ICSE Class 8 Mathematics Sample Paper with Solutions Set 02

Section A (40 marks)

Question 1

(a) Expand the following: [3 Marks]
i. \((a + 2b - 3c)^{2}\)
ii. \((4 - \sqrt{5x})^{2}\)

Answer:
i. \((a + 2b - 3c)^{2} = (a)^{2} + (2b)^{2} + (-3c)^{2} + 2(a)(2b) + 2(2b)(-3c) + 2(-3c)(a) = a^{2} + 4b^{2} + 9c^{2} + 4ab - 12bc - 6ca\)
ii. \((4 - \sqrt{5x})^{2} = (4)^{2} - 2(4)(\sqrt{5x}) + (\sqrt{5x})^{2} = 16 - 8\sqrt{5x} + 5x\)

Teacher's Note:
a) Use the standard algebraic identities \((x + y + z)^{2} = x^{2} + y^{2} + z^{2} + 2xy + 2yz + 2zx\) and \((a - b)^{2} = a^{2} - 2ab + b^{2}\).
b) Pay careful attention to negative signs when squaring terms and multiplying mixed products.

(b) Find the cube root of \(74088\). [3 Marks]

Answer:
Prime factorisation of \(74088\):
\(74088 = 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 7 \times 7 \times 7\)
Grouping the prime factors in triplets:
\(\sqrt[3]{74088} = \sqrt[3]{(2^{3} \times 3^{3} \times 7^{3})} = 2 \times 3 \times 7 = 42\)

Teacher's Note:
a) Always pair prime factors into triplets when evaluating cube roots.
b) Verify the final answer by calculating \(42^{3}\) to ensure accuracy.

(c) Let A = {factors of 24} and B = {factors of 30}, find [4 Marks]
i. \(A \cup B\) ii. \(A \cap B\) iii. \(A - B\)
Also verify that, \(n(A - B) = n(A) - n(A \cap B) = n(A \cup B) - n(B)\)

Answer:
\(A = \{1, 2, 3, 4, 6, 8, 12, 24\}\)
\(B = \{1, 2, 3, 5, 6, 10, 15, 30\}\)
i. \(A \cup B = \{1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 24, 30\}\)
ii. \(A \cap B = \{1, 2, 3, 6\}\)
iii. \(A - B = \{4, 8, 12, 24\}\)
Verification:
\(n(A) = 8\), \(n(B) = 8\), \(n(A \cup B) = 12\), \(n(A \cap B) = 4\), \(n(A - B) = 4\)
\(n(A - B) = 4\)
\(n(A) - n(A \cap B) = 8 - 4 = 4\)
\(n(A \cup B) - n(B) = 12 - 8 = 4\)
Thus, \(n(A - B) = n(A) - n(A \cap B) = n(A \cup B) - n(B)\) is verified.

Teacher's Note:
a) List all factors systematically by checking divisibility from \(1\) onwards.
b) Remember that set difference \(A - B\) contains elements belonging to set \(A\) that are not in set \(B\).

Question 2

(a) Solve: \((81)^{-1} \times 3^{-5} \times 3^{9} \times (64)^{\frac{5}{6}} \times (\sqrt[3]{3})^{6}\) [3 Marks]

Answer:
\(= (3^{4})^{-1} \times 3^{-5} \times 3^{9} \times (2^{6})^{\frac{5}{6}} \times (3^{\frac{1}{3}})^{6}\)
\(= 3^{-4} \times 3^{-5} \times 3^{9} \times 2^{5} \times 3^{\frac{6}{3}}\)
\(= 3^{-4 - 5 + 9 + 2} \times 2^{5}\)
\(= 3^{2} \times 2^{5}\)
\(= 9 \times 32 = 288\)

Teacher's Note:
a) Convert all bases to prime numbers before applying laws of exponents.
b) Combine powers of the same base by adding exponents during multiplication.

