ICSE Class 7 Mathematics Sample Paper with Solutions Set 02

SECTION A (40 marks)

Question 1

a) Three metallic cubes with edges 3 cm, 4 cm and 5 cm were melted and recast into a new cube. What will be the length of the edge of the new cube? [3 Marks]

[Figure: Three separate cubes of edge lengths 3 cm, 4 cm, and 5 cm melting into a single larger cube]

Answer:
Volume of the first cube = \( 3^3 = 27 \text{ cm}^3 \)
Volume of the second cube = \( 4^3 = 64 \text{ cm}^3 \)
Volume of the third cube = \( 5^3 = 125 \text{ cm}^3 \)
Total volume of the new cube = \( 27 + 64 + 125 = 216 \text{ cm}^3 \)
Edge of the new cube = \( \sqrt[3]{216} = 6 \text{ cm} \).

Teacher's Note:
a) The total volume remains conserved when solid shapes are melted and recast into a new shape.
b) Students must remember to take the cube root of the total volume to find the edge length, not the sum of the edges.

b) Draw a Venn-Diagram to represent A = {2, 4, 6, 8, 10} and B = {4, 8, 12, 16, 20}. Find A ∪ B and A ∩ B. [4 Marks]

[Figure: A Venn diagram showing two intersecting circles labeled A and B. Circle A contains elements 2, 6, 10 in the non-intersecting part; Circle B contains 12, 16, 20 in the non-intersecting part; and the intersection contains elements 4, 8.]

Answer:
\( A \cup B = \{2, 4, 6, 8, 10, 12, 16, 20\} \)
\( A \cap B = \{4, 8\} \)

Teacher's Note:
a) Elements common to both sets must be placed in the overlapping region of the Venn diagram.
b) Union includes all elements from both sets without repetition, while intersection includes only the common elements.

c) Simplify: \( \left(\frac{y}{6} + \frac{2y}{3}\right) \div \left(2y - \frac{3y - 2}{2}\right) \) [3 Marks]

Answer:
\( = \left(\frac{y + 4y}{6}\right) \div \left(\frac{4y - (3y - 2)}{2}\right) \)
\( = \left(\frac{5y}{6}\right) \div \left(\frac{4y - 3y + 2}{2}\right) \)
\( = \left(\frac{5y}{6}\right) \div \left(\frac{y + 2}{2}\right) \)
\( = \frac{5y}{6} \times \frac{2}{y + 2} \)
\( = \frac{5y}{3(y + 2)} = \frac{5y}{3y + 6} \)

Teacher's Note:
a) Simplify inside the brackets first by finding common denominators.
b) Be careful with the negative sign when opening brackets in the numerator of the second term: \( -(3y - 2) \) becomes \( -3y + 2 \).

Question 2

a) In the given figure, find \( m\angle CGF \) and \( m\angle DGF \). [3 Marks]

[Figure: Two parallel lines AB and CD cut by a transversal EH at points F and G respectively. Angle AFG is given as \( 70^{\circ} \). Angles around point G are labeled: angle DGF as 'b' and angle CGF adjacent to it.]

Answer:
Since AB and CD are parallel lines and EH is a transversal, alternate interior angles are equal.
Therefore, \( m\angle DGF = m\angle AFG = 70^{\circ} \), so \( b = 70^{\circ} \).
Angles \( \angle CGF \) and \( \angle DGF \) form a linear pair on the straight line CD.
Therefore, \( m\angle CGF + m\angle DGF = 180^{\circ} \)
\( m\angle CGF + 70^{\circ} = 180^{\circ} \)
\( m\angle CGF = 180^{\circ} - 70^{\circ} = 110^{\circ} \).

