Oswal Model Specimen Papers ICSE Class 10 Mathematics Solutions (Specimen Paper - 7)

Section-A

1. (i) (d) ₹256.50

Explanation :

We have,

P = ₹ 200, n = 18 months, r = 9% p.a.

$$\therefore\space\text{Interest earned =}\\\space\text{P}×\frac{n(n+1)}{2×12}×\frac{r}{100}\\= 200×\frac{18×19}{24}×\frac{9}{100}$$

= ₹ 256.50.

(ii) (b) 4, –2

Explanation :

x² – 2x – 8 = x² – 4x + 2x – 8

= x(x – 4) + 2(x – 4)

= (x – 4) (x + 2)

Therefore, x = 4, –2

(iii) (c) (x – 5) (2x + 3)

Explanation :

Let, p(x) = 2x² – 7x – 15

= 2x² – 10x + 3x – 15

= 2x (x – 5) + 3(x – 5)

= (x – 5) (2x + 3)

(iv) (c) Square matrix

Explanation :

If a matrix has equal number of columns and rows, then it is said to be a “square matrix”.

(v) (d) 2B

Explanation :

Given: Sum of first n terms, of an AP is

S_n = An + Bn²

If n = 1 First term = S₁ = A + B

If n = 2

⇒ Second term (a₂) = S₂ – S₁ = A + 3B

∴ Common difference (d) = a₂ – a₁ = 3B – B = 2B

(vi) (d) (2, – 3)

Explanation :

Using section formula,

$$x =\frac{m_{1}x_{2} + m_{2}x_{1}}{m_{1} + m_{2}}\\=\frac{2×(\normalsize-5) + 7×4}{2 + 7}\\=\frac{-10 + 28}{9}=\frac{18}{9}= 2\\\text{and}\space\\ y =\frac{m_{1}y_{2} + m_{2}y_{1}}{m_{1} + m_{2}}\\=\frac{2×11 +7×(\normalsize-7)}{2 + 7}\\=\frac{22 - 49}{9}=\frac{-27}{9}=-3$$

Hence, the co-ordinates of P are (2, – 3).

$$\text{(vii)\space(c)}\space m = \frac{1}{\sqrt{3}}, c = 2$$

Explanation :

Equation of line,

$$3y =\sqrt{3x} + 6\\\text{or}\space y =\frac{\sqrt{3x}}{3} + \frac{6}{3}\\\Rarr\space y =\frac{1}{\sqrt{3}}x + 2$$

Comparing with y = mx + c, we get

$$m =\frac{1}{\sqrt{3}}\text{and c = 2}$$

(viii) (a) 91·6 cm³

Explanation :

According to question,

2πr = 12 cm

$$r =\frac{12×7}{2×22}=\frac{21}{11}\text{cm}$$

Now, volume of cylinder = πr²h

$$=\frac{22}{7}×\frac{21}{11}×\frac{21}{11}×8$$

= 91·6 cm³.

$$\text{(ix)}\space (a)\space\begin{Bmatrix}x: -3\leq x \lt\frac{7}{5}, x\epsilon \text{R}\end{Bmatrix}$$

Explanation :

We have, 2x – 5 ≤ 5x + 4 < 11

⇒ 2x – 5 ≤ 5x + 4 ; 5x + 4 < 11

⇒ 2x – 5x ≤ 4 + 5 ; 5x < 11 – 4

⇒ – 3x ≤ 9 ; 5x < 7

$$\Rarr\space x\geq-3; x\lt\frac{7}{5}\\\Rarr\space -3\leq x\lt\frac{7}{5}\\\therefore\space\text{Solution set =}\\\begin{Bmatrix}x : -3\leq x\lt\frac{7}{5}, x\epsilon R\end{Bmatrix}$$

(x) (c) 2 sec²A

Explanation :

(1 + tan A)² + (1 – tan A)²

⇒ = 2(1² + tan²A) [∵ (a + b)² + (a – b)² = 2(a² + b²)]

= 2(1 + tan²A) = 2 sec²A (Using identity)

(xi) (c) m × p

Explanation :

$$\text{(xii)\space (c)\space}-\frac{2}{3}$$

Explanation :

Given, 2y = 3x + 2

$$\Rarr\space y =\bigg(\frac{3}{2}\bigg)x +1\\\therefore\space \text{slope m}_{1}=\frac{3}{2}$$

and y = ax + 5

∴ slope m₂ = a

But as the lines are perpendicular,

∴ m₁m₂ = – 1

$$\therefore\space\bigg(\frac{3}{2}\bigg)a = -1\\\Rarr\space a =-\frac{2}{3}.$$

(xiii) (b) 65°

Explanation :

Since, Angle subtended by an arc at the centre is twice the angle subtended in the remaining part of the circle.

