ICSE Class 8 Mathematics Sample Paper with Solutions Set 01

Section A (40 marks)

Question 1

(a) Do the ratios 30 cm to 4 m and 20 sec to 6 minutes form a proportion? [3 Marks]

Answer:
First ratio = \( 30 \text{ cm} : 4 \text{ m} = 30 : 4 \times 100 = 30 : 400 = 3 : 40 \)
Second ratio = \( 20 \text{ sec} : 6 \text{ minutes} = 20 : 6 \times 60 = 20 : 360 = 1 : 18 \)
Since \( 3 : 40 \neq 1 : 18 \), the given ratios do not form a proportion.

Teacher's Note:
a) Always convert quantities to the same units before comparing or finding ratios.
b) Two ratios form a proportion only if they are equal when expressed in simplest form.

(b) If RO is perpendicular to PT, find the measure of angles 1 and 2 in the figure below: [3 Marks]

[Figure: A straight line PT with point O on it. Rays OQ, OR, and OS emanate from O upwards. RO is perpendicular to PT, making angle ROT and angle POR right angles (\( 90^{\circ} \)). Angle POQ is \( 52^{\circ} \), angle 1 is between OQ and OR, angle ROS is \( 67^{\circ} \), and angle 2 is between OS and OT.]

Answer:
Since \( RO \perp PT \), both \( \angle POR \) and \( \angle ROT \) are \( 90^{\circ} \).
\( \angle POQ + \angle 1 = 90^{\circ} \)
\( 52^{\circ} + \angle 1 = 90^{\circ} \)
\( \angle 1 = 90^{\circ} - 52^{\circ} = 38^{\circ} \)
Also, \( \angle ROS + \angle 2 = 90^{\circ} \)
\( 67^{\circ} + \angle 2 = 90^{\circ} \)
\( \angle 2 = 90^{\circ} - 67^{\circ} = 23^{\circ} \)

Teacher's Note:
a) Use the property of perpendicular lines which form right angles (\( 90^{\circ} \)).
b) Ensure proper subtraction from \( 90^{\circ} \) for complementary angles.

(c) Simplify: \( \dfrac{7\sqrt{3}}{\sqrt{6} - \sqrt{3}} - \dfrac{2\sqrt{5}}{\sqrt{8} + \sqrt{2}} \) [4 Marks]

Answer:
Rationalising factor of \( (\sqrt{6} - \sqrt{3}) \) is \( (\sqrt{6} + \sqrt{3}) \), and of \( (\sqrt{8} + \sqrt{2}) \) is \( (\sqrt{8} - \sqrt{2}) \).
\( = \dfrac{7\sqrt{3}(\sqrt{6} + \sqrt{3})}{(\sqrt{6} - \sqrt{3})(\sqrt{6} + \sqrt{3})} - \dfrac{2\sqrt{5}(\sqrt{8} - \sqrt{2})}{(\sqrt{8} + \sqrt{2})(\sqrt{8} - \sqrt{2})} \)
\( = \dfrac{7\sqrt{18} + 7(3)}{6 - 3} - \dfrac{2\sqrt{40} - 2\sqrt{10}}{8 - 2} \)
\( = \dfrac{21\sqrt{2} + 21}{3} - \dfrac{4\sqrt{10} - 2\sqrt{10}}{6} \)
\( = \dfrac{21(\sqrt{2} + 1)}{3} - \dfrac{2\sqrt{10}}{6} \)
\( = 7(\sqrt{2} + 1) - \dfrac{\sqrt{10}}{3} \)
\( = \dfrac{21(\sqrt{2} + 1) - \sqrt{10}}{3} \)

Teacher's Note:
a) Rationalise the denominators separately by multiplying numerator and denominator by the conjugate.
b) Simplify surds carefully (e.g., \( \sqrt{18} = 3\sqrt{2} \)) before combining terms.

Question 2

(a) The sum of two numbers is 55 and their H.C.F. and L.C.M. are 5 and 120 respectively, then, find the sum of the reciprocals of the numbers. [3 Marks]

Answer:
Let the two numbers be \( a \) and \( b \).
Then, \( a + b = 55 \)
We know that \( a \times b = \text{H.C.F.} \times \text{L.C.M.} = 5 \times 120 = 600 \)
Required sum of reciprocals = \( \dfrac{1}{a} + \dfrac{1}{b} = \dfrac{a + b}{ab} \)
Substitute the values: \( \dfrac{55}{600} = \dfrac{11}{120} \)

Teacher's Note:
a) Use the fundamental relation: Product of two numbers equals the product of their H.C.F. and L.C.M.
b) Simplify the final fraction to its lowest terms.

(b) If the product of two positive consecutive even integers is 168, find the integers. [3 Marks]

Answer:
Let the first positive even integer be \( x \).
The other consecutive even integer is \( x + 2 \).
According to the given condition:
\( x(x + 2) = 168 \)
\( x^2 + 2x - 168 = 0 \)
\( x^2 + 14x - 12x - 168 = 0 \)
\( x(x + 14) - 12(x + 14) = 0 \)
\( (x + 14)(x - 12) = 0 \)
\( x = -14 \) or \( x = 12 \)
Since \( x \) is a positive even integer, \( x = 12 \).
The other integer is \( 12 + 2 = 14 \).
Hence, the required integers are 12 and 14.

Teacher's Note:
a) Formulate the quadratic equation correctly using consecutive even integers \( x \) and \( x + 2 \).
b) Reject the negative value since the problem specifies positive integers.

(c) A's income is 60% more than that of B. By what percent is B's income less than A's? [4 Marks]

Answer:
Let B's income be Rs. 100.
Then A's income = Rs. 160.
Difference in income = \( 160 - 100 = \text{Rs. } 60 \).
Percentage by which B's income is less than A's = \( \left(\dfrac{60}{160} \times 100\right)\% \)
\( = \left(\dfrac{3}{8} \times 100\right)\% = \left(\dfrac{300}{8}\right)\% = 37.5\% \)

Teacher's Note:
a) Base the percentage decrease on A's income since the question asks "less than A's".
b) Avoid common errors of taking the base as B's income.

Question 3

(a) In a parallelogram ABCD, if its area is \( 20 \text{ cm}^2 \), find the area of \( \Delta ABC \) and the distance between the sides AB and CD, if AB = 5 cm. [3 Marks]

Answer:
Area of parallelogram ABCD = \( 20 \text{ cm}^2 \).
Since a diagonal divides a parallelogram into two triangles of equal area, Area of \( \Delta ABC = \dfrac{1}{2} \times \text{Area of parallelogram ABCD} \)
\( \text{Area of } \Delta ABC = \dfrac{1}{2} \times 20 = 10 \text{ cm}^2 \).
Also, Area of \( \Delta ABC = \dfrac{1}{2} \times \text{base} \times \text{height} \)
\( 10 = \dfrac{1}{2} \times AB \times \text{height} \)
\( 10 = \dfrac{1}{2} \times 5 \times \text{height} \)
\( \text{height} = \dfrac{10 \times 2}{5} = 4 \text{ cm} \).
Thus, the distance between the parallel sides AB and CD is 4 cm.

Teacher's Note:
a) Recall that a diagonal bisects a parallelogram into two equal triangles.
b) The distance between two parallel sides is the perpendicular height of the parallelogram.

(b) Given: A = {1, 2, 3}, B = {3, 4}, C = {4, 5, 6}, find \( (A \times B) \cap (B \times C) \). [3 Marks]

Answer:
\( A \times B = \{1, 2, 3\} \times \{3, 4\} = \{(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)\} \)
\( B \times C = \{3, 4\} \times \{4, 5, 6\} = \{(3, 4), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6)\} \)
\( (A \times B) \cap (B \times C) = \{(3, 4)\} \)

Teacher's Note:
a) Cartesian product of two sets consists of all ordered pairs where the first element belongs to the first set and the second to the second set.
b) Intersection means finding the common ordered pairs present in both Cartesian products.

(c) Simplify: \( \dfrac{(2x^2y^3)^5 \times (2x^2y^2)^3}{(5x^4y)^6} \) [4 Marks]

Answer:
Numerator = \( (2^5 \cdot x^{10} \cdot y^{15}) \times (2^3 \cdot x^6 \cdot y^6) = 2^8 \cdot x^{16} \cdot y^{21} \)
Denominator = \( 5^6 \cdot x^{24} \cdot y^6 \)
Expression = \( \dfrac{2^8 \cdot x^{16} \cdot y^{21}}{5^6 \cdot x^{24} \cdot y^6} \)
\( = \dfrac{2^8}{5^6} \cdot x^{16 - 24} \cdot y^{21 - 6} \)
\( = \dfrac{256}{15625} x^{-8} y^{15} \) or \( \dfrac{2^8 y^{15}}{5^6 x^8} \)

Teacher's Note:
a) Apply laws of exponents: \( (a^m)^n = a^{mn} \) and \( a^m \times a^n = a^{m+n} \).
b) Simplify variables by subtracting exponents for division.

Question 4

(a) Find the square root of \( 5\dfrac{19}{25} \). [3 Marks]

Answer:
\( 5\dfrac{19}{25} = \dfrac{25 \times 5 + 19}{25} = \dfrac{144}{25} \)
Square root = \( \sqrt{\dfrac{144}{25}} = \dfrac{\sqrt{144}}{\sqrt{25}} \)
\( = \dfrac{12}{5} = 2\dfrac{2}{5} \)

Teacher's Note:
a) Convert mixed fractions into improper fractions before finding the square root.
b) Find the square roots of the numerator and denominator separately.

(b) Find the fraction which becomes \( \dfrac{1}{2} \) when its numerator is increased by 6 and is equal to \( \dfrac{1}{3} \) when its denominator is increased by 7. Find the fraction. [3 Marks]

Answer:
Let the numerator be \( x \) and denominator be \( y \text{ (fraction } \dfrac{x}{y}\text{)} \).
According to the first condition:
\( \dfrac{x + 6}{y} = \dfrac{1}{2} \implies 2x + 12 = y \implies 2x - y = -12 \) ---(1)
According to the second condition:
\( \dfrac{x}{y + 7} = \dfrac{1}{3} \implies 3x = y + 7 \implies 3x - y = 7 \) ---(2)
Subtracting equation (1) from equation (2):
\( (3x - y) - (2x - y) = 7 - (-12) \)
\( x = 19 \)
Substitute \( x = 19 \) in equation (1):
\( 2(19) - y = -12 \implies 38 - y = -12 \implies y = 50 \)
Hence, the required fraction is \( \dfrac{19}{50} \).

Teacher's Note:
a) Formulate two linear equations in two variables based on the given word problem statements.
b) Use the elimination or substitution method to solve for \( x \) and \( y \).

(c) The table below classifies the days of the months of June, July and August according to the rainfall received in a locality. [4 Marks]

Draw a histogram for this data.

Answer:
[Figure: A histogram with Rain (mm) on the X-axis and Days (frequency) on the Y-axis. Rectangles are drawn contiguously for each class interval with heights corresponding to the given frequencies: 8, 10, 14, 20, 15, 8, 7, 6, and 4 respectively.]

Teacher's Note:
a) Ensure the X-axis has uniform class intervals and the Y-axis has an appropriate scale for frequencies.
b) Since class intervals are continuous, draw adjacent rectangular bars without gaps.

Section B (40 Marks)

Question 5

(a) Evaluate: \( 5 + 6 - 3 \times (2 + 70) - \dfrac{50}{2} + (3 + 7 \times 2 - 9) \) [3 Marks]

Answer:
\( = 5 + 6 - 3 \times 72 - 25 + (3 + 14 - 9) \)
\( = 5 + 6 - 3 \times 72 - 25 + 8 \)
\( = 5 + 6 - 216 - 25 + 8 \)
\( = 19 - 216 - 25 + 8 \)
\( = -222 \)

Teacher's Note:
a) Strictly follow the BODMAS rule (Brackets, Orders, Division, Multiplication, Addition, Subtraction).
b) Perform multiplications and divisions before additions and subtractions.

(b) Draw a circle of radius 2.5 cm. Show and define minor and major segments. [3 Marks]

Answer:
[Figure: A circle with centre O and radius 2.5 cm. A chord PQ divides the circle into two parts. The smaller region containing the minor arc is shaded and labeled 'Minor segment', and the larger region containing the center and major arc is labeled 'Major segment'.]
Definition: A chord of a circle divides the circular region into two parts called segments. The segment containing the minor arc is the minor segment, and the segment containing the major arc is the major segment.

Teacher's Note:
a) Draw the circle accurately using a compass with radius set to 2.5 cm.
b) Clearly demarcate and label both minor and major segments with respect to the chord.

