ICSE Class 10 Mathematics Solved Paper 2024

(c) 12%

Explanation:

CGST% = \frac{\text{CGST}}{\text{Marked Price}} × 100

⇒ CGST% = \frac{120}{2000} × 100

⇒ CGST% = 6%

For an intra-state sale:

GST% = 2 × CGST%

⇒ GST% = 2 × 6%

⇒ GST% = 12%

Therefore, the rate of GST is 12%.

(d) 31

Explanation:

Let k be the number to be subtracted.

The resulting polynomial is x³ + x² − 2x + 1 − k.

For it to be divisible by (x − 3), put x = 3:

3³ + 3² − 2(3) + 1 − k = 0

⇒ 27 + 9 − 6 + 1 − k = 0

⇒ 31 − k = 0

⇒ k = 31

Therefore, 31 must be subtracted.

(b) q² = 4pr

Explanation:

For a quadratic equation ax² + bx + c = 0, the roots are real and equal when the discriminant is zero.

D = b² − 4ac

Here, a = p, b = −q and c = r.

Therefore,

D = (−q)² − 4pr

⇒ D = q² − 4pr

For equal roots:

q² − 4pr = 0

⇒ q² = 4pr

(c) 8

Explanation:

A² = A × A

A² = \begin{bmatrix}2 & 2 \\ 0 & 2\end{bmatrix}\begin{bmatrix}2 & 2 \\ 0 & 2\end{bmatrix}

= \begin{bmatrix}(2\times2)+(2\times0) & (2\times2)+(2\times2) \\ (0\times2)+(2\times0) & (0\times2)+(2\times2)\end{bmatrix}

= \begin{bmatrix}4 & 8 \\ 0 & 4\end{bmatrix}

Now, \begin{bmatrix}4 & 8 \\ 0 & 4\end{bmatrix} = \begin{bmatrix}4 & x \\ 0 & 4\end{bmatrix}

x = 8

(c) 62

Explanation:

Number of observations, n = 10.

Since n is even, the median is the mean of the 5th and 6th observations.

5th observation = x

6th observation = x + 4

Median = \frac{x+(x+4)}{2}

Given that the median is 64:

\frac{2x+4}{2} = 64

⇒ x + 2 = 64

⇒ x = 64 − 2

⇒ x = 62

(d) 7 − 3x

Explanation:

Since A, B and C are collinear:

Slope of AB = Slope of BC

\frac{-2-y}{3-x}=\frac{-5-(-2)}{4-3}

⇒ \frac{-2-y}{3-x}=\frac{-3}{1}

⇒ −2 − y = −3(3 − x)

⇒ −2 − y = −9 + 3x

⇒ −y = −7 + 3x

⇒ y = 7 − 3x

(d) x = 3, y = 150

Explanation:

The train moves with uniform speed.

Speed = \frac{\text{Distance}}{\text{Time}}

Using the first pair of values:

Speed = \frac{60}{2} = 30 m/s

For a distance of 90 m:

\frac{90}{x} = 30

⇒ 90 = 30x

⇒ x = \frac{90}{30} = 3 s

For a time of 5 s:

\frac{y}{5} = 30

⇒ y = 30 × 5

⇒ y = 150 m

Therefore, x = 3 and y = 150.

(a) \frac{1}{a}+6

Explanation:

First term = \frac{1}{a}

Common difference = 1

The nth term of an A.P. is:

Tₙ = first term + (n − 1) × common difference

For n = 7:

T₇ = \frac{1}{a} + (7 − 1) × 1

⇒ T₇ = \frac{1}{a} + 6

(d) ₹900

Explanation:

Nominal value of one share = ₹75

Discount = 20%

Discount on one share = \frac{20}{100} × 75

⇒ Discount on one share = ₹15

Market value of one share = Nominal value − Discount

⇒ Market value of one share = ₹75 − ₹15

⇒ Market value of one share = ₹60

Number of shares purchased = 15

Total investment = 15 × ₹60

⇒ Total investment = ₹900

(b) At an equal distance from the three vertices of the triangle.

