Oswal Model Specimen Papers ICSE Class 10 Mathematics Solutions (Specimen Paper - 8)

Section-A

1. (i) (c) ₹ 1,050

Explanation :

Let the price of the article be ₹ x.

So, x + GST = 1,239

⇒ x + 18% of x = 1,239

$$\Rarr\space x + \frac{18 x}{100} = 1,239\\\Rarr\space \frac{118x}{100} = 1,239$$

⇒ x = ₹ 1,050

$$\textbf{(ii)\space}\text{(b)\space -2}\sqrt{6}$$

Explanation :

3x² + mx + 2 = 0

$$\sqrt{\frac{2}{3}}\space\text{is a solution of the equation.}$$

So,

$$3×\bigg(\sqrt{\frac{2}{3}}\bigg)^{2} + m×\sqrt{\frac{2}{3}} +2 =0\\ 3×\frac{2}{3} +m\sqrt{\frac{2}{3}}+2 = 0\\ m\sqrt{\frac{2}{3}} = \normalsize-4\\\Rarr\space m =\frac{-4\sqrt{3}}{\sqrt{2}}\\\Rarr\space m =-2\sqrt{6}$$

$$\text{(iii)\space (c)}\frac{1}{3}$$

Explanation :

Let, p(x) = x³ – ax² + 2x + a – 1

∵ (x – a) is a factor of p(x).

∴ a is the zero of p(x).

∴ p(x = a) = 0

⇒ (a³) – a(a²) + 2a + a – 1 = 0

⇒ a³ – a³ + 3a – 1 = 0

⇒ 3a – 1 = 0

$$\Rarr\space a =\frac{1}{3}$$

$$\text{(iv)\space (a)}\begin{bmatrix}3 &3\\\normalsize-1 &0\end{bmatrix}$$

Explanation :

Given,

$$\text{A =}\begin{bmatrix}5 &3\\\normalsize-1 &2\end{bmatrix}\\\text{We know}\\\text{I =}\begin{bmatrix}1 &0\\0 &1\end{bmatrix}$$

∴ Required matrix = A – 2I

$$=\begin{bmatrix}5 &3\\\normalsize-1 &2\end{bmatrix}- 2\begin{bmatrix}1 &0\\0 &1\end{bmatrix}\\=\begin{bmatrix}5 &3\\\normalsize-1 &2\end{bmatrix}-\begin{bmatrix}2 &0\\\normalsize0 &2\end{bmatrix}\\=\begin{bmatrix}5-2 &3-0\\-1-0 &2-2\end{bmatrix}\\\text{A - 2I}=\begin{bmatrix}3 &3\\\normalsize-1 &0\end{bmatrix}$$

(v) (c) 945

Explanation :

List of first 14 natural numbers, where each number is divisible by 9 is :

9, 18, 27, .................. .

Clearly, this list forms an A. P.

∴ First term of A.P. (a) = 9

Common difference (d) = 9

We know,

$$\text{S}_{n}=\frac{n}{2}[2a +(n-1)d]\\\text{S}_{14}=\frac{14}{2}[2×9 + (14-1)×9]$$

= 7[18 + 117]

= 945

∴ Required sum = 945

(vi) (d) (2, – 1)

Explanation :

(vii) (b) 6 cm

Explanation :

In ΔADE and ΔABC,

∠ADE = ∠ABC [Corresponding angles]

∠A = ∠A [Common angle]

∴ ΔADE ~ ΔABC [AA similarity rule]

$$\therefore\space \frac{\text{AD}}{\text{AB}}=\frac{\text{AE}}{\text{AC}}\\\Rarr\space \frac{2}{6}=\frac{\text{AE}}{9}$$

⇒ AE = 3 cm

∴ CE = AC – AE

= 9 – 3 = 6 cm.

(viii) (a) 2

Explanation :

$$\text{Inequality 7 – 3x}\geq\bigg(\frac{\normalsize-1}{2}\bigg)\\\Rarr\space -3x\geq\frac{\normalsize-1}{2}-7\\\Rarr\space -3x\geq\frac{\normalsize-15}{2}\\\Rarr\space x\leq\frac{15}{6}$$

⇒ x ≤ 2.5

∴ Greatest value of x = 2

(ix) (d) 8

Explanation :

No. of possible outcomes n(S) = n(HHH, HHT, HTH, HTT, THH, THT, TTH, TTT)

= 8.

