CBSE Class 10 Mathematics Real Numbers, Polynomials and Pair of Linear Equations Worksheet Set 03

Answer: It is used to compute the HCF (highest common factor) of the two integers.

Teacher's Note:
1. The algorithm repeats division until the remainder is zero.
2. The last non-zero divisor is the HCF.

Answer: 124 = 24 × 5 + 4. 24 = 4 × 6 + 0. The remainder is zero, so HCF = 4.

Teacher's Note:
1. Divide the larger number by the smaller one.
2. Stop when the remainder becomes zero.

Answer: Every composite number can be expressed as a product of primes, and this factorisation is unique apart from the order in which the prime factors occur.

Teacher's Note:
1. Uniqueness means different orders give the same set of primes.
2. The theorem is the basis for finding HCF and LCM by prime factorisation.

Answer: The blank is LCM(a, b). LCM = \( \dfrac{306 \times 657}{9} \) = 34 × 657 = 22338.

Teacher's Note:
1. The product of two numbers equals the product of their HCF and LCM.
2. Divide before multiplying to keep the numbers small.

Answer: A number that cannot be written in the form \( \dfrac{p}{q} \), with p, q integers and q ≠ 0, is irrational. Examples are √2 and √3.

Teacher's Note:
1. Irrational numbers have non-terminating, non-repeating decimals.
2. The square root of a non-perfect-square prime is irrational.

Answer: Any positive integer is 3q, 3q + 1 or 3q + 2. Their squares are 9q² = 3(3q²), 9q² + 6q + 1 = 3(3q² + 2q) + 1 and 9q² + 12q + 4 = 3(3q² + 4q + 1) + 1. So the square is 3m or 3m + 1.

Teacher's Note:
1. Divide by 3 to get three possible remainders.
2. Square each form and regroup as a multiple of 3 plus a remainder.

Answer: 12 = 2² × 3, 15 = 3 × 5, 21 = 3 × 7. HCF = 3 (common factor with the least power). LCM = 2² × 3 × 5 × 7 = 420.

Teacher's Note:
1. For HCF take the smallest power of each common prime.
2. For LCM take the greatest power of every prime that appears.

Answer: No. 6ⁿ = 2ⁿ × 3ⁿ has no factor 5. A number ends in 0 only if it is divisible by both 2 and 5.

Teacher's Note:
1. A last digit of 0 needs the prime factors 2 and 5.
2. Use the prime factorisation of the base.

Answer: A polynomial of the form ax² + bx + c, where a, b, c are real numbers and a ≠ 0, is a quadratic polynomial. Its degree is 2.

Teacher's Note:
1. The degree is the highest power of the variable.
2. If a = 0 the polynomial would be linear.

Answer: The zeroes are the values of x for which p(x) = 0. Geometrically they are the x-coordinates of the points where the graph of y = p(x) meets the x-axis.

Teacher's Note:
1. A zero is also called a root of the equation p(x) = 0.
2. A quadratic polynomial can have at most two zeroes.

Answer: The graph is a parabola. It opens upward when the coefficient of x² is positive and downward when it is negative.

Teacher's Note:
1. The sign of a decides the direction.
2. A parabola that does not cross the x-axis has no real zeroes.

Answer: A polynomial of degree n has at most n zeroes. The graph of ax + b is a straight line that meets the x-axis at exactly one point, namely x = \( -\dfrac{b}{a} \).

Teacher's Note:
1. A cubic polynomial can have at most three zeroes.
2. A non-horizontal line cuts the x-axis once.

Answer: Sum = −7 and product = 10. x² + 7x + 10 = (x + 2)(x + 5), so the zeroes are −2 and −5, with sum −7 and product 10.

Teacher's Note:
1. Sum = −b/a and product = c/a.
2. Factorisation gives an independent check.

Answer: The polynomial is k[x² − (sum)x + product] = k\( \left(x^2 - \dfrac{1}{4}x - 1\right) \). Taking k = 4, we get 4x² − x − 4.

Teacher's Note:
1. Multiply by a suitable constant to clear fractions.
2. Any non-zero multiple is also correct.

Answer: x² − 2x − 8 = (x − 4)(x + 2), so the zeroes are 4 and −2. Sum = 2 = −(−2)/1 and product = −8 = −8/1, so the relations hold.

Teacher's Note:
1. Split −2x as −4x + 2x.
2. Compare with −b/a and c/a.

Answer: Since x − 1 is a factor, the value at x = 1 is zero. 3 + 2 − 3 + k = 0, so k = −2.

Teacher's Note:
1. Substitute x = 1 because x − 1 = 0 there.
2. The result must equal zero for a factor.

Answer: (i) \( \dfrac{a_1}{a_2} \ne \dfrac{b_1}{b_2} \). (ii) \( \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \ne \dfrac{c_1}{c_2} \). (iii) \( \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2} \).

Teacher's Note:
1. Unique solution means the lines intersect, inconsistent means parallel, and dependent means coincident.
2. Compare ratios of coefficients, not the coefficients themselves.

Answer: No solution needs parallel lines: \( \dfrac{3}{2} = \dfrac{2k}{5} \ne \dfrac{-2}{1} \). So 4k = 15 and k = \( \dfrac{15}{4} \). The third ratio is −2, which is different from \( \dfrac{3}{2} \).

Teacher's Note:
1. Write both equations in the form ax + by + c = 0 first.
2. Confirm the third ratio is different.

Answer: Let width = x m, so length = x + 4. Then x + x + 4 = 36, giving x = 16. Width = 16 m and length = 20 m.

Teacher's Note:
1. Half the perimeter equals length plus width.
2. Check: 16 + 20 = 36.

Answer: Multiply the first by 3 and the second by 2: 9x + 6y = 33 and 4x + 6y = 8. Subtracting gives 5x = 25, so x = 5. Then 15 + 2y = 11, so y = −2. Solution: x = 5, y = −2.

Teacher's Note:
1. Equalise the coefficients of y and then subtract.
2. Verify in the second equation: 10 − 6 = 4.

Answer: \( \dfrac{5}{10} = \dfrac{7}{14} = \dfrac{1}{2} \) but \( \dfrac{8}{4} = 2 \). The first two ratios are equal but the third differs, so the lines are parallel and not coincident.

Teacher's Note:
1. For coincident lines all three ratios must be equal.
2. Here the pair has no solution.

Answer: \( \dfrac{k}{12} = \dfrac{3}{k} = \dfrac{k-3}{k} \). From the last two, k − 3 = 3, so k = 6. The first ratio then gives \( \dfrac{6}{12} = \dfrac{3}{6} \), which is true. k = 6.

Teacher's Note:
1. Use the ratio that gives a linear equation.
2. Check the value in the remaining ratio.