Oswal 61 Sample Question Papers ICSE Class 10 Maths Solutions

Section-A

Answer 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

x+18x/100=1,239

118x/100=1,239

⇒ x = ₹ 1,050

(ii) (b) –2 √ 6

Explanation :

√2/3 is a solution of the equation

So, 3 ×(√2/3)² + m × √(2/3)+2=0

3 ×2/3 + m × √(2/3)+2=0

m × √(2/3)=-4

m =-4√3/√2=0

m =-2√6

(iii) (c) 1/3

Explanation :

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

⇒ a = 1/3

Explanation :

$$\text{Given, \space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space}\text{A =}\begin{bmatrix}5&3\\-1&2\\\end{bmatrix}\\\text{}\\\text{We Know, \space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space}\text{I = }\begin{bmatrix}1&0\\0&1\\\end{bmatrix}\\\text{}\\\therefore\text{\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space}\text{Required matrix = A – 2I}\\\text{}\\\text{\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space}\text{=}\begin{bmatrix}5&3\\-1&2\end{bmatrix}-2\begin{bmatrix}1&0\\0&1\end{bmatrix}\\\text{}\\\text{\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space}\text{=}\begin{bmatrix}5&3\\-1&2\end{bmatrix}-\begin{bmatrix}2&0\\0&2\end{bmatrix}\\\text{}\\\text{\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space}\text{=}\begin{bmatrix}5-2&3-0\\-1-0&2-2\end{bmatrix}\\\text{}\\\text{\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space}\text{A – 2I =}\begin{bmatrix}3&3\\-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,

S_n = n/2 [2a + (n – 1)d]

S_n = 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 DABC

∠ADE = ∠ABC [Corresponding angles]

∠A = ∠A [Common angle]

∠ADE ~ ∠ABC [AA similarity rule]

AD/AB=AE/AC

⇒ 2/6=AE/9

⇒ AE = 3 cm

CE = AC – AE

= 9 – 3 = 6 cm.

(viii) (b) 2.5

Explanation :

Inequality 7 – 3x ≥ (-1/2)

⇒ – 3x ≥ (-1/2) - 7

⇒ – 3x ≥ (-15/2)

⇒ x ≤ 15/6

⇒ x ≤ 2.5

Greatest value of x = 2.5

(ix) (d) 8

Explanation :

No. of possible outcomes n(S) = n(HHH, HHT, HTH, HTT, THH, THT, TTH, TTT)
= 8.

(x) (b) 48π cm²

Explanation :

Let r = 2a and h = 3a Given Volume = πr²h = 96π cm3 ⇒ π × (2a)² × 3a = 96π ⇒ 12a³ = 96 ⇒ a³ = 8 ⇒ a = 2 r = 2 × 2 = 4 cm and h = 2 × 3 = 6 cm Hence, Curved surface area = 2πrh = 2 ×π × 4 × 6 = 48π cm2.

(xi) (d) 2

Explanation :

$$\begin{bmatrix}x-2y & 5\\ 3 & y\end{bmatrix} = \begin{bmatrix}6 & 5\\3 & -2\end{bmatrix}\\\text{}\\\text{Comparing, we get}\\\text{}\\\text{y = -2}$$

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) ∠A = ∠B

Explanation :

The angles at the circumference subtended by the same arc are equal.

(xiv) (c) n(n+1)/√2

Explanation :

Given: series is, √2+√8+√18+√32+.........

⇒ √2+2√2+3√2+4√2+......+n√2

⇒√2(1+2+3+4+.....+n)

We know that,

$$\sum^n_{i=1}k = \text{1+2+3+…..+n.}\\\text{}\\=\frac{\text{n(n+1)}}{\text{2}}\\\text{}\\=\sqrt{2}\bigg[{\frac{n(n+1)}{2}}\bigg]\text{ We know that,}\\\text{}\\ \therefore \text{}=\sqrt{2}\text{(1+2+3+…..+n)}\\ \text{}\\ = \frac{\text{n(n+1)}}{\sqrt{2}}$$

