ICSE Class 7 Mathematics Sample Paper with Solutions Set 01

SECTION A (40 marks)

Question 1

a) If the marks of Rohit, Ajay and Vipul are in ratio of 4 : 5 : 6, and Ajay got 75 marks, then find the marks of Rohit and Vipul? [2 Marks]

Answer:
Let the marks of Rohit, Ajay and Vipul be \( 4x \), \( 5x \) and \( 6x \) respectively.
Given that Ajay's marks = \( 75 \)
\( \Rightarrow 5x = 75 \)
\( \Rightarrow x = \frac{75}{5} = 15 \)
Marks of Rohit = \( 4x = 4 \times 15 = 60 \)
Marks of Vipul = \( 6x = 6 \times 15 = 90 \)
Thus, the marks of Rohit and Vipul are \( 60 \) and \( 90 \) respectively.

Teacher's Note:
a) Use a common variable multiplier such as \( x \) to split the given ratio into individual algebraic quantities.
b) Ensure both required quantities are calculated and stated clearly at the end.

b) A train 270 m long is running at 40.5 km/hr. How much time will it take to cross the tree? [3 Marks]

Answer:
Length of the train = \( 270\text{ m} = \frac{270}{1000}\text{ km} = 0.27\text{ km} \)
Speed of the train = \( 40.5\text{ km/hr} \)
Time taken to cross the tree = \( \frac{\text{Distance}}{\text{Speed}} = \frac{0.27}{40.5}\text{ hr} \)
\( = \frac{27}{4050}\text{ hr} = \frac{1}{150}\text{ hr} \)
\( = \frac{1}{150} \times 3600\text{ sec} = 24\text{ sec} \)
Thus, the train will take \( 24 \) seconds to cross the tree.

Teacher's Note:
a) The distance covered by a train to cross a stationary point like a tree is equal to the length of the train itself.
b) Convert units carefully between meters, kilometers, hours, and seconds before performing division.

c) Express the following as a rational number:
\( \left[\left(\frac{2}{3}\right)^{2}\right]^{3} \times \left(\frac{1}{3}\right)^{-2} \times 3^{-1} \times \frac{1}{6} \) [3 Marks]

Answer:
\( = \left(\frac{2}{3}\right)^{2 \times 3} \times 3^{2} \times 3^{-1} \times \frac{1}{6} \)
\( = \left(\frac{2}{3}\right)^{6} \times 3^{2 - 1} \times \frac{1}{6} \)
\( = \frac{2^{6}}{3^{6}} \times 3^{1} \times \frac{1}{6} \)
\( = \frac{64}{729} \times 3 \times \frac{1}{6} \)
\( = \frac{64}{729} \times \frac{1}{2} \)
\( = \frac{32}{729} \)

Teacher's Note:
a) Apply the laws of exponents such as \( (a^{m})^{n} = a^{m \times n} \) and negative exponent rule \( a^{-n} = \frac{1}{a^{n}} \) step by step.
b) Reduce fractions fully at the final multiplication stage to avoid arithmetic errors.

d) State whether True or False: [2 Marks]
i. If an object looks exactly the same after a rotation, then it has a rotational symmetry.
ii. If a transversal cuts a pair of parallel lines, then the alternate angles formed are congruent.

Answer:
i. True
ii. True

Teacher's Note:
a) Rotational symmetry is defined as the property where an object fits onto itself more than once during a full \( 360^{\circ} \) turn.
b) Alternate interior and exterior angles formed by a transversal with parallel lines are always equal in measure (congruent).

Question 2

a) The given data shows the marks obtained by 20 students of a class in a Math test.
31, 9, 8, 20, 8, 7, 30, 31, 24, 20, 13, 13, 28, 26, 19, 27, 13, 12, 25, 21
Represent the data in a frequency distribution table and find the mean. [4 Marks]

Answer:

Sum of all observations = \( 7 + 8 + 8 + 9 + 12 + 13 + 13 + 13 + 19 + 20 + 20 + 21 + 24 + 25 + 26 + 27 + 28 + 30 + 31 + 31 = 300 \)
Mean marks = \( \frac{\text{Sum of observations}}{\text{Total number of observations}} = \frac{300}{20} = 15 \)

