Long Answer Questions 2

(a)

For pen :

No. of pens purchased = 12

Price of 12 pens = ₹5 × 12 = ₹60

Discount (D) : 5% on dozen pens.

$D = \dfrac{5}{100} \times 60 = 3.$

Price after discount = ₹ 60 - ₹ 3 = ₹ 57.

GST = 18% = $\dfrac{18}{100} \times 57$ = ₹ 10.26

Total price : ₹ 57 + ₹ 10.26 = ₹ 67.26

For pencil :

No. of pencils purchased = 20

Price of 20 pencils = ₹ 7 × 20 = ₹ 140

Discount (D) : 10% on 20 pencils.

$D = \dfrac{10}{100} \times 140 = ₹ 14.$

Price after discount = ₹ 140 - ₹ 14 = ₹ 126.

GST = 12% = $\dfrac{12}{100} \times 126$ = ₹ 15.12

Total price : ₹ 126 + ₹ 15.12 = ₹ 141.12

For notebook :

No. of notebooks purchased = 5

Price of 5 notebooks = ₹ 200 × 5 = ₹ 1000

Premium (P) : ₹ 50 on each notebook

Total premium : ₹ 50 × 5 = ₹ 250.

Price after premium = ₹ 1000 + ₹ 250 = ₹ 1250.

GST = 12% = $\dfrac{12}{100} \times 1250$ = ₹ 150.

Total price : ₹ 1250 + ₹ 150 = ₹ 1400.

(i) Total selling price of all items = ₹ 57 + ₹ 126 + ₹ 1250 = ₹ 1433.

Hence, total selling price = ₹ 1433.

(ii) Total price paid (including GST) = ₹ 67.26 + ₹ 141.12 + ₹ 1400 = ₹ 1608.38

Hence, total price paid (including GST) = ₹ 1608.38

(iii) On applying incorrect GST rates :

On pen :

Price after discount = ₹ 57.

GST = 18% = $\dfrac{18}{100} \times 57$ = ₹ 10.26

Total price : ₹ 57 + ₹ 10.26 = ₹ 67.26

On pencil :

Price after discount = ₹ 126.

GST = 18% = $\dfrac{18}{100} \times 126$ = ₹ 22.68

Total price : ₹ 126 + ₹ 22.68 = ₹ 148.68

For notebook :

Price after premium = ₹ 1250.

GST = 18% = $\dfrac{18}{100} \times 1250$ = ₹ 225.

Total price : ₹ 1250 + ₹ 225 = ₹ 1475.

When the GST rate is applied as 18% uniformly, bill need to be paid by Chetan :

= ₹ 1475 + ₹ 148.68 + ₹ 67.26

= ₹ 1690.94

Discount given on total bill = 2%

$\dfrac{2}{100} \times 1690.94$ = 33.82

Bill = ₹ 1690.94 - ₹ 33.82 = ₹ 1657.12

Hence, bill = ₹ 1657.12

(b) The bill should be of ₹ 1608.38 but instead it was of ₹ 1657.12

⇒ ₹ 1657.12 - ₹ 1608.38 = ₹ 48.78

Hence, the shopkeeper overcharged an amount of ₹ 48.78

2x + 3 = 0

⇒ 2x = -3

⇒ x = $-\dfrac{3}{2}$

Substituting value of x in equation 2x² + 11x + 12, we get :

$\Rightarrow 2 \times \Big(-\dfrac{3}{2}\Big)^2 + 11 \times \Big(-\dfrac{3}{2}\Big) + 12 \\[1em] \Rightarrow 2 \times \dfrac{9}{4} - \dfrac{33}{2} + 12 \\[1em] \Rightarrow \dfrac{9}{2} - \dfrac{33}{2} + 12 \\[1em] \Rightarrow \dfrac{9 - 33 + 24}{2} \\[1em] \Rightarrow \dfrac{0}{2} \\[1em] \Rightarrow 0.$

Since, remainder = 0.

∴ 2x + 3 is a factor of the polynomial 2x² + 11x + 12.

