ICSE Class 9 Mathematics Sample Paper with Solutions Set 05

ICSE EXAMINATION

Sample Question Paper - 4

Mathematics

Time: 2 ½ Hours    Total Marks: 80

General Instructions:
1. Answers to this Paper must be written on the paper provided separately.
2. You will not be allowed to write during the first 15 minutes. This time is to be spent in reading the question paper.
3. The time given at the head of this Paper is the time allowed for writing the answers.
4. Attempt all questions from Section A and any four questions from Section B.
5. The intended marks for questions or parts of questions are given in brackets [ ].

Section A

(Attempt all questions from this section.)

Question 1
Choose the correct answers to the questions from the given options. [15]

(i) Which of the following is a rational number? [1 Mark]
(a) \((3+\sqrt{3})(3-\sqrt{3})\)
(b) \(\sqrt{23}-6+\sqrt{36}\)
(c) \(\sqrt{64}-2\sqrt{8}\)
(d) \(4\sqrt{6}-2\sqrt{6}\)

Answer: (a) \((3+\sqrt{3})(3-\sqrt{3})\)

\((3+\sqrt{3})(3-\sqrt{3}) = 3^{2}-(\sqrt{3})^{2} = 9-3 = 6\), which is a rational number.

Teacher's Note:
a) A rational number is a number that can be expressed as the quotient or fraction \(\frac{p}{q}\) of two integers.
b) Always simplify surds completely before determining whether the resulting expression is rational or irrational.

(ii) What will be the amount on Rs. 10000 invested for 1 year at the rate of 8% per annum compounded annually? [1 Mark]
(a) Rs. 11664
(b) Rs. 10800
(c) Rs. 10000
(d) Rs. 800

Answer: (b) Rs. 10800

Amount = \(10000 \left(1 + \frac{8}{100}\right)^{1} = \text{Rs. } 10800\).

Teacher's Note:
a) When interest is compounded annually for 1 year, the compound amount is equal to the simple amount.
b) Ensure proper substitution of Principal, Rate of interest, and Time period in the compound interest formula.

(iii) If \(a - b = 1\) and \(ab = 6\), what is the value of \((a + b)\)? [1 Mark]
(a) 7
(b) 6
(c) 5
(d) 1

Answer: (c) 5

\((a + b)^{2} = (a - b)^{2} + 4ab = 1^{2} + 4(6) = 1 + 24 = 25 \implies a + b = \pm 5\). Since options provide 5, the correct choice is (c).

Teacher's Note:
a) Use the algebraic identity \((a + b)^{2} = (a - b)^{2} + 4ab\) to connect sum, difference, and product.
b) Be careful with signs when taking the square root of both sides.

(iv) Factors of \(3ax - 6ay - 8by + 4bx\) are: [1 Mark]
(a) \((x - 2y)(3a + 4b)\)
(b) \((x + 2y)(3a - 4b)\)
(c) \((3x - 4y)(a + b)\)
(d) \((3x + 4y)(a - b)\)

Answer: (a) \((x - 2y)(3a + 4b)\)

\(3ax - 6ay - 8by + 4bx = 3a(x - 2y) + 4b(x - 2y) = (x - 2y)(3a + 4b)\).

Teacher's Note:
a) Factorization by grouping terms requires careful rearrangement to find a common binomial factor.
b) Verify the expansion of factors to check the correctness of signs.

(v) The cost of 11 pens and 19 pencils is Rs. 502 and the cost of 19 pens and 11 pencils is Rs. 758. Which of the following linear equations represent the given situation? [1 Mark]
(a) \(11x + 19y = 502\), \(19x + 11y = 758$
(b) \(19x + 11y = 502\), \(11x + 19y = 758$
(c) \(11x + 19y = 758\), \(19x + 11y = 502$
(d) \(11x - 19y = 502\), \(19x - 11y = 758$
*(Note: let the cost of one pen be Rs. \(x\) and one pencil be Rs. \(y\))

Answer: (a) \(11x + 19y = 502\), \(19x + 11y = 758\)

Translating word statements directly yields \(11x + 19y = 502\) and \(19x + 11y = 758\).

Teacher's Note:
a) Assign variables clearly to unknown quantities as instructed.
b) Match the coefficients with respective items carefully.

(vi) If \((81)^{x} = 3^{12}\), then the value of \(x\) is: [1 Mark]
(a) 1
(b) 2
(c) 4
(d) 3

Answer: (d) 3

\((3^{4})^{x} = 3^{12} \implies 3^{4x} = 3^{12} \implies 4x = 12 \implies x = 3\).

Teacher's Note:
a) Express both sides of an exponential equation with the same base before equating powers.
b) Remember that \((a^{m})^{n} = a^{m \times n}\).

(vii) Which of the following is not a condition for congruence of triangles? [1 Mark]
(a) SAS
(b) ASA
(c) SSS
(d) AAA

Answer: (d) AAA

AAA only proves similarity, not congruence, because triangles of different sizes can have identical corresponding angles.

Teacher's Note:
a) Valid congruence criteria are SSS, SAS, ASA, AAS, and RHS.
b) Students must note that AAA guarantees similar shapes, not equal sizes.

