CBSE Class 9 Mathematics Revision Assignments Set 02

Question 1. Find the value of \(a\) and \(b\) if \(\frac{\sqrt{2} + \sqrt{3}}{3\sqrt{2} - 2\sqrt{3}} = a + b\sqrt{6}\)
Answer: Rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator, which is \(3\sqrt{2} + 2\sqrt{3}\):
\(\text{LHS} = \frac{(\sqrt{2} + \sqrt{3})(3\sqrt{2} + 2\sqrt{3})}{(3\sqrt{2} - 2\sqrt{3})(3\sqrt{2} + 2\sqrt{3})}\)
Expand the numerator:
\(= \frac{\sqrt{2}(3\sqrt{2} + 2\sqrt{3}) + \sqrt{3}(3\sqrt{2} + 2\sqrt{3})}{(3\sqrt{2})^2 - (2\sqrt{3})^2}\)
\(= \frac{3(2) + 2\sqrt{6} + 3\sqrt{6} + 2(3)}{18 - 12}\)
\(= \frac{6 + 5\sqrt{6} + 6}{6}\)
\(= \frac{12 + 5\sqrt{6}}{6}\)
\(= 2 + \frac{5}{6}\sqrt{6}\)
Comparing this with \(a + b\sqrt{6}\), we get:
\(a = 2\), \(b = \frac{5}{6}\)
In simple words: Multiply top and bottom by the conjugate to remove roots from the denominator, then compare coefficients.

Exam Tip: Always look for the conjugate of the binomial denominator to simplify surds.

Question 2. If \(a\) and \(b\) are rational numbers and \(\frac{\sqrt{11} - \sqrt{7}}{\sqrt{11} + \sqrt{7}} = a - b\sqrt{77}\), then find the values of \(a\) and \(b\).
Answer: Rationalize the denominator using the conjugate \(\sqrt{11} - \sqrt{7}\):
\(\text{LHS} = \frac{(\sqrt{11} - \sqrt{7})(\sqrt{11} - \sqrt{7})}{(\sqrt{11} + \sqrt{7})(\sqrt{11} - \sqrt{7})}\)
\(= \frac{(\sqrt{11} - \sqrt{7})^2}{(\sqrt{11})^2 - (\sqrt{7})^2}\)
\(= \frac{11 + 7 - 2\sqrt{77}}{11 - 7}\)
\(= \frac{18 - 2\sqrt{77}}{4}\)
\(= \frac{18}{4} - \frac{2}{4}\sqrt{77}\)
\(= \frac{9}{2} - \frac{1}{2}\sqrt{77}\)
Comparing with \(a - b\sqrt{77}\), we get:
\(a = \frac{9}{2}\), \(b = \frac{1}{2}\)
In simple words: Use the formula \((x-y)^2 = x^2 + y^2 - 2xy\) in the numerator and the difference of squares in the denominator.

Exam Tip: Be careful with the negative sign when matching coefficients.

Question 3. In a survey, it was found that 9 out of every 11 households are donating some amount of their income to an orphanage or old age home or institutions for physically handicapped.
I. What fraction of households is not donating?
II. Write it in decimal form and find what type of decimal expansion it has?
III. What values of society are depicted in this question?
Answer: I. Fraction of households donating = \(\frac{9}{11}\)
Fraction of households not donating = \(1 - \frac{9}{11} = \frac{2}{11}\)
II. Converting \(\frac{2}{11}\) to decimal form using long division: \(2 \div 11 = 0.181818... = 0.\overline{18}\).
The type of decimal expansion is non-terminating repeating (recurring).
III. The values depicted are empathy, social responsibility, charity, kindness, and helpfulness towards the needy and underprivileged in society.
In simple words: Subtract the donating fraction from 1 to find the non-donating fraction, then divide to get the decimal.

Exam Tip: Represent recurring decimals with a bar over the repeating block of digits.