Solving polynomial, 2x² + 11x + 12, we get :

⇒ 2x² + 8x + 3x + 12

⇒ 2x(x + 4) + 3(x + 4)

⇒ (2x + 3)(x + 4).

Solving polynomial, x² + 3x - 4, we get :

⇒ x² + 4x - x - 4

⇒ x(x + 4) - 1(x + 4)

⇒ (x - 1)(x + 4).

∴ (x - 1) and (x + 4) are factors of x² + 3x - 4.

∴ On multiplying polynomial, 2x² + 11x + 12 by (x - 1) it will be divisible by x² + 3x - 4.

⇒ (2x² + 11x + 12)(x - 1)

⇒ 2x³ - 2x² + 11x² - 11x + 12x - 12

⇒ 2x³ + 9x² + x - 12.

Hence, the resulting polynomial = 2x³ + 9x² + x - 12.

(a) Difference between common terms : 16 - 9 = 9 - 2 = 7.

Since, difference between common terms are equal.

Hence, the given sequence is an A.P.

(b) By formula,

a_n = a + (n - 1)d

a₂₀ = 2 + (20 - 1) × 7

= 2 + 19 × 7

= 2 + 133

= 135.

Hence, 20th term of the sequence = 135.

(c) By formula,

S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

Substituting values we get :

Sum upto 22 terms :

$S_{22} = \dfrac{22}{2}[2 \times 2 + (22 - 1) \times 7] \\[1em] = 11 \times [4 + 21 \times 7] \\[1em] = 11 \times [4 + 147] \\[1em] = 11 \times 151 \\[1em] = 1661.$

Sum upto 25 terms :

$S_{25} = \dfrac{25}{2}[2 \times 2 + (25 - 1) \times 7] \\[1em] = \dfrac{25}{2} \times [4 + 24 \times 7] \\[1em] = \dfrac{25}{2} \times [4 + 168] \\[1em] = \dfrac{25}{2} \times 172 \\[1em] = 25 \times 86 \\[1em] = 2150.$

S₂₅ - S₂₂ = 2150 - 1661 = 489.

Hence, difference between the sum of its first 22 and 25 terms = 489.

(d) Let n^th term be 102.

⇒ a_n = a + (n - 1)d

⇒ 102 = 2 + 7(n - 1)

⇒ 102 = 2 + 7n - 7

⇒ 102 = 7n - 5

⇒ 102 + 5 = 7n

⇒ 7n = 107

⇒ n = $\dfrac{107}{7} = 15\dfrac{2}{7}$.

Since, n cannot be in fraction.

Hence, 102 is not the term of the sequence.

(e) If k is added to each term.

Sequence : 2 + k, 9 + k, 16 + k,.............

The common difference between terms is still equal to 7.

Hence, sequence is in A.P.

(a) L1 : x - y = 1

⇒ y = x - 1

Comparing above equation with y = mx + c, we get :

⇒ m = 1.

We know that,

Slope of parallel lines are equal.

∴ Slope of L3 = 1.

Given,

L3 has y-intercept = 3.

∴ y = mx + c

⇒ y = 1.x + 3

⇒ y = x + 3.

Hence, equation of line L3 : y = x + 3.

(b) L2 : x + y = 5 and L3 : y = x + 3

⇒ x + y = 5 .........(1)

⇒ y = x + 3 .........(2)

Substituting value of y from equation (2) in (1), we get :

⇒ x + (x + 3) = 5

⇒ 2x + 3 = 5

⇒ 2x = 5 - 3

⇒ 2x = 2

⇒ x = $\dfrac{2}{2}$

⇒ x = 1.

Substituting value of x in equation (2), we get :

⇒ y = 1 + 3 = 4.

⇒ (x, y) = (1, 4)

∴ P(k, 4) = (1, 4)

Hence, value of k = 1.

(c) Given,

L2 meets x-axis at point R.

At point on x-axis, y-coordinate = 0.

L2 : x + y = 5

⇒ x + 0 = 5

⇒ x = 5

⇒ R = (x, y) = (5, 0).