Teacher's Note:
a) Always arrange raw data in ascending order before constructing a frequency distribution table to avoid missing values.
b) Verify that the sum of all frequencies matches the total number of students given in the problem statement.

b) Add : i) \( 4x^{2} + 3x + y \) and \( 5x - 3y \)
ii) \( 9a^{2} + 4b - 4c \) and \( -5a^{2} - 5b \) [3 Marks]

Answer:
i) \( (4x^{2} + 3x + y) + (5x - 3y) \)
\( = 4x^{2} + 3x + y + 5x - 3y \)
\( = 4x^{2} + 3x + 5x + y - 3y \)
\( = 4x^{2} + 8x - 2y \)

ii) \( (9a^{2} + 4b - 4c) + (-5a^{2} - 5b) \)
\( = 9a^{2} + 4b - 4c - 5a^{2} - 5b \)
\( = 9a^{2} - 5a^{2} + 4b - 5b - 4c \)
\( = 4a^{2} - b - 4c \)

Teacher's Note:
a) Combine only like terms by grouping their coefficients together.
b) Pay careful attention to signs when removing parentheses containing negative terms.

c) Find the area of the shaded region? [3 Marks]

[Figure: A rectangle of total width 12 cm (4 cm + 4 cm + 4 cm) and height 8 cm. Inside the rectangle, there is an unshaded square of side 4 cm placed on the left side, and a right-angled triangle on the right side with base 4 cm and height 8 cm. The remaining region is shaded pink.]

Answer:
The given figure is a rectangle with length \( 12\text{ cm} \) (\( 4\text{ cm} + 4\text{ cm} + 4\text{ cm} \)) and height \( 8\text{ cm} \).
The shaded region can be divided into two smaller rectangles and one right-angled triangle.
Area of 1st rectangle = \( 4\text{ cm} \times 8\text{ cm} = 32\text{ cm}^{2} \)
Area of 2nd rectangle = \( 4\text{ cm} \times 6\text{ cm} = 24\text{ cm}^{2} \)
Area of the right triangle = \( \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2\text{ cm} \times 4\text{ cm} = 4\text{ cm}^{2} \)
Total area of the shaded region = \( 32 + 24 + 4 = 60\text{ cm}^{2} \)

Teacher's Note:
a) Break complex composite figures into standard geometric shapes like rectangles and triangles to calculate areas easily.
b) Ensure all dimensions used for individual sub-shapes are correctly deduced from the main figure diagram.

Question 3

a) \( \frac{5}{6} \)th of the cake was eaten by 5 friends. The next day 3 other friends ate \( \frac{1}{2} \) of what was left. How much of the cake is left? [3 Marks]

Answer:
Cake left on day 1 = \( 1 - \frac{5}{6} = \frac{6 - 5}{6} = \frac{1}{6} \)
Cake eaten on day 2 = \( \frac{1}{2} \) of remaining cake = \( \frac{1}{2} \times \frac{1}{6} = \frac{1}{12} \)
Cake left at the end = \( \frac{1}{6} - \frac{1}{12} = \frac{2 - 1}{12} = \frac{1}{12} \)
Hence, \( \frac{1}{12} \) of the cake is remaining.

Teacher's Note:
a) Always subtract the consumed fraction from whole unity (\( 1 \)) to find the remaining fraction.
b) Pay close attention to phrasing like "half of what was left" versus "half of the total cake".

b) Sanket’s monthly expenditure is Rs. 15000. He spends 25% on house rent, 40% on food and groceries, 5% each on travelling and entertainment and the rest on education. Calculate the amount he spends on each. [4 Marks]

Answer:
Total monthly expenditure = Rs. 15,000
Amount spent on rent = \( \frac{25}{100} \times 15,000 = \text{Rs. } 3,750 \)
Amount spent on food and groceries = \( \frac{40}{100} \times 15,000 = \text{Rs. } 6,000 \)
Amount spent on travelling = \( \frac{5}{100} \times 15,000 = \text{Rs. } 750 \)
Amount spent on entertainment = \( \frac{5}{100} \times 15,000 = \text{Rs. } 750 \)
Amount spent on education = \( 15,000 - (3,750 + 6,000 + 750 + 750) \)
\( = 15,000 - 11,250 = \text{Rs. } 3,750 \)

Teacher's Note:
a) Calculate percentages of a given total by converting each percentage into a fraction or decimal multiplier.
b) Verify that the sum of all individual expense amounts equals the total monthly expenditure of Rs. 15,000.

c) In the given figure, \( l \parallel m \) find \( x \). [3 Marks]

[Figure: Two parallel lines \( l \) and \( m \) intersected by a transversal line \( t \). An interior angle is marked as \( 105^{\circ} \) adjacent to angle 1 on line \( l \), and angle \( x \) is marked as an interior angle on line \( m \).]