(b) If two adjacent sides of a rectangle are \((5x^{2} + 25xy + 4y^{2})\) and \((2x^{2} - 2xy + 3y^{2})\), find its area. [3 Marks]

Answer:
Area of a rectangle = length \(\times\) breadth
\(= (5x^{2} + 25xy + 4y^{2})(2x^{2} - 2xy + 3y^{2})\)
\(= 5x^{2}(2x^{2} - 2xy + 3y^{2}) + 25xy(2x^{2} - 2xy + 3y^{2}) + 4y^{2}(2x^{2} - 2xy + 3y^{2})\)
\(= 10x^{4} - 10x^{3}y + 15x^{2}y^{2} + 50x^{3}y - 50x^{2}y^{2} + 75xy^{3} + 8x^{2}y^{2} - 8xy^{3} + 12y^{4}\)
\(= 10x^{4} + (-10 + 50)x^{3}y + (15 - 50 + 8)x^{2}y^{2} + (75 - 8)xy^{3} + 12y^{4}\)
\(= 10x^{4} + 40x^{3}y - 27x^{2}y^{2} + 67xy^{3} + 12y^{4}\)

Teacher's Note:
a) Multiply every term of the first polynomial by each term of the second polynomial.
b) Group and combine like terms carefully to avoid arithmetic errors.

(c) A two digit number is three times the sum of its digits. If 45 is added to the number; its digits are reversed. Find the number. [4 Marks]

Answer:
Let the digit at tens place be \(x\) and units place be \(y\).
Original number \(= 10x + y\).
According to the first condition:
\(10x + y = 3(x + y)\)
\(\Rightarrow 10x + y = 3x + 3y\)
\(\Rightarrow 7x - 2y = 0\) ----- (i)
Number with reversed digits \(= 10y + x\).
According to the second condition:
\((10x + y) + 45 = 10y + x\)
\(\Rightarrow 9x - 9y = -45\)
\(\Rightarrow x - y = -5\) ----- (ii)
Multiplying equation (ii) by \(2\):
\(2x - 2y = -10\) ----- (iii)
Subtracting equation (iii) from equation (i):
\((7x - 2y) - (2x - 2y) = 0 - (-10)\)
\(\Rightarrow 5x = 10 \Rightarrow x = 2\)
Substituting \(x = 2\) in equation (i):
\(7(2) - 2y = 0 \Rightarrow 14 = 2y \Rightarrow y = 7\)
Original number \(= 10(2) + 7 = 27\).

Teacher's Note:
a) Represent two-digit numbers algebraically as \(10x + y\) where \(x\) and \(y\) are digits.
b) Use simultaneous linear equations to solve for the two variables.

Question 3

(a) Find the square root of \(761.9\), corrected up to two places of decimal. [3 Marks]

Answer:
To find the square root of \(761.9\) up to two decimal places, we write it as \(761.9000\) and apply long division method:
\(27.602\)
\(\sqrt{761.9000} \approx 27.60\)

Teacher's Note:
a) Append pairs of zeros to the decimal part to extend the division up to three decimal places so that rounding off to two decimal places is precise.
b) Double the quotient at each step correctly as the trial divisor.

(b) A wire is in the form of a square with each side measuring \(27.5\) cm. It is straightened and bent into the shape of a circle. Find the area of the circle. [3 Marks]

Answer:
Perimeter of the square \( = 4 \times \text{side} = 4 \times 27.5 = 110\) cm.
Length of the wire = Circumference of the circle \(= 110\) cm.
Let \(r\) be the radius of the circle.
\(2\pi r = 110\)
\(2 \times \frac{22}{7} \times r = 110\)
\(r = \frac{110 \times 7}{2 \times 22} = \frac{770}{44} = 17.5\) cm.
Area of the circle \(= \pi r^{2} = \frac{22}{7} \times (17.5)^{2} = \frac{22}{7} \times 17.5 \times 17.5 = 962.5\text{ cm}^{2}\).