Explanation:

The circumcentre is the point of intersection of the perpendicular bisectors of the sides of a triangle.

Every point on the perpendicular bisector of a side is equidistant from the endpoints of that side.

Therefore, the circumcentre is at an equal distance from all three vertices of the triangle.

(a) Only (i)

Explanation:

The fundamental trigonometric identity is:

sin² θ + cos² θ = 1

Therefore, Statement (i) is true.

Another identity is:

cosec² θ − cot² θ = 1

Statement (ii) uses a plus sign, so it is false.

Hence, only Statement (i) is valid.

(d) 50°

Explanation:

PS = PT because tangents drawn from the same external point are equal.

Therefore, △PST is isosceles.

Let ∠PST = ∠PTS = a.

Using the angle-sum property of △PST:

a + a + 80° = 180°

⇒ 2a = 100°

⇒ a = 50°

Since SQ ∥ PT, ∠QST = ∠STP by alternate interior angles.

Therefore, ∠QST = 50°.

(b) A is false, R is true.

Explanation:

The complete sample space when a die is thrown is {1, 2, 3, 4, 5, 6}.

The favourable outcomes for an even number are {2, 4, 6}.

Number of favourable outcomes = 3

Total number of outcomes = 6

Probability of getting an even number = \frac{3}{6} = \frac{1}{2}

Therefore, the assertion is false, while the reason is true.

(a) 1 : 1

Explanation:

For the first cylinder:

Height, h = 7 cm

Circumference = 11 cm

⇒ 2πr = 11

⇒ r = \frac{11}{2\pi} cm

For the second cylinder:

Height, H = 11 cm

Circumference = 7 cm

⇒ 2πR = 7

⇒ R = \frac{7}{2\pi} cm

Required ratio = 2πrh : 2πRH

⇒ rh : RH

⇒ \frac{11}{2\pi}\times7:\frac{7}{2\pi}\times11

⇒ 77 : 77

⇒ 1 : 1

(b) △ABD ∼ △PQS

Explanation:

Given, △ABC ∼ △PQR.

Therefore, ∠A = ∠P and ∠B = ∠Q.

Since AD and PS bisect ∠A and ∠P respectively:

∠BAD = \frac{1}{2}∠A

and ∠QPS = \frac{1}{2}∠P

Since ∠A = ∠P:

∠BAD = ∠QPS

Also, ∠ABD = ∠PQS because ∠B = ∠Q.

Hence, △ABD ∼ △PQS by the AA criterion.

Given, AB = C.

\begin{bmatrix}x & 0 \\ 1 & 1\end{bmatrix}\begin{bmatrix}4 & 0 \\ y & 1\end{bmatrix}=\begin{bmatrix}4 & 0 \\ x & 1\end{bmatrix}

Multiplying the matrices:

\begin{bmatrix}(x\times4)+(0\times y) & (x\times0)+(0\times1) \\ (1\times4)+(1\times y) & (1\times0)+(1\times1)\end{bmatrix}=\begin{bmatrix}4 & 0 \\ x & 1\end{bmatrix}

\begin{bmatrix}4x & 0 \\ 4+y & 1\end{bmatrix}=\begin{bmatrix}4 & 0 \\ x & 1\end{bmatrix}

Equating corresponding elements:

4x = 4

⇒ x = 1

Also, 4 + y = x

⇒ 4 + y = 1

⇒ y = 1 − 4

⇒ y = −3

Therefore, x = 1 and y = −3.

Diameter (d) = 7 cm

Radius (r) = d/2 = 7/2 = 3.5 cm

Height (h) = 10 cm

(a) When the cylinder is cut along its height, two new rectangular surfaces are formed.

Total surface area of both halves = Total surface area of the cylinder + Area of two rectangles

⇒ Total surface area = 2πr(h + r) + 2(h × d)

⇒ Total surface area = 2 × \frac{22}{7} × 3.5 × (10 + 3.5) + 2 × 10 × 7

⇒ Total surface area = 22 × 13.5 + 140

⇒ Total surface area = 297 + 140

⇒ Total surface area = 437 cm²

Therefore, the total surface area of both halves is 437 cm².