(x) (c) 154 cm²

Explanation :

Area of circular base = πr²

$$=\frac{22}{7}×(7)^{2}$$

= 22 × 7

= 154 cm².

(xi) (d) 2

Explanation :

$$\begin{bmatrix}x-2y &5\\3 &y\end{bmatrix}=\begin{bmatrix}6 &5\\3 &\normalsize-2\end{bmatrix}$$

Comparing, we get

y = –2

and x – 2y = 6

⇒ x – 2(–2) = 6

⇒ x + 4 = 6

⇒ x = 6 – 4 = 2

(xii) (b) y = 2x

Explanation :

Let the equation be y = mx + c

As the slope is 2, y = 2x + c

As it is passing through (1,2), substituting this in the given equation

2 = 2 × 1 + c

⇒ c = 0 Hence, the equation is y = 2x

(xiii) (b) 120°

Explanation :

reflex ∠AOC = 360° – 120° = 240°

∵ since, the angle subtended at the centre by an arc of a circle is double the angle which this arc subtends at any remaining part of the circumference.

∴ reflex ∠AOC = 2 × ∠ABC

$$\Rarr\space \angle\text{ABC}=\frac{1}{2}\text{reflex}\space\angle\text{AOC}\\=\frac{1}{2}×240\degree =1 20\degree.$$

(xiv) (c) Mode

(xv) (c) Both A and B are true

Explanation :

First 100 natural numbers divisible by 5 are 5, 10, 15, 20, ...

The above series is an A.P. with a = 5, d = 10 – 5 = 5 and n = 100.

$$\text{S}_{n}=\frac{n}{2}[2a + (n-1)d]\\=\frac{100}{2}[2×5 + (100-1)× 5]$$

= 50 × 505 = 25250.

2. (i) Since, money deposited = ₹1,000 per month i.e. P = ₹1,000 and number of months = 2 12 = 24 i.e., n = 24 and r = 6%

(i) Interest earned in 2 years

$$=\text{P}×\frac{n(n+1)}{2×12}×\frac{r}{100}\\= 1000×\frac{24(24+1)}{2×12}×\frac{6}{100}\\=₹ 1,500.\space\textbf{Ans.}$$

(ii) Matured value = Sum deposited + Interest

= ₹ (1,000 × 24) + ₹1,500

= ₹ 25,500. Ans.

$$\text{(ii)}\space (a)\space\text{Given,}\space\frac{7m + 2n}{7m - 2n}=\frac{5}{3}$$

Using componendo and dividendo,

$$\frac{(7m + 2n) + (7m - 2n)}{(7m + 2n)-(7m - 2n)}\\=\frac{5 + 3}{5-3}\\\Rarr\space \frac{7m + 7m}{2n + 2n}=\frac{8}{2}\\\Rarr\space \frac{14m}{4m}=\frac{4}{1}\\\Rarr\space\frac{7m}{2n} =\frac{4}{1}\\\Rarr\space \frac{m}{n}=\frac{4}{1}×\frac{2}{7}$$

⇒ m : n = 8 : 7 Ans.

$$\text{(b)}\space\frac{m}{n}=\frac{8}{7}\\\Rarr\space \frac{m^{2}}{n^{2}}=\frac{64}{49}$$