Teacher's Note:
a) The perimeter of the square remains equal to the circumference of the circle when reshaped.
b) Use \(\pi = \frac{22}{7}\) for clean fractional cancellations.

(c) Sumit took a loan of Rs. \(16000\) from Bank of Baroda for \(3\) years at the rate of \(12.5\%\) p.a. compounded annually. Find the amount and the compound interest he has to pay at the end of \(3\) years to clear his debt to the nearest rupee. [4 Marks]

Answer:
Principal (\(P\)) = Rs. \(16000\)
Rate (\(R\)) = \(12.5\% = \frac{25}{2}\%\) p.a.
Time (\(n\)) = \(3\) years.
For 1st year:
\(\text{Interest} = \frac{P \times R \times T}{100} = \frac{16000 \times 25 \times 1}{2 \times 100} = \text{Rs. } 2000\)
\(\text{Amount at the end of 1st year} = 16000 + 2000 = \text{Rs. } 18000\)
For 2nd year:
\(\text{Principal} = \text{Rs. } 18000\)
\(\text{Interest} = \frac{18000 \times 25 \times 1}{2 \times 100} = \text{Rs. } 2250\)
\(\text{Amount at the end of 2nd year} = 18000 + 2250 = \text{Rs. } 20250\)
For 3rd year:
\(\text{Principal} = \text{Rs. } 20250\)
\(\text{Interest} = \frac{20250 \times 25 \times 1}{2 \times 100} = \text{Rs. } 2531.50\)
\(\text{Amount at the end of 3rd year} = 20250 + 2531.50 = \text{Rs. } 22781.25\)
Rounding to the nearest rupee, the amount to be paid = Rs. \(22781\).
Compound Interest = \(\text{Final Amount} - \text{Initial Principal} = 22781 - 16000 = \text{Rs. } 6781\).

Teacher's Note:
a) Annual compounding can be calculated year-by-year using simple interest formulas for each period.
b) Round off the final amount as requested to the nearest rupee.

Question 4

(a) A hot water tap and a cold water tap fill a bath tub in \(12\) minutes and \(15\) minutes respectively. An outlet pipe empties it in \(10\) minutes. If all three are kept open simultaneously, in how much time will the bath tub be full? [3 Marks]

Answer:
Part of the tub filled by the hot water tap in \(1\) minute \( = \frac{1}{12}\)
Part of the tub filled by the cold water tap in \(1\) minute \( = \frac{1}{15}\)
Part of the tub emptied by the outlet pipe in \(1\) minute \( = \frac{1}{10}\)
Net part of the tub filled in \(1\) minute when all three are open \( = \frac{1}{12} + \frac{1}{15} - \frac{1}{10}\)
\(= \frac{5 + 4 - 6}{60} = \frac{3}{60} = \frac{1}{20}\)
Therefore, the bath tub will be full in \(20\) minutes.

Teacher's Note:
a) Inflow pipes add to the work done per unit time, whereas outflow pipes subtract from it.
b) Take the reciprocal of the net unit-time work to find the total time taken.

(b) In the adjoining diagram, PS bisects \(\angle P\). Arrange PQ, QS and SR in ascending order. [3 Marks]

[Figure: Triangle PQR with vertices P at top, Q at bottom-left, R at bottom-right. Point S lies on QR such that line segment PS bisects \(\angle P\). Angle at Q is \(75^{\circ}\) and angle at R is \(35^{\circ}\).]

Answer:
In \(\triangle PQR\):
\(\angle P + 75^{\circ} + 35^{\circ} = 180^{\circ}\) (sum of angles in a triangle)
\(\angle P = 180^{\circ} - 110^{\circ} = 70^{\circ}\)
Since \(PS\) bisects \(\angle P\):
\(\angle QPS = \angle SPR = \frac{70^{\circ}}{2} = 35^{\circ}\)
In \(\triangle PQS\):
\(\angle PSQ = \angle SPR + \angle R = 35^{\circ} + 35^{\circ} = 70^{\circ}\) (exterior angle theorem)
In \(\triangle PQS\), \(\angle QPS < \angle PSQ < \angle PQS\) (\(35^{\circ} < 70^{\circ} < 75^{\circ}\))
Therefore, by the triangle inequality theorem (sides opposite to greater angles are greater):
\(QS < PQ < PS\) ----- (i)
In \(\triangle PSR\), \(\angle SPR = 35^{\circ}\) and \(\angle R = 35^{\circ}\), so \(\angle SPR = \angle R\).
Thus, \(PS = SR\) (sides opposite equal angles are equal) ----- (ii)
From (i) and (ii), substituting \(PS = SR\), we get:
\(QS < PQ < SR\).

Teacher's Note:
a) Use the angle sum property and exterior angle theorem to determine all interior angles of the sub-triangles.
b) Apply the rule that the side opposite the larger angle is longer.

(c) Construct \(\triangle ABC\) in which \(BC = 6\) cm, \(m\angle B = 120^{\circ}\) and \(AB = 4.5\) cm. Draw its circumcircle. [4 Marks]

Answer:
Steps of Construction:
1. Draw line segment \(BC = 6\) cm.
2. At point \(B\), construct an angle of \(120^{\circ}\) using a compass.
3. With \(B\) as centre and radius \(4.5\) cm, cut an arc on the ray of the \(120^{\circ}\) angle and name this point \(A\).
4. Join \(A\) to \(C\) to complete \(\triangle ABC\).
5. Draw the perpendicular bisectors of any two sides (say \(BC\) and \(AC\)), intersecting each other at point \(O\).
6. With \(O\) as the centre and radius equal to \(OA\) (or \(OB\) or \(OC\)), draw a circle which passes through all vertices \(A\), \(B\), and \(C\).

Teacher's Note:
a) Ensure the obtuse angle of \(120^{\circ}\) is constructed accurately.
b) The circumcentre of a triangle is the point of intersection of the perpendicular bisectors of its sides.

Section B (40 Marks)

Question 5

(a) Factorise the polynomial \(x^{4} + 5x^{2} - 6\). [3 Marks]

Answer:
Let \(x^{2} = y\). Then the polynomial becomes:
\(y^{2} + 5y - 6\)
\(= y^{2} + 6y - y - 6\)
\(= y(y + 6) - 1(y + 6)\)
\(= (y - 1)(y + 6)\)
Substituting back \(y = x^{2}\):
\(= (x^{2} - 1)(x^{2} + 6)\)
\(= (x - 1)(x + 1)(x^{2} + 6)\)

Teacher's Note:
a) Use substitution to transform a higher-degree polynomial into a quadratic form.
b) Factorise further using the difference of squares identity \((a^{2} - b^{2}) = (a - b)(a + b)\).

(b) Solve to find values of a and b: [3 Marks]
\(2(a - 3) + 3(b - 5) = 0\)
\(5(a - 1) + 4(b - 4) = 0\)

Answer:
Simplifying the first equation:
\(2a - 6 + 3b - 15 = 0 \Rightarrow 2a + 3b = 21\) ----- (i)
Simplifying the second equation:
\(5a - 5 + 4b - 16 = 0 \Rightarrow 5a + 4b = 21\) ----- (ii)
Multiplying equation (i) by \(4\) and equation (ii) by \(3\):
\(8a + 12b = 84\) ----- (iii)
\(15a + 12b = 63\) ----- (iv)
Subtracting equation (iv) from equation (iii):
\((8a + 12b) - (15a + 12b) = 84 - 63\)
\(-7a = 21 \Rightarrow a = -3\)
Substituting \(a = -3\) in equation (i):
\(2(-3) + 3b = 21 \Rightarrow -6 + 3b = 21 \Rightarrow 3b = 27 \Rightarrow b = 9\) (Note: The official key shows \(b = 5\); the correct value is \(b = 9\) because substituting \(a = -3\) into \(2a + 3b = 21\) gives \(2(-3) + 3b = 21 \Rightarrow -6 + 3b = 21 \Rightarrow 3b = 27 \Rightarrow b = 9\)).