∵ ∠POQ = 2∠ORP

$$\Rarr\space\angle\text{ORP}=\frac{1}{2}\angle\text{POQ}\\=\frac{1}{2}×130\degree = 65\degree$$

∵ OP = OR (same radius)

⇒ ∠OPR = ∠ORP = 65°.

(xiv) (c) 52·4

Explanation :

Σf = 8 + 10 + 10 + 12 + 10 = 50

Σfx = 80 + 300 + 500 + 840 + 900

= 2620

$$\therefore\space \text{Average =}\frac{\Sigma fx}{\Sigma f}\\=\frac{2620}{50} = 52.4.$$

(xv) (b) Assertion is incorrect but Reason is correct.

Explanation :

For the given A.P., we have

a = 9, d = 17 – 9 = 8 and S_n = 636

We know,

$$\text{S}_{n}=\frac{n}{2}[2a +(n-1)d]\\\Rarr\space 636=\frac{n}{2}[2×9 + (n-1)×8]$$

⇒ 1272 = n[8n + 10]

⇒ 4n² + 5n – 636 = 0

Using quadratic formula,

$$n =\frac{-5\pm\sqrt{(5)^{2} -4×4×(-636)}}{2×4}\\=\frac{-5\pm\sqrt{10201}}{8}\\=\frac{-5\pm 101}{8}=\frac{96}{8},\\\frac{-106}{8}= 12,\frac{-53}{4}$$

Since, number of terms cannot be negative.

∴ n = 12.

2. (i) Cost of medicines = ₹950

GST on medicines = 5% of 950

$$=\frac{5}{100}×950$$

= ₹47.50

Cost of a pair of shoes = ₹3000

GST on shoes = 18% of ₹3000

$$=\frac{18}{100}×3000$$

= ₹ 540

Cost of laptop bag = ₹1000

Discount on bag = 30% of 1000

$$=\frac{30}{100}×1000$$

= ₹300

∴ Cost of laptop bag after discount

= ₹(1000 – 300)

= ₹700

GST on laptop bag = 18% of ₹700

$$=\frac{18}{100}×700$$

= ₹126

∴ Total GST on all items

= ₹ (47.50 + 540 + 126)

= ₹ 713.50

(ii) Total bill including GST = cost of (medicines + shoes + laptop bag) + Total GST on all items.

= ₹(950 + 3000 + 700) + ₹ 713.50

= ₹ (4650 + 713.50)

= ₹ 5363.50

(ii) Given, a, b, c, d are in continued proportion.

$$\therefore \frac{a}{b}=\frac{b}{c}= \frac{c}{d}= k$$

(say)

⇒ c = kd, b = kc = k²d, a = kb = k³d

Now, L.H.S. = (a² + b² + c²) (b² + c² + d²)

= (d²k⁶ + d²k⁴ + d²k²)(d²k⁴ + d²k² + d²)

= d²k²(k⁴ + k² + 1)·d²(k⁴ + k² + 1)

= d⁴k² (k⁴ + k²+ 1)²

R.H.S. = (ab + bc + cd)² = (dk³ · dk² + dk²·dk + dk·d)²

= (d²k⁵ + d²k³ + d²k)² = d⁴k²(k⁴ + k² + 1)²

∴ L.H.S. = R.H.S. Hence Proved. Ans.

$$\text{(iii)\space}\frac{\text{cot B - tan A}}{\text{cot A - tan B}}\\=\text{tan A cot B}\\\therefore\space\text{L.H.S} =\frac{\text{cot B - tan A}}{\frac{1}{\text{tan A}} -\frac{1}{\text{cot B}}}\\=\frac{\text{cot B - tanA}}{\frac{\text{cot B - tan A}}{\text{tan A cot B}}}\\=\text{cot B - tan A}×\\\frac{\text{tan A cot B}}{\text{cot B - tan A}}$$

= tan A cot B = R.H.S. Hence Proved.