Question 4. If \((4)^{2x-1} - (16)^{x-1} = 384\), then find the value of \(x\).
Answer: Express all terms with base 4:
\((4)^{2x-1} - (4^2)^{x-1} = 384\)
\((4)^{2x-1} - (4)^{2x-2} = 384\)
Factor out \((4)^{2x-2}\):
\((4)^{2x-2} (4^1 - 1) = 384\)
\((4)^{2x-2} (3) = 384\)
Divide both sides by 3:
\((4)^{2x-2} = \frac{384}{3}\)
\((4)^{2x-2} = 128\)
Convert both sides to base 2:
\((2^2)^{2x-2} = 2^7\)
\((2)^{4x-4} = 2^7\)
Equate the exponents since bases are equal:
\(4x - 4 = 7\)
\(4x = 11\)
\(x = \frac{11}{4}\)
In simple words: Convert bases to the same number, factor out the lowest power, and equate exponents.

Exam Tip: Exponent laws like \((a^m)^n = a^{mn}\) are crucial for solving exponential equations.

Question 5. Assuming that \(x, y, z\) are positive real numbers and the exponents are all rational numbers, show that \(\left(\frac{x^a}{x^b}\right)^{a+b} \left(\frac{x^b}{x^c}\right)^{b+c} \left(\frac{x^c}{x^a}\right)^{c+a} = 1\)
Answer: Consider the left-hand side (LHS):
\(\text{LHS} = (x^{a-b})^{a+b} \cdot (x^{b-c})^{b+c} \cdot (x^{c-a})^{c+a}\)
Using the power rule \((x^m)^n = x^{mn}\):
\(= x^{(a-b)(a+b)} \cdot x^{(b-c)(b+c)} \cdot x^{(c-a)(c+a)}\)
Using the algebraic identity \((m-n)(m+n) = m^2 - n^2\):
\(= x^{a^2 - b^2} \cdot x^{b^2 - c^2} \cdot x^{c^2 - a^2}\)
Using the product rule \(x^m \cdot x^n = x^{m+n}\):
\(= x^{(a^2 - b^2 + b^2 - c^2 + c^2 - a^2)}\)
\(= x^0\)
\(= 1 = \text{RHS}\)
In simple words: Apply laws of exponents to simplify powers into products, then use algebraic identities to cancel out terms.

Exam Tip: Memorize basic algebraic identities as they frequently simplify complex index problems.

Question 6. Evaluate \(\frac{40}{\sqrt{5} + \sqrt{20} + \sqrt{40} - \sqrt{5}}\) when it is given \(\sqrt{10} = 3.162\) and \(\sqrt{5} = 2.236\)
Answer: Simplify the denominator expression:
\(\sqrt{5} + \sqrt{20} + \sqrt{40} - \sqrt{5}\)
\(= \sqrt{20} + \sqrt{40}\)
\(= \sqrt{4 \times 5} + \sqrt{4 \times 10}\)
\(= 2\sqrt{5} + 2\sqrt{10}\)
Now substitute this back into the fraction:
\(\frac{40}{2\sqrt{5} + 2\sqrt{10}} = \frac{20}{\sqrt{5} + \sqrt{10}}\)
Rationalize the denominator by multiplying numerator and denominator by \((\sqrt{10} - \sqrt{5})\):
\(= \frac{20(\sqrt{10} - \sqrt{5})}{(\sqrt{10} + \sqrt{5})(\sqrt{10} - \sqrt{5})}\)
\(= \frac{20(\sqrt{10} - \sqrt{5})}{10 - 5}\)
\(= \frac{20(\sqrt{10} - \sqrt{5})}{5}\)
\(= 4(\sqrt{10} - \sqrt{5})\)
Substitute the given values \(\sqrt{10} = 3.162\) and \(\sqrt{5} = 2.236\):
\(= 4(3.162 - 2.236)\)
\(= 4(0.926)\)
\(= 3.704\)
In simple words: Simplify the surds first, rationalize the denominator, and then substitute the decimal values.

Exam Tip: Simplify individual surds completely before performing addition or subtraction.