Teacher's Note:
a) Identify parallel line properties such as alternate interior angles correctly.
b) Angles on a straight line always add up to \( 180^{\circ} \) (linear pair).

b) Represent the inequation on a number line: \( 3x + 14 \geq 8, x \in I \) [3 Marks]

[Figure: A number line from -5 to 5 with dark solid dots at -2, -1, 0, 1, 2, 3, 4, 5 and an arrow extending to the right from -2]

Answer:
\( 3x + 14 \geq 8 \)
\( 3x \geq 8 - 14 \)
\( 3x \geq -6 \)
\( x \geq -2 \)
Since \( x \in I \), the solution set is \( \{-2, -1, 0, 1, 2, 3, \dots\} \), represented on the number line by solid dots at -2 and all integers to its right with an arrow.

Teacher's Note:
a) Solve linear inequations using the same rules as linear equations, keeping in mind that reversing the inequality sign occurs only when multiplying or dividing by a negative number.
b) For integers (\( I \)), use distinct solid dots on the number line rather than a continuous shaded line.

c) Solve for a: \( a + b + 27 = 45 \) [2 Marks]

Answer:
\( a + b + 27 = 45 \)
\( a + b = 45 - 27 \)
\( a + b = 18 \)
\( a = 18 - b \)

Teacher's Note:
a) Transpose constant terms to the right-hand side to isolate terms containing variables.
b) Since there are two variables and one equation, the value of 'a' is expressed in terms of 'b'.

d) Calculate the interest earned on a sum of Rs. 18,000, lent for 3 years at 6% per annum. [2 Marks]

Answer:
\( \text{Simple Interest} = \frac{P \times R \times T}{100} \)
\( \text{S.I.} = \frac{18000 \times 3 \times 6}{100} \)
\( \text{S.I.} = 180 \times 18 = \text{Rs. } 3240 \)

Teacher's Note:
a) State the standard simple interest formula clearly before substitution.
b) Ensure units (Rupees) are explicitly written in the final answer.

Question 3

a) Convert as instructed: i. 36 km/hr to m/sec ii. 75 cm/sec to m/sec [3 Marks]

Answer:
i. \( 36 \text{ km/hr} = 36 \times \frac{5}{18} = 10 \text{ m/sec} \)
ii. \( 75 \text{ cm/sec} = \frac{75}{100} = 0.75 \text{ m/sec} \)

Teacher's Note:
a) To convert km/hr to m/sec, multiply the given value by \( \frac{5}{18} \).
b) To convert cm/sec to m/sec, divide the value by 100.

b) Solve: i. \( \{5 + (5 \times 8) \div 2 - 3\} \div (-11) \) ii. \( \{63 \div (-15 + 8)\} - (-3 \times 7) \) [4 Marks]

Answer:
i. \( \{5 + (5 \times 8) \div 2 - 3\} \div (-11) \)
\( = \{5 + 40 \div 2 - 3\} \div (-11) \)
\( = \{5 + 20 - 3\} \div (-11) \)
\( = 22 \div (-11) = -2 \)
ii. \( \{63 \div (-15 + 8)\} - (-3 \times 7) \)
\( = \{63 \div (-7)\} - (-21) \)
\( = -9 + 21 = 12 \)

Teacher's Note:
a) Strictly follow the BODMAS rule: evaluate innermost brackets first, followed by division, multiplication, addition, and subtraction.
b) Pay close attention to integer signs, especially when dividing or subtracting negative numbers.

c) In the figure, AD = DC and AB = BC. Prove that \( \Delta ABD \cong \Delta CBD \). [3 Marks]

[Figure: A kite-shaped quadrilateral ABCD divided by diagonal BD into two triangles \( \Delta ABD \) and \( \Delta CBD \), with sides AD and DC marked equal, and sides AB and BC marked equal.]

Answer:
In \( \Delta ABD \) and \( \Delta CBD \):
\( AD = DC \) (given)
\( AB = BC \) (given)
\( BD = BD \) (common side)
Therefore, by SSS congruence criterion, \( \Delta ABD \cong \Delta CBD \).

Teacher's Note:
a) Clearly state the three corresponding pairs of equal sides or angles when proving triangle congruence.
b) Mention the correct congruence criterion (SSS test) at the conclusion.