Answer:
Let the angle vertically opposite or adjacent to \( 105^{\circ} \) forming a linear pair on line \( l \) be \( \angle 1 \).
\( m\angle 1 + 105^{\circ} = 180^{\circ} \) (linear pair)
\( \therefore m\angle 1 = 180^{\circ} - 105^{\circ} = 75^{\circ} \)
Since \( l \parallel m \), corresponding angles are equal.
\( x = m\angle 1 = 75^{\circ} \)

Teacher's Note:
a) Use angle properties such as linear pairs and corresponding angles for parallel lines cut by a transversal.
b) Clearly mention the geometric theorem used at each step of angle calculation.

Question 4

a) The population of a town increases by 6% annually. If the present population is 17490, what was it a year ago? [4 Marks]

Answer:
Let the population of the town a year ago be \( x \).
Then, its present population = \( 106\% \) of \( x = 17,490 \)
\( \frac{106}{100} \times x = 17,490 \)
\( \frac{53}{50} \times x = 17,490 \)
\( x = \frac{17,490 \times 50}{53} = 16,500 \)
Hence, the population of the town a year ago was \( 16,500 \).

Teacher's Note:
a) When finding a past value given percentage growth, set up the equation based on the original unknown base value \( x \), not the current value.
b) Cross-check the answer by calculating a \( 6\% \) increase on \( 16,500 \), which gives \( 16,500 + 990 = 17,490 \).

b) Simplify: \( \left(\frac{a^{3}}{b^{4}}\right)^{2} \times \left(\frac{b^{2}}{a^{3}}\right)^{3} \) [3 Marks]

Answer:
\( = \frac{(a^{3})^{2}}{(b^{4})^{2}} \times \frac{(b^{2})^{3}}{(a^{3})^{3}} \)
\( = \frac{a^{6}}{b^{8}} \times \frac{b^{6}}{a^{9}} \)
\( = \frac{a^{6}}{a^{9}} \times \frac{b^{6}}{b^{8}} \)
\( = a^{6 - 9} \times b^{6 - 8} \)
\( = a^{-3}b^{-2} \) or \( \frac{1}{a^{3}b^{2}} \)

Working Notes:
1. Apply exponent power rule \( (x^{m})^{n} = x^{m \times n} \).
2. Group like bases and apply quotient rule \( \frac{x^{m}}{x^{n}} = x^{m - n} \).

Teacher's Note:
a) Expand powers of products and quotients carefully before combining terms.
b) Negative exponents can be left as such or expressed in positive denominator form depending on standard preference.

c) The angles of a triangle are in the ratio 2 : 3 : 5. Find the angles. [3 Marks]

Answer:
Let the angles of the triangle be \( 2x \), \( 3x \) and \( 5x \).
Sum of angles of a triangle = \( 180^{\circ} \)
\( 2x + 3x + 5x = 180^{\circ} \)
\( 10x = 180^{\circ} \)
\( x = \frac{180^{\circ}}{10} = 18^{\circ} \)
Thus, the angles are:
\( 2x = 2 \times 18^{\circ} = 36^{\circ} \)
\( 3x = 3 \times 18^{\circ} = 54^{\circ} \)
\( 5x = 5 \times 18^{\circ} = 90^{\circ} \)
Hence, the angles are \( 36^{\circ} \), \( 54^{\circ} \) and \( 90^{\circ} \).

Teacher's Note:
a) Use the angle sum property of triangles to form a linear algebraic equation in terms of multiplier \( x \).
b) Verify that the sum of the three calculated angles equals \( 180^{\circ} \).

SECTION B (40 marks)

Instruction: Solve any four questions from Section B.