(b) Cost of painting = Area × Rate

⇒ Cost = 437 × ₹30

⇒ Cost = ₹13,110

Therefore, the total cost of painting is ₹13,110.

Given G.P.: 15, 30, 60, 120, …

First term, a = 15

Common ratio, r = 30/15 = 2

(a) The nth term of a G.P. is:

Tₙ = arⁿ⁻¹

⇒ Tₙ = 15 × 2ⁿ⁻¹

Therefore, the nth term is 15 × 2ⁿ⁻¹.

(b) Let the required number of terms be n.

Sum of n terms of a G.P. = \frac{a(r^n-1)}{r-1}

Given that the sum is 945:

945 = \frac{15(2^n-1)}{2-1}

⇒ 945 = 15(2ⁿ − 1)

⇒ 2ⁿ − 1 = 945/15

⇒ 2ⁿ − 1 = 63

⇒ 2ⁿ = 64

⇒ 2ⁿ = 2⁶

⇒ n = 6

Therefore, 6 terms give the sum 945.

Solution:

Using a³ + b³ = (a + b)(a² − ab + b²):

sin³ θ + cos³ θ

= (sin θ + cos θ)(sin² θ − sin θ cos θ + cos² θ)

Since sin² θ + cos² θ = 1:

sin³ θ + cos³ θ = (sin θ + cos θ)(1 − sin θ cos θ) … (1)

Now consider the left-hand side of the identity:

\frac{\sin^3\theta+\cos^3\theta}{\sin\theta+\cos\theta}+\sin\theta\cos\theta

Substituting from (1):

\frac{(\sin\theta+\cos\theta)(1-\sin\theta\cos\theta)}{\sin\theta+\cos\theta}+\sin\theta\cos\theta

= 1 − sin θ cos θ + sin θ cos θ

= 1

Therefore, LHS = RHS. Hence proved.

(a) The radius is perpendicular to the tangent at the point of contact.

Therefore, ∠OQP = 90°.

∠OQM = ∠OQP − ∠PQM

⇒ ∠OQM = 90° − 42°

⇒ ∠OQM = 48°

(b) By the alternate segment theorem:

∠QNM = ∠PQM = 42°

and ∠SNM = ∠PSM = 25°

Therefore, ∠QNS = ∠QNM + ∠SNM

⇒ ∠QNS = 42° + 25°

⇒ ∠QNS = 67°

(c) The angle subtended by an arc at the centre is twice the angle subtended by it at the circumference.

∠QOS = 2∠QNS

⇒ ∠QOS = 2 × 67°

⇒ ∠QOS = 134°

(d) QMSN is a cyclic quadrilateral, so its opposite angles are supplementary.

∠QMS + ∠QNS = 180°

⇒ ∠QMS + 67° = 180°

⇒ ∠QMS = 113°

(a) Plot A(0, 3), B(2, 1) and C(4, −1) on the graph.

(b) Reflection in the y-axis changes (x, y) to (−x, y).

Therefore:

B(2, 1) → B′(−2, 1)

C(4, −1) → C′(−4, −1)

(c) The line BB′ passes through B(2, 1) and B′(−2, 1), so its equation is y = 1.

Reflecting A(0, 3) in y = 1 places its image the same perpendicular distance below the line.

Distance of A from y = 1 is 3 − 1 = 2 units.

Therefore, A′ is 2 units below y = 1.

A′ = (0, −1)

(d) The required coordinates are:

A′(0, −1), B′(−2, 1) and C′(−4, −1).

(e) On joining A, B, A′ and B′, all four sides are equal and adjacent sides are perpendicular.

Therefore, ABA′B′ is a square.

Monthly deposit (P) = ₹2000

Rate of interest (r) = 8% per annum

Interest (I) = ₹1040

Let the account be held for n months.