(xv) (c) Mode

Explanation :

$$\text{(i) Given:P=₹ 200, M.V. = ₹ 5150, r = 7}\%\\\text{}\\\text{Let the number of instalments be n.}\\\text{}\\\therefore\text{ }\text{ I = p}×\frac{\text{n(n+1)}}{2×12}×\frac{r}{100}-200×\frac{\text{n}^2+\text{n}}{24}×\frac{7}{100}\\\text{}\\=\frac{7}{12}(\text{n}^2+\text{n})\\\text{}\\ \therefore\text{ }\text{ M.V. = }\text{P}_\text{n}\text{+I=}\text{200}_\text{n}+\frac{7}{12}(\text{n}^2+\text{n)} = \frac{\text{2400n}+\text{7n}^{2}+\text{7n}}{12}\\\text{}\text{ }\\=\frac{ \text{7n}^2+2407n}{12}\\ \text{According to the question.}\\\text{}\\ \therefore \text{}\frac{\text{7n}^\text{2}+\text{2407n}\\\text{}}{12}=\text{5150}\\\text{}\\ \Rightarrow \text{ }\text{7n}^2+2407\text{n}=61800\\\text{}\\ \Rightarrow \text{ }\text{7n}^{2}+2575\text{n}-168\text{n}-61800 = 0 \\\text{}\\ \Rightarrow \text{}\text{n}(\text{7n}+2575)-24(\text{7n}+2575) = 0\\\text{}\\ \Rightarrow \text{}(7\text{n}+2575)(\text{n}-24) = 0\\\text{}\\ \therefore \text{n}=24 \text{ }[\because \text{n cannot be negative}]\\ \text{}\\ \therefore \text{The required number of instalments = 24.}\text{ Ans.}$$

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

$$L.H.S. = (cosec\space \theta -sin\space \theta)(sec\space \theta -cos\space \theta) (tan\space \theta+cot\space \theta) \\ =\left(\frac{1}{sin\space \theta}-sin\space \theta)\right)\left(\frac{1}{cos\space \theta}-cos\space \theta)\right)\left(\frac{sin \space \theta}{cos\space \theta}+\frac{cos\space \theta}{sin\space \theta}\right) \\ = \left(\frac{1-sin^2 \space \theta}{sin\space \theta}\right)\left(\frac{1-cos^2 \space \theta}{cos\space \theta}\right)\left(\frac{sin^2\space \theta+cos^2 \space \theta}{sin\space \theta \space cos\space \theta}\right) \\ = \frac{cos^2\space \theta}{sin\space \theta}×\frac{sin^2\space \theta}{cos\space \theta}× \frac{1}{sin\space \theta cos\space \theta}\space\space [\because sin^2 \theta + cos^2\theta = 1] \\ = \frac{sin^2 \space \theta\space cos^2 \space \theta}{sin^2 \space \theta\space cos^2 \space \theta} \\ = 1= R.H.S. \space\space Hence. Proved.$$

Answer 3.

(i) For cylinder 1 :

Diameter = 24 cm

Radius, R = 24/2 = 12 cm

Height, H = 220 cm

For cylinder 2 :

Radius, r = 8 cm

Height, h = 60 cm

Total volume of iron in two cylinders = πR²H + πr²h

= π(144 × 220 + 64 × 60) cm3

= 3.14 × 35520 = 111532.8 cm3

Total mass of iron = 111532.8 × 8 g

=111532.8 ×8/1000 kg

892.26 kg (approximately). Ans.

(ii) For Point P :

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

$$\therefore \text{Using section formula}\\\text{}\\\text{P =}\bigg(\frac{m_1x_2+m_2x_1}{m_1+m_2},\frac{m_1y_2+m_2y_1}{m_1+m_2}\bigg)\\\text{}\\=\bigg(\frac{2×3+3×(-2)}{2+3},\frac{2×(-4)+3×6}{2+3}\bigg)\\\text{}\\ = \bigg(\frac{6-6}{5},\frac{-8+18}{5}\bigg)\\\text{}\\ = \bigg(0,\frac{10}{5}\bigg)=(0,2)\\\text{}\\ \text{For the required line :}\\\text{}\\m=\frac{3}{2}\text{and}(x_1,y_1)=\text{P}(0,2)\\\text{}\\\text{Equation of line is given as,}\\\text{}\\ y-y_1 = m(x – x_1)\\\text{}\\ \Rightarrow y-2=\frac{3}{2}(x-0)\\\text{}\\ \Rightarrow 2y-4=3x \space i.e.\space \space 3x-2y+4 = 0\text{Ans.}$$

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

$$(ii)\space A \space(3,5) \xrightarrow{R_x}A'(3,-5)$$

$$(iii) B(-2,-4)\xrightarrow{R_y}(2,-4)\xrightarrow{R_O}B'(-2,4)$$

(iv) AA′BB′ is an isosceles trapezium.