Question 5

a) Write the following sets in the roster method. [2 Marks]
i. M is a set of first 5 multiples of 3.
ii. V is a set of the vowels in ‘DISJOINT’.

Answer:
i. \( M = \{3, 6, 9, 12, 15\} \)
ii. \( V = \{I, O\} \)

Teacher's Note:
a) Roster form requires listing all elements within curly brackets separated by commas.
b) In set notation, duplicate elements must not be repeated (e.g., vowel 'I' appears twice in DISJOINT but written once).

b) The figure below is made of 3 squares with sides 5 cm. What is the perimeter of the figure? [2 Marks]

[Figure: A geometric figure composed of 3 adjoining squares arranged in a row or diagonal touching pattern with individual side length 5 cm, showing an outer boundary outline.]

Answer:
Perimeter = Sum of all outer boundary sides of the 3 squares forming the figure.
Counting the outer boundary segments of length \( 5\text{ cm} \):
Perimeter = \( (5 + 5 + 5 + 5 + 5 + 5 + 5 + 5)\text{ cm} = 40\text{ cm} \)

Teacher's Note:
a) Perimeter is the total length of the outer boundary enclosing the shape; internal shared edges are not counted.
b) Count the exact number of exposed outer segments carefully from the diagram.

c) The score of 8 members of a team is 360 and that of 7 members of another team is 322. Which team scored better? [3 Marks]

Answer:
Total score of 1st team = \( 360 \)
Total members of 1st team = \( 8 \)
Average score of 1st team = \( \frac{360}{8} = 45 \)
Total score of 2nd team = \( 322 \)
Total members of 2nd team = \( 7 \)
Average score of 2nd team = \( \frac{322}{7} = 46 \)
Since the average score of the 2nd team (\( 46 \)) is greater than the average score of the 1st team (\( 45 \)), the 2nd team scored better.

Teacher's Note:
a) Compare teams with different numbers of members by calculating their respective mean (average) scores.
b) Clearly state the comparison conclusion based on the calculated averages.

d) The cost of 1 L milk is Rs. 22.50. What is the cost of 40.3 L of milk? [3 Marks]

Answer:
Cost of \( 1\text{ L} \) of milk = Rs. \( 22.50 \)
Cost of \( 40.3\text{ L} \) of milk = \( 40.3 \times 22.50 = \text{Rs. } 906.75 \)
Thus, the cost of \( 40.3\text{ L} \) of milk is Rs. \( 906.75 \).

Teacher's Note:
a) Use direct multiplication of decimal numbers to find the total cost.
b) Count decimal places correctly in the product (one decimal place from \( 40.3 \) and two from \( 22.50 \) making three, rounded to currency two decimal places).

Question 6

a) Solve: i) \( (-48) \times 24 \times (-10) + 100 \)
ii) \( (-56) + 27 - 45 - 17 + 19 \) [4 Marks]

Answer:
i) \( (-48) \times 24 \times (-10) + 100 \)
\( = -1152 \times (-10) + 100 \)
\( = 11520 + 100 \)
\( = 11620 \)

ii) \( (-56) + 27 - 45 - 17 + 19 \)
\( = -29 - 45 - 17 + 19 \)
\( = -74 - 17 + 19 \)
\( = -91 + 19 \)
\( = -72 \)

Teacher's Note:
a) Follow standard BODMAS rules and handle multiplication of negative integers (- multiplied by - gives +).
b) For addition and subtraction of integers, perform operations sequentially from left to right.

b) If \( 5x - \frac{3}{4} = 2x - \frac{2}{3} \), then find the value of \( x \). [3 Marks]

Answer:
\( 5x - \frac{3}{4} = 2x - \frac{2}{3} \)
\( 5x - 2x = \frac{3}{4} - \frac{2}{3} \)
\( 3x = \frac{9 - 8}{12} \)
\( 3x = \frac{1}{12} \)
\( x = \frac{1}{12} \div 3 \)
\( x = \frac{1}{12} \times \frac{1}{3} \)
\( x = \frac{1}{36} \)

Teacher's Note:
a) Group all variable terms on one side and constant terms on the other side of the equation.
b) Take the LCM of denominators ( \( 12 \) for \( 4 \) and \( 3 \) ) to subtract fractions accurately.