For a recurring deposit:

I = P × \frac{n(n+1)}{2}\times\frac{r}{12\times100}

Substituting the values:

1040 = 2000\times\frac{n(n+1)}{2}\times\frac{8}{1200}

⇒ 1040 = \frac{2000n(n+1)}{300}

⇒ n(n + 1) = \frac{1040\times300}{2000}

⇒ n(n + 1) = 156

⇒ n² + n − 156 = 0

⇒ n² + 13n − 12n − 156 = 0

⇒ n(n + 13) − 12(n + 13) = 0

⇒ (n − 12)(n + 13) = 0

⇒ n = 12 or n = −13

The number of months cannot be negative, so n = 12 months.

12 months = 1 year.

Therefore, the account was held for 1 year.

Class width, i = 10. Take the assumed mean A = 135.

Σf = 50 and Σfu = −38

Mean = A + \frac{\Sigma fu}{\Sigma f}\times i

⇒ Mean = 135 + \frac{-38}{50}\times10

⇒ Mean = 135 − 7.6

⇒ Mean = 127.4 minutes

Therefore, the mean duration is 127.4 minutes.

(a) Given:

\frac{(a+b)^3}{(a-b)^3}=\frac{64}{27}

\left(\frac{a+b}{a-b}\right)^3=\frac{4^3}{3^3}

Taking the cube root of both sides:

\frac{a+b}{a-b}=\frac{4}{3}

(b) Using the above proportion:

3(a + b) = 4(a − b)

⇒ 3a + 3b = 4a − 4b

⇒ 4a − 3a = 3b + 4b

⇒ a = 7b

⇒ a/b = 7/1

Therefore, a : b = 7 : 1.

(a) Reading the heights of the bars gives the following frequency table:

(b) The greatest frequency is 12, corresponding to the class 60–70 cm.

Therefore, the modal class is 60–70 cm.

(c) From the graph, the mode of the distribution is 64 cm.

(d) The frequency corresponding to 80–90 cm is 3.

Therefore, 3 plants have heights between 80 cm and 90 cm.

Let C be the foot and D the top of the tree.

CD = 100 m

In right-angled △ACD:

tan 52° = CD/AC

⇒ 1.28 = 100/AC

⇒ AC = 100/1.28

⇒ AC = 78.125 m

In right-angled △BCD:

tan 45° = CD/BC

⇒ 1 = 100/BC

⇒ BC = 100 m

Since A and B lie on opposite sides of the tree:

AB = AC + BC

⇒ AB = 78.125 + 100

⇒ AB = 178.125 m

To the nearest metre, AB = 178 m.

Comparing 2x² − 10x + 5 = 0 with ax² + bx + c = 0:

a = 2, b = −10 and c = 5

Using the quadratic formula:

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

Substituting the values:

x=\frac{-(-10)\pm\sqrt{(-10)^2-4(2)(5)}}{2(2)}

⇒ x=\frac{10\pm\sqrt{100-40}}{4}

⇒ x=\frac{10\pm\sqrt{60}}{4}

⇒ x=\frac{10\pm2\sqrt{15}}{4}

⇒ x=\frac{5\pm\sqrt{15}}{2}

Since √15 ≈ 3.87298:

x = (5 + 3.87298)/2 or x = (5 − 3.87298)/2

⇒ x ≈ 4.43649 or x ≈ 0.563508

Correct to three significant figures:

x = 4.44 or x = 0.564

Given, T_n = 6(7 − n).

(a) Put n = 1:

T₁ = 6(7 − 1)

⇒ T₁ = 6 × 6

⇒ T₁ = 36

Therefore, the first term a = 36.

Put n = 2:

T₂ = 6(7 − 2)

⇒ T₂ = 6 × 5

⇒ T₂ = 30

Common difference, d = T₂ − T₁

⇒ d = 30 − 36

⇒ d = −6

(b) The sum of n terms of an A.P. is:

Sₙ = \frac{n}{2}[2a+(n-1)d]

For n = 25:

S₂₅ = \frac{25}{2}[2(36)+(25-1)(-6)]

⇒ S₂₅ = \frac{25}{2}[72+24(-6)]

⇒ S₂₅ = \frac{25}{2}[72-144]

⇒ S₂₅ = \frac{25}{2}(-72)

⇒ S₂₅ = 25 × (−36)

⇒ S₂₅ = −900

Therefore, the first term is 36, the common difference is −6, and the sum of the first 25 terms is −900.