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

Section-B

Answer 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

IGST = 18% of 25,000 = ₹ 4,500

= 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

(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 = 0 4k + 3 = 0

k = 3 k = − 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

⇒ x = 3/2, 3/2

For k = – 3/4 , the equation will be

(-3+1/4)x²-4(-3/4)x+9=0

⇒ 1x²/4 + 3x + 9 = 0

⇒ x² + 12x + 36 = 0

⇒ x² + 6x + 6x + 36 = 0

⇒ x(x + 6) + 6(x + 6) = 0

⇒ (x + 6)² = 0

⇒ x = – 6, – 6.

Hence, x = 3 /2 , − 6.

(iii)

Class Interval	Frequency
0 – 5	2
5 – 10	7
10 – 15	18
20 – 25	8
25 – 30	5
15 – 20	10

∴ Mode = 13 (using histogram).

Answer 4.

$$\text{(i) We have}\\\text{}\\ \text{x+y}=\begin{bmatrix}5&2\\0&9\end{bmatrix}\\ \text{}\\ \text{and }\text{X-Y=}\begin{bmatrix}3&6\\0&-1\end{bmatrix}\\ \text{}\\ \text{Now }\text{(X+Y)+(X-Y)=}\begin{bmatrix}5&2\\0&9\end{bmatrix}+\begin{bmatrix}3&6\\0&-1\end{bmatrix}\\ \text{}\\ \Rightarrow 2X = \begin{bmatrix}8&8\\0&8\end{bmatrix}\\ \\ \text{}\\ \Rightarrow X = \frac{1}{2} \begin{bmatrix}8&8\\0&8\end{bmatrix}= \begin{bmatrix}4&4\\0&4\end{bmatrix}\\\text{}\\\text{}\text{Also }\text{(X+Y)-(X-Y)=}\begin{bmatrix}5&2\\0&9\end{bmatrix}-\begin{bmatrix}3&6\\0&-1\end{bmatrix}\\\text{}\\ \Rightarrow \text{ }\text{2Y}=\begin{bmatrix}2&-4\\0&10\end{bmatrix}\\ \text{}\\ \Rightarrow \text{ }\text{Y=}\frac{1}{2}\begin{bmatrix}2&-4\\0&10\end{bmatrix}=\begin{bmatrix}1&-2\\0&5\end{bmatrix}\\\text{}\\ \text{Thus, } X = \begin{bmatrix}4&4\\0&4\end{bmatrix}\text{and Y =}\begin{bmatrix}1&-2\\0&5\end{bmatrix}\text{ Ans.}$$

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

∠TSR = 90° [Angle between the radius and tangent]

∴ ∠SRT + ∠STR = ∠TSR = 90°

⇒ ∠STR = ∠TSR – ∠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),

$$\space\space\space\space\space\space\space\space\space\space\space(2x^2+3x+1)\\x-2)\overline{2x^3-x^2-5x-2}\\\space\space\space\space\space\space\space\space\space\space\space 2x^3-4x^2\\ \space\space\space\space\space\space\space\space\space\space -\space\space\space\space\space+\\\space\space\space\space\space\space\space\space\space\overline{\space\space\space\space\space\space\space\space\space\space\space3x^2-5x}\\\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space3x^2-6x\\\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space-\space\space\space\space\space+\\\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\overline{\space\space\space\space\space\space\space\space\space\space\space\space\space\space x-2}\\\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space x-2\\\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space-\space\space\space+\\\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\overline{\space\space\space\space\space ×\space\space\space\space\space\space}$$

∴ 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.

Answer 6.

(i) Given, A(7, – 3) and B(1, 9).

(a) Slope of AB = y2-y1/x2-x1

= 9-(-3) /1-7

=12/-6

=-2

(b) Slope of perpendicular bisector of AB (m) = -1/(-2)=1/2

Mid-point of AB =((x1+x2)/2,(y1+y2)/2)

=((7+1)2,(-3+9)/2)=(4,3)

Let the equation of required line be

y – y1 = m(x – x1)

⇒ y – 3 = 1/2 (x – 4)

⇒ 2y – 6 = x – 4

⇒ x – 2y – 4 + 6 = 0

⇒ x – 2y + 2 = 0, which is the required line. Ans.

(c) Since (– 2, p) lies on x – 2y + 2 = 0

Putting x = 2, y = p, we get

– 2 – 2p + 2 = 0

⇒ 2p = 0

⇒ p = 0 Ans.

$$\text{(ii) To prove},\frac{1}{\text{cos }\theta+\text{sin} \theta}+\frac{1}{\text{sin }\theta-\text{cos} \theta}=\frac{2\text{sin } \theta}{\text{1-2cos}^2 \theta}\\ \text{}\\ \text{L.H.S = }\frac{1}{\text{cos }\theta+\text{sin }\theta}+\frac{1}{\text{sin}\theta-\text{cos}\theta}\\\text{}\\=\frac{\text{sin }\theta-\text{cos }\theta+\text{cos }\theta+\text{sin }\theta}{(\text{cos }\theta+\text{sin }\theta )(\text{sin }\theta-\text{cos }\theta )}\\\text{}\\=\frac{2\text{sin} \theta}{\text{sin}^2 \theta-\text{cos}^2\theta}\\\text{}\\=\frac{2\text{sin}\theta}{1-\text{cos}^2\theta-\text{cos}^2 \theta}\text{ }[\because \text{sin}^2 \theta -1-\text{cos}^2 \theta]\\ \text{}\\=\frac{2\text{sin} \theta}{1-2\text{cos}^2\theta}\\\text{}\\=\text{R.H.S }\text{Hence Proved}$$

(iii) Let the first term of A.P. be a and the common difference be d.
Given,

$$\text{Given}, \text{\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space}\text{T}_{24}-\text{2T}_{10}\\ \Rightarrow \text{\space\space\space\space\space\space\space\space} \{\text{a+(24-1)d}\}=2\{\text{a+(10-1)d}\}\\ \Rightarrow \text{\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space} \text{a+23d}=2\text{a+18d}\\ \Rightarrow \text{\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space} \text{a}=5\text{d}\text{\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space …(i)}\\ \text{Now},\text{\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space} \text{T}_{15}=\text{\{a+(15-1)d\}}\\ \Rightarrow \text{\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space} \text{T}_{15}=\text{a+14d}\\ \\ \Rightarrow \text{\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space} \text{T}_{15}=\text{5d+14d}\text{\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space}[\because \text{a=5d}]\\ \\ \Rightarrow \text{\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space} \text{T}_{15}=\text{19d}\\ \text{and}\text{\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space} \text{T}_{72}=\text{\{a+(72-1)d\}}\\ \Rightarrow \text{\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space} \text{T}_{72}=\text{a+71d}\\ \Rightarrow \text{\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space} \text{T}_{72}=\text{5d+71d}\\ \Rightarrow \text{\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space} \text{T}_{72}=\text{76d-4(19d)}\\ \Rightarrow \text{\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space} \text{T}_{72}=\text{4T}_{15.}\text{\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space}\text{Hence Proved}$$

Answer 7.

(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

∴ P (a king) = 3/49. Ans.

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

∴ No. of favourable outcomes = 10

∴ P (a club) = 10/49. Ans.

(c) There is only one ‘10’ of hearts

∴ No. of favourable outcomes = 1

∴ P (a ‘10’ of hearts) = 1/49. Ans.

(ii) According to the question,

2/3 (Volume of hemisphere) = Volume of cone

2/3(2 πr³/3) = -1πr²h/3

4/9 (3.5)³=1/3(3.5)².h

h =4 × 3.5 × 3.5 × 3.5 × 3/3.5 × 3.5 × 9

42.0/9

14m/3 = 4.67 m

Slant height of cone, l = √r²+h²

=√(3.5)²+(4.67)²

35m/6

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

= πrl + 2πr²

= pr(l + 2r)

22/7 × 3.5 ((35/6)+2 × 3.5)

= 11 × (5.83 + 7)

= 11 × 12.83

= 141.13 m2 Ans.

(iii) Given, AB is a diameter.

and ∠BAC = 40°.

In ΔABC

∴ ∠ABC + ∠ACB + ∠BAC = 180° [Sum of angles in a D is 180°]

⇒ ∠ABC + 90° + 40° = 180°

∠ABC = 180° – 130° = 50°

Now, ∠ACT = ∠ABC [Angles in alternate segment]

⇒ ∠ACT = 50°. Ans.

Answer 8.

$$\text{(i) We have-}\frac{x}{3}\le\frac{x}{2}-1\frac{1}{3}<\frac{1}{6},\text{x}\isin\text{R}\\\text{}\\\text{Now}, \text{\space\space\space\space\space\space} -\frac{x}{3}\le\frac{x}{2}-1\frac{1}{3} \text{\space\space\space\space\space\space}\\-\frac{x}{3}\le\frac{x}{2}-\frac{4}{3}\\\text{}\\\text{\space\space\space\space\space\space} \frac{4}{3}\le\frac{x}{2}+\frac{x}{3}\\ \text{}\\\text{\space\space\space\space\space\space\space}\frac{4}{3}\le\frac{5x}{6}\\\text{}\\\text{\space\space\space\space\space\space}\frac{6}{5}×\frac{4}{3}\le\text{x}\\\text{}\\\text{\space\space\space\space\space\space}\frac{8}{5}\le\text{x}\text{ \space\space\space\space\space\space…(i)}\\\text{}\\\text{Thus}, \space\space\space\space\space\space\space\frac{8}{5}\le\text{x}<3\\\text{}\\\text{or}\space\space\space\space\space\space\text{1.6}\le\text{x<3}\\\text{}\\ \therefore\text{ Solution set = \{1 x : 1.6} \le \text{x <3, x } \epsilon \text{ R\}}.$$