Using componendo and dividendo,

$$\frac{m^{2} + n^{2}}{m^{2} - n^{2}}=\frac{64 + 49}{64 -49}\\=\frac{113}{15}$$

(iii) (cosec θ – sin θ)(sec θ – cos θ) (tan θ + cot θ) = 1

∴ L.H.S. = (cosec θ – sin θ)(sec θ – cos θ) (tan θ + cot θ)

$$=\bigg(\frac{1}{\text{sin}\space\theta} -\text{sin}\theta\bigg)\bigg(\frac{1}{\text{cos}\theta}-\text{cos}\theta\bigg)\\\bigg(\frac{\text{sin}\space\theta}{\text{cos}\space\theta} + \frac{\text{cos}\space\theta}{\text{sin}\space\theta}\bigg)\\=\bigg(\frac{\text{1 - sin}^{2}\space\theta}{\text{sin}\space\theta}\bigg)\bigg(\frac{\text{1 - cos}^{2}\theta}{\text{cos}\theta}\bigg)\\\bigg(\frac{\text{sin}^{2}\theta +\text{cos}^{2}\theta}{\text{sin}\theta \text{cos}\theta}\bigg)\\=\frac{\text{cos}^{2}\theta}{\text{sin}\theta}×\frac{\text{sin}^{2}\theta}{\text{cos}\theta}×\frac{1}{\text{sin}\theta cos\theta}\\\lbrack\because\space \text{sin}^{2}\theta +\text{cos}^{2}\theta =1\rbrack\\=\frac{\text{sin}^{2}\theta\text{cos}^{2}\theta}{\text{sin}^{2}\theta\text{cos}^{2}\theta}$$

= 1 = R.H.S. Hence Proved.

3. (i) Given,

Radius of sphere = 10 cm = R

Radius of cylinder = 5 cm = r

∴ Height of cylinder = 10 cm = h

Volume of shaded region = Volume of sphere – Volume of cylinder

$$\text{Volume of sphere =}\frac{4}{3}\pi R^{3}\\=\frac{4}{3}\pi(10)^{3}$$

= 1333·33 π cm³

Volume of cylinder = πr²h

= π (5)² × 20

= 500π cm³

Now, Volume of shaded region

= 1333·33 π – 500 π

= 833·33 π

= 833·33 × 3·14

= 2616·65 cm³ Ans.

Hence, the volume of the shaded region is 2616·65 cm³. Ans.

(ii) For Point P :

(x₁, y₁) = A(– 2, 6), (x₂, y₂) = B(3, – 4) and m₁ : m₂ = 2 : 3.

∴ Using section formula,

$$\text{P} =\begin{pmatrix}\frac{m_{1}x_{2} + m_{2}x_{1}}{m_{1} + m_{2}},\frac{m_{1}y_{2} + m_{2}y_{1}}{m_{1} + m_{2}}\end{pmatrix}\\=\begin{pmatrix}\frac{2×3 + 3×(\normalsize-2)}{2 + 3},\frac{2×(\normalsize-4 + 3×6)}{2 +3}\end{pmatrix}\\=\bigg(\frac{6-6}{5},\frac{-8 + 18}{5}\bigg)\\=\bigg(0,\frac{10}{5}\bigg)$$

= (0, 2)

For the required line :

$$m =\frac{3}{2}\space\text{and (x}_{1},y_{1})\\=\text{P(0,2)}$$

Equation of a line is given as,

y – y₁ = m(x – x₁)

$$\Rarr\space y-2 =\frac{3}{2}(x-0)$$

⇒ 2y – 4 = 3x i.e., 3x – 2y + 4 = 0 Ans.

(iii) Steps of construction :

(i) Draw a line AB = 5 cm.

(ii) Mark a point C on AB such that AC = 3 cm.

(iii) Draw a perpendicular bisector of AC.

(iv) Mark a point O perpendicular bisector from A of length 2·5 cm.

(v) Taking O as centre and OA as radius draw a circle, which is the required circle.

(vi) Join O and B.

(vii) Draw a circle with OB as diameter which cuts the given circle at points P and Q.

(viii) Join PB and QB, which are the required tangents. QB = PB = 3 cm.

Here, length of tangents is 3 cm.

Section-B

4. (i) When the product is sold from Agra to Kanpur (intra - state transaction) :

For dealer in Agra :

S.P. = ₹20,000

CGST = 9% of ₹20,000 = ₹1,800

SGST = 9% of ₹20,000 = ₹1,800

When product is sold from Kanpur to Jaipur (inter- state transaction):

For the dealer in Kanpur :

Input tax credit = ₹1,800 + ₹1,800 = ₹3,600

C.P. = ₹20,000 and Profit = ₹5,000

S.P. = ₹20,000 + ₹5,000 = ₹25,000

IGST = 18% of 25,000 = ₹4,500

Net GST paid by the dealer at Kanpur

= Output GST – Input GST

= 4,500 – 3,600 = ₹900

The cost of goods/services at Jaipur

= S.P. in Kanpur + IGST

= 25,000 + 18% of 25,000

= 25,000 + 4,500 = ₹ 29,500 Ans.