(viii) Assertion (A): The sides measuring \(4\text{ cm}\), \(7\text{ cm}\), \(9\text{ cm}\) can form a right-angled triangle.
Reason (R): In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. [1 Mark]
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
(c) Assertion (A) is true but Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is true.

Answer: (d) Assertion (A) is false and Reason (R) is true.

For the sides \(4\), \(7\), and \(9\), \(4^{2} + 7^{2} = 16 + 49 = 65 \neq 9^{2}\ (81)\), so Assertion is false, while Reason states Pythagoras theorem correctly.

Teacher's Note:
a) Verify Pythagorean triplets using \(a^{2} + b^{2} = c^{2}\) where \(c\) is the longest side.
b) Always test both assertion and reason independently.

(ix) If chord \(AB = 8\text{ cm}\) of a circle is at a distance of \(3\text{ cm}\) from the centre, then the radius of the circle is: [1 Mark]
(a) \(3\text{ cm}\)
(b) \(4\text{ cm}\)
(c) \(5\text{ cm}\)
(d) \(6\text{ cm}\)

Answer: (c) \(5\text{ cm}\)

Perpendicular from centre bisects chord, so half chord \( = 4\text{ cm}\). Radius \(r = \sqrt{4^{2} + 3^{2}} = \sqrt{25} = 5\text{ cm}\).

Teacher's Note:
a) The perpendicular drawn from the centre of a circle to a chord bisects the chord.
b) Apply Pythagoras theorem in the right-angled triangle formed by the radius, distance from centre, and half-chord.

(x) The median of observations \(7, 3, 5, 1, 9, 2, 4\) is: [1 Mark]
(a) 2
(b) 3
(c) 5
(d) 4

Answer: (d) 4

Arranging in ascending order: \(1, 2, 3, 4, 5, 7, 9\). Middle observation (\(n = 7\)) is the \(4^{\text{th}}\) term, which is 4.

Teacher's Note:
a) Always arrange data in ascending or descending order before finding the median.
b) For an odd number of observations, median is the value of the \(\left(\frac{n+1}{2}\right)^{\text{th}}\) term.

(xi) The class mark of the class interval \(60 - 70\) is: [1 Mark]
(a) 60
(b) 70
(c) 65
(d) 55

Answer: (c) 65

Class Mark \( = \frac{\text{Lower Limit} + \text{Upper Limit}}{2} = \frac{60 + 70}{2} = 65\).

Teacher's Note:
a) Class mark represents the middle value of a class interval.
b) Use the standard formula sum of limits divided by two.

(xii) The cost of making a closed cardboard box of total surface area \(5.45\text{ m}^{2}\) at the rate of Rs. 20 per \(\text{m}^{2}\) is: [1 Mark]
(a) Rs. 100
(b) Rs. 100.50
(c) Rs. 108
(d) Rs. 109

Answer: (d) Rs. 109

Cost = \(5.45 \times 20 = \text{Rs. } 109\).

Teacher's Note:
a) Total cost is calculated by multiplying total surface area by rate per unit area.
b) Perform decimal multiplication accurately.

(xiii) If \(\sec(90^{\circ} - x) = 2\), then the value of \(x\) is: [1 Mark]
(a) \(60^{\circ}\)
(b) \(30^{\circ}\)
(c) \(45^{\circ}\)
(d) \(90^{\circ}\)

Answer: (b) \(30^{\circ}\)

\(\sec(90^{\circ} - x) = \csc x = 2 \implies \csc x = \csc 30^{\circ} \implies x = 30^{\circ}\).

Teacher's Note:
a) Use complementary angle relation \(\sec(90^{\circ} - \theta) = \csc\theta\).
b) Recall standard trigonometric table values for cosecant.

(xiv) If \((3x + 1, 2y - 7) = (10, -11)\), then the values of \(x\) and \(y\) are: [1 Mark]
(a) \(x = 3, y = 2$
(b) \(x = -3, y = 2$
(c) \(x = 3, y = -2$
(d) \(x = -3, y = -2\)

Answer: (c) \(x = 3, y = -2\)

\(3x + 1 = 10 \implies 3x = 9 \implies x = 3\). Also \(2y - 7 = -11 \implies 2y = -4 \implies y = -2\).

Teacher's Note:
a) Two ordered pairs are equal if and only if their corresponding first and second components are equal.
b) Solve simple linear equations for each variable independently.

(xv) Assertion (A): The distance between points \(A(-9, -6)\) and \(B(-12, -4)\) is \(17\) units.
Reason (R): The distance between points \((x_{1}, y_{1})\) and \((x_{2}, y_{2})\) is given by \(\sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}\). [1 Mark]
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
(c) Assertion (A) is true but Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is true.

Answer: (d) Assertion (A) is false and Reason (R) is true.

\(AB = \sqrt{(-12 - (-9))^{2} + (-4 - (-6))^{2}} = \sqrt{(-3)^{2} + 2^{2}} = \sqrt{9 + 4} = \sqrt{13}\text{ units} \neq 17\). Thus Assertion is false, while Reason states the distance formula correctly.

Teacher's Note:
a) Apply the distance formula correctly keeping track of negative coordinate signs.
b) Verify arithmetic calculations under square roots carefully.

Question 2

(i) Find the sum of money which amounts to Rs. 216 more in 2 years at 12% per annum compound interest than when it is put at simple interest for the same time and at the same rate. [3 Marks]

Answer:
Let the principal sum be \(P\).
1. Compound Interest (C.I.) for \(n = 2\) years at \(R = 12\%\) per annum is given by:
\(\text{C.I.} = P \left(1 + \frac{12}{100}\right)^{2} - P = P \left(\frac{112}{100}\right)^{2} - P = \frac{12544P}{10000} - P = \frac{2544P}{10000}\).
2. Simple Interest (S.I.) for 2 years at 12% is:
\(\text{S.I.} = \frac{P \times 12 \times 2}{100} = \frac{24P}{100}\).
3. Given that \(\text{C.I.} - \text{S.I.} = 216\):
\(\frac{2544P}{10000} - \frac{2400P}{10000} = 216 \implies \frac{144P}{10000} = 216 \implies P = \frac{216 \times 10000}{144} = \text{Rs. } 15000\).

Teacher's Note:
a) Formulate both compound interest and simple interest expressions in terms of principal \(P\).
b) Take care during algebraic simplification of fractions to avoid calculation errors.

(ii) Solve the following simultaneous linear equations:
\(\frac{3}{4}x - \frac{2}{3}y = 1\)
\(\frac{3}{8}x - \frac{1}{6}y = 1\) [3 Marks]

Answer:
1. Simplifying the first equation by taking LCM of denominators (12):
\(9x - 8y = 12 \implies x = \frac{12 + 8y}{9} \quad \text{.....(i)}\).
2. Simplifying the second equation by taking LCM of denominators (24):
\(9x - 4y = 24 \quad \text{.....(ii)}\).
3. Substituting \(x\) from (i) into (ii):
\(9\left(\frac{12 + 8y}{9}\right) - 4y = 24 \implies 12 + 8y - 4y = 24 \implies 4y = 12 \implies y = 3\).
4. Substituting \(y = 3\) into equation (i):
\(x = \frac{12 + 8(3)}{9} = \frac{36}{9} = 4\).
Hence, \(x = 4\) and \(y = 3\).

Teacher's Note:
a) Clearing fractions first simplifies simultaneous equations significantly.
b) Always substitute the obtained value of one variable back into an expression to find the other.

(iii) Given: In \(\triangle ABC\), \(\angle B = 90^{\circ}\) and \(P\) is the mid-point of \(AC\).
To prove: \(BP = \frac{1}{2}AC\) [3 Marks]
[Figure: Right-angled triangle ABC with \(\angle B = 90^{\circ}\), AC as hypotenuse, P as midpoint of AC, and a line segment drawn from P parallel to BC meeting AB at Q.]

Answer:
1. Construction: Draw a straight line through \(P\) parallel to \(BC\) meeting \(AB\) at \(Q\).
2. Proof: Since \(PQ \parallel BC\) and \(P\) is the mid-point of \(AC\), by converse of mid-point theorem, \(Q\) is the mid-point of \(AB\) (\(\implies AQ = QB\)).
3. In \(\triangleAQP\) and \(\triangleBQP\):
\(AQ = QB\) (\(Q\) is mid-point of \(AB\))
\(\angleAQP = \angleBQP = 90^{\circ}\) (\(PQ \parallel BC\) and \(\angle B = 90^{\circ}\))
\(QP = QP\) (Common)
Therefore, \(\triangleAQP \cong \triangleBQP\) (by SAS congruence).
4. Hence, \(AP = BP\) (by c.p.c.t.).
Since \(P\) is the mid-point of \(AC\), \(AP = \frac{1}{2}AC\).
Therefore, \(BP = \frac{1}{2}AC\).

Teacher's Note:
a) Use standard geometric theorems such as the mid-point theorem and triangle congruence criteria.
b) Clearly state construction steps and corresponding reasons for every logical deduction.

Question 3

(i) In \(\triangle ABC\), \(\angle ACB = 90^{\circ}\), \(CD\) is perpendicular to \(AB\), \(AC = a\), \(BC = b\), \(AB = c\) and \(CD = p\). Prove that \(\frac{1}{p^{2}} = \frac{1}{a^{2}} + \frac{1}{b^{2}}\). [3 Marks]

Answer:
1. In right-angled \(\triangle ABC\), by Pythagoras theorem: \(c^{2} = a^{2} + b^{2} \quad \text{.....(i)}\).
2. Area of \(\triangle ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times c \times p = \frac{1}{2} \times a \times b\).
3. Therefore, \(cp = ab \implies c = \frac{ab}{p}\).
4. Substituting \(c = \frac{ab}{p}\) into equation (i):
\(\left(\frac{ab}{p}\right)^{2} = a^{2} + b^{2} \implies \frac{a^{2}b^{2}}{p^{2}} = a^{2} + b^{2}\).
5. Dividing both sides by \(a^{2}b^{2}\):
\(\frac{1}{p^{2}} = \frac{a^{2} + b^{2}}{a^{2}b^{2}} = \frac{1}{b^{2}} + \frac{1}{a^{2}}\).
Hence proved.

Teacher's Note:
a) Relate the sides and altitude of a right-angled triangle using area expressions.
b) Apply algebraic manipulation carefully to arrive at the reciprocal form.

(ii) Given: In parallelogram \(ABCD\), \(AC = BD\).
To prove: Parallelogram \(ABCD\) is a rectangle. [3 Marks]
[Figure: Parallelogram ABCD with diagonals AC and BD intersecting, showing equal diagonals.]

Answer:
1. In \(\triangle ABC\) and \(\triangle BAD\):
\(AB = BA\) (Common)
\(BC = AD\) (Opposite sides of a parallelogram)
\(AC = BD\) (Given)
Therefore, \(\triangle ABC \cong \triangle BAD\) (by SSS congruence).
2. Consequently, \(\angle ABC = \angle BAD\) (by c.p.c.t.).
3. Since \(ABCD\) is a parallelogram, consecutive interior angles are supplementary: \(\angle ABC + \angle BAD = 180^{\circ}\).
4. Therefore, \(2 \angle BAD = 180^{\circ} \implies \angle BAD = 90^{\circ}\).
5. Since one interior angle is \(90^{\circ}\) and opposite sides are equal and parallel, parallelogram \(ABCD\) is a rectangle.

Teacher's Note:
a) To prove a parallelogram is a rectangle, it is sufficient to prove that one of its interior angles is a right angle.
b) Triangle congruence is an effective tool for establishing angle equality.

(iii) Evaluate without using trigonometric tables:
\(\frac{6.67 \times 6.67 \times 6.67 + 5.33 \times 5.33 \times 5.33}{6.67 \times 6.67 - 6.67 \times 5.33 + 5.33 \times 5.33}\) [4 Marks]

OR

Evaluate: \(\frac{(18.5)^{2} - (6.5)^{2}}{18.5 + 6.5}\) [4 Marks]

Answer:
Using algebraic identity \(a^{3} + b^{3} = (a + b)(a^{2} - ab + b^{2})\):
Numerator \( = (6.67)^{3} + (5.33)^{3} = (6.67 + 5.33)[(6.67)^{2} - 6.67 \times 5.33 + (5.33)^{2}]\).
Denominator \( = (6.67)^{2} - 6.67 \times 5.33 + (5.33)^{2}\).
Expression \( = \frac{(6.67 + 5.33)[(6.67)^{2} - 6.67 \times 5.33 + (5.33)^{2}]}{(6.67)^{2} - 6.67 \times 5.33 + (5.33)^{2}} = 6.67 + 5.33 = 12\).

Teacher's Note:
a) Recognize standard algebraic identities to simplify complex numerical expressions.
b) Avoid direct multiplication when numbers fit algebraic formulas.

OR

Evaluate: \(\frac{(18.5)^{2} - (6.5)^{2}}{18.5 + 6.5}\) [4 Marks]

Answer:
Using identity \(a^{2} - b^{2} = (a - b)(a + b)\):
Numerator \( = (18.5 - 6.5)(18.5 + 6.5)\).
Expression \( = \frac{(18.5 - 6.5)(18.5 + 6.5)}{18.5 + 6.5} = 18.5 - 6.5 = 12\).

Teacher's Note:
a) Apply the difference of squares identity \(a^{2} - b^{2} = (a - b)(a + b)\).
b) Cancel common factors in numerator and denominator to simplify calculations.

Section B

(Attempt any four questions from this Section.)

Question 4

(i) Find three rational numbers between \(-\frac{3}{8}\) and \(\frac{1}{4}\). [3 Marks]

Answer:
1. LCM of denominators 8 and 4 is 8.
2. Expressing the numbers with a common denominator:
\(-\frac{3}{8}\) and \(\frac{1}{4} = \frac{1 \times 2}{4 \times 2} = \frac{2}{8}\).
3. Three rational numbers between \(-\frac{3}{8}\) and \(\frac{2}{8}\) can be chosen as:
\(-\frac{2}{8}, -\frac{1}{8}, \text{and } \frac{1}{8}\) (or any other appropriate values such as \(-\frac{1}{4}, -\frac{1}{8}, 0\)).

Teacher's Note:
a) Make denominators equal to easily find rational numbers between two fractions.
b) Multiple valid answers exist for finding rational numbers between given limits.

(ii) During every financial year, the value of a machine depreciates by 10%. Find the original value (cost) of the machine which depreciates by Rs. 2952 during the second year (Without using formula). [3 Marks]

Answer:
1. Let the original cost of the machine be Rs. 100.
2. Depreciation during the 1st year = 10% of 100 = Rs. 10.
Value at the beginning of the 2nd year = \(100 - 10 = \text{Rs. } 90\).
3. Depreciation during the 2nd year = 10% of 90 = Rs. 9.
4. When depreciation during the 2nd year is Rs. 9, original cost is Rs. 100.
When depreciation is Rs. 2952, original cost = \(\frac{100}{9} \times 2952 = \text{Rs. } 32800\).

Teacher's Note:
a) Calculate successive depreciation step by step as requested without direct formulas.
b) Use unitary method to scale up from the second-year depreciation to the original cost.