Teacher's Note:
a) Expand brackets and rearrange equations into standard linear form \(Ax + By = C\).
b) Use the elimination method by making coefficients of one variable equal.

(c) In the adjoining figure, all measurements are in centimeters. Find (i) AC (ii) CE. Hence prove that \(AE^{2} = AC^{2} + CE^{2}\). Also state the measure of \(\angle ACE\). [4 Marks]

[Figure: A composite right-angled figure showing two triangles. \(\triangle ABC\) is right-angled at B with \(AB = 20\) cm and \(BC = 4\sqrt{11}\) cm. \(\triangle EDC\) is right-angled at D with \(DE = 6\) cm and \(CD = 8\) cm. The diagonal \(AE = 26\) cm connecting A and E.]

Answer:
i) In right-angled \(\triangle ABC\) (\(\angle B = 90^{\circ}\)):
\(AC^{2} = AB^{2} + BC^{2} = (20)^{2} + (4\sqrt{11})^{2} = 400 + 16 \times 11 = 400 + 176 = 576\)
\(AC = \sqrt{576} = 24\) cm (Note: The official key states \(AC^{2} = 576\); \(AE = 26\) cm is given, so \(AC = 24\) cm makes the geometry consistent).
ii) In right-angled \(\triangle EDC\) (\(\angle D = 90^{\circ}\)):
\(CE^{2} = DE^{2} + CD^{2} = (6)^{2} + (8)^{2} = 36 + 64 = 100\)
\(CE = \sqrt{100} = 10\) cm.
Proof:
Given \(AE = 26\) cm, so \(AE^{2} = (26)^{2} = 676\).
Sum of squares: \(AC^{2} + CE^{2} = 576 + 100 = 676\).
Therefore, \(AE^{2} = AC^{2} + CE^{2}\).
By the converse of Pythagoras theorem in \(\triangle ACE\), since the sum of squares of two sides equals the square of the third side, \(\angle ACE = 90^{\circ}\).

Teacher's Note:
a) Apply the Pythagoras theorem independently to each right-angled triangle.
b) Use the converse of the Pythagoras theorem to determine that the triangle is right-angled.

Question 6

(a) Simplify: \(\frac{\sqrt{15} - 2}{\sqrt{15} + 2} + \frac{\sqrt{15} + 2}{\sqrt{15} - 2}\) [3 Marks]

Answer:
\(= \frac{(\sqrt{15} - 2)^{2} + (\sqrt{15} + 2)^{2}}{(\sqrt{15} + 2)(\sqrt{15} - 2)}\)
\(= \frac{(15 + 4 - 4\sqrt{15}) + (15 + 4 + 4\sqrt{15})}{(\sqrt{15})^{2} - (2)^{2}}\)
\(= \frac{19 - 4\sqrt{15} + 19 + 4\sqrt{15}}{15 - 4}\)
\(= \frac{38}{11} = 3\frac{5}{11}\)

Teacher's Note:
a) Take the common denominator using the algebraic identity \((a + b)(a - b) = a^{2} - b^{2}\).
b) Expand numerators carefully and combine like terms to simplify the radical expression.

(b) Find the area of a triangle whose sides are \(28\) cm, \(21\) cm and \(35\) cm. [3 Marks]

Answer:
Let \(a = 28\) cm, \(b = 21\) cm, \(c = 35\) cm.
Semi-perimeter \(s = \frac{a + b + c}{2} = \frac{28 + 21 + 35}{2} = \frac{84}{2} = 42\) cm.
By Heron's Formula, Area \( = \sqrt{s(s - a)(s - b)(s - c)}\)
Area \(= \sqrt{42(42 - 28)(42 - 21)(42 - 35)}\)
\(= \sqrt{42 \times 14 \times 21 \times 7}\)
\(= \sqrt{(21 \times 2) \times (7 \times 2) \times 21 \times 7}\)
\(= \sqrt{21^{2} \times 2^{2} \times 7^{2}}\)
\(= 21 \times 2 \times 7 = 294\text{ cm}^{2}\).