c) Simplify: \( \frac{m}{5} - \frac{m - 2}{3} + m \) [3 Marks]

Answer:
\( = \frac{m}{5} - \frac{m - 2}{3} + \frac{m}{1} \)
Taking LCM of denominators \( 5 \) and \( 3 \), which is \( 15 \):
\( = \frac{m \times 3 - (m - 2) \times 5 + m \times 15}{15} \)
\( = \frac{3m - (5m - 10) + 15m}{15} \)
\( = \frac{3m - 5m + 10 + 15m}{15} \)
\( = \frac{13m + 10}{15} \)

Teacher's Note:
a) Be extremely careful with the negative sign outside the fraction numerator when expanding \( -(5m - 10) \).
b) Combine all like variable terms in the numerator correctly before writing the final expression.

Question 7

a) Find \( x \): \( x + z - 15 = 65 \) [2 Marks]

Answer:
\( x + z - 15 = 65 \)
\( x + z = 65 + 15 \)
\( x + z = 80 \)
\( x = 80 - z \)

Teacher's Note:
a) Transpose constant terms to the right side of the equation.
b) Since there are two variables, express \( x \) in terms of \( z \) as the final answer.

b) What is the square root of i) 1.44 and ii) 289 [3 Marks]

Answer:
i) \( \sqrt{1.44} = \sqrt{\frac{144}{100}} = \frac{\sqrt{144}}{\sqrt{100}} = \frac{12}{10} = 1.2 \)
ii) \( \sqrt{289} = \sqrt{17 \times 17} = 17 \)

Teacher's Note:
a) For decimals, convert to fraction form with powers of ten or use decimal place root rules.
b) Memorize standard square roots up to 30 for speed and accuracy in exams.

c) Solve the inequation: \( -14 - 5x \ge 3x + 2 \), \( x \in \text{integers} \) [2 Marks]

Answer:
\( -14 - 5x \ge 3x + 2 \)
\( -14 - 2 \ge 3x + 5x \)
\( -16 \ge 8x \)
\( \frac{-16}{8} \ge x \)
\( -2 \ge x \)
\( x \le -2 \)
Solution set: \( x = \{-2, -3, -4, \dots\} \)

Teacher's Note:
a) Isolate the variable term by transposing terms across the inequality sign.
b) Clearly write the solution set representing integers satisfying the inequality condition.

d) In the figure, AD bisects \( \angle A \) and \( \angle AD \perp BC \). Show that \( \triangle ADB \cong \triangle ADC \). [3 Marks]

[Figure: Triangle ABC with altitude AD perpendicular to BC, and line AD bisecting angle A into two equal angles.]

Answer:
Given: AD bisects \( \angle A \) and \( \angle AD \perp BC \).
In \( \triangle ADB \) and \( \triangle ADC \):
\( \angle BAD = \angle CAD \) (given, AD bisects \( \angle A \))
\( AD = AD \) (common side)
\( \angle ADB = \angle ADC = 90^{\circ} \) (given, \( AD \perp BC \))
Thus, by ASA (Angle-Side-Angle) congruence criterion, \( \triangle ADB \cong \triangle ADC \).

Teacher's Note:
a) Identify three corresponding equal parts (two angles and the included side) to establish triangle congruence.
b) State the congruence test criterion clearly at the end of the proof.

Question 8

a) A person weighing 60 kg on the Earth weighs 9.9 kg on the Moon and 141.8 kg on Jupiter. How much will another person weighing 75 kg weigh on the Moon and on Jupiter? [4 Marks]

Answer:
Weight of a person on Earth = \( 60\text{ kg} \)
Weight of a person on the Moon = \( 9.9\text{ kg} \)
Weight of a person on Jupiter = \( 141.8\text{ kg} \)
If the weight of a person on Earth = \( 75\text{ kg} \), then:
Weight on the Moon = \( \frac{75 \times 9.9}{60} = \frac{742.5}{60} = 12.375 \approx 12.4\text{ kg} \)
Weight on Jupiter = \( \frac{75 \times 141.8}{60} = \frac{10635}{60} = 177.25 \approx 177.3\text{ kg} \)