(c) In \( \Delta ABC \), BE and CF are altitudes on the sides AC and AB respectively such that BE = CF. Prove that AB = AC. [4 Marks]

[Figure: Triangle ABC with altitudes BE on AC and CF on AB such that BE = CF.]

Answer:
In right-angled triangles BEC and CFB:
\( \angle BEC = \angle CFB = 90^{\circ} \)
\( BE = CF \) (Given)
\( BC = BC \) (Common hypotenuse)
\( \therefore \Delta BEC \cong \Delta CFB \) (RHS congruence criterion)
\( \therefore \angle BCE = \angle CBF \) (Corresponding parts of congruent triangles)
Since \( \angle BCE \) is same as \( \angle BCA \) and \( \angle CBF \) is same as \( \angle CBA \):
\( \angle BCA = \angle CBA \)
In \( \Delta ABC \), sides opposite to equal angles are equal, so \( AB = AC \).

Teacher's Note:
a) Prove the congruence of triangles BEC and CFB using RHS criterion.
b) Use the property that sides opposite to equal angles in a triangle are equal.

Question 6

(a) Raj covered a certain distance in 6 hours. He covered some part of the journey by bus at 30 km/h and the remaining part of the journey by train at 50 km/h. Find the distance covered for the entire journey. [3 Marks]

Answer:
Speed by bus = \( 30 \text{ km/hr} \)
Speed by train = \( 50 \text{ km/hr} \)
Average speed = \( \dfrac{30 + 50}{2} = \dfrac{80}{2} = 40 \text{ km/hr} \)
Distance covered = Time \( \times \) Average speed
Distance = \( 6 \times 40 = 240 \text{ km} \)
Thus, the total distance covered by Raj is 240 km.

Teacher's Note:
a) When time taken for equal halves of speed is considered, average speed is the arithmetic mean of the two speeds.
b) Multiply average speed by total time to get the total distance.

(b) Simplify: \( \dfrac{x^2 - 3x - 10}{x^2 - x - 20} \times \dfrac{x^2 - 2x + 4}{x^3 + 8} \) [3 Marks]

Answer:
Factorize each part:
\( x^2 - 3x - 10 = (x - 5)(x + 2) \)
\( x^2 - x - 20 = (x - 5)(x + 4) \)
\( x^3 + 8 = (x + 2)(x^2 - 2x + 4) \)
Substitute into the expression:
\( = \dfrac{(x - 5)(x + 2)}{(x - 5)(x + 4)} \times \dfrac{x^2 - 2x + 4}{(x + 2)(x^2 - 2x + 4)} \)
Cancel out common factors \( (x - 5) \), \( (x + 2) \), and \( (x^2 - 2x + 4) \):
\( = \dfrac{1}{x + 4} \)

Answer:

Teacher's Note:
a) Factorize all quadratic and cubic expressions completely using identities like sum of cubes \( (a^3 + b^3) \).
b) Cancel common factors across numerators and denominators carefully.

(c) Draw triangle according to the following measures: \( \Delta DEF: l(DE) = l(DF) = 6 \text{ cm}, m\angle D = 40^{\circ} \) [4 Marks]

[Figure: Triangle DEF with DE = DF = 6 cm and included angle D = 40 degrees.]

Answer:
Steps of construction:
1. Draw segment DE of length 6 cm.
2. At vertex D, construct an angle of \( 40^{\circ} \) using a protractor, say \( \angle XDE = 40^{\circ} \).
3. With D as centre and radius 6 cm, draw an arc to cut the ray DX at point F.
4. Join points F and E.
\( \Delta DEF \) is the required triangle.

Teacher's Note:
a) Use a ruler and protractor accurately to construct the given side-angle-side measurements.
b) Label all vertices and dimensions clearly on the constructed figure.

Question 7

(a) How much compound interest is earned on Rs. 18,000 at 7% interest rate for 1 year? [3 Marks]

Answer:
For the first year, simple interest and compound interest are equal.
Principal \( (P) = \text{Rs. } 18,000 \), Rate \( (R) = 7\% \), Time \( (T) = 1 \text{ year} \)
\( I = \dfrac{P \times R \times T}{100} \)
\( I = \dfrac{18000 \times 7 \times 1}{100} = 180 \times 7 = \text{Rs. } 1260 \)
Thus, the compound interest earned is Rs. 1,260.