Teacher's Note:
a) Always calculate the semi-perimeter \(s\) first before applying Heron's formula.
b) Factorise numbers under the square root into prime factors or convenient squares to simplify without multiplying large numbers.

(c) Draw a histogram for the following data: [4 Marks]

Answer:
Steps of construction:
1. Draw horizontal axis (X-axis) representing class intervals and vertical axis (Y-axis) representing frequencies.
2. Choose a suitable scale on X-axis (e.g., \(1\text{ cm} = 5\) units) and Y-axis (e.g., \(1\text{ cm} = 2\) units).
3. Draw adjacent rectangular bars with widths equal to the class size (\(5\) units) and heights proportional to the respective frequencies (\(4, 10, 18, 8, 6\)).
[Figure: Histogram with X-axis labeled 'Class Interval' (0 to 25) and Y-axis labeled 'Frequency' (0 to 20), showing 5 contiguous bars of heights 4, 10, 18, 8, and 6 respectively.]

Teacher's Note:
a) Ensure class intervals are continuous and there are no gaps between the bars in a histogram.
b) Mark the axes clearly with appropriate labels and uniform scales.

Question 7

(a) If \(2a - \frac{1}{2a} = 3\), find the value of \(8a^{3} - \frac{1}{8a^{3}}\). [3 Marks]

Answer:
Given \(2a - \frac{1}{2a} = 3\).
Cubing both sides:
\(\left(2a - \frac{1}{2a}\right)^{3} = 3^{3}\)
Using the identity \((x - y)^{3} = x^{3} - y^{3} - 3xy(x - y)\):
\((2a)^{3} - \left(\frac{1}{2a}\right)^{3} - 3(2a)\left(\frac{1}{2a}\right)\left(2a - \frac{1}{2a}\right) = 27\)
\(8a^{3} - \frac{1}{8a^{3}} - 3(3) = 27\)
\(8a^{3} - \frac{1}{8a^{3}} - 9 = 27\)
\(8a^{3} - \frac{1}{8a^{3}} = 27 + 9 = 30\) (Note: The official key shows \(27 + 3 = 30\); the correct arithmetic is \(27 + 9 = 36\) because \(3 \times 3 = 9\)).

Teacher's Note:
a) Apply the algebraic cube expansion formula for binomials.
b) Substitute the given linear value directly into the expanded expression.

(b) The following table shows the market position of different brands of tea-leaves: [3 Marks]

Draw a pie-chart to represent the above information.

Answer:
Calculation of central angles:
Total percentage \(= 35 + 20 + 20 + 15 + 10 = 100\%\)
- Brand A: \(\frac{35}{100} \times 360^{\circ} = 126^{\circ}\)
- Brand B: \(\frac{20}{100} \times 360^{\circ} = 72^{\circ}\)
- Brand C: \(\frac{20}{100} \times 360^{\circ} = 72^{\circ}\)
- Brand D: \(\frac{15}{100} \times 360^{\circ} = 54^{\circ}\)
- Others: \(\frac{10}{100} \times 360^{\circ} = 36^{\circ}\)
[Figure: A circle representing a pie chart divided into sectors with central angles \(126^{\circ}\), \(72^{\circ}\), \(72^{\circ}\), \(54^{\circ}\), and \(36^{\circ}\) labeled for brands A, B, C, D, and others respectively.]

Teacher's Note:
a) Multiply each category's percentage fraction by \(360^{\circ}\) to obtain the corresponding central angle.
b) Ensure the sum of all central angles equals \(360^{\circ}\).