Teacher's Note:
a) Use direct proportion methods since planetary weights are proportional to Earth weight.
b) Round off decimal answers appropriately to one decimal place if required.

b) What is the sum of the interior angles of a polygon with:
i) 12 sides and ii) 25 sides [3 Marks]

Answer:
Formula for the sum of interior angles of an \( n \)-sided polygon = \( (n - 2) \times 180^{\circ} \)
i) For a polygon with \( 12 \) sides (\( n = 12 \)):
Sum = \( (12 - 2) \times 180^{\circ} = 10 \times 180^{\circ} = 1800^{\circ} \)
ii) For a polygon with \( 25 \) sides (\( n = 25 \)):
Sum = \( (25 - 2) \times 180^{\circ} = 23 \times 180^{\circ} = 4140^{\circ} \)

Teacher's Note:
a) State the formula \( (n - 2) \times 180^{\circ} \) clearly before substituting values.
b) Verify calculations by double-checking multiplication steps.

c) In the given figure, O is the centre of the circle and \( m\angle CAB = 35^{\circ} \). Calculate the measure of \( \angle ABC \). [3 Marks]

[Figure: A circle with centre O and diameter AB. Point C is on the circumference. Angle CAB is marked as \( 35^{\circ} \).]

Answer:
O is the centre of the circle and AB passes through O, hence AB is the diameter.
\( m\angle BCA = 90^{\circ} \) (angle subtended in a semi-circle is a right angle)
In \( \triangle ABC \):
\( m\angle ABC + m\angle BCA + m\angle CAB = 180^{\circ} \)
\( m\angle ABC + 90^{\circ} + 35^{\circ} = 180^{\circ} \)
\( m\angle ABC + 125^{\circ} = 180^{\circ} \)
\( m\angle ABC = 180^{\circ} - 125^{\circ} = 55^{\circ} \)

Teacher's Note:
a) Recall the geometric theorem that the angle in a semi-circle is always \( 90^{\circ} \).
b) Use the angle sum property of triangles to find the remaining angle.

Question 9

a) Find the supplementary and complementary angle for an angle measuring \( 39^{\circ} \). [3 Marks]

Answer:
Complementary angle for \( 39^{\circ} = 90^{\circ} - 39^{\circ} = 51^{\circ} \)
Supplementary angle for \( 39^{\circ} = 180^{\circ} - 39^{\circ} = 141^{\circ} \)

Teacher's Note:
a) Complementary angles sum up to \( 90^{\circ} \), whereas supplementary angles sum up to \( 180^{\circ} \).
b) Keep subtraction calculations accurate.

b) The ratio of the ages of two brothers is 3 : 2. If the elder brother’s age is 21 years, how old is the younger brother? Find the ratio of their ages after 7 years. [4 Marks]

Answer:
Ratio of the ages of the two brothers = \( 3 : 2 \)
Elder brother’s age = \( 21\text{ years} \)
Let the younger brother’s age be \( x \) years.
\( 3 : 2 :: 21 : x \)
\( 3x = 2 \times 21 \)
\( 3x = 42 \)
\( x = \frac{42}{3} = 14\text{ years} \)
Hence, the younger brother is \( 14 \) years old.
Elder brother’s age after \( 7 \) years = \( 21 + 7 = 28\text{ years} \)
Younger brother’s age after \( 7 \) years = \( 14 + 7 = 21\text{ years} \)
Ratio of their ages after \( 7 \) years = \( 28 : 21 = 4 : 3 \)

Teacher's Note:
a) Set up a proportion equation using the given age ratio to find the unknown age.
b) Add \( 7 \) years to both brothers' ages before computing the new ratio.

c) Write the co-ordinates for the following points when reflected on the y-axis:
i) \( A(3, 6) \) ii) \( B(-4, 8) \) iii) \( C(4, -7) \) [3 Marks]

Answer:
When a point is reflected across the y-axis, its x-coordinate changes sign while its y-coordinate remains unchanged:\
i) \( A(3, 6) \to (-3, 6) \)
ii) \( B(-4, 8) \to (4, 8) \)
iii) \( C(4, -7) \to (-4, -7) \)

Teacher's Note:
a) Reflection across the y-axis transforms \( (x, y) \) to \( (-x, y) \).
b) Double-check signs of coordinates carefully after reflection.