(a) In △APD and △BPC:

∠APD = ∠BPC because they are vertically opposite angles.

∠ADP = ∠BCP = 90°.

Therefore, △APD ∼ △BPC by the AA criterion.

(b) In right-angled △ADB, by the Pythagoras theorem:

AB² = AD² + BD²

⇒ 10² = 6² + BD²

⇒ 100 = 36 + BD²

⇒ BD² = 64

⇒ BD = √64

⇒ BD = 8 cm

Since BD = BP + PD:

PB = BD − PD

⇒ PB = 8 − 4.5

⇒ PB = 3.5 cm

(c) In right-angled △APD:

AP² = AD² + DP²

⇒ AP² = 6² + 4.5²

⇒ AP² = 36 + 20.25

⇒ AP² = 56.25

⇒ AP = √56.25

⇒ AP = 7.5 cm

(d) The ratio of the areas of similar triangles equals the square of the ratio of corresponding sides.

\frac{\text{Area of }\triangle APD}{\text{Area of }\triangle BPC}=\frac{AD^2}{BC^2}

⇒ Area of △APD : Area of △BPC = 6² : 2.4²

⇒ 36 : 5.76

Multiplying both terms by 100:

⇒ 3600 : 576

Dividing by 144:

⇒ 25 : 4

Therefore, area of △APD : area of △BPC = 25 : 4.

(a) Since OM is perpendicular to chord AC, it bisects both the chord and the angle subtended by the chord at the centre.

Therefore, ∠AOM = ∠COM = 65°.

∠AOC = ∠AOM + ∠COM

⇒ ∠AOC = 65° + 65°

⇒ ∠AOC = 130°

The angle subtended by an arc at the centre is twice the angle subtended by the same arc at the circumference.

∠AOC = 2∠ABC

⇒ ∠ABC = 130°/2

⇒ ∠ABC = 65°

(b) Since △ABC is isosceles, AB = AC.

Therefore, ∠ACB = ∠ABC = 65°.

Using the angle-sum property of △ABC:

∠BAC + 65° + 65° = 180°

⇒ ∠BAC = 180° − 130°

⇒ ∠BAC = 50°

(c) By the alternate segment theorem, the angle between tangent CQ and chord CB equals the angle in the alternate segment.

Therefore, ∠BCQ = ∠BAC = 50°.

Given:

−3 + x ≤ \frac{7x}{2} + 2 < 8 + 2x

Solving the left-hand part:

−3 + x ≤ \frac{7x}{2} + 2

⇒ \frac{7x}{2} − x ≥ −3 − 2

⇒ \frac{7x-2x}{2} ≥ −5

⇒ \frac{5x}{2} ≥ −5

⇒ 5x ≥ −10

⇒ x ≥ −2 … (1)

Solving the right-hand part:

\frac{7x}{2} + 2 < 8 + 2x

⇒ \frac{7x}{2} − 2x < 8 − 2

⇒ \frac{7x-4x}{2} < 6

⇒ \frac{3x}{2} < 6

⇒ 3x < 12

⇒ x < 4 … (2)

Combining (1) and (2):

−2 ≤ x < 4

Since x ∈ I, the solution set is:

{−2, −1, 0, 1, 2, 3}

On the number line, mark filled points at −2, −1, 0, 1, 2 and 3.

(a) From the graph:

A = (0, 6) and D = (−3, 0).

(b) The centroid of a triangle with vertices (x₁, y₁), (x₂, y₂) and (x₃, y₃) is:

\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right)

For A(0, 6), B(4, −4) and C(−4, −2):

Centroid = \left(\frac{0+4-4}{3},\frac{6-4-2}{3}\right)

⇒ Centroid = (0, 0)

(c) D divides AC in the ratio k : 1.