Question 7. Find the area of a triangle whose two sides are \(24\text{ cm}\) and \(10\text{ cm}\) and the perimeter of the triangle is \(62\text{ cm}\).
Answer: Let the three sides of the triangle be \(a = 24\text{ cm}\), \(b = 10\text{ cm}\), and \(c\).
Perimeter \(= a + b + c = 62\text{ cm}\)
\(24 + 10 + c = 62\)
\(34 + c = 62\)
\(c = 62 - 34 = 28\text{ cm}\)
Semi-perimeter \(s = \frac{62}{2} = 31\text{ cm}\)
Using Heron's formula, Area \(= \sqrt{s(s-a)(s-b)(s-c)}\):
Area \(= \sqrt{31(31-24)(31-10)(31-28)}\)
\(= \sqrt{31 \times 7 \times 21 \times 3}\)
\(= \sqrt{31 \times 7 \times (7 \times 3) \times 3}\)
\(= \sqrt{31 \times 7^2 \times 3^2}\)
\(= 7 \times 3 \times \sqrt{31}\)
\(= 21\sqrt{31}\text{ cm}^2\)
In simple words: Find the third side using the perimeter, calculate the semi-perimeter, and apply Heron's formula.

Exam Tip: Factorize numbers inside the square root to easily pull out pairs without doing heavy multiplication.

Question 8. A field is in the shape of a trapezium whose parallel sides are \(25\text{ cm}\) and \(10\text{ cm}\). The non-parallel sides are \(14\text{ cm}\) and \(13\text{ cm}\). Find the area of the field.
Answer: Let the trapezium be ABCD with parallel sides \(AB = 25\text{ cm}\) and \(DC = 10\text{ cm}\), and non-parallel sides \(AD = 13\text{ cm}\) and \(BC = 14\text{ cm}\).
Draw a line \(CE \parallel AD\) meeting \(AB\) at \(E\). Then AECD is a parallelogram.
\(AE = DC = 10\text{ cm}\), and \(CE = AD = 13\text{ cm}\).
\(EB = AB - AE = 25 - 10 = 15\text{ cm}\).
For triangle CEB, the sides are \(13\text{ cm}\), \(14\text{ cm}\), and \(15\text{ cm}\).
Semi-perimeter of \(\Delta CEB\) is \(s = \frac{13 + 14 + 15}{2} = 21\text{ cm}\).
Area of \(\Delta CEB = \sqrt{21(21-13)(21-14)(21-15)} = \sqrt{21 \times 8 \times 7 \times 6}\)
\(= \sqrt{(3 \times 7) \times (4 \times 2) \times 7 \times (2 \times 3)} = \sqrt{3^2 \times 7^2 \times 4^2} = 3 \times 7 \times 4 = 84\text{ cm}^2\).
Also, Area of \(\Delta CEB = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 15 \times h = 84\)
\(h = \frac{84 \times 2}{15} = \frac{168}{15} = \frac{56}{5} = 11.2\text{ cm}\).
Area of trapezium \(= \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}\)
\(= \frac{1}{2} \times (25 + 10) \times \frac{56}{5} = \frac{1}{2} \times 35 \times \frac{56}{5} = 7 \times 28 = 196\text{ cm}^2\).
In simple words: Convert the trapezium into a parallelogram and a triangle, find the triangle's area and height, then use the trapezium area formula.

Exam Tip: Drawing a parallel line to one of the non-parallel sides converts a trapezium problem into a standard triangle problem.

Question 9. If ABC is an isosceles triangle with \(AB = AC\). The perimeter of the triangle is \(36\text{ cm}\) and \(AB = 10\text{ cm}\). What is the area of triangle?
Answer: Since \(\Delta ABC\) is isosceles with \(AB = AC = 10\text{ cm}\):
Perimeter \(= AB + AC + BC = 36\text{ cm}\)
\(10 + 10 + BC = 36\)
\(20 + BC = 36\)
\(BC = 36 - 20 = 16\text{ cm}\)
Sides of the triangle are \(a = 10\text{ cm}\), \(b = 10\text{ cm}\), and \(c = 16\text{ cm}\).
Semi-perimeter \(s = \frac{36}{2} = 18\text{ cm}\).
Using Heron's formula:
Area \(= \sqrt{18(18-10)(18-10)(18-16)}\)
\(= \sqrt{18 \times 8 \times 8 \times 2}\)
\(= \sqrt{(9 \times 2) \times 8 \times 8 \times 2}\)
\(= \sqrt{9 \times 2^2 \times 8^2}\)
\(= 3 \times 2 \times 8 = 48\text{ cm}^2\)
In simple words: Find the third side from the perimeter, then apply Heron's formula with the three side lengths.
Exam Tip: For an isosceles triangle, dropping an altitude to the unequal side also works by splitting it into two congruent right-angled triangles.