(iii) In a circle of radius \(5\text{ cm}\), \(AB\) and \(CD\) are two parallel chords of lengths \(8\text{ cm}\) and \(6\text{ cm}\), respectively. Calculate the distance between the chords if they are on the:
A. Same side of the centre
B. Opposite sides of the centre [4 Marks]

Answer:
1. Let radius \(OA = OC = 5\text{ cm}\).
Half of chord \(AB\) (\(AL\)) \( = \frac{8}{2} = 4\text{ cm}\).
Half of chord \(CD\) (\(CM\)) \( = \frac{6}{2} = 3\text{ cm}\).
2. Distance of chord \(AB\) from centre (\(OL\)) \( = \sqrt{5^{2} - 4^{2}} = \sqrt{25 - 16} = 3\text{ cm}\).
3. Distance of chord \(CD\) from centre (\(OM\)) \( = \sqrt{5^{2} - 3^{2}} = \sqrt{25 - 9} = 4\text{ cm}\).
A. If chords are on the same side of the centre:
Distance between chords \( = OM - OL = 4 - 3 = 1\text{ cm}\).
B. If chords are on opposite sides of the centre:
Distance between chords \( = OM + OL = 4 + 3 = 7\text{ cm}\).

Teacher's Note:
a) Perpendicular from the centre bisects the chord in a circle.
b) Consider both relative positions (same side and opposite sides) when chords are parallel.

Question 5

(i) If \(a + b = 6\) and \(ab = 5\), what is the value of \(a^{3} + b^{3}\)? [3 Marks]

Answer:
Using the identity \(a^{3} + b^{3} = (a + b)(a^{2} - ab + b^{2}) = (a + b)[(a + b)^{2} - 3ab]\):
\(a^{3} + b^{3} = 6 \times [6^{2} - 3(5)] = 6 \times [36 - 15] = 6 \times 21 = 126\).

Teacher's Note:
a) Express sum of cubes in terms of sum and product of numbers.
b) Substitute given values systematically.

(ii) Factorise: \(x^{4} - 14x^{2}y^{2} - 51y^{4}\). [3 Marks]

Answer:
1. Splitting the middle term \(-14\) into \(-17\) and \(+3\) (\(\because -17 \times 3 = -51\)):
\(x^{4} - 17x^{2}y^{2} + 3x^{2}y^{2} - 51y^{4}\)
2. Grouping terms:
\(= x^{2}(x^{2} - 17y^{2}) + 3y^{2}(x^{2} - 17y^{2})\)
\(= (x^{2} - 17y^{2})(x^{2} + 3y^{2})\)
3. Further factoring using surds if applicable, or leaving as \((x - \sqrt{17}y)(x + \sqrt{17}y)(x^{2} + 3y^{2})\).

Teacher's Note:
a) Treat expressions quadratic in form (\(x^{2}\)) as standard quadratic polynomials for middle-term splitting.
b) Factorize further using the difference of squares formula where possible.

(iii) The mean of 5 numbers is 20. If one number is excluded, the mean of the remaining numbers becomes 23. Find the excluded observation. [4 Marks]

Answer:
1. Mean of 5 numbers = 20.
Sum of all 5 observations \( = 20 \times 5 = 100\).
2. When one number is excluded, number of observations = 4, and new mean = 23.
Sum of remaining 4 observations \( = 23 \times 4 = 92\).
3. Excluded observation \( = \text{Sum of 5 numbers} - \text{Sum of remaining 4 numbers} = 100 - 92 = 8\).

Teacher's Note:
a) Total sum is obtained by multiplying mean by the number of observations.
b) The difference between total sums gives the value of the excluded observation.

Question 6

(i) Solve using the method of elimination by equating coefficients:
\(23x - 29y = 98\)
\(29x - 23y = 110\) [3 Marks]

Answer:
\(23x - 29y = 98 \quad \text{.....(i)}\)
\(29x - 23y = 110 \quad \text{.....(ii)}\)
1. Adding (i) and (ii):
\(52x - 52y = 208 \implies x - y = 4 \quad \text{.....(iii)}\).
2. Subtracting (ii) from (i):
\(-6x - 6y = -12 \implies x + y = 2 \quad \text{.....(iv)}\).
3. Adding (iii) and (iv):
\(2x = 6 \implies x = 3\).
4. Substituting \(x = 3\) in (iv):
\(3 + y = 2 \implies y = -1\).
Hence, \(x = 3\) and \(y = -1\).

Teacher's Note:
a) For symmetric linear equations with swapped coefficients, adding and subtracting equations simplifies them greatly.
b) Solve the resulting simpler simultaneous pair to find values.

(ii) Prove that: \(\frac{x + y + z}{x^{-1}y^{-1} + y^{-1}z^{-1} + z^{-1}x^{-1}} = xyz\) [3 Marks]

Answer:
L.H.S. \( = \frac{x + y + z}{\frac{1}{xy} + \frac{1}{yz} + \frac{1}{zx}} = \frac{x + y + z}{\frac{z + x + y}{xyz}} = (x + y + z) \times \frac{xyz}{x + y + z} = xyz = \text{R.H.S.}\).
Hence proved.

Teacher's Note:
a) Convert negative exponents into fractions using \(a^{-1} = \frac{1}{a}\).
b) Take the LCM of denominators in the denominator expression to simplify the complex fraction.