(ii) Given equation is,

(k + 1)x² – 4kx + 9 = 0

Here, a = (k + 1), b = – 4k, c = 9

For real and equal roots,

D = b² – 4ac = 0

⇒ (– 4k)² – 4(k + 1)9 = 0

⇒ 16k² – 36(k + 1) = 0

⇒ 16k² – 36k – 36 = 0

Dividing both sides by 4, we get

4k² – 9k – 9 = 0

⇒ 4k² – 12k + 3k – 9 = 0

⇒ 4k(k – 3) + 3(k – 3) = 0

⇒ (k – 3)(4k + 3) = 0

⇒ k = 3 or

$$k =-\frac{3}{4}$$

For k = 3, the equation will be

4x² – 12x + 9 = 0

⇒ 4x² – 6x – 6x + 9 = 0

⇒ 2x(2x – 3) – 3(2x – 3) = 0

⇒ (2x – 3)² = 0

$$\Rarr\space x =\frac{3}{2},\frac{3}{2}\\\text{For k = –}\frac{3}{4},\\\space\text{the equation will be}$$

$$\bigg(-\frac{3}{4} + 1\bigg)x^{2} -\\ 4\bigg(\frac{\normalsize-3}{4}\bigg)x + 9 = 0\\\Rarr\space \frac{1}{4}x^{2} +3x+ 9 = 0$$

⇒ x² + 12x + 36 = 0

⇒ x(x + 6) + 6(x + 6) = 0

⇒ (x + 6)² = 0

⇒ x = – 6, – 6.

$$\text{Hence}\space x =\frac{3}{2},-6\space\textbf{Ans.}$$

(iii) On graph

Marks	0–10	10–20	20–30	30–40	40–50	50–60	60–70	70–80	80–90	90–100
f	5	11	10	20	28	37	40	29	14	6
c.f.	5	16	26	46	74	111	151	80	194	200

$$\text{(i)\space Median} =\bigg(\frac{n}{2}\bigg)^{\text{th}}\space\\\text{observation}\\=\bigg(\frac{200}{2}\bigg)^{\text{th}}\space\text{observation}$$

= 100^th observation = 57 Ans.

(ii) No. of students who failed = 46 Ans.

(iii) No. of students who secured grade one = 200 – 186 = 14 Ans.

5. (i) We have

$$\text{X + Y} =\begin{bmatrix}5 &2\\0 &9\end{bmatrix}\\\text{and}\space\text{X - Y}=\begin{bmatrix}3 &6\\0 &\normalsize-1\end{bmatrix}$$

Now (X + Y) + (X – Y) =

$$\begin{bmatrix}5 &2\\0 &9\end{bmatrix} + \begin{bmatrix}3 &6\\0 &\normalsize-1\end{bmatrix}\\\Rarr\space 2x =\begin{bmatrix}8 &8\\0 &8\end{bmatrix}\\\Rarr\space x =\frac{1}{2}\begin{bmatrix}8 &8\\0 &8\end{bmatrix}=\begin{bmatrix}4 &4\\0 &4\end{bmatrix}$$

Also (X + Y) – (X – Y) =

$$\begin{bmatrix}5 &2\\0 &9\end{bmatrix}-\begin{bmatrix}3 &6\\0 &\normalsize-1\end{bmatrix}\\\Rarr\space 2Y =\begin{bmatrix}2 &\normalsize-4\\0 &10\end{bmatrix}\\\Rarr\space \text{Y}=\frac{1}{2}\begin{bmatrix}2 &\normalsize-4\\0 &10\end{bmatrix} =\begin{bmatrix}1 &\normalsize-2\\0 &5\end{bmatrix}$$

Thus,

$$\text{x}=\begin{bmatrix}4 &4\\0 &4\end{bmatrix}\text{and Y}=\begin{bmatrix}1 &\normalsize-2\\0 &5\end{bmatrix}.$$ Ans.