Question 4

a) By selling a dress for Rs. 729, a shopkeeper experienced a loss of 10%. Find the cost price of the dress? [3 Marks]

Answer:
\( \text{Selling Price (SP)} = \text{Rs. } 729 \)
\( \text{Loss}\% = 10\% \)
\( \text{Cost Price (CP)} = \left[\frac{100}{100 - \text{Loss}\%}\right] \times \text{SP} \)
\( \text{CP} = \left[\frac{100}{100 - 10}\right] \times 729 \)
\( \text{CP} = \frac{100}{90} \times 729 = \text{Rs. } 810 \)

Teacher's Note:
a) Use the standard formula relating cost price, selling price, and loss percentage.
b) Ensure proper simplification of fractions to avoid calculation errors.

b) Aryan has a rectangular garden whose length is double its width. The area of the garden is 450 square cm. What is the length of the garden? [2 Marks]

Answer:
Let the width of the garden be \( w \) cm.
Then, the length of the garden = \( 2w \) cm.
\( \text{Area} = \text{length} \times \text{breadth} \)
\( 450 = 2w \times w \)
\( 2w^2 = 450 \)
\( w^2 = 225 \)
\( w = \sqrt{225} = 15 \text{ cm} \)
Length = \( 2w = 2 \times 15 = 30 \text{ cm} \).

Teacher's Note:
a) Set up the algebraic equation based on the given relation between length and width.
b) Do not forget to multiply the width by 2 at the end to find the required length.

c) Find the value of x in the given figure if AOB is a straight line. [2 Marks]

[Figure: A straight line AOB with a ray originating from point O, dividing the straight angle into two adjacent angles labeled \( 3x + 5^{\circ} \) and \( 2x - 25^{\circ} \).]

Answer:
Since AOB is a straight line, the sum of adjacent angles forming a linear pair is \( 180^{\circ} \).
\( (3x + 5) + (2x - 25) = 180^{\circ} \)
\( 5x - 20 = 180^{\circ} \)
\( 5x = 180 + 20 \)
\( 5x = 200^{\circ} \)
\( x = 40^{\circ} \)

Teacher's Note:
a) Use the linear pair property where adjacent angles on a straight line sum to \( 180^{\circ} \).
b) Combine like terms carefully before transposing constants.

d) Each interior angle of a polygon is \( 140^{\circ} \). Find the number of sides. [3 Marks]

Answer:
Let the number of sides be \( n \).
Each interior angle = \( \frac{(n - 2) \times 180^{\circ}}{n} \)
\( \frac{(n - 2) \times 180^{\circ}}{n} = 140^{\circ} \)
\( \frac{n - 2}{n} = \frac{140}{180} = \frac{7}{9} \)
\( 9(n - 2) = 7n \)
\( 9n - 18 = 7n \)
\( 9n - 7n = 18 \)
\( 2n = 18 \)
\( n = 9 \)
Thus, the polygon has 9 sides.

Teacher's Note:
a) Alternatively, find the exterior angle first: \( 180^{\circ} - 140^{\circ} = 40^{\circ} \), then use \( n = \frac{360^{\circ}}{40^{\circ}} = 9 \).
b) Both methods are accepted, but the exterior angle method is often quicker.

SECTION B (40 marks)

Question 5

a) Subtract the sum of \( (8a^3 + 4a + 5c^2) \) and \( (4a^2 + 8b - 4c^2) \) from \( (-2a^3 + 9a - 5c^2 + 8a^2 + 6) \) [3 Marks]

Answer:
Sum of the two expressions:
\( (8a^3 + 4a + 5c^2) + (4a^2 + 8b - 4c^2) = 8a^3 + 4a^2 + c^2 + 8b + 4a \)
Now, subtracting this sum from \( (-2a^3 + 9a - 5c^2 + 8a^2 + 6) \):
\( (-2a^3 + 8a^2 + 9a - 5c^2 + 6) - (8a^3 + 4a^2 + 4a + c^2 + 8b) \)
\( = -2a^3 - 8a^3 + 8a^2 - 4a^2 + 9a - 4a - 5c^2 - c^2 - 8b + 6 \)
\( = -10a^3 + 4a^2 + 5a - 6c^2 - 8b + 6 \)