Using the section formula:

D=\left(\frac{k(-4)+1(0)}{k+1},\frac{k(-2)+1(6)}{k+1}\right)

Since D = (−3, 0):

-3=\frac{-4k}{k+1}

⇒ −3(k + 1) = −4k

⇒ −3k − 3 = −4k

⇒ k = 3

Checking the y-coordinate:

0=\frac{-2k+6}{k+1}

⇒ −2k + 6 = 0

⇒ k = 3

(d) The line BD passes through B(4, −4) and D(−3, 0).

Slope = \frac{0-(-4)}{-3-4}=\frac{4}{-7}=-\frac{4}{7}

Using the point-slope form with B(4, −4):

y − (−4) = -\frac{4}{7}(x − 4)

⇒ 7(y + 4) = −4(x − 4)

⇒ 7y + 28 = −4x + 16

⇒ 4x + 7y + 12 = 0

Therefore, the equation of BD is 4x + 7y + 12 = 0.

Since (x − 1) is a factor, x = 1 makes the polynomial zero.

3(1)³ + 8(1)² − 15(1) + k = 0

⇒ 3 + 8 − 15 + k = 0

⇒ −4 + k = 0

⇒ k = 4

The resulting polynomial is:

3x³ + 8x² − 15x + 4

On dividing (3x³ + 8x² – 15x + 4) by (x – 1), we get :

3x³ + 8x² − 15x + 4 = (x − 1)(3x² + 11x − 4)

Splitting the middle term:

3x² + 11x − 4 = 3x² + 12x − x − 4

⇒ 3x(x + 4) − 1(x + 4)

⇒ (3x − 1)(x + 4)

Hence:

3x³ + 8x² − 15x + 4 = (x − 1)(3x − 1)(x + 4)

The cards contain {A, D, M, N, O, S, U, Y}.

Total number of cards = 8.

(a) The letters of MONDAY present on the cards are {M, O, N, D, A, Y}.

Number of favourable cards = 6.

Required probability = \frac{6}{8}

⇒ Required probability = \frac{3}{4}

(b) The cards whose letters do not appear in MONDAY are {S, U}.

Number of favourable cards = 2.

Required probability = \frac{2}{8}

⇒ Required probability = \frac{1}{4}

(c) The letters that appear in both SUNDAY and MONDAY are {N, D, A, Y}.

Number of favourable cards = 4.

Required probability = \frac{4}{8}

⇒ Required probability = \frac{1}{2}

Radius of the spherical vessel, r = 28 cm.

Volume of the spherical vessel = \frac{4}{3}\pi r^3

The oil occupies \frac{3}{4} of the vessel.

Volume of oil = \frac{3}{4}\times\frac{4}{3}\pi r^3

⇒ Volume of oil = πr³

⇒ Volume of oil = π × 28³ cm³

Radius of the cylindrical vessel, R = 21 cm.

Let the height of oil in the cylinder be h cm.

Volume of oil in the cylinder = πR²h

Equating the two volumes:

π × 28³ = π × 21² × h

⇒ 28³ = 21²h

⇒ h = \frac{28^3}{21^2}

⇒ h = \frac{21952}{441}

⇒ h ≈ 49.78 cm

Correct to the nearest centimetre, h = 50 cm.

Radius, OA = OB = 9 cm.

Therefore, diameter AB = OA + OB.

⇒ AB = 9 + 9

⇒ AB = 18 cm

Since PA = 24 cm:

PB = PA − AB

⇒ PB = 24 − 18

⇒ PB = 6 cm

By the tangent-secant theorem:

PQ² = PB × PA

⇒ PQ² = 6 × 24

⇒ PQ² = 144

⇒ PQ = √144

⇒ PQ = 12 cm

Therefore, the length of tangent PQ is 12 cm.