3. (i) (i) Total area of the internal surface, including base

= 2πrh + 2πr²

= 2πr (h + r)

$$= 2×\frac{22}{7}×3.5×(7 + 3.5)\space m^{2}\\=\bigg(2×\frac{22}{7}×3.5×10.5\bigg)m^{2}$$

= 231 m². Ans.

(ii) Internal volume of the container

$$=\pi r^{2}h + \frac{2}{3}\pi r^{3}\\=\frac{1}{3}×\frac{22}{7}×3.5×3.5×(21 + 7)m^{3}\\=\bigg(\frac{1}{3}×\frac{22}{7}×3.5×3.5×28\bigg)m^{3}$$

= 359·33 m³. Ans.

(ii) Given points are A (– 4, 1) and B (17, 10) and ratio = 1 : 2.

Let the coordinates of P be (x, y)

$$\therefore\space x =\frac{m_{1}x_{2} + m_{2}x_{1}}{m_{1} + m_{2}}\\= \frac{1×17 + 2×(-4)}{1 + 2}\\=\frac{17-8}{3};\\=\frac{9}{3}$$

x = 3

$$y =\frac{m_{1}y_{2} + m_{2}y_{1}}{m_{1} + m_{2}}\\=\frac{1×10 + 2×1}{1 + 2}\\=\frac{10 +2}{3}\\=\frac{12}{3}$$

y = 4

∴ The coordinates of P = (3, 4).

$$\therefore\space \text{OP}=\sqrt{x^{2} + y^{2}}\\=\sqrt{3^{2} + 4^{2}}\\=\sqrt{9 + 16} = \sqrt{25}= 5\space\text{units}.$$

Now, Let Y-axis divide the line AB in the ratio k : 1 at the point (0, y).

$$\therefore\space x =\frac{m_{1}x_{2} + m_{2}x_{1}}{m_{1} + m_{2}}\\\Rarr\space 0 =\frac{k×17 + 1 ×(\normalsize-4)}{k+1}$$

⇒ 0 = 17k – 4

⇒ 17k = 4

$$\Rarr\space k =\frac{4}{17}\\\therefore\space k:1 = \frac{4}{17}:1$$

= 4 : 17

∴ The required ratio = 4 : 17. Ans.

(iii) (a) A (3, 5), B (– 2, – 4) are plotted on graph.

$$\text{(b)}\space\text{A(3,5)}\xrightarrow{\text{Rx}}\text{A'(3, \normalsize-5)}\\\text{(c)\space \text{B}(-2-4)}\xrightarrow{\text{RY}}(2,-4)\\\xrightarrow{\text{R}_{\text{O}}}\text{B'}(-2,4)$$

(d) AA′BB′ is an isosceles trapezium.

(e) Two invariant points under reflection in X-axis are C (3, 0) and D (– 2, 0).

Section-B

4. (a) Since, number of months (n) = 24 and rate of interest (r) = 6%

$$\text{I} =\text{P}×\frac{m(n+1)}{2×12}×\frac{r}{100}\\\Rarr\space 1200 =\text{P}×\frac{24(24 + 1)}{2×12}×\frac{6}{100}\\\Rarr\space \text{P}=\frac{1,200×24×100}{6×24×25}$$

= ₹ 800

∴ Monthly instalment = ₹800 Ans.

(ii) Sum deposited = ₹ 800 × 24

= ₹ 19200

Amount on maturity = ₹ 19,200 + ₹ 1,200

= ₹ 20,400 Ans.

(ii) Given quadratic equation is, 2x² + px – 15 = 0

Since, – 5 is its root,

∴ 2 (– 5)² + p(– 5) – 15 = 0

⇒ 50 – 5p – 15 = 0

⇒ 5p = 35

⇒ p = 7

Now, putting p = 7 in p(x² + x) + k = 0, we have

7(x² + x) + k = 0

⇒ 7x² + 7x + k = 0

Hence, a = 7, b = 7, c = k.

∴ Discriminant = b² – 4ac = 7² – 4 × 7 × k = 49 – 28k.

Since, the roots are equal,

49 – 28k = 0

⇒ 28k = 49

$$\Rarr\space k=\frac{49}{28} =\frac{7}{4}.\\\textbf{Ans.}$$

(iii) (a)

Tomatoes per Plant	Mid-Point (x)	Number of Plants (f)	fx
1 – 5	3	20	60
6 – 10	8	50	400
11 – 15	13	46	598
16 – 20	18	22	396
21 – 25	23	12	276
Total		150	1730

$$\text{Mean} =\frac{\Sigma f{x}}{\Sigma f}=\frac{1730}{150}\\=\frac{173}{15}= 11.53\space\textbf{Ans.}$$

(b) The modal class is 6 – 10. Ans.