Teacher's Note:
a) Group like terms carefully after removing parentheses.
b) Remember to change the signs of all terms inside the subtrahend expression.

b) Solve: \( (2^{-1} \times 3^{-1})^2 \times \left(\frac{-3}{8}\right)^{-1} \) [2 Marks]

Answer:
\( = \left(\frac{1}{2} \times \frac{1}{3}\right)^2 \times \left(-\frac{8}{3}\right) \)
\( = \left(\frac{1}{6}\right)^2 \times \left(-\frac{8}{3}\right) \)
\( = \frac{1}{36} \times \left(-\frac{8}{3}\right) \)
\( = -\frac{8}{108} = -\frac{2}{27} \)

Teacher's Note:
a) Apply laws of exponents: \( a^{-1} = \frac{1}{a} \) and \( (ab)^m = a^m b^m \).
b) Simplify the final fraction to its lowest terms.

c) Simplify: \( \frac{\frac{1}{3}\left(\frac{1}{2} + \frac{1}{5}\right)}{\frac{1}{5}\left(\frac{1}{2} + \frac{1}{3}\right)} \) [2 Marks]

Answer:
Numerator = \( \frac{1}{3}\left(\frac{5 + 2}{10}\right) = \frac{1}{3} \times \frac{7}{10} = \frac{7}{30} \)
Denominator = \( \frac{1}{5}\left(\frac{3 + 2}{6}\right) = \frac{1}{5} \times \frac{5}{6} = \frac{1}{6} \)
Expression = \( \frac{\frac{7}{30}}{\frac{1}{6}} = \frac{7}{30} \times \frac{6}{1} = \frac{7}{5} \)

Teacher's Note:
a) Simplify the numerator and denominator fractions independently before dividing them.
b) Dividing by a fraction is equivalent to multiplying by its reciprocal.

d) Find x, y and z in the following triangle. [3 Marks]

[Figure: A triangle with exterior and interior angles. An interior angle at one vertex is \( 80^{\circ} \) (with a vertical angle also \( 80^{\circ} \)), another interior angle is \( 45^{\circ} \) with an adjacent angle \( 20^{\circ} \), and exterior angles z and y at the base extensions.]

Answer:
\( x = 180^{\circ} - 80^{\circ} = 100^{\circ} \) (linear pair)
\( z = 180^{\circ} - (x + 45^{\circ}) = 180^{\circ} - (100^{\circ} + 45^{\circ}) = 180^{\circ} - 145^{\circ} = 35^{\circ} \) (angle sum property of a triangle)
\( y = x + 45^{\circ} = 100^{\circ} + 45^{\circ} = 145^{\circ} \) (exterior angle property of a triangle)

Teacher's Note:
a) Use standard geometric properties such as linear pairs, angle sum property, and exterior angle theorem.
b) Check calculations by verifying that interior angles and corresponding exterior angles sum to \( 180^{\circ} \).

Question 6

a) Find the perimeter of the following plot (all measures are in m). [3 Marks]

[Figure: An L-shaped or stepped rectilinear polygon plot with marked side dimensions: horizontal top edge 1, vertical steps 1, 1, 4, 5, and bottom/side extensions.]

Answer:
Perimeter = Sum of all outer boundary lengths
\( = 1 + 4 + 5 + 1 + 1 + 4 + 5 + 4 + 4 + 1 + 1 + 1 = 32 \text{ m} \)

Teacher's Note:
a) Ensure all missing outer boundary lengths are accounted for by summing parallel horizontal and vertical segments.
b) Perimeter is the total boundary distance around a 2D shape.

b) Divide 104 pens between 3 friends in the ratio \( \frac{1}{2} : \frac{1}{3} : \frac{1}{4} \) [4 Marks]