Money invested = ₹33,000

Nominal value of each share = ₹100

Premium = 10%

Premium on each share = \frac{10}{100} × ₹100

⇒ Premium on each share = ₹10

Market value of each share = Nominal value + Premium

⇒ Market value = ₹100 + ₹10

⇒ Market value = ₹110

(a) Number of shares = Money invested/Market value per share

⇒ Number of shares = \frac{33000}{110}

⇒ Number of shares = 300

(b) Dividend on one share = \frac{12}{100} × ₹100

⇒ Dividend on one share = ₹12

Annual dividend = 300 × ₹12

⇒ Annual dividend = ₹3,600

Therefore, 300 shares were purchased and the annual dividend is ₹3,600.

Use the upper class boundaries and cumulative frequencies to plot the less-than ogive.

Take 2 cm = 5 years on the x-axis and 2 cm = 10 policy holders on the y-axis.

Plot the points:

(20, 0), (25, 2), (30, 6), (35, 18), (40, 38), (45, 66), (50, 88), (55, 96) and (60, 100).

Join the plotted points with a smooth free-hand curve.

(a) Total number of policy holders, n = 100.

Median corresponds to the n/2th observation.

⇒ Median position = 100/2

⇒ Median position = 50th observation

From 50 on the cumulative-frequency axis, draw a horizontal line to meet the ogive. From that point, draw a vertical line to the age axis.

The graph gives the median age as approximately 42 years.

(b) From 52 years on the age axis, draw a vertical line to meet the ogive. From that point, draw a horizontal line to the cumulative-frequency axis.

The graph gives approximately 91 policy holders aged 52 years or below.

Number aged above 52 years = 100 − 91

⇒ Number aged above 52 years = 9

For laddus:

Cost before GST = ₹500 × 2

⇒ Cost before GST = ₹1,000

GST = \frac{5}{100} × ₹1,000

⇒ GST on laddus = ₹50

Amount including GST = ₹1,000 + ₹50

⇒ Amount including GST = ₹1,050

For pastries:

Cost before GST = ₹100 × 12

⇒ Cost before GST = ₹1,200

GST = \frac{18}{100} × ₹1,200

⇒ GST on pastries = ₹216

Amount including GST = ₹1,200 + ₹216

⇒ Amount including GST = ₹1,416

(a) Total GST paid = ₹50 + ₹216

⇒ Total GST paid = ₹266

(b) Total bill amount = ₹1,050 + ₹1,416

⇒ Total bill amount = ₹2,466

(a) For the first line:

kx − y + 4 = 0

⇒ y = kx + 4

Therefore, its slope m₁ = k.

For the second line:

2y = 6x + 7

⇒ y = 3x + 7/2

Therefore, its slope m₂ = 3.

For perpendicular lines:

m₁m₂ = −1

⇒ k × 3 = −1

⇒ k = −1/3

(b) A line parallel to 2y = 6x + 7 has slope 3.

Using the point-slope form through (−1, 1):

y − 1 = 3[x − (−1)]

⇒ y − 1 = 3(x + 1)

⇒ y − 1 = 3x + 3

⇒ y = 3x + 4

Therefore, k = −1/3 and the required parallel line is y = 3x + 4.

1. Draw the line segment BC = 8 cm.

2. At B, construct ∠ABC = 90°.

3. On the perpendicular ray, mark A such that AB = 6 cm.

(a) The locus of points equidistant from B and C is the perpendicular bisector of BC.

Construct the perpendicular bisector of BC and name it XY.

(b) The locus of points equidistant from A and B is the perpendicular bisector of AB.

Construct the perpendicular bisector of AB and name it PQ.

(c) Mark the intersection of XY and PQ as O.

A point moving at a fixed distance OA from O traces a circle.

With O as centre and OA as radius, draw a circle. This circle is the required locus.

(d) The locus of points equidistant from the intersecting lines BA and BC is their angle bisector.

Construct the angle bisector of ∠ABC and name it BZ.

Therefore:

• XY is the locus equidistant from B and C.

• PQ is the locus equidistant from A and B.

• The circle with centre O and radius OA is the fixed-distance locus.

• BZ is the locus equidistant from BA and BC.