5. (i) Given :

$$\text{A} =\begin{bmatrix}1 &4\\0 &\normalsize-1\end{bmatrix},\text{B}=\begin{bmatrix}2 &x\\0 &-\frac{1}{2}\end{bmatrix}$$

∵ AB = BA

$$\Rarr\space\begin{bmatrix}1 &4\\0 &\normalsize-1\end{bmatrix}\begin{bmatrix}2 &x\\0 &-\frac{1}{2}\end{bmatrix}\\=\begin{bmatrix}2 &x\\0 &-\frac{1}{2}\end{bmatrix}\begin{bmatrix}1 &4\\0 &\normalsize-1\end{bmatrix}\\\Rarr\space \begin{bmatrix}2 &x-2\\0 &\frac{1}{2}\end{bmatrix}=\begin{bmatrix}2 &8-x\\0 &\frac{1}{2}\end{bmatrix}$$

On comparing both sides, we get

x – 2 = 8 – x

⇒ 2x = 10

⇒ x = 5. Ans.

(ii) Given : AC = BD and ∠AOC = 72°

(a)∵ Q Angle subtended by an arc at the centre is twice the angle subtended by the same arc at any point on the remaining part of the circle.

∴ ∠AOC = 2∠ABC

$$\Rarr\space\angle\text{ABC} =\frac{1}{2}\angle\text{AOC}\\=\frac{1}{2}×72\degree = 36\degree$$

Ans.

(b) Since BD = AC and equal chords subtend equal angles at the centre.

So, ∠BOD = ∠AOC = 72°

and,

$$\angle\text{BAD} =\frac{1}{2}\angle\text{BOD}\\=\frac{1}{2}×72\degree = 36\degree$$

Ans.

∠BAD + ∠ABD + ∠ADB = 180°

⇒ 36° + ∠ABD + 90° = 180°

[·.· ∠ADB is in semicircle and angles of a triangle adds upto 180°]

⇒ ∠ABD = 180° – 126°

= 54° Ans.

(iii) Let f (x) = x³ + 5x² – ax + 6 be divided by (x – 1).

Then, Remainder = 2a

⇒ f (1) = 2a

⇒ 1³ + 5.1² – a.1 + 6 = 2a

⇒ 1 + 5 – a + 6 = 2a

⇒ – a – 2a = – 12

⇒ – 3a = – 12

$$\Rarr\space a =\frac{\normalsize-12}{\normalsize-3}$$

∴ a = 4. Ans.

6. (i) Given : A (– 1, 3), B (4, 2), C (3, – 2)

(a) Coordinates of centroid G =

$$\begin{pmatrix}\frac{x_{1} + x_{2} + x_{3}}{3},\\\frac{y_{1} + y_{2} + y_{3}}{3}\end{pmatrix}\\=\begin{pmatrix}\frac{-1 + 4 + 3}{3},\\\frac{3 + 2-2}{3}\end{pmatrix}\\=\begin{pmatrix}\frac{6}{3},\frac{3}{3}\end{pmatrix} =(2,1) \textbf{Ans.}$$

$$(b)\space \text{Slope of AC =}\frac{y_{2}-y_{1}}{x_{2} - x_{1}}\\=\frac{-2-3}{3-(\normalsize-1)}=-\frac{5}{4}$$

$$\therefore\space\text{Slope of required line (m) = }-\frac{5}{4}$$

Let the equation of the line through G, be

y – y₁ = m(x – x₁)

$$\Rarr\space y-1=-\frac{5}{4}(x-2)$$

⇒ 4y – 4 = – 5x + 10

⇒ 5x + 4y – 14 = 0 which is the required line. Ans.