(ii) Given, ∠SRT = 65° and SP is tangent

∠TSR = 90°

[Angle between the radius and tangent]

∴ ∠SRT + ∠STR = ∠TSR = 90°

⇒ ∠STR = 90°– ∠SRT

∴ ∠STR = x° = 90° – 65° = 25°

∠y° = 2∠x°

[Angle subtended at the centre is double that of the angle subtended by the arc at same circle]

∴ ∠y° = 2 × 25° = 50°

and ∠SPO = ∠OSP – ∠SOP

∴ ∠SPO = ∠z = 90° – 50° = 40°

Hence, ∠x = 25°, ∠y = 50° and ∠z = 40°. Ans.

(iii) Given expression is 2x³ – x² – px – 2 and x – 2 is the factor.

(a) Put x – 2 = 0 ⇒ x = 2 in expression, we get

2(2)³ – (2)² – p(2) – 2 = 0

⇒ 16 – 4 – 2p – 2 = 0

⇒ 10 – 2p = 0

⇒ p = 5 Ans.

(b) Putting the value of p, the expression becomes 2x³ – x² – 5x – 2.

Dividing the expression by (x – 2),

∴ 2x³ – x² – 5x – 2 = (x – 2)(2x² + 3x + 1)

= (x – 2) {2x² + (2 + 1) x + 1}

= (x – 2) (2x² + 2x + x + 1)

= (x – 2) {2x (x + 1) +1 (x + 1)}

= (x – 2) (2x + 1) (x + 1) Ans.

6. (i) Given, A(7, – 3) and B(1, 9).

$$\text{(a) Slope of AB =}\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\=\frac{9-(\normalsize-3)}{1-7}=\frac{12}{\normalsize-6} =-2\\\textbf{Ans.}$$

(b) Slope of perpendicular bisector of AB (m) =

$$-\frac{1}{(\normalsize-2)}=\frac{1}{2}$$

$$\therefore\space\text{Mid-point of AB = }\\\bigg(\frac{x_{1} + x_{2}}{2},\frac{y_{1} + y_{2}}{2}\bigg)\\=\bigg(\frac{7+1}{2},\frac{-3 +9}{2}\bigg) =(4,3)$$

Let the equation of required line be

y – y₁ = m(x – x₁)

$$\Rarr\space y-3=\frac{1}{2}(x-4)$$

⇒ 2y – 6 = x – 4

⇒ x – 2y – 4 + 6 = 0

⇒ x – 2y + 2 = 0,

which is the required line. Ans.

Putting x = – 2, y = p, we get

– 2 – 2p + 2 = 0

⇒ 2p = 0

⇒ p = 0 Ans.

(ii) To prove,

$$\frac{1}{\text{cos}\space\theta + \text{sin}\space\theta} + \frac{1}{\text{sin}\space\theta - \text{cos}\space\theta}\\=\frac{2\text{sin}\theta}{\text{1 - 2 cos}^{2}\theta}$$

L.H.S. =

$$\frac{1}{\text{cos}\space\theta + \text{sin}\space\theta} +\frac{1}{\text{sin}\space\theta - \text{cos}\space\theta}\\=\frac{\text{sin}\space\theta -\text{cos}\space\theta + \text{cos}\theta +\text{sin}\space\theta}{(\text{cos}\space\theta + \text{sin}\space\theta)(\text{sin}\space\theta - \text{cos}\space\theta)}\\=\frac{2\space\text{sin}\space\theta}{\text{sin}^{2}\theta - \text{cos}^{2}\theta}\\=\frac{2\text{sin}\space\theta}{1 -\text{cos}^{2}\theta - \text{cos}^{2}\theta}\\\lbrack\because\space \text{sin}^{2}\theta = 1 -\text{cos}^{2}\theta\rbrack\\=\frac{2\text{sin}\space\theta}{\text{1 - 2}\text{cos}^{2}\theta}$$

= R.H.S. Hence Proved.