Teacher's Note:
a) For a time period of 1 year compounded annually, compound interest equals simple interest.
b) Apply the simple interest formula directly to save time.

(b) Make d as the subject of the formula: \( S = \dfrac{n}{2}\{2a + (n - 1)d\} \) [3 Marks]

Answer:
\( S = \dfrac{n}{2}\{2a + (n - 1)d\} \)
Multiply both sides by 2: \( 2S = n\{2a + (n - 1)d\} \)
Expand the bracket: \( 2S = 2an + n(n - 1)d \)
Rearrange to isolate the term containing \( d \):
\( n(n - 1)d = 2S - 2an \)
\( d = \dfrac{2(S - an)}{n(n - 1)} \)

Teacher's Note:
a) Clear the fraction first by multiplying both sides by the denominator.
b) Isolate the term with the target variable before dividing by its coefficient.

(c) The surface area of a cuboidal wooden box is \( 470 \text{ cm}^2 \). If its length and breadth are 15 cm and 8 cm respectively, find its height. [4 Marks]

Answer:
Given: Length \( (l) = 15 \text{ cm} \), breadth \( (b) = 8 \text{ cm} \)
Let height be \( h \text{ cm} \).
Total Surface Area of cuboid = \( 2(lb + bh + hl) \)
\( 470 = 2(15 \times 8 + 8h + 15h) \)
\( 470 = 2(120 + 23h) \)
\( 235 = 120 + 23h \)
\( 23h = 235 - 120 = 115 \)
\( h = \dfrac{115}{23} = 5 \text{ cm} \)
Thus, the height of the cuboidal box is 5 cm.

Teacher's Note:
a) Substitute the given values of length, breadth, and total surface area into the standard cuboid formula.
b) Solve the resulting simple linear equation for height \( h \).

Question 8

(a) Simplify: \( 20x - [15x^3 + 5x^2 - \{8x^2 - (4 - 2x - x^3) - 5x^3\} - 2x] \) [3 Marks]

Answer:
Remove innermost brackets first:
\( = 20x - [15x^3 + 5x^2 - \{8x^2 - 4 + 2x + x^3 - 5x^3\} - 2x] \)
Combine like terms inside braces:
\( = 20x - [15x^3 + 5x^2 - \{8x^2 - 4 + 2x - 4x^3\} - 2x] \)
Remove braces by changing signs:
\( = 20x - [15x^3 + 5x^2 - 8x^2 + 4 - 2x + 4x^3 - 2x] \)
Simplify inside the square brackets:
\( = 20x - [19x^3 - 3x^2 + 4 - 4x] \)
Remove square brackets and combine all like terms:
\( = 20x - 19x^3 + 3x^2 - 4 + 4x \)
\( = -19x^3 + 3x^2 + 24x - 4 \)

Teacher's Note:
a) Follow the order of bracket removal: parentheses (), braces {}, then square brackets [].
b) Be extremely careful with negative signs when opening brackets.

(b) Write down the co-ordinates of the images for the points plotted in the graph.
i. Point A and Point D reflected in the x-axis. [3 Marks]
ii. Point B and Point C reflected in the y-axis.

[Figure: Cartesian plane with points A(4, 5), B(-2, -2), C(-7, -6), D(-4, 8) plotted along with their respective reflected images A'(4, -5), B'(2, -2), C'(7, -6), D'(-4, -9).]

Answer:
i. Reflection in the x-axis changes the sign of the y-coordinate \( (x, y) \to (x, -y) \):
- Image of Point A(4, 5) is \( A'(4, -5) \).
- Image of Point D(-4, 8) is \( D'(-4, -8) \).
ii. Reflection in the y-axis changes the sign of the x-coordinate \( (x, y) \to (-x, y) \):
- Image of Point B(-2, -2) is \( B'(2, -2) \).
- Image of Point C(-7, -6) is \( C'(7, -6) \).

Teacher's Note:
a) Recall reflection rules: reflection across the x-axis negates the y-coordinate, and across the y-axis negates the x-coordinate.
b) Read coordinates accurately from the provided graph.

(c) The marks obtained by the students in a class test are given below: [4 Marks]
31, 12, 28, 45, 32, 16, 49, 12, 18, 26, 34, 39, 29, 28, 25, 46, 32, 13, 14, 26, 25, 34, 23, 23, 25, 45, 33, 22, 18, 37, 26, 19, 20, 30, 28, 38, 42, 21, 36, 19, 20, 40, 48, 15, 46, 26, 23, 33, 47, 40.
Taking class intervals 10-15, 15-20, ...... 45-50; construct a frequency table.

Answer:

Teacher's Note:
a) Use exclusive class intervals where the lower limit is included and the upper limit is excluded.
b) Cross-check the sum of frequencies to ensure it equals the total number of observations (50).

Question 9

(a) Find the sum of the interior angles of a polygon of: [3 Marks]
i. 6 sides
ii. 8 sides
iii. 13 sides

Answer:
Formula for the sum of interior angles of an \( n \)-sided polygon = \( (2n - 4) \times 90^{\circ} \)
i. For 6 sides (\( n = 6 \)):
\( (2 \times 6 - 4) \times 90^{\circ} = (12 - 4) \times 90^{\circ} = 8 \times 90^{\circ} = 720^{\circ} \)
ii. For 8 sides (\( n = 8 \)):
\( (2 \times 8 - 4) \times 90^{\circ} = (16 - 4) \times 90^{\circ} = 12 \times 90^{\circ} = 1080^{\circ} \)
iii. For 13 sides (\( n = 13 \)):
\( (2 \times 13 - 4) \times 90^{\circ} = (26 - 4) \times 90^{\circ} = 22 \times 90^{\circ} = 1980^{\circ} \)

Teacher's Note:
a) Memorize the standard polygon angle sum formula \( (2n - 4) \times 90^{\circ} \) or \( (n - 2) \times 180^{\circ} \).
b) Substitute the number of sides correctly for each sub-part.

(b) The area of a trapezium is \( 105 \text{ cm}^2 \) and its height is 7 cm. If one of the parallel sides is longer than the other by 6 cm. Find the two parallel sides. [3 Marks]

Answer:
Let the shorter parallel side be \( x \text{ cm} \).
Then the longer parallel side is \( (x + 6) \text{ cm} \).
Area of trapezium = \( \dfrac{1}{2} \times (\text{sum of parallel sides}) \times \text{height} \)
\( 105 = \dfrac{1}{2} \times (x + x + 6) \times 7 \)
\( 105 = \dfrac{1}{2} \times (2x + 6) \times 7 \)
\( \dfrac{105 \times 2}{7} = 2x + 6 \)
\( 30 = 2x + 6 \)
\( 2x = 24 \implies x = 12 \text{ cm} \)
Shorter side = 12 cm, Longer side = \( 12 + 6 = 18 \text{ cm} \).

Teacher's Note:
a) Set up the equation using the formula for the area of a trapezium.
b) Solve for \( x \) to find both parallel side lengths.

(c) A and B are two sets such that \( n(A - B) = 32 + x \), \( n(B - A) = 5x \) and \( n(A \cap B) = x \). Illustrate the information by means of a Venn-diagram. [4 Marks]
Given that \( n(A) = n(B) \), calculate
i. the value of x.
ii. \( n(A \cup B) \)

[Figure: A Venn diagram showing two intersecting sets A and B within a universal set, with region A - B labeled as 32 + x, intersection A intersect B labeled as x, and region B - A labeled as 5x.]

Answer:
i. From the Venn diagram:
\( n(A) = n(A - B) + n(A \cap B) = (32 + x) + x = 32 + 2x \)
\( n(B) = n(B - A) + n(A \cap B) = 5x + x = 6x \)
Given that \( n(A) = n(B) \):
\( 32 + 2x = 6x \)
\( 4x = 32 \implies x = 8 \)
ii. \( n(A \cup B) = n(A - B) + n(B - A) + n(A \cap B) \)
\( n(A) = 32 + 2(8) = 32 + 16 = 48 \)
\( n(B) = 6(8) = 48 \)
\( n(A \cap B) = 8 \)
\( n(A \cup B) = n(A) + n(B) - n(A \cap B) = 48 + 48 - 8 = 88 \)

Teacher's Note:
a) Understand set relations: \( n(A) = n(A - B) + n(A \cap B) \).
b) Apply the inclusion-exclusion principle for union of two sets.