(ii) Consider,

$$\text{L.H.S}=\frac{\text{sin}\space\theta - 2\text{sin}^{3}\theta}{2\text{cos}^{3}\theta -\text{cos}\theta}\\=\frac{\text{sin}\theta(1 - 2 sin^{2}\theta)}{\text{cos}\theta(2 \text{cos}^{2}\theta - 1)}\\=\frac{\text{sin}\space\theta[1 - 2 (1 - \text{cos}^{2}\theta)]}{\text{cos}\theta(2 \text{cos}^{2}\theta - 1)}\\\lbrack\because\space \text{1 - cos}^{2}\theta = \text{sin}^{2}\theta\rbrack\\=\frac{\text{sin}\theta(1 -2 + 2 \text{cos}^{2}\theta)}{\text{cos}\theta(2\text{cos}^{2}\theta-1)}\\=\frac{\text{sin}\space\theta(2\text{cos}^{2}\theta - 1)}{\text{cos}\space\theta(2\text{cos}^{2}\theta - 1)}\\=\frac{\text{sin}\space\theta}{\text{cos}\space\theta}=\text{tan\space}\theta =\text{R.H.S}$$

Hence Proved.

(iii) Given, The first term = 98

Last term = 1001

Common difference = 7

(a) Number of terms(n) = ?

T_n = a + (n – 1)d

l = a + (n – 1)d

1001 = 98 + (n – 1)7

1001 = 98 + 7n – 7 = 7n + 91

7n = 1001 – 91 = 910

$$n =\frac{910}{7}= 130$$

⇒ n = 130

(b) Sum of the 'n' terms

$$\text{S}_{n} =\frac{n}{2}(a + l)\\=\frac{130}{2}(98 +1001)$$

= 65 × (1099)

= 71435. Ans.

7. (i) Let the no. of rows be x.

∴ No. of seats in each row = x

∴ Total no. of seats = x × x = x²

After rearrangement,

No. of seats in each row = x – 10,

No. of rows = 2x.

∴ Total no. of seats = 2x(x – 10)

According to question,

2x(x – 10) = x² + 300

⇒ 2x² – 20x – x² – 300 = 0

⇒ x² – 20x – 300 = 0

⇒ x² – 30x + 10x – 300 = 0

⇒ x(x – 30) + 10(x – 30) = 0

⇒ (x – 30)(x + 10) = 0

⇒ x – 30 = 0 or x + 10 = 0

⇒ x = 30 or x = – 10.

Since x cannot be negative,

∴ x = 30

(a) The required no. of rows = 30. Ans.

(b) The no. of seats after rearrangement = 2x(x – 10)

= 2 × 30 (30 – 10) = 60 × 20 = 1200. Ans.

(ii)

Scores	f	c.f
0 – 10	5	5
10 – 20	9	14
20 – 30	16	30
30 – 40	22	52
40 – 50	26	78
50 – 60	18	96
60 – 70	11	107
70 – 80	6	113
80 – 90	4	117
90 – 100	3	120
	N = 120

Scale : On X-axis : 2 cm = 10 scores

On Y-axis : 1 cm = 10 shooters

$$\text{(a)}\space\text{Median}\bigg(\frac{\text{N}}{2}\bigg)^{\text{th}}\space\text{term}\\=\bigg(\frac{120}{2}\bigg)^{\text{th}}\space\text{term = 60}^{\text{th}}\text{term}$$

Through mark for 60 on Y-axis, draw a horizontal line which meets the ogive drawn at point A.

Through point A, draw a vertical line which meets the X-axis at the mark of 43.

∴ Median = 43

(b) Since, Number of terms = 120

$$\text{Lower quartile (Q}_1) =\bigg(\frac{120}{4}\bigg)^{\text{th}}\text{term}\\= 30^{\text{th}}\text{term} = 30$$

$$\text{Upper quartile (Q}_3) =\frac{3}{4}(120)^{\text{th}}\space\text{term}\\= 90^{\text{th}}\space\text{term} = 57$$

∴ Interquartile range = Q₃ – Q₁ = 57 – 30 = 27.

(c) On X-axis, mark 75 and through this mark draw a vertical line to meet the ogive curve at B. Through B, draw a horizontal line which meets the Y-axis at the mark of 110.