(iii) (a) b + c, c + a, a + b will be in A.P.

if(c + a) – (b + c)=(a + b) – (c + a)

i.e., if a – b = b – c

i.e., if 2b = a + c

i.e., if a, b, c are in A.P.

Thus, if a, b, c are in A.P.

⇒ b + c, c + a, a + b are in A.P.

Hence Proved.

(b) Given, a, b, c are in A.P.

$$\Rarr\space\frac{a}{abc},\frac{b}{abc},\frac{c}{abc}\\\text{are in A.P.}\\\lbrack\text{on dividing each term by abc}\rbrack\\\Rarr\space\frac{1}{bc},\frac{1}{ca},\frac{1}{ab}\text{are in A.P. }\\\textbf{Hence Proved.}$$

$$\text{(c)}\space\frac{1}{\sqrt{b} + \sqrt{c}},\frac{1}{\sqrt{c} + \sqrt{a}},\\\frac{1}{\sqrt{a} + \sqrt{b}}\space\text{will be in A.P.}$$

$$\text{if}\space\frac{1}{\sqrt{c} + \sqrt{a}}-\frac{1}{\sqrt{b} +\sqrt{c}}\\=\frac{1}{\sqrt{a} + \sqrt{b}}-\frac{1}{\sqrt{c} +\sqrt{a}}\\\text{i.e., if}\frac{\sqrt{b}- \sqrt{a}}{(\sqrt{c} + \sqrt{a})(\sqrt{b} + \sqrt{c})}\\=\frac{\sqrt{c}-\sqrt{b}}{(\sqrt{a} + \sqrt{b})(\sqrt{c} + \sqrt{a})}\\\text{i.e., if}\space\frac{\sqrt{b} -\sqrt{a}}{\sqrt{b} + \sqrt{c}}\\=\frac{\sqrt{c} - \sqrt{b}}{\sqrt{b} + \sqrt{a}}$$

i.e., if b – a = c – b

i.e., if 2b = a + c

i.e., if a, b, c are in A.P. Hence Proved.

7. (i) Given, Distance = 400 km.

Car A travels x km/litre.

Car B travels (x + 5) km/litre.'

(i) Number of litre used by car A =

$$\frac{\text{Distance}}{\text{Speed of car A}}=\frac{400}{x}\text{litre}$$

Number of litre used by car B =

$$\frac{\text{Distance}}{\text{Speed of car B}}=\frac{400}{x + 5}\text{litre}.$$

(ii) Car A uses 4 litre more than car B

$$\therefore\space\frac{400}{x} - \frac{400}{x + 5} = 4$$

⇒ 400 (x + 5) – 400x = 4x(x + 5)

⇒ 400x + 2000 – 400x = 4x² + 20x

⇒ 4x² + 20x – 2000 = 0

⇒ 4 (x² + 5x – 500) = 0

⇒ x² + 25x – 20x – 500 = 0

⇒ x (x + 25) – 20 (x + 25) = 0

⇒ (x + 25) (x – 20) = 0

∴ x = – 25 (not acceptable)

or x = 20

Number of litre of petrol used by car B

$$=\frac{400}{20 +5} =\frac{400}{25}=16$$

Ans.

(ii)

Marks	No. of Students	c.f.
0 – 10	5	5
10 – 20	11	16
20 – 30	10	26
30 – 40	20	46
40 – 50	28	74
50 – 60	37	111
60 – 70	40	151
70 – 80	29	180
80 – 90	14	194
90 – 100	6	200

∴ N = 200 [even]

$$\text{(a)\space}\text{Median} =\frac{\text{N}}{2}\text{th value}$$

= 100th value

= 57 marks (from ogive) Ans.

