factors other than 1 and 29. Hence, 23 and 29 are prime numbers.

To find when they meet again, we need the LCM of 45 and 60. Prime factorization of 45: 45 = 3 × 3 × 5 = 3² × 5¹. Prime factorization of 60: 60 = 2 × 2 × 3 × 5 = 2² × 3¹ × 5¹. LCM is the product of the greatest power of each prime factor involved. LCM(45, 60) = 2² × 3² × 5¹ = 4 × 9 × 5 = 180. They will meet again after 180 minutes (or 3 hours).

The prime factorisation of the denominator 441 is 3^2 * 7^2. Since it contains prime factors other than 2 and 5, the decimal expansion of 125/441 is non-terminating repeating.

First, find the prime factorization of each number: 12 = 2² × 3¹ 15 = 3¹ × 5¹ 21 = 3¹ × 7¹ HCF is the product of the smallest power of each common prime factor. The only common prime factor is 3, with the smallest power being 3¹. So, HCF(12, 15, 21) = 3. LCM is the product of the greatest power of each prime factor involved. The primes involved are 2, 3, 5, and 7. Greatest powers: 2² (from 12), 3¹ (common), 5¹ (from 15), 7¹ (from 21). LCM(12, 15, 21) = 2² × 3¹ × 5¹ × 7¹ = 4 × 3 × 5 × 7 = 420. So, HCF is 3 and LCM is 420.

If 6ⁿ ends with the digit 0, it must be divisible by 10. For a number to be divisible by 10, its prime factorization must contain both the prime factors 2 and 5. Let us find the prime factorization of 6ⁿ. 6 = 2 × 3 Therefore, 6ⁿ = (2 × 3)ⁿ = 2ⁿ × 3ⁿ. The prime factors of 6ⁿ are only 2 and 3. By the Fundamental Theorem of Arithmetic, the prime factorization is unique. Since the prime factor 5 is not present in the factorization of 6ⁿ, 6ⁿ cannot be divisible by 5, and consequently, not divisible by 10. Therefore, 6ⁿ cannot end with the digit 0 for any natural number n.

Prime factorization of the numbers: 336 = 2 × 168 = 2² × 84 = 2³ × 42 = 2⁴ × 21 = 2⁴ × 3¹ × 7¹ 54 = 2 × 27 = 2¹ × 3³ HCF is the product of the smallest power of common prime factors. Common primes are 2 and 3. Smallest power of 2 is 2¹. Smallest power of 3 is 3¹. HCF(336, 54) = 2¹ × 3¹ = 6. LCM is the product of the greatest power of each prime factor involved. Greatest power of 2 is 2⁴. Greatest power of 3 is 3³. Greatest power of 7 is 7¹. LCM(336, 54) = 2⁴ × 3³ × 7¹ = 16 × 27 × 7 = 432 × 7 = 3024. Verification: Product of numbers = 336 × 54 = 18,144. Product of HCF and LCM = 6 × 3024 = 18,144. The relationship HCF × LCM = Product of numbers is verified.

We perform prime factorisation of 3825. 3825 ends in 5, so it is divisible by 5. 3825 ÷ 5 = 765. 765 ÷ 5 = 153. 153 is divisible by 3 (sum of digits 1+5+3=9). 153 ÷ 3 = 51. 51 ÷ 3 = 17. 17 is a prime number. So, 3825 = 5 × 5 × 3 × 3 × 17. In exponential form, arranging primes in ascending order: 3² × 5² × 17.

Step 1: 210 = 55 × 3 + 45. Step 2: 55 = 45 × 1 + 10. Step 3: 45 = 10 × 4 + 5. Step 4: 10 = 5 × 2 + 0. HCF is 5. Working backwards: 5 = 45 - 10 × 4. 5 = 45 - (55 - 45 × 1) × 4 = 45 × 5 - 55 × 4. 5 = (210 - 55 × 3) × 5 - 55 × 4 = 210 × 5 - 55 × 19. So x = 5, y = -19.