Question 10. ABCD is a rhombus whose one side is \(10\text{ cm}\) and one of its diagonals is \(16\text{ cm}\). What is the area of rhombus?
Answer: Let the diagonals of rhombus ABCD intersect at point O. All sides of a rhombus are equal, so \(AB = 10\text{ cm}\).
The diagonals of a rhombus bisect each other at right angles (\(90^\circ\)).
Let diagonal \(d_1 = AC = 16\text{ cm}\). Then \(AO = \frac{16}{2} = 8\text{ cm}\).
In right-angled triangle AOB: \(AB^2 = AO^2 + BO^2\)
\(10^2 = 8^2 + BO^2\)
\(100 = 64 + BO^2\)
\(BO^2 = 100 - 64 = 36\)
\(BO = 6\text{ cm}\)
Therefore, the other diagonal \(d_2 = BD = 2 \times BO = 2 \times 6 = 12\text{ cm}\).
Area of rhombus \(= \frac{1}{2} \times d_1 \times d_2\)
\(= \frac{1}{2} \times 16 \times 12 = 8 \times 12 = 96\text{ cm}^2\)
In simple words: Use the right-angled triangle formed by the intersecting diagonals to find half of the second diagonal, then apply the rhombus area formula.

Exam Tip: Remember that diagonals of a rhombus bisect each other perpendicularly.

Question 11. A triangle and a parallelogram have the same area. If the sides of the triangle are \(26\text{ cm}\), \(28\text{ cm}\) and \(30\text{ cm}\) and the parallelogram stands on the base \(28\text{ cm}\), then find the height of the parallelogram.
Answer: First, find the area of the triangle with sides \(a = 26\text{ cm}\), \(b = 28\text{ cm}\), and \(c = 30\text{ cm}\).
Semi-perimeter \(s = \frac{26 + 28 + 30}{2} = \frac{84}{2} = 42\text{ cm}\).
Area of triangle \(= \sqrt{42(42-26)(42-28)(42-30)}\)
\(= \sqrt{42 \times 16 \times 14 \times 12}\)
\(= \sqrt{(6 \times 7) \times 16 \times 14 \times (2 \times 6)}\)
\(= \sqrt{6^2 \times 7 \times 16 \times (2 \times 7) \times 2}\)
\(= \sqrt{6^2 \times 7^2 \times 16 \times 2^2}\)
\(= 6 \times 7 \times 4 \times 2 = 336\text{ cm}^2\).
Since the area of the parallelogram equals the area of the triangle, Area of parallelogram \(= 336\text{ cm}^2\).
Base of parallelogram \(= 28\text{ cm}\).
Area of parallelogram \(= \text{base} \times \text{height}\)
\(336 = 28 \times h\)
\(h = \frac{336}{28} = 12\text{ cm}\).
In simple words: Calculate the triangle's area using Heron's formula, equate it to the parallelogram's area, and solve for height.

Exam Tip: Area of a parallelogram is simply base multiplied by height.

Question 12. The sides of a triangle are \(x\), \(x+1\), \(2x-1\) and its area is \(x\sqrt{10}\). Find the value of \(x\).
Answer: Sides of the triangle are \(a = x\), \(b = x+1\), \(c = 2x-1\).
Semi-perimeter \(s = \frac{x + (x+1) + (2x-1)}{2} = \frac{4x}{2} = 2x\).
Using Heron's formula, Area \(= \sqrt{s(s-a)(s-b)(s-c)}\):
\(s - a = 2x - x = x\)
\(s - b = 2x - (x+1) = x - 1\)
\(s - c = 2x - (2x-1) = 1\)
Area \(= \sqrt{2x \cdot x \cdot (x-1) \cdot 1} = \sqrt{2x^2(x-1)} = x\sqrt{2(x-1)}\).
Given that Area \(= x\sqrt{10}\):
\(x\sqrt{2(x-1)} = x\sqrt{10}\)
Dividing both sides by \(x\) (\(x \neq 0\)):
\(\sqrt{2(x-1)} = \sqrt{10}\)
Square both sides:
\(2(x-1) = 10\)
\(x - 1 = 5\)
\(x = 6\)
In simple words: Find terms for Heron's formula in terms of \(x\), simplify the radical expression, and equate it to the given area.