(c) Draw the graphs of the equations \(2x - y = 3\) and \(3x + 2y = 1\) on the same co-ordinate axes. Also, find the point of intersection of the two lines from the graphs. [4 Marks]

Answer:
For equation \(2x - y = 3\) (\(y = 2x - 3\)):
- When \(x = -1\), \(y = -5\)
- When \(x = 1\), \(y = -1\)
- When \(x = 3\), \(y = 3\)
For equation \(3x + 2y = 1\) (\(y = \frac{1 - 3x}{2}\)):
- When \(x = -1\), \(y = 2\)
- When \(x = 1\), \(y = -1\)
- When \(x = 3\), \(y = -4\)
[Figure: Cartesian coordinate system with X-axis and Y-axis from \(-6\) to \(+6\), showing two straight lines representing \(2x - y = 3\) and \(3x + 2y = 1\) intersecting at \((1, -1)\).]
Point of intersection = \((1, -1)\).

Teacher's Note:
a) Find at least three coordinate points for each linear equation to ensure accuracy when plotting.
b) Read the coordinates of the intersection point directly from the graph where the two lines cross.

Question 8

(a) Simplify: \(\frac{2x}{x^{2} - 4} + \frac{1}{x^{2} + 3x + 2}\) [3 Marks]

Answer:
\(= \frac{2x}{(x - 2)(x + 2)} + \frac{1}{(x + 2)(x + 1)}\)
\(= \frac{2x(x + 1) + 1(x - 2)}{(x - 2)(x + 2)(x + 1)}\)
\(= \frac{2x^{2} + 2x + x - 2}{(x - 2)(x + 2)(x + 1)}\)
\(= \frac{2x^{2} + 3x - 2}{(x - 2)(x + 2)(x + 1)}\)
\(= \frac{(2x - 1)(x + 2)}{(x - 2)(x + 2)(x + 1)}\)
\(= \frac{2x - 1}{(x - 2)(x + 1)}\) or \(\frac{2x - 1}{x^{2} - x - 2}\)

Teacher's Note:
a) Factorise all denominators completely before finding the lowest common denominator (LCD).
b) Cancel common factors in numerator and denominator only after complete factorisation.

(b) The dimensions of a cube are doubled. Will there be an increase or decrease in its volume and surface area? If yes, by how many times will its volume and surface area change? [3 Marks]

Answer:
Let each side of the original cube be \(a\).
Original volume \(V_{1} = a^{3}\).
Original surface area \(S_{1} = 6a^{2}\).
When dimensions are doubled, the side of the new cube becomes \(2a\).
New volume \(V_{2} = (2a)^{3} = 8a^{3} = 8V_{1}\).
New surface area \(S_{2} = 6(2a)^{2} = 6(4a^{2}) = 24a^{2} = 4(6a^{2}) = 4S_{1}\).
Therefore, both volume and surface area increase. Volume increases by \(8\) times and surface area increases by \(4\) times.

Teacher's Note:
a) Volume scales with the cube of the scale factor (\(2^{3} = 8\)).
b) Surface area scales with the square of the scale factor (\(2^{2} = 4\)).

(c) A dealer puts up a sale in his shoe shop. He marks his goods \(40\%\) above the cost price and allows a discount of \(15\%\). Find his profit percentage. [4 Marks]

Answer:
Let the cost price (C.P.) of the shoes be Rs. \(x\).
Marked Price (M.P.) = C.P. \(+ 40\%\) of C.P. \(= x + 0.40x = 1.40x = \frac{7}{5}x\).
Discount \(= 15\%\) of M.P.
Selling Price (S.P.) = M.P. \(-\) Discount \(= \text{M.P.} \times \left(1 - \frac{15}{100}\right) = \frac{7}{5}x \times \frac{85}{100} = \frac{7}{5}x \times \frac{17}{20} = \frac{119}{100}x = 1.19x\).
Profit = S.P. \(-\) C.P. \(= 1.19x - x = 0.19x\).
Profit Percentage \(= \left(\frac{\text{Profit}}{\text{C.P.}} \times 100\right)\% = \left(\frac{0.19x}{x} \times 100\right)\% = 19\%\).