(iii) The daily wages of 50 workers in a factory are given below:

Construct a histogram to represent the above frequency distribution. [4 Marks]

Answer:
1. Take suitable scales: on the x-axis, represent daily wages (in Rs.), and on the y-axis, represent number of workers.
2. Construct rectangular bars with class intervals as bases and corresponding frequencies as heights.
3. Since the scale on the x-axis starts at 140 instead of zero, indicate a kink (break) or zigzag curve near the origin.

[Figure: Histogram showing rectangular bars for class intervals 140-180 (height 16), 180-220 (height 9), 220-260 (height 12), 260-300 (height 2), 300-340 (height 7), and 340-380 (height 4) with a kink on the x-axis.]

Teacher's Note:
a) A kink on the axis is mandatory when the scale does not start from zero.
b) Ensure adjacent bars touch each other as it is a continuous frequency distribution histogram.

Question 7

(i) In triangle \(ABC\), \(AD\) is perpendicular to \(BC\). \(\sin B = 0.6\), \(BD = 8\text{ cm}\) and \(\tan C = 1\). Find the length of \(AB\), \(AD\), \(AC\) and \(DC\). [5 Marks]

Answer:
1. In right-angled \(\triangle ABD\):
\(\sin B = \frac{AD}{AB} = 0.6 = \frac{3}{5}\).
Let \(AD = 3x\) and \(AB = 5x\).
By Pythagoras theorem, \(BD^{2} = AB^{2} - AD^{2} = (5x)^{2} - (3x)^{2} = 16x^{2} \implies BD = 4x\).
Given \(BD = 8\text{ cm} \implies 4x = 8 \implies x = 2\text{ cm}\).
Therefore, \(AB = 5 \times 2 = 10\text{ cm}\) and \(AD = 3 \times 2 = 6\text{ cm}\).
2. In right-angled \(\triangle ADC\):
\(\tan C = \frac{AD}{DC} = 1 \implies AD = DC\).
Since \(AD = 6\text{ cm}\), therefore \(DC = 6\text{ cm}\).
3. Hypotenuse \(AC = \sqrt{AD^{2} + DC^{2}} = \sqrt{6^{2} + 6^{2}} = \sqrt{72} = 6\sqrt{2}\text{ cm}\).

Teacher's Note:
a) Apply trigonometric ratios in right-angled triangles separately.
b) Use the Pythagorean theorem to find missing side lengths from proportional ratio parts.

(ii) Find the perimeter and area of a quadrilateral \(ABCD\) in which \(BC = 12\text{ cm}\), \(CD = 9\text{ cm}\), \(BD = 15\text{ cm}\), \(DA = 17\text{ cm}\) and \(\angle ABD = 90^{\circ}\). [5 Marks]
[Figure: Quadrilateral ABCD divided into two triangles ABD and BCD by diagonal BD = 15 cm, with AB, BC = 12 cm, CD = 9 cm, DA = 17 cm, and \(\angle ABD = 90^{\circ}\).]

Answer:
1. In right-angled \(\triangle ABD\), by Pythagoras theorem:
\(AB^{2} = AD^{2} - BD^{2} = 17^{2} - 15^{2} = 289 - 225 = 64 \implies AB = 8\text{ cm}\).
2. Perimeter of quadrilateral \(ABCD = AB + BC + CD + DA = 8 + 12 + 9 + 17 = 46\text{ cm}\).
3. Area of \(\triangle ABD = \frac{1}{2} \times AB \times BD = \frac{1}{2} \times 8 \times 15 = 60\text{ cm}^{2}\).
4. In \(\triangle BCD\), sides are \(a = 12\text{ cm}\), \(b = 9\text{ cm}\), \(c = 15\text{ cm}\).
Semi-perimeter \(s = \frac{12 + 9 + 15}{2} = 18\text{ cm}\).
Area of \(\triangle BCD = \sqrt{18(18 - 12)(18 - 9)(18 - 15)} = \sqrt{18 \times 6 \times 9 \times 3} = \sqrt{2916} = 54\text{ cm}^{2}\).
5. Total area of quadrilateral \(ABCD = 60 + 54 = 114\text{ cm}^{2}\).

Teacher's Note:
a) Split the quadrilateral into two triangles along the diagonal.
b) Use right triangle area formula for one part and Heron's formula for the other.

Question 8

(i) In the figure, equilateral triangle \(EDC\) surmounts square \(ABCD\). If \(\angle DEB = x\), then find the value of \(x\). [3 Marks]
[Figure: Square ABCD with equilateral triangle EDC built on top side DC. Line segment EB connects vertex E to B, forming angle \(\angle DEB = x\).]

Answer:
1. \(\triangle EDC\) is equilateral, so \(\angle DEC = \angle CDE = \angle DCE = 60^{\circ}\).
2. In square \(ABCD\), interior angles are \(90^{\circ}\), so \(\angle BCD = 90^{\circ}\).
Therefore, \(\angle BCE = \angle BCD + \angle DCE = 90^{\circ} + 60^{\circ} = 150^{\circ}\).
3. In isosceles \(\triangle BCE\) (\(BC = CE\)), base angles \(\angle CBE = \angle CEB\):
\(\angle CEB = \frac{180^{\circ} - 150^{\circ}}{2} = \frac{30^{\circ}}{2} = 15^{\circ}\).
4. Also, \(\angle DEC = 60^{\circ}\), so \(\angle DEB = \angle DEC - \angle CEB = 60^{\circ} - 15^{\circ} = 45^{\circ}\).
Hence, \(x = 45^{\circ}\).