P = (1, 4), Q = (3, 2) and R = (5, 0)

Let PQ : QR = k : 1

By section formula,

$\Rightarrow (x, y) = \Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big) \\[1em] \Rightarrow (3, 2) = \Big(\dfrac{k \times 5 + 1 \times 1}{k + 1}, \dfrac{k \times 0 + 1 \times 4}{k + 1}\Big) \\[1em] \Rightarrow (3, 2) = \Big(\dfrac{5k + 1}{k + 1}, \dfrac{4}{k + 1}\Big) \\[1em] \Rightarrow \dfrac{5k + 1}{k + 1} = 3 \text{ and } 2 = \dfrac{4}{k + 1} \\[1em] \Rightarrow 5k + 1 = 3(k + 1) \text{ and } 4 = 2(k + 1) \\[1em] \Rightarrow 5k + 1 = 3k + 3 \text{ and } 4 = 2k + 2 \\[1em] \Rightarrow 5k - 3k = 3 - 1 \text{ and } 2k = 4 - 2 \\[1em] \Rightarrow 2k = 2 \text{ and } 2k = 2 \\[1em] \Rightarrow k = \dfrac{2}{2} = 1.$

∴ PQ : QR = 1 : 1.

Hence, R = (5, 0) and PQ : QR = 1 : 1.

(a) In △ ACD and △ FCE,

⇒ ∠ACD = ∠FCE (Common angles)

⇒ ∠ADC = ∠FEC (Corresponding angles are equal)

∴ △ ACD ~ △ FCE (By A.A. axiom)

We know that,

Ratio of corresponding sides of similar triangle are proportional.

$\therefore \dfrac{FC}{AC} = \dfrac{EC}{CD} \\[1em] \Rightarrow \dfrac{FC}{AC} = \dfrac{EC}{EC + ED} \\[1em] \Rightarrow \dfrac{FC}{AC} = \dfrac{3}{3 + 18} \\[1em] \Rightarrow \dfrac{FC}{AC} = \dfrac{3}{21} \\[1em] \Rightarrow \dfrac{FC}{AC} = \dfrac{1}{7}.$

Let FC = x and AC = 7x.

From figure,

AF = AC - FC = 7x - x = 6x.

$\therefore \dfrac{AF}{FC} = \dfrac{6x}{x} = \dfrac{6}{1}$ = 6 : 1.

Hence, AF : FC = 6 : 1.

(b) Since, △ ACD ~ △ FCE

$\therefore \dfrac{EF}{AD} = \dfrac{EC}{CD} \\[1em] \Rightarrow \dfrac{EF}{35} = \dfrac{3}{21} \\[1em] \Rightarrow EF = \dfrac{3}{21} \times 35 \\[1em] \Rightarrow EF = \dfrac{105}{21} = 5 \text{ cm}.$

Hence, EF = 5 cm.

(c) We know that,

The ratio of the area of two similar triangles is equal to the square of the ratio of any pair of the corresponding sides of the similar triangles.

$\therefore \dfrac{\text{Area of △ ADC}}{\text{Area of △ FCE}} = \Big(\dfrac{DC}{EC}\Big)^2 \\[1em] \Rightarrow \dfrac{\text{Area of △ ADC}}{\text{Area of △ FCE}} = \Big(\dfrac{21}{3}\Big)^2 \\[1em] \Rightarrow \dfrac{\text{Area of △ ADC}}{\text{Area of △ FCE}} = \dfrac{441}{9} \\[1em] \Rightarrow \dfrac{\text{Area of △ ADC}}{\text{Area of △ FCE}} = \dfrac{49}{1}.$

Let area of △ ADC = 49a and area of △ FCE = a.

Area of trapezium ADEF = Area of △ ADC - Area of △ FCE = 49a - a = 48a.

$\therefore \dfrac{\text{Area of trapezium ADEF}}{\text{Area of △ FCE}} = \dfrac{48a}{a} = \dfrac{48}{1}.$

Hence, area of trapezium ADEF : area of △ EFC = 48 : 1.