Exam Tip: Always simplify the terms \(s-a\), \(s-b\), and \(s-c\) first before substituting into the radical.

Question 13. Sides of triangle are in the ratio \(12 : 17 : 25\) and its perimeter is \(540\text{ cm}\). Find the area.
Answer: Let the sides of the triangle be \(12x\), \(17x\), and \(25x\).
Perimeter \(= 12x + 17x + 25x = 540\text{ cm}\)
\(54x = 540\)
\(x = 10\)
Therefore, the sides are:
\(a = 12 \times 10 = 120\text{ cm}\)
\(b = 17 \times 10 = 170\text{ cm}\)
\(c = 25 \times 10 = 250\text{ cm}\)
Semi-perimeter \(s = \frac{540}{2} = 270\text{ cm}\).
Using Heron's formula:
Area \(= \sqrt{270(270-120)(270-170)(270-250)}\)
\(= \sqrt{270 \times 150 \times 100 \times 20}\)
\(= \sqrt{(27 \times 10) \times (15 \times 10) \times 100 \times (2 \times 10)}\)
\(= \sqrt{(9 \times 3) \times (5 \times 3) \times 10 \times 10 \times 100 \times 2 \times 10}\)
\(= \sqrt{3^2 \times 3^2 \times 5^2 \times 10^2 \times 10^2}\)
\(= 3 \times 3 \times 5 \times 10 \times 10 = 4500\text{ cm}^2\)
In simple words: Use variables with the given ratio to find the exact side lengths, then apply Heron's formula.

Exam Tip: When sides are given in ratios, always introduce a common variable \(x\).

Question 14. Find the zeroes of the polynomial \(3x^2 + x - 2\).
Answer: Let \(p(x) = 3x^2 + x - 2\).
To find the zeroes, set \(p(x) = 0\):
\(3x^2 + x - 2 = 0\)
Factorize by splitting the middle term (product = \(3 \times (-2) = -6\), sum = 1):
\(3x^2 + 3x - 2x - 2 = 0\)
\(3x(x + 1) - 2(x + 1) = 0\)
\((3x - 2)(x + 1) = 0\)
Setting each factor to zero:
\(3x - 2 = 0 \implies x = \frac{2}{3}\)
\(x + 1 = 0 \implies x = -1\)
Thus, the zeroes of the polynomial are \(\frac{2}{3}\) and \(-1\).
In simple words: Factor the quadratic expression by splitting the middle term and find the values of \(x\) that make it zero.

Exam Tip: Check your factorization by expanding the factors back out.

Question 15. If \(ax^3 + bx + x - 6\) has \((x + 2)\) as a factor and leaves remainder 4, when divided by \((x - 2)\), then find the values of \(a\) and \(b\).
Answer: Let \(p(x) = ax^3 + bx + x - 6 = ax^3 + (b+1)x - 6\).
Since \((x + 2)\) is a factor, by the Factor Theorem, \(p(-2) = 0\):
\(a(-2)^3 + (b+1)(-2) - 6 = 0\)
\(-8a - 2b - 2 - 6 = 0\)
\(-8a - 2b - 8 = 0\)
\(4a + b = -4 \quad \text{--- (Equation 1)}\)
When divided by \((x - 2)\), the remainder is 4. By the Remainder Theorem, \(p(2) = 4\):
\(a(2)^3 + (b+1)(2) - 6 = 4\)
\(8a + 2b + 2 - 6 = 4\)
\(8a + 2b - 4 = 4\)
\(8a + 2b = 8\)
\(4a + b = 4 \quad \text{--- (Equation 2)}\)
Comparing Equation 1 and Equation 2 yields a contradiction (\(-4 = 4\)), which implies a sign typo in the original prompt's expression \(bx + x\). Let us treat the polynomial as \(p(x) = ax^3 + bx^2 + x - 6\):
With \(p(x) = ax^3 + bx^2 + x - 6\):
\(p(-2) = a(-2)^3 + b(-2)^2 + (-2) - 6 = -8a + 4b - 8 = 0 \implies -2a + b = 2\)
\(p(2) = a(2)^3 + b(2)^2 + (2) - 6 = 8a + 4b - 4 = 4 \implies 8a + 4b = 8 \implies 2a + b = 2\)
Adding the two equations: \(4b = 4 \implies b = 1\).
Substituting \(b = 1\): \(-2a + 1 = 2 \implies -2a = 1 \implies a = -\frac{1}{2}\).
In simple words: Apply the Factor Theorem (\(p(-2)=0\)) and Remainder Theorem (\(p(2)=4\)) to set up a system of linear equations, then solve for \(a\) and \(b\).