Answer:
Given ratio = \( \frac{1}{2} : \frac{1}{3} : \frac{1}{4} \)
LCM of denominators (2, 3, 4) = 12
Multiply each term by 12 to convert to whole numbers:
\( \frac{1}{2} \times 12 : \frac{1}{3} \times 12 : \frac{1}{4} \times 12 = 6 : 4 : 3 \)
Total ratio parts = \( 6 + 4 + 3 = 13 \)
Number of pens for 1st friend = \( \frac{6}{13} \times 104 = 48 \)
Number of pens for 2nd friend = \( \frac{4}{13} \times 104 = 32 \)
Number of pens for 3rd friend = \( \frac{3}{13} \times 104 = 24 \)
Thus, the 3 friends get 48, 32, and 24 pens respectively.

Teacher's Note:
a) Always convert fractional ratios into simplified whole-number ratios by multiplying by the LCM of denominators.
b) Verify the answer by adding the distributed parts to ensure the total equals 104.

c) Factorise: \( 49(2x + y)^2 - 64(x - 3y)^2 \) [3 Marks]

Answer:
\( = [7(2x + y)]^2 - [8(x - 3y)]^2 \)
Using identity \( a^2 - b^2 = (a + b)(a - b) \):
\( = \{[7(2x + y)] + [8(x - 3y)]\}\{[7(2x + y)] - [8(x - 3y)]\} \)
\( = \{14x + 7y + 8x - 24y\}\{14x + 7y - 8x + 24y\} \)
\( = (22x - 17y)(6x + 31y) \)

Teacher's Note:
a) Recognize the expression as a difference of two squares.
b) Expand and collect like terms carefully within each set of curly brackets.

Question 7

a) The average of 6 numbers is 36.5. If 5 of the six numbers are 25, 29, 33, 37 and 51, find the \( 6^{\text{th}} \) number. [3 Marks]

Answer:
Total number of observations = 6
Average = 36.5
Sum of 6 numbers = \( 36.5 \times 6 = 219 \)
Sum of the 5 given numbers = \( 25 + 29 + 33 + 37 + 51 = 175 \)
\( 6^{\text{th}} \) number = Sum of 6 numbers - Sum of 5 numbers
\( = 219 - 175 = 44 \)

Teacher's Note:
a) Total sum is found by multiplying the average by the count of numbers.
b) Subtract the sum of the known numbers from the total sum to find the missing number.

b) Find x: \( 4x + 3 - x + 3 = 60 \) [2 Marks]

Answer:
\( 4x - x + 3 + 3 = 60 \)
\( 3x + 6 = 60 \)
\( 3x = 60 - 6 \)
\( 3x = 54 \)
\( x = 18 \)

Teacher's Note:
a) Group like terms on the left-hand side before solving.
b) Divide by the coefficient of x to find the final value.

c) Divide 0.00945 by 0.315. [2 Marks]

Answer:
\( \frac{0.00945}{0.315} = \frac{945 / 100000}{315 / 1000} \)
\( = \frac{945}{100000} \times \frac{1000}{315} \)
\( = \frac{3}{100} = 0.03 \)

Teacher's Note:
a) Convert decimals to fractions or shift decimal points equally in both numerator and denominator.
b) \( 945 \div 315 = 3 \), and adjust the powers of 10 accordingly.

d) Find the co-ordinates of the image of the following points under reflection in the origin: i. A(2, -3) ii. B(-3, -6) iii. C(0, 9) [3 Marks]

Answer:
i. Reflection of \( A(2, -3) \) in the origin is \( (-2, 3) \).
ii. Reflection of \( B(-3, -6) \) in the origin is \( (3, 6) \).
iii. Reflection of \( C(0, 9) \) in the origin is \( (0, -9) \).

Teacher's Note:
a) Reflection of a point \( (x, y) \) in the origin changes the signs of both coordinates: \( (-x, -y) \).
b) Zero remains unchanged in sign.