(d) In △ AGF and △ ABC,

⇒ ∠AGF = ∠ABC (Corresponding angles are equal)

⇒ ∠GAF = ∠BAC (Common angles)

∴ △ AGF ~ △ ABC (By A.A. axiom)

We know that,

Ratio of corresponding sides of similar triangle are proportional.

$\therefore \dfrac{AC}{AF} = \dfrac{BC}{GF}$

From part (a),

⇒ FC = x and AC = 7x

⇒ AF = AC - FC = 7x - x = 6x.

$\Rightarrow \dfrac{BC}{GF} = \dfrac{7x}{6x} = \dfrac{7}{6}.$

Hence, BC : GF = 7 : 6.

We know that,

Locus of a point at a fixed distance from a fixed point is the circumference of the circle with fixed point as center and fixed distance as radius.

Steps of construction :

1. Mark point O.

2. With center O and radius = 4.5 cm draw a circle.

3. Use a ruler to mark point A.

4. From point A, measure 6 cm along a straight line and mark point B.

5. Draw a straight line connecting points A and B.
   We know that, locus of points equidistant from two points is the perpendicular bisector of the line joining the two points.

6. Draw perpendicular bisector of AB.

7. Mark the points P and Q where the perpendicular bisector intersects circle.
   We know that, locus of points equidistant from two lines is the angle bisector of the angle between the lines.

8. Join PA.

9. Draw AX, the angle bisector of ∠A.

10. Mark point R as the intersection point of AX on the circumference of the circle.

11. Measure AR.

Hence, AR = 4.8 cm.

We know that each angle in a regular hexagon = 120°.

1. Draw a line segment AB = 4.3 cm.

2. At A and B draw rays making an angle of 120° each and cut off AF = BC = 4.3 cm.

3. At F and C, draw rays making angle of 120° each and cut off EF = CD = 4.3 cm.

4. Join ED. Hence, ABCDEF is the required hexagon.

5. Draw the perpendicular bisector of AB and AF. Let these bisectors meet at the point O.

6. With O as center and radius equal to OA or OB draw a circle which passes through the vertices of the hexagon. This is the required circumcircle of hexagon ABCDEF.

7. Draw the radius OB and OC.

8. At point B, construct a line perpendicular to OB. This line is the tangent at B.

9. At point C, construct a line perpendicular to OC. This line is the tangent at C.

10. The two tangents will intersect at point P.

11. Measure ∠BPC.

Hence, ∠BPC = 120°.

First, a sphere of radius (r) 7 cm is formed.

Volume of sphere = $\dfrac{4}{3}πr^3$

$= \dfrac{4}{3} \times \dfrac{22}{7} \times 7^3 \\[1em] = \dfrac{4}{3} \times 22 \times 7^2 \\[1em] = \dfrac{4312}{3} \text{ cm}^3.$

Next, a right circular cone with radius (r₁) 14 cm is formed. Let height of cone be h cm.

Since same amount of clay is used to make cone and sphere.

∴ Volume of cone = Volume of sphere

$\Rightarrow \dfrac{1}{3}πr_1^2h = \dfrac{4312}{3} \\[1em] \Rightarrow πr_1^2h = 4312 \\[1em] \Rightarrow \dfrac{22}{7} \times 14^2 \times h = 4312 \\[1em] \Rightarrow 22 \times 2 \times 14 \times h = 4312 \\[1em] \Rightarrow h = \dfrac{4312}{22 \times 2 \times 14} \\[1em] \Rightarrow h = \dfrac{4312}{616} = 7 \text{ cm}.$

Given,

The same clay is used to make a right circular cylinder of height (h₁) $\dfrac{7}{3}$ cm. Let its radius be r₂.

Since same amount of clay is used to make cylinder and sphere.