(b) No. of students scoring less than 40 marks = 45. Ans.

$$\textbf{8.}\space\text{We have}-\frac{x}{3}\leq\frac{x}{2}-1\frac{1}{3}\lt\frac{1}{6},\\x\epsilon\text{R}$$

Now,

$$-\frac{x}{3}\leq\frac{x}{2}-1\frac{1}{3}\\-\frac{x}{3}\leq\frac{x}{2}-\frac{4}{3}\\\frac{4}{3}\leq\frac{x}{2} + \frac{x}{3}\\\frac{4}{3}\leq\frac{5x}{6}\\\frac{6}{5}×\frac{4}{3}\leq x\\\frac{8}{5}\leq x\space\text{...(i)}$$

$$\frac{x}{2}-1\frac{1}{3}\lt\frac{1}{6}\\\frac{x}{2}\lt \frac{1}{6} + \frac{4}{3}\\\frac{x}{2}\lt\frac{1 + 8}{6}\\x\lt\frac{9×2}{6}$$

x < 3 ...(ii)

$$\text{Thus,}\space \frac{8}{5}\leq x\lt 3$$

$$\text{or}\space 1.6\leq x\lt 3\\\therefore\space\text{Solution set =}\\\lbrace x: 1.6\leq x\leq 3, x\epsilon\ R\rbrace.$$

(ii)

Marks : Less than	Classes	Mid Values (x)	No. of students (c.f.)	Frequency (f)	fx
10	0 – 10	5	2	2	10
20	10 – 20	15	7	5	75
30	20 – 30	25	15	8	200
40	30 – 40	35	18	3	105
50	40 – 50	45	20	2	90
				Σf = 20	Σfx = 480

∴ Arithmetic mean =

$$\frac{\Sigma fx}{\Sigma f}=\frac{480}{20}= 24.$$ Ans.

(iii) Given, ∠ABC = ∠DAC = x (say)

AB = 8 cm, AC = 4 cm, AD = 5 cm.

(a) In ΔACD and ΔBCA

∠ABC = ∠DAC [Given]

∠ACD = ∠BCA [Common]

∴ DACD ~ DBCA [AA axiom]

Hence ΔACD is similar to ΔBCA.

Hence Proved.

(b) ∵ ΔACD ~ ΔBCA

$$\therefore\space \frac{\text{AC}}{\text{BC}} =\frac{\text{CD}}{\text{CA}}=\frac{\text{AD}}{\text{BA}}$$

[Corresponding parts of similar triangles]

$$\Rarr\space\frac{4}{\text{BC}} =\frac{\text{CD}}{4}=\frac{5}{8}\\\Rarr\space\frac{4}{\text{BC}}=\frac{5}{8}\\\Rarr\space\text{BC}= \frac{8×4}{5}=\frac{32}{5}= 6.4\space\text{cm}\\\text{and}\space \frac{\text{CD}}{4}=\frac{5}{8}\\\Rarr\space \text{CD}=\frac{5×4}{8}$$

⇒ CD = 2.5 cm. Ans.

9. (i) After removing king, queen and jack of clubs, 49 cards are left in the deck.

∴ Total number of possible outcomes = 49

(a) There are 3 kings left in the deck.

∴ Number of favourable outcomes = 3

$$\therefore\space\text{P (a king) =}\frac{3}{49}.\space\textbf{Ans.}$$

(b) After removing king, queen and jack of clubs, number of clubs cards left = 13 – 3 = 10.

∴ No. of favourable outcomes = 10

$$\therefore\space\text{P (a club) =}\frac{10}{49}.\space\textbf{Ans.}$$

∴ No. of favourable outcomes = 1

∴ P (a ‘10’ of hearts) =

$$\frac{1}{49}.\space\textbf{Ans.}$$

(ii) According to the question,

$$\frac{2}{3}\text{(Volume of hemisphere)}\\=\text{Volume of cone}\\\Rarr\space\frac{2}{3}\bigg(\frac{2}{3}\pi r^{3}\bigg)=\frac{1}{3}\pi r^{2}h\\\Rarr\space \frac{4}{9}(3.5)^{3} =\frac{1}{3}(3.5)^{2}.h\\\Rarr\space h =\frac{4×3.5×3.5×3.5×3}{3.5×3.5×9}\\=\frac{42.0}{9}$$

$$=\frac{14}{3}m = 4.67 m\\\text{Slant height of cone, l} =\\\sqrt{r^{2} + h^{2}}\\=\sqrt{(3.5)^{2} + (4.67)^{2}}\\=\frac{35}{6}m$$

Now, Surface area of buoy = Surface area of right cone + Surface area of hemisphere

= πrl + 2πr²

= πr(l + 2r)

$$= \frac{22}{7}×3.5\bigg(\frac{35}{6} + 2×3.5\bigg)$$

= 11 × (5.83 + 7)

= 11 × 12.83

= 141.13 m² Ans.