Exam Tip: Factor Theorem states if \((x-c)\) is a factor, \(p(c)=0\). Remainder Theorem states division by \((x-c)\) leaves remainder \(p(c)\).

Question 16. If the polynomial \(p(x) = 2x^3 + bx^2 + 3x - 5\) and \(q(x) = x^3 + x^2 - 4x - b\) leave the same remainder, when divided by \(x - 2\), then prove that \(b = -\frac{13}{5}\).
Answer: By the Remainder Theorem, when \(p(x)\) is divided by \((x-2)\), the remainder is \(p(2)\).
\(p(2) = 2(2)^3 + b(2)^2 + 3(2) - 5\)
\(= 2(8) + 4b + 6 - 5\)
\(= 16 + 4b + 1 = 4b + 17\)
Similarly, the remainder when \(q(x)\) is divided by \((x-2)\) is \(q(2)\):
\(q(2) = (2)^3 + (2)^2 - 4(2) - b\)
\(= 8 + 4 - 8 - b = 4 - b\)
Since both leave the same remainder, set \(p(2) = q(2)\):
\(4b + 17 = 4 - b\)
\(4b + b = 4 - 17\)
\(5b = -13\)
\(b = -\frac{13}{5}\)
Hence proved.
In simple words: Find the remainder for both polynomials by substituting \(x = 2\), equate them, and solve for \(b\).

Exam Tip: "Leaves the same remainder" means you simply calculate \(p(c)\) and \(q(c)\) and equate them directly.

Question 17. Solve: \(x^4 - 625\)
Answer: Factor the expression using the difference of squares formula \(a^2 - b^2 = (a-b)(a+b)\):
\(x^4 - 625 = (x^2)^2 - (25)^2\)
\(= (x^2 - 25)(x^2 + 25)\)
Further factor the term \((x^2 - 25)\):
\(= (x^2 - 5^2)(x^2 + 25)\)
\(= (x - 5)(x + 5)(x^2 + 25)\)
Setting the expression to zero to solve: \((x - 5)(x + 5)(x^2 + 25) = 0\)
Real solutions are \(x = 5\) and \(x = -5\) (since \(x^2 + 25 = 0\) gives non-real complex roots \(x = \pm 5i\)).
In simple words: Treat \(x^4\) as \((x^2)^2\) and repeatedly apply the difference of squares formula.

Exam Tip: Sum of squares like \(x^2 + 25\) cannot be factored further using real numbers.

Question 18. Solve: \(5(3x + y)^2 + 6(3x + y) - 8\)
Answer: Let \(u = 3x + y\). Substitute \(u\) into the given expression:
\(5u^2 + 6u - 8\)
Factorize by splitting the middle term (product = \(5 \times (-8) = -40\), sum = 6):
\(5u^2 + 10u - 4u - 8 = 0\)
\(5u(u + 2) - 4(u + 2) = 0\)
\((5u - 4)(u + 2) = 0\)
Substitute back \(u = 3x + y\):
\((5(3x + y) - 4)(3x + y + 2) = 0\)
\((15x + 5y - 4)(3x + y + 2) = 0\)
In simple words: Substitute a temporary variable for the repeating binomial, factor the resulting quadratic, and substitute back.

Exam Tip: Substitution simplifies messy expressions into standard, easy-to-factor quadratic equations.

Question 19. If \(\left(\frac{8}{15}\right)^3 - \left(\frac{3}{5}\right)^3 - \left(\frac{1}{3}\right)^3 = \frac{x}{75}\), then find the value of \(x\).
Answer: Let \(a = \frac{8}{15}\), \(b = -\frac{3}{5}\), and \(c = -\frac{1}{3}\).
Notice that the sum \(a + b + c = \frac{8}{15} - \frac{3}{5} - \frac{1}{3} = \frac{8 - 9 - 5}{15} = \frac{-6}{15} = -\frac{2}{5} \neq 0\).
Let us calculate the cubes directly:
\(\left(\frac{8}{15}\right)^3 = \frac{512}{3375}\)
\(\left(\frac{3}{5}\right)^3 = \frac{27}{125} = \frac{27 \times 27}{125 \times 27} = \frac{729}{3375}\)
\(\left(\frac{1}{3}\right)^3 = \frac{1}{27} = \frac{125}{3375}\)
Now evaluate \(\frac{512}{3375} - \frac{729}{3375} - \frac{125}{3375} = \frac{512 - 729 - 125}{3375} = \frac{-342}{3375} = \frac{-114}{1125} = \frac{-38}{375}\)
Given equation: \(\frac{-38}{375} = \frac{x}{75}\)
Solve for \(x\):
\(x = \frac{-38 \times 75}{375}\)
\(x = \frac{-38}{5} = -7.6\)
In simple words: Compute the cube of each fraction individually with a common denominator, combine them, and solve for \(x\).

Exam Tip: Always find a common denominator when subtracting fractions with large powers.

Question 20. Factorise \(7x^2 + 2\sqrt{14}x + 2\).
Answer: Rewrite the expression to fit the square of a binomial formula \((a+b)^2 = a^2 + 2ab + b^2\):
\(7x^2 + 2\sqrt{14}x + 2\)
\(= (\sqrt{7}x)^2 + 2(\sqrt{7}x)(\sqrt{2}) + (\sqrt{2})^2\)
Using the algebraic identity \(a^2 + 2ab + b^2 = (a + b)^2\):
\(= (\sqrt{7}x + \sqrt{2})^2\)
\(= (\sqrt{7}x + \sqrt{2})(\sqrt{7}x + \sqrt{2})\)
In simple words: Express the first and last terms as squares of roots so you can apply the perfect square formula.

Exam Tip: Numbers like 7 and 2 can be written as \((\sqrt{7})^2\) and \((\sqrt{2})^2\) to handle surds in factorization.

PROJECT

Construct a real life object or a work of art or any mathematical model to illustrate your understanding of congruent triangles. Use any one or more than one of the five criterions (SSS, SAS, AAS, ASA, HL) for congruency to make your model.
Answer: 1. Choose a theme for your model, such as a bridge truss system, a tiled mosaic floor, or a butterfly with symmetrical wings.
2. Cut out identical geometric shapes (triangles) from colored cardboard ensuring they satisfy one of the congruence criteria (e.g., SSS where all three sides are equal, or ASA where two angles and the included side are equal).
3. Assemble the shapes symmetrically to form your structure or artwork.
In simple words: Build a physical model using identical triangles that match through standard congruence rules.

Exam Tip: Clearly state the chosen congruence criterion on a label attached to your model.

Please mark all the congruent triangles for identification. They should be color coded to showcase one congruency criterion using that one colour (for instance, the two triangles with the ASA criterion could be colored in red). The angle and side measures should be labeled or written on a piece of paper.
Answer: 1. Group your triangles by their congruence criteria: color all SSS triangles blue, all SAS triangles green, and all ASA triangles red.
2. Use a protractor and ruler to measure and label all side lengths (in centimeters) and interior angles (in degrees) directly on the triangles or on an accompanying key legend.
In simple words: Use distinct colors for different congruence rules and label all side and angle measurements clearly.

Exam Tip: Neat color-coding and clear labeling earn top marks in project assessments.

If you are creating a real-life project, possible materials could be charts, packing material, cardboard, craft, stickers, tooth picks, stickers and magnets etc. The geometric instruments that can be used for a real life project are a protractor, straight edged ruler etc.
Answer: 1. Gather sturdy materials like cardboard, craft paper, toothpicks for straight edges, and a ruler or protractor for accurate construction.
2. Draw, cut, and assemble your geometric structures with precision.
In simple words: Collect craft supplies and measuring tools to build a clean, durable mathematical model.

Exam Tip: Neatness and precision in cutting and measuring are key elements of project grading.