∴ Volume of cylinder = Volume of sphere

$\Rightarrow πr_2^2h_1 = \dfrac{4312}{3} \\[1em] \Rightarrow \dfrac{22}{7} \times r_2^2 \times \dfrac{7}{3} = \dfrac{4312}{3} \\[1em] \Rightarrow 22 \times r_2^2 = 4312 \\[1em] \Rightarrow r_2^2 = \dfrac{4312}{22} \\[1em] \Rightarrow r_2^2 = 196 \\[1em] \Rightarrow r_2 = \sqrt{196} = 14 \text{ cm}.$

Total surface area of sphere = 4πr²

Total surface area of cylinder = 2πr₂(r₂ + h₁)

$\Rightarrow \dfrac{\text{TSA sphere}}{\text{TSA cylinder}} = \dfrac{4πr^2}{2πr_2(r_2 + h_1)} \\[1em] = \dfrac{2r^2}{r_2(r_2 + h_1)} \\[1em] = \dfrac{2 \times 7^2}{14(14 + \dfrac{7}{3})} \\[1em] = \dfrac{98}{14 \times \dfrac{49}{3}} \\[1em] = \dfrac{98 \times 3}{14 \times 49} \\[1em] = \dfrac{3}{7}.$

Hence, height of cone = 7 cm, radius of cylinder = 14 cm and Total surface area of sphere : Total surface area of cylinder = 3 : 7.

(a) From figure,

tan 30° = $\dfrac{AB}{AR}$

Substituting values we get :

$\Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{AB}{150} \\[1em] \Rightarrow AB = \dfrac{150}{\sqrt{3}} = \dfrac{150}{1.732} \\[1em] = 86.6 \text{ m}.$

Hence, height of the building = 86.6 m.

(b)

1. From figure,

tan 60° = $\dfrac{AB}{AN}$

Substituting values we get :

$\Rightarrow \sqrt{3} = \dfrac{\dfrac{150}{\sqrt{3}}}{AN} \\[1em] \Rightarrow AN = \dfrac{150}{\sqrt{3}} \times \dfrac{1}{\sqrt{3}} \\[1em] \Rightarrow AN = \dfrac{150}{3} = 50\text{ m}.$

Hence, distance between Neha and foot of the building = 50 m.

2. From figure,

RN = AR - AN = 150 - 50 = 100 m.

Hence, distance between Rohit and Neha = 100 m.

3. From figure,

tan 60° = $\dfrac{ST}{NT}$

Substituting values we get :

$\Rightarrow \sqrt{3} = \dfrac{30}{NT} \\[1em] \Rightarrow NT = \dfrac{30}{\sqrt{3}} \\[1em] \Rightarrow NT = \dfrac{30}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}} \\[1em] \Rightarrow NT = \dfrac{30\sqrt{3}}{3} \\[1em] \Rightarrow NT = 10\sqrt{3} \\[1em] \Rightarrow NT = 10 \times 1.732 = 17.32 \text{ m}.$

Hence, distance between Neha and the tree = 17.32 m.

(iv) From figure,

AT = AN - NT = 50 - 17.32 = 32.68 m

Hence, distance between building and tree = 32.68 m.

(a) Given,

Total no. of policy holders = 100

∴ 7 + 12 + 15 + 22 + f + 14 + 8 + 4 = 100

⇒ 82 + f = 100

⇒ f = 100 - 82 = 18.

Hence, f = 18.

(b) Cumulative frequency distribution table :

Since, n = 100, which is even.

Median = $\dfrac{n}{2}\text{ th term} = \dfrac{100}{2}$ = 50th term.

Steps of construction :

1. Plot class interval on x-axis and cumulative frequency on y-axis.

2. Mark points (20, 7), (25, 19), (30, 34), (35, 56), (40, 74), (45, 88), (50, 96) and (55, 100).

3. Draw a free hand curve passing through the points marked and starting from the lower limit of first class and terminating at upper limit of the last class.

4. From point A = 50 draw a line parallel to x-axis touching the graph at point B. From point B draw a line parallel to y-axis touching x-axis at C.

From graph,

C = 33.75, which lies between 30-35.

Hence, median class = 30-35.