$$\frac{x}{2}-1\frac{1}{3}<\frac{1}{6}\\\text{}\\\text{\space\space\space\space\space\space\space\space\space\space}\frac{x}{2}<\frac{1}{6}+\frac{4}{3}\\\text{}\\\text{\space\space\space\space\space\space\space\space\space\space}\frac{x}{2}<\frac{1+8}{6}\\\text{}\\\text{\space\space\space\space\space\space\space\space\space\space}\text{x}<\frac{9×2}{6}\\\text{}\\\text{\space\space\space\space\space\space\space\space\space\space}\text{x}<\text{3}\text{\space\space\space\space\space\space\space}\text{…(ii)}$$

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 = Σfx/Σf=480/20=24

(iii) Given, ∠ABC = ∠DAC = x (say)

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

(a) In ΔACD andΔBCA

∠ABC = ∠DAC [Given]

∠ACD = ∠BCA [Common]

∴ ΔACD ~ ΔBCA [AA axiom]

Hence ΔACD is similar to ΔBCA. Hence Proved.

(b) ΔACD ~ ΔBCA

∴ AC/BC=CD/CA=AD/BA [Corresponding parts of similar triangles]

⇒ 4/BC=CD/4=5/8

⇒ 4/BC = 5/8

⇒ BC = 8 x 4/5=32/5=6.4 CM

and CD/4 = 5/8

⇒ CD = 5 x 4 /8

⇒ CD = 2.5 cm. Ans.

Answer 9.

(i) The original cost of each book = ₹ x.

Total cost = ₹ 960

No. of books purchased = 960/x

If the price was ₹ 8 less, the cost of each book = ₹ (x – 8)water.

No. of books purchased = 960/x -8

According to question,

(960/x-8)-(960/x) = 4

960(1/x-8)-(1/x) = 4

960 x (x-x+8)/x(x-8)=4

⇒ 960 × 8 = 4x (x – 8)

⇒ 4x² – 32x = 7680

⇒ x² – 8x – 1920 = 0

⇒ x² – 48x + 40x – 1920 = 0

⇒ x (x – 48) + 40 (x – 48) = 0

⇒ (x – 48) (x + 40) = 0

⇒ x – 48 = 0 or x + 40 = 0

⇒ x = 48 or x = – 40 (not possible)

x = 48

The original cost of each book = ₹ 48.

(ii)

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

∴ N = 200 [even]

(a) Median = Nth/2 value

= 100th value

= 57 marks (from ogive) Ans.

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

Answer 10.

(i) We have,

(x⁴+1)/2x²=17/8

(x⁴+1+2x²)/(x⁴+1-2x²)=17+8/17+8

⇒( (x²)²+2.x².1+1)/( (x²)²-2.x².1+1)=25/9

⇒ (x² +1)²/(x² -1)² =25/9

⇒ (x²+1)/(x² - 1) =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 :

(a) Draw DABC with each side 5 cm.

(b) Bisect ∠ABC and ∠ACB.

(c) Let the bisectors meet at O.

(d) Draw a perpendicular OD from O to BC.

(e) Taking OD as radius, draw the required incircle.

(iii) Let the aeroplane be at position A and BC be the river. Drop a perpendicular from A on BC. Let it intersect BC at D.

In DADB, tan 60° = AD/BD

$$\sqrt{3} = \frac{250}{x}\\\text{}\\\text{x = }\frac{250}{\sqrt{3}}\text{m}\\\text{}\\\text{In }\Delta\text{ADC},\text{ tan 45\textdegree} = \frac{\text{AD}}{\text{DC}}\\\text{}\\\text{1}=\frac{\text{250}}{\text{y}}\\\text{}\\\text{y }=\text{ 250m}\\\text{}\\ Thus,\space width\space of\space the\space river \space = 250+\frac{250}{\sqrt{3}} = 250\left(1+\frac{1}{\sqrt{3}}\right) = 250 \left(\frac{\sqrt{3}+1}{\sqrt{3}}\right)\\ \text{= 250}×\frac{2.732}{1.732}\\\text{}\\\text{= 394.34 = 394m}\text{ \space\space\space\space Ans.}$$

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