Teacher's Note:
a) Combine properties of squares and equilateral triangles to find compound angles.
b) Recognize isosceles triangles formed by equal sides of the square and equilateral triangle.

(ii) The lengths of two parallel chords of a circle are \(6\text{ cm}\) and \(8\text{ cm}\). If the smaller chord is at a distance \(4\text{ cm}\) from the centre, what is the distance of the other chord from the centre? [3 Marks]

Answer:
1. Radius \(r\) of the circle can be found using the smaller chord of length \(6\text{ cm}\) at distance \(4\text{ cm}\):
\(r^{2} = 4^{2} + \left(\frac{6}{2}\right)^{2} = 16 + 9 = 25 \implies r = 5\text{ cm}\).
2. For the larger chord of length \(8\text{ cm}\), half-chord length is \(\frac{8}{2} = 4\text{ cm}\).
3. Distance of the larger chord from the centre (\(d\)) is:
\(d = \sqrt{r^{2} - (\text{half-chord})^{2}} = \sqrt{5^{2} - 4^{2}} = \sqrt{25 - 16} = \sqrt{9} = 3\text{ cm}\).

Teacher's Note:
a) First determine the radius of the circle using information from one chord.
b) Apply the radius to find the distance of the second chord from the centre.

(iii) A wooden bookshelf has external dimensions as follows:
Height = \(110\text{ cm}\), breadth = \(25\text{ cm}\), length = \(85\text{ cm}\).
The thickness of the plank is \(5\text{ cm}\) everywhere. The external faces are to be polished, and the inner faces are to be painted. If the rate of polishing is 20 paise per \(\text{cm}^{2}\) and the rate of painting is 10 paise per \(\text{cm}^{2}\), find the total expenses required for polishing and painting the surface of the bookshelf. [4 Marks]
[Figure: Wooden bookshelf of height 110 cm, breadth 25 cm, length 85 cm with open front and shelves.]

Answer:
1. External surface area to be polished (excluding front open face):
\(\text{External Area} = l h + 2(bh + lb) = 85 \times 110 + 2(25 \times 110 + 85 \times 25) = 9350 + 2(2750 + 2125) = 19100\text{ cm}^{2}\).
Area of the front frame opening \( = (85 \times 110 - 75 \times 100 + 2(75 \times 5)) = 2600\text{ cm}^{2}\).
Total area to be polished \( = 19100 + 2600 = 21700\text{ cm}^{2}\).
Polishing cost \( = 21700 \times \text{Rs. } 0.20 = \text{Rs. } 4340\).
2. Inner faces to be painted (3 rows):
Area of inner faces per row \( = 2(l + h)b + lh = 6450\text{ cm}^{2}\).
Total painting area for 3 rows \( = 3 \times 6450 = 19350\text{ cm}^{2}\).
Painting cost \( = 19350 \times \text{Rs. } 0.10 = \text{Rs. } 1935\).
3. Total expense = Rs. \(4340 + 1935 = \text{Rs. } 6275\).

Teacher's Note:
a) Distinguish between external surfaces to be polished and internal compartments to be painted.
b) Pay careful attention to plank thickness when calculating inner dimensions and areas.

Question 9

(i) The bisectors of \(\angle B\) and \(\angle C\) of an isosceles triangle with \(AB = AC\) intersect each other at a point \(O\). \(BO\) is produced to meet \(AC\) at a point \(M\). Prove that \(\angle MOC = \angle ABC\). [3 Marks]

Answer:
1. In \(\triangle ABC\), since \(AB = AC\), \(\angle ABC = \angle ACB\) (Angles opposite to equal sides).
2. \(BO\) and \(CO\) are angle bisectors, so \(\angle OBC = \frac{1}{2}\angle ABC\) and \(\angle OCB = \frac{1}{2}\angle ACB\), which implies \(\angle OBC = \angle OCB\).
3. In \(\triangle OBC\), exterior angle \(\angle MOC\) is equal to the sum of interior opposite angles:\br />\(\angle MOC = \angle OBC + \angle OCB = 2 \angle OBC\).
Since \(\angle OBC = \frac{1}{2}\angle ABC\), we get \(\angle MOC = 2 \left(\frac{1}{2}\angle ABC\right) = \angle ABC\).
Hence proved.

Teacher's Note:
a) Use the exterior angle property of a triangle to relate exterior and interior angles.
b) Apply angle bisector definitions correctly.

(ii) In the given figure, \(O\) is a point in the interior of square \(ABCD\) such that \(\triangle OAB\) is an equilateral triangle. Show that \(\triangle OCD\) is an isosceles triangle. [3 Marks]
[Figure: Square ABCD with interior point O connected to vertices A, B, C, D, forming equilateral triangle OAB inside.]

Answer:
1. Since \(\triangle OAB\) is equilateral, \(OA = OB = AB\) and \(\angle OAB = \angle OBA = 60^{\circ}\).
2. Since \(ABCD\) is a square, \(AD = AB = BC\) and each interior angle is \(90^{\circ}\).
Therefore, \(\angle DAB = 90^{\circ} \implies \angle DAO = 90^{\circ} - 60^{\circ} = 30^{\circ}\).
Similarly, \(\angle CBO = 30^{\circ}\).
3. In \(\triangle OAD\) and \(\triangle OBC\):
\(AD = BC\) (Sides of square)
\(\angle DAO = \angle CBO = 30^{\circ}\)
\(OA = OB\) (Sides of equilateral \(\triangle OAB\))
Therefore, \(\triangle OAD \cong \triangle OBC$ (by SAS congruence).
4. Hence, \(OD = OC\) (by c.p.c.t.).
Thus, \(\triangle OCD\) is an isosceles triangle.

Teacher's Note:
a) Use properties of squares (equal sides and \(90^{\circ}\) angles) along with equilateral triangle properties.
b) Prove triangle congruence to establish equal side lengths for \(\triangle OCD\).

(iii) A rectangular lawn, \(75\text{ m}\) by \(60\text{ m}\), has two roads, each road \(4\text{ m}\) wide, running through the middle of the lawn, one parallel to the length and the other parallel to the breadth, as shown in the figure. Find the cost of gravelling the roads at Rs. 50 per \(\text{m}^{2}\). [4 Marks]
[Figure: Rectangular lawn PQRS measuring 75 m by 60 m with two intersecting cross roads of width 4 m running parallel to length and breadth, with central intersection EFGH.]

Answer:
1. Area of the road running parallel to length = length \(\times\) width \(= 75 \times 4 = 300\text{ m}^{2}\).
2. Area of the road running parallel to breadth = length \(\times\) width \(= 60 \times 4 = 240\text{ m}^{2}\).
3. Area of the central square intersection common to both roads = \(4 \times 4 = 16\text{ m}^{2}\).
4. Total area of roads for gravelling = \(300 + 240 - 16 = 524\text{ m}^{2}\).
5. Cost of gravelling \(524\text{ m}^{2}\) road \( = 524 \times \text{Rs. } 50 = \text{Rs. } 26,200\).

Teacher's Note:
a) Remember to subtract the area of the central intersection once to avoid double counting.
b) Multiply the net road area by the rate per square meter to find total cost.

Question 10

(i) Find the area of the four walls and the ceiling of a room whose length is \(10\text{ m}\), breadth is \(8\text{ m}\) and height is \(5\text{ m}\). Also find the cost of whitewashing the walls and ceiling at the rate of Rs. 15 per \(\text{m}^{2}\). [3 Marks]

Answer:
1. Area of four walls and ceiling = Lateral surface area of cuboid + Area of the top (ceiling):
\(\text{Area} = 2h(l + b) + (l \times b) = 2 \times 5(10 + 8) + (10 \times 8) = 10(18) + 80 = 180 + 80 = 260\text{ m}^{2}\).
2. Cost of whitewashing \(260\text{ m}^{2}\) area at Rs. 15 per \(\text{m}^{2}\):
\(\text{Cost} = 260 \times 15 = \text{Rs. } 3900\).

Teacher's Note:
a) Include only the four walls and the ceiling as specified, omitting the floor.
b) Verify dimensions and formula substitutions carefully.

(ii) Find the point on the x-axis which is equidistant from the points \(A(2, -5)\) and \(B(-2, 9)\). [3 Marks]

Answer:
1. Let the required point on the x-axis be \(C(x, 0)\).
2. Given that \(AC = BC \implies AC^{2} = BC^{2}\):
\((x - 2)^{2} + (0 - (-5))^{2} = (x - (-2))^{2} + (0 - 9)^{2}\).
\(x^{2} - 4x + 4 + 25 = x^{2} + 4x + 4 + 81\).
3. Simplifying both sides:
\(-4x + 29 = 4x + 85 \implies 8x = 29 - 85 = -56 \implies x = -7\).
4. Therefore, the required point on the x-axis is \((-7, 0)\).

Teacher's Note:
a) Any point on the x-axis has a y-coordinate equal to zero, represented as \((x, 0)\).
b) Squaring both sides of the distance equation eliminates square roots and simplifies solving for \(x\).

(iii) Solve the following simultaneous equations using the graphical method:
\(2x + 3y = 2$
\(x - 2y = 8\) [4 Marks]

Answer:
1. For equation \(2x + 3y = 2 \implies x = \frac{2 - 3y}{2}\):
When \(y = 2, x = -2\)
When \(y = 0, x = 1\)
When \(y = -2, x = 4\)
Points: \((-2, 2), (1, 0), (4, -2)\).
2. For equation \(x - 2y = 8 \implies x = 8 + 2y\):
When \(y = -3, x = 2\)
When \(y = -4, x = 0\)
When \(y = 0, x = 8\)
Points: \((2, -3), (0, -4), (8, 0)\).
3. Plotting these points on graph paper and drawing the straight lines, the lines intersect at \((4, -2)\).
Hence, the solution is \(x = 4, y = -2\).

[Figure: Cartesian plane showing two intersecting straight lines representing equations \(2x + 3y = 2\) and \(x - 2y = 8\) intersecting at point \((4, -2)\).]

Teacher's Note:
a) Plot at least three coordinate points for each linear equation to ensure graph accuracy.
b) The coordinates of the intersection point represent the simultaneous solution of the equations.