∴ Number of shooters who obtained more than 75% score

= 120 – 110

= 10

8. (i) Given inequation is,

$$-8\frac{1}{2}\lt-\frac{1}{2}-4x\leq7\frac{1}{2}, x\epsilon 1\\\Rarr\space -\frac{17}{2}\lt\frac{-1-8x}{2}\leq\frac{15}{2}\\\Rarr\space-\frac{17}{2}×2\lt\frac{-1-8x}{2}×2\leq\frac{15}{2}×2$$

⇒ – 17 < – 1 – 8x < 15

⇒ – 17 + 1 < –1 + 1 – 8x < 15 + 1

$$\Rarr\space -16\lt-8x\leq 16\\\Rarr\space \frac{\normalsize-16}{\normalsize-8}\gt\frac{\normalsize-8x}{\normalsize-8}\geq\frac{16}{8}$$

⇒ 2 > x > – 2

⇒ – 2 < x < 2

∴ Solution set = {– 2, – 1, 0, 1}.

Ans.

(ii)

Height (in cm)	No. of students	Cumulative Frequency
151	6	6
152	4	10
153	11	21
154	9	30
155	16	46
156	12	58
157	2	60

∴ n = 60

$$(a)\space\text{Median = }\\\frac{\frac{n}{2}\text{th observation} + \bigg(\frac{n}{2}+1\bigg)\text{th observation}}{2}\\=\\\frac{30 \text{th observation} + 31\text{st observation}}{2}\\=\frac{154 + 155}{2} = 154.5\space\text{cm}\space\textbf{Ans.}$$

(b) Lower quartile (Q₁) =

$$\frac{n}{4}\text{th observation}\\=\frac{60}{4}\text{th observation}$$

= 15th observation = 153 cm. Ans.

$$\frac{3n}{4}\text{th observation}\\= 3×\frac{60}{4}\text{th observation}$$

(d) Interquartile range = Q₃ – Q₁ = 155 – 153 = 2 cm. Ans.

(iii) Given : DE || BC, AD : DB = 5 : 4.

$$\Rarr\space \frac{\text{AD}}{\text{AD + DB}}=\frac{5}{5 + 4}\\\Rarr\space\frac{\text{AD}}{\text{AB}} =\frac{5}{9}$$

(a) In ∆ADE and ∆ABC,

∠ADE = ∠ABC [Corresponding angles, DE || BC]

∠A = ∠A [Common angle]

⇒ ∆ADE ~ ∆ABC [AA axiom]

$$\Rarr\space\frac{\text{DE}}{\text{BC}} =\frac{\text{AD}}{\text{AB}}=\frac{5}{9}$$

⇒ DE : BC = 5 : 9. Ans.

(b) In ∆DOE and ∆COB,

∠OED = ∠OBC [DE || BC, Alternate angles]

∠DOE = ∠BOC [Vertically opposite angles]

∴ ∆DOE ~ ∆COB [AA axiom]

$$\Rarr\space \frac{\text{DO}}{\text{OC}} =\frac{\text{DE}}{\text{BC}}=\frac{5}{9}\\\Rarr\space \frac{\text{DO}}{\text{DO + OC}}=\frac{5}{5 + 9}\\\Rarr\space\frac{\text{DO}}{\text{DC}}=\frac{5}{14}$$

⇒ DO : DC = 5 : 14. Ans.

$$\therefore\space \frac{\text{Area of} \Delta \text{DOE}}{\text{Area of}\Delta\text{DCE}}=\frac{\frac{1}{2}×\text{DO × EF}}{\frac{1}{2}×\text{DC × EF}}$$

$$=\frac{\text{DO}}{\text{DC}}=\frac{5}{14}.\space\textbf{Ans.}$$

(d) ∵ ∆DOE ~ ∆COB

$$\therefore\space\frac{\text{Area of} \Delta \text{DOE}}{\text{Area of}\Delta \text{COB}}=\frac{\text{DE}^{2}}{\text{BC}^{2}}\\=\bigg(\frac{\text{DE}}{\text{BC}}\bigg)^{2} =\bigg(\frac{5}{9}\bigg)^{2}=\frac{25}{81}.$$

Ans.

9. (i) Tickets are numbered from 1 to 20.

∴ S = {1, 2, 3, 4, ....., 19, 20}

∴ n(S) = 20

Multiple of 3 = 3, 6, 9, 12, 15, 18

Multiple of 7 = 7, 14

Let E be the event of getting a multiple of 3 or 7.

∴ E = {3, 6, 7, 9, 12, 14, 15, 18}

∴ n(E) = 8

$$\therefore\space\text{P(E)}=\frac{\text{n(E)}}{\text{n(S)}}=\frac{8}{20}=\frac{2}{5}\\\textbf{Ans.}$$

(ii) Given, radius of sphere (r₁) = 6 cm

and height of cylinder (h) = 32 cm

$$\therefore\text{Volume of sphere (V}_1) =\frac{4}{3}\pi r_{1}^{3}\\=\frac{4}{3}\pi×(6)^{3}\space\text{cm}^{3}$$

Let radius of cylinder be r₂.

∴ Volume of cylinder, (V₂)=

$$\pi r_{2}^{2}h\\=\pi r^{2}_{2}× 32\space\text{cm}^{3}\\\because\space\text{V}_{1} =\text{V}_{2}\\\Rarr\space \frac{4}{3}\pi×6^{3}\\=\pi r^{2}_{2}×32\\\Rarr\space r^{2}_{2} =\frac{4×\pi×63}{3×\pi×32}$$

$$\Rarr\space r^{2}_{2} = 9\\\Rarr\space r_{2} = 3$$ Ans.

(i) Radius of the cylinder, (r2) = 3 cm Ans.

(ii) Curved surface area of the cylinder

= 2πr₂h

= 2 × 3.1 × 3 × 32

= 595.2 cm² Ans.

(iii) Given, ∠DAE = 70°

(a) ∠BAD + ∠DAE = 180° [Linear pair]

⇒ ∠BAD = 180° – 70° = 110°

Now, ∠BCD + ∠BAD = 180° [Sum of opposite angles of cyclic quadrilateral is 180°]

⇒ ∠BCD = 180° – 110°

= 70°

(b) ∠BOD = 2 ∠BCD Ans.

[Angle at centre is twice the angle at circumference]

= 2 × 70°

= 140° Ans.

$$\lbrack\because\space \text{OB = OD = radius}\rbrack$$

In ∆OBD,

∠OBD + ∠ODB + ∠BOD = 180° [Sum of angles in a triangle is 180°]

⇒ ∠OBD + ∠OBD + 140° = 180°

[∵ ∠OBD = ∠OBD]

⇒ 2 ∠OBD = 180° – 140°

$$\Rarr\space\angle\text{OBD}=\frac{40\degree}{2}=20\degree$$

Ans.

10. (i) We have

$$\frac{x^{3} + 12x}{6x^{2} + 8}=\frac{y^{3} + 27y}{9y^{2} +27}$$

Applying componendo and dividendo, we get

$$\frac{x^{3} + 12x +6x^{2} + 8}{x^{3} +12x-6x^{2}-8}\\=\frac{y^{3} + 27y + 9y^{2}+ 27}{y^{3} +27y -9y^{2}-27}\\\Rarr\space\frac{(x + 2)^{3}}{(x - 2)^{3}}=\frac{(y +3)^{3}}{(y-3)^{3}}\\\Rarr\space \frac{x+2}{x-2} =\frac{y +3}{y-3}$$

[on taking cube root on both sides]

Again using componendo and dividendo, we get

$$\frac{x + 2 +x-2}{x + 2-x + 2}\\=\frac{y +3 + y-3}{y + 3-y + 3}\\\Rarr\space \frac{2x}{4}=\frac{2y}{6}\\\frac{x}{y}=\frac{2}{3}$$

x : y = 2 : 3. Ans.

(ii) Steps of construction :

(a) Construct ∆ABC with given measurement.

(b) Construct perpendicular bisectors of BC and AC which meet at O.

(iii) We have, AB = 60 m.

Let BD = x m, CD = y m

In ∆ABD,

$$\text{tan\space}60\degree =\frac{\text{AB}}{\text{BD}}\\\Rarr\space \sqrt{3}=\frac{60}{x}\\\Rarr\space x-\frac{60}{\sqrt{3}}×\frac{\sqrt{3}}{\sqrt{3}}\\=\frac{60\sqrt{3}}{3}= 20\sqrt{3}$$

= 20 × 1.732 = 34.64 m.

In ∆ACE,

$$\text{tan 30}\degree=\frac{\text{AE}}{\text{CE}}\\\Rarr\space \frac{1}{\sqrt{3}}=\frac{\text{AE}}{20\sqrt{3}}\\\Rarr\space \text{AE}=\frac{20\sqrt{3}}{\sqrt{3}}= 20\space m$$

Now, BE = CD = y = AB – AE = 60 – 20 = 40 m.

(a) Distance between AB and CD = x = 34.64 m. Ans.

(b) Height of lamp-post (CD) = y = 40 m. Ans.