Question 8

a) Find the square and cube of: i. 15 and ii. 1.5 [3 Marks]

Answer:
i. For 15:
Square = \( 15^2 = 15 \times 15 = 225 \)
Cube = \( 15^3 = 15 \times 15 \times 15 = 3375 \)
ii. For 1.5:
Square = \( (1.5)^2 = 1.5 \times 1.5 = 2.25 \)
Cube = \( (1.5)^3 = 1.5 \times 1.5 \times 1.5 = 3.375 \)

Teacher's Note:
a) Multiply the number by itself for the square, and three times for the cube.
b) Pay attention to decimal placement when cubing numbers like 1.5.

b) In a poultry farm, 300 eggs are produced every day. If 100 birds produce an average of 75 eggs each day, how many birds are there in the poultry farm? Also find how many more birds should be bought if 375 eggs are needed every day. [4 Marks]

Answer:
Eggs produced by 100 birds per day = 75
Number of birds needed to produce 300 eggs per day = \( \frac{100 \times 300}{75} = 400 \) birds.
Number of birds needed to produce 375 eggs per day = \( \frac{100 \times 375}{75} = 500 \) birds.
Number of more birds required = \( 500 - 400 = 100 \) birds.

Teacher's Unit:
a) Use unitary method proportions to solve for the number of birds.
b) Subtract the current bird count from the target bird count to find the additional birds needed.

c) Construct \( \Delta XYZ \) in which XZ = 5 cm, YZ = 3.6 cm and XY = 5.4 cm. [3 Marks]

[Figure: A triangle XYZ with side lengths labeled: XY = 5.4 cm, XZ = 5 cm, and YZ = 3.6 cm]

Answer:
Steps of construction:
1. Draw line segment \( XY = 5.4 \text{ cm} \).
2. With X as center and radius 5 cm, draw an arc.
3. With Y as center and radius 3.6 cm, draw another arc to intersect the previous arc at point Z.
4. Join XZ and YZ to complete \( \Delta XYZ \).

Teacher's Note:
a) Use a ruler and compass for accurate geometric construction.
b) Clearly label all vertices and side lengths in the final diagram.

Question 9

a) The temperatures of 10 days in a city are given below. Plot a line graph to represent the data. [4 Marks]

[Figure: A line graph with Days on the horizontal axis (Day 1 to Day 10) and Temperature in \( ^{\circ}\text{C} \) on the vertical axis (from 36 to 45), showing connected data points corresponding to the table values]

Answer:
A line graph is plotted with Days along the x-axis and Temperature (\( ^{\circ}\text{C} \)) along the y-axis, plotting points (Day 1, 41), (Day 2, 40.5), (Day 3, 42), (Day 4, 39), (Day 5, 40), (Day 6, 43), (Day 7, 44), (Day 8, 41.5), (Day 9, 38), and (Day 10, 40), and connecting them consecutively with straight line segments.

Teacher's Note:
a) Ensure appropriate scaling on both axes so the graph is clear and legible.
b) Label axes with proper titles and units.

b) The H.C.F. of two numbers is 18 and their L.C.M. is 108. One of the numbers is 54. Find the other number. [3 Marks]

Answer:
We know that for any two numbers, Product of numbers = \( \text{HCF} \times \text{LCM} \)
\( 54 \times x = 18 \times 108 \)
\( x = \frac{18 \times 108}{54} \)
\( x = 1 \times 36 = 36 \)
Thus, the other number is 36.

Teacher's Note:
a) Apply the standard relation: \( \text{Product of two numbers} = \text{HCF} \times \text{LCM} \).
b) Simplify before multiplying to make calculations easier.

c) Multiply: \( 57 \times 63 \) [3 Marks]

Answer:
\( 57 = (60 - 3) \)
\( 63 = (60 + 3) \)
Using algebraic identity \( (a - b)(a + b) = a^2 - b^2 \):
\( (60 - 3)(60 + 3) = 60^2 - 3^2 \)
\( = 3600 - 9 = 3591 \)

Teacher's Note:
a) Using algebraic identities like \( (a+b)(a-b) \) simplifies mental or manual arithmetic.
b) Alternatively, standard multiplication can be used to verify the result.