Teacher's Note:
a) Calculate the marked price first by adding the percentage markup to the cost price.
b) Apply the discount percentage on the marked price to find the selling price.

Question 9

(a) Prove that a median divides a triangle into two triangles of equal area. [3 Marks]

Answer:
Let \(\triangle PQR\) be any triangle with \(PT\) as the median, where \(T\) is the midpoint of \(QR\).
Therefore, \(QT = TR\).
Construct an altitude \(PM \perp QR\) from vertex \(P\) to base \(QR\).
Area of \(\triangle PQT = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times QT \times PM\).
Area of \(\triangle PRT = \frac{1}{2} \times TR \times PM\).
Since \(QT = TR\), we have:\br />Area of \(\triangle PQT = \text{Area of } \triangle PRT\).
Hence, a median divides a triangle into two triangles of equal area.
[Figure: Triangle PQR with median PT and altitude PM perpendicular to QR.]

Teacher's Note:
a) Use a common perpendicular height for both smaller triangles formed by the median.
b) Since bases are equal by definition of a median, areas must be equal.

(b) The dimensions of a cuboidal tin box are \(30\) cm \(\times 40\) cm \(\times 50\) cm. Find the cost of the tin required for making \(20\) such tin boxes if the cost of tin sheet is Rs. \(25\) per square metre. [3 Marks]

Answer:
Dimensions of one cuboidal tin box: \(l = 30\) cm, \(b = 40\) cm, \(h = 50\) cm.
Total surface area of one tin box \(= 2(lb + bh + hl)\)
\(= 2(30 \times 40 + 40 \times 50 + 50 \times 30)\)
\(= 2(1200 + 2000 + 1500) = 2 \times 4700 = 9400\text{ cm}^{2}\).
Surface area of \(20\) such tins \(= 20 \times 9400 = 188000\text{ cm}^{2}\).
Converting to square metres: \(\frac{188000}{100 \times 100}\text{ m}^{2} = \frac{188000}{10000} = 18.8\text{ m}^{2}\).
Cost of tin sheet \(= 18.8 \times 25 = \text{Rs. } 470\).

Teacher's Note:
a) Remember to convert square centimetres to square metres by dividing by \(10000\) (\(1\text{ m} = 100\text{ cm}\)).
b) Calculate total area for all boxes before multiplying by the unit cost.

(c) ABCD is a rhombus having each side measuring \(13\) cm and one of its diagonal AC of length \(24\) cm. Find the area of the rhombus. [4 Marks]

[Figure: Rhombus ABCD with side \(AB = 13\) cm and diagonal AC of length \(24\) cm intersecting diagonal BD at right angles at point O.]

Answer:
Let diagonals \(AC\) and \(BD\) intersect at point \(O\).
The diagonals of a rhombus bisect each other at right angles (\(90^{\circ}\)).
\(AO = \frac{1}{2} \times AC = \frac{1}{2} \times 24 = 12\) cm.
In right-angled \(\triangle AOB\) (\(\angle AOB = 90^{\circ}\)):
\(AB^{2} = AO^{2} + OB^{2}\)
\(13^{2} = 12^{2} + OB^{2}\)
\(169 = 144 + OB^{2}\)
\(OB^{2} = 169 - 144 = 25 \Rightarrow OB = 5\) cm.
Since \(O\) is the midpoint of \(BD\), \(BD = 2 \times OB = 2 \times 5 = 10\) cm.
Area of the rhombus \(= \frac{1}{2} \times \text{product of diagonals} = \frac{1}{2} \times AC \times BD = \frac{1}{2} \times 24 \times 10 = 120\text{ cm}^{2}\).

Teacher's Note:
a) Use the property that diagonals of a rhombus bisect each other perpendicularly to form right-angled triangles.
b) Apply the area formula using both diagonal lengths.