(iii) Given,

AB = 24 cm, ON = 12 cm, OM = 5 cm.

(i) In ∠AOM, OA² = OM² + AM²

= (5)² + (12)²

= 25 + 144 = 169

∴ OA = 13 cm

Thus, radius of the circle is 13 cm.

(ii) In ∠CON, OC² = ON² + CN²

(13)² = (12)² + CN²

[∵ OC = OA = 13 (Radius)]

⇒ 169 – 144 = CN²

⇒ CN² = 25

⇒ CN = 5

Thus, length of chord CD = 2 CN

= 2 × 5

= 10 cm.

10. (i) We have,

$$\frac{x^{4} +1}{2x^{2}} =\frac{17}{8}$$

Using componendo and dividendo, we have

$$\frac{x^{4} + 1 + 2x^{2}}{x^{4} + 1 - 2x^{2}}=\frac{17 + 8}{17 -8}$$

$$\Rarr\space\frac{(x^{2})^{2} + 2.x^{2}.1 +1}{(x^{2})^{2} - 2.x^{2}.1 +1}=\frac{25}{9}\\\Rarr\space \frac{(x^{2}+1)^{2}}{(x^{2}-1)^{2}} =\frac{25}{9}\\\Rarr\space \frac{x^{2}+1}{x^{2}-1}=\frac{5}{3}$$

⇒ 5(x² – 1) = 3(x² + 1)

⇒ 5x² – 5 = 3x² + 3

⇒ 5x² – 3x² = 3 + 5

⇒ 2x² = 8

⇒ x² = 4

⇒ x = ± 2. Ans.

(ii) Steps of construction :

(i) Take BC = 10 cm.

(ii) Make ∠ABC = 45° and with centre B, cut the arc BA = 9 cm.

(iii) Join AC, so ABC is the required triangle.

(iv) With A as centre and radius = 2·5 cm, draw a circle. It will pass through D.

(v) Draw perpendicular from D which cuts BC at E.

(vi) Draw the angle bisector of BED which cut BD at O.

(vii) Taking radius = OD, draw a circle which touches the first circle at D and also touches the line BC at F.

(viii) This is the required circle. The radius OD = 2·7 cm.

(iii) Let AB be the cliff and CD be the tower.

Also, let DB = CE = x m and AB = h m

(a) In ΔABD,

$$\text{tan}\space 60\degree =\frac{\text{AB}}{\text{DB}}\\\Rarr\space\sqrt{3}=\frac{h}{x}\\\Rarr\space h = x\sqrt{3}\space\text{...(i)}$$

And, in ΔACE,

$$\text{tan}\space 45\degree=\frac{\text{AE}}{\text{CE}}\\\Rarr\space 1 =\frac{\text{AB - BE}}{x}\\\Rarr\space 1 =\frac{h-20}{x}$$

⇒ x = h – 20 ...(ii)

Putting the value of x in equation (i), we get

$$h = (h-20)\sqrt{3}\\\Rarr\space h =\sqrt{3}h - 20\sqrt{3}\\\Rarr\space \sqrt{3}h-h = 20\sqrt{3}\\\Rarr\space h(\sqrt{3}-1) = 20\sqrt{3}\\\Rarr\space h =\frac{20\sqrt{3}}{\sqrt{3}-1}×\frac{\sqrt{3}+1}{\sqrt{3}+1}\\=\frac{20\sqrt{3}(\sqrt{3}+1)}{3-1}\\= 10(3 + \sqrt{3})$$

= 10(3 + 1.732)

= 10 × 4.732

= 47.32 m

Hence, the height of cliff is 47.32 m. Ans.

(b) Putting the value of h in equation (ii), we get

x = h – 20 = 47.32 – 20

= 27.32

Hence, the distance between the cliff and the tower is 27.32 m. Ans.