Short Answer Questions 1

M.P. of pressure cooker = ₹ 1800.

Let after discount the price of cooker be ₹ x.

GST on pressure cooker = 12%

According to question,

$\therefore x + \dfrac{12}{100} \times x = 1792 \\[1em] \Rightarrow x + 0.12x = 1792 \\[1em] \Rightarrow 1.12x = 1792 \\[1em] \Rightarrow x = \dfrac{1792}{1.12} = \text{₹ 1600}.$

Discount given by shopkeeper = ₹ 1800 - ₹ 1600 = ₹ 200.

Discount % = $\dfrac{\text{Discount}}{\text{M.P.}} \times 100$

$= \dfrac{200}{1800} \times 100 \\[1em] = \dfrac{1}{9} \times 100 \\[1em] = 11.11$

Hence, discount offered = 11.11 %.

Given,

Time (n) = 2 years or 24 months

Rate = r% (let)

P = ₹ x/month

I = ₹ x

By formula,

I = $\dfrac{P \times n(n + 1)}{2 \times 12} \times \dfrac{r}{100}$

Substituting values we get :

$\Rightarrow x = \dfrac{x \times 24 \times (24 + 1)}{2 \times 12} \times \dfrac{r}{100} \\[1em] \Rightarrow x = \dfrac{x \times 24 \times 25}{24} \times \dfrac{r}{100} \\[1em] \Rightarrow 1 = \dfrac{r}{4} \\[1em] \Rightarrow r = 4$

Hence, rate of interest = 4%.

Given,

Rate of dividend = 14%

Dividend on ₹ 50 = $\dfrac{14}{100} \times 50$ = ₹ 7.

Given,

Return percent from the shares is 10%.

∴ Interest on ₹ 100 = $\dfrac{10}{100} \times 100$ = ₹ 10.

∴ ₹ 7 will be interest on $\dfrac{100}{10} \times 7$ = ₹ 70.

Hence, market value of each share = ₹ 70.

(a) Given,

$\dfrac{1}{2}(2x - 1) \le 2x + \dfrac{1}{2} \le 5\dfrac{1}{2} + x$

Solving L.H.S. of the inequation, we get :

$\Rightarrow \dfrac{1}{2}(2x - 1) \le 2x + \dfrac{1}{2} \\[1em] \Rightarrow x - \dfrac{1}{2} \le 2x + \dfrac{1}{2} \\[1em] \Rightarrow 2x - x \ge -\dfrac{1}{2} - \dfrac{1}{2} \\[1em] \Rightarrow x \ge -1 \text{ ...........(1)}$

Solving R.H.S. of the inequation, we get :

$\Rightarrow 2x + \dfrac{1}{2} \le 5\dfrac{1}{2} + x \\[1em] \Rightarrow 2x + \dfrac{1}{2} \le \dfrac{11}{2} + x \\[1em] \Rightarrow 2x - x \le \dfrac{11}{2} - \dfrac{1}{2} \\[1em] \Rightarrow x \le \dfrac{10}{2} \\[1em] \Rightarrow x \le 5 \text{ ...........(2)}$

From equation (1) and (2), we get :

-1 ≤ x ≤ 5 and x ∈ R.

Hence, minimum and maximum value of x is -1 and 5 respectively.

(b) If x ∈ W.

Then, minimum value = 0 and maximum value = 5.

Hence, minimum and maximum value of x is 0 and 5 respectively, when x is a whole number.

Given,

$\dfrac{5}{x} + 4\sqrt{3} = \dfrac{2\sqrt{3}}{x^2}$

Substituting $\dfrac{1}{x}$ = t in the given equation, we get :

$\Rightarrow 5t + 4\sqrt{3} = 2\sqrt{3}t^2 \\[1em] \Rightarrow 2\sqrt{3}t^2 - 5t - 4\sqrt{3} = 0 \\[1em] \Rightarrow 2\sqrt{3}t^2 - 8t + 3t - 4\sqrt{3}= 0 \\[1em] \Rightarrow 2t(\sqrt{3}t - 4) + \sqrt{3}(\sqrt{3}t - 4) = 0 \\[1em] \Rightarrow (2t + \sqrt{3})(\sqrt{3}t - 4) = 0 \\[1em] \Rightarrow 2t + \sqrt{3} = 0 \text{ or } \sqrt{3}t - 4 = 0 \\[1em] \Rightarrow 2t = -\sqrt{3} \text{ or } \sqrt{3}t = 4 \\[1em] \Rightarrow t = -\dfrac{\sqrt{3}}{2} \text{ or } t = \dfrac{4}{\sqrt{3}} \\[1em] \Rightarrow \dfrac{1}{x} = -\dfrac{\sqrt{3}}{2} \text{ or } \dfrac{1}{x} = \dfrac{4}{\sqrt{3}} \\[1em] \Rightarrow x = -\dfrac{2}{\sqrt{3}} \text{ or } x = \dfrac{\sqrt{3}}{4}.$

Hence, x = $-\dfrac{2}{\sqrt{3}} \text{ or } x = \dfrac{\sqrt{3}}{4}.$

Given,

M.P. of toy = ₹ x

G.S.T. charged = x%

Price including G.S.T. = ₹ 24

$\therefore 24 = x + \dfrac{x}{100} \times x \\[1em] \Rightarrow 24 = x + \dfrac{x^2}{100} \\[1em] \Rightarrow 24 = \dfrac{100x + x^2}{100} \\[1em] \Rightarrow 2400 = x^2 + 100x \\[1em] \Rightarrow x^2 + 100x - 2400 = 0 \\[1em] \Rightarrow x^2 + 120x - 20x - 2400 = 0 \\[1em] \Rightarrow x(x + 120) - 20(x + 120) = 0 \\[1em] \Rightarrow (x - 20)(x + 120) = 0 \\[1em] \Rightarrow x - 20 = 0 \text{ or } x + 120 = 0 \\[1em] \Rightarrow x = 20 \text{ or } x = -120$

Since, M.P. cannot be negative.

∴ M.P. = ₹ 20 and G.S.T. = 20%.

Hence, the marked price of toy = ₹ 20 and G.S.T. = 20%.

Let the numbers be a and b.

Given,

Mean proportion between the numbers is 6.

$\therefore \dfrac{a}{6} = \dfrac{6}{b} \\[1em] \Rightarrow ab = 36 \\[1em] \Rightarrow a = \dfrac{36}{b}\text{ .........(1)}$

Third proportion of the numbers is 48.

$\therefore \dfrac{a}{b} = \dfrac{b}{48} \\[1em] \Rightarrow b^2 = 48a \\[1em] \Rightarrow b^2 = 48 \times \dfrac{36}{b} \\[1em] \Rightarrow b^3 = 1728 \\[1em] \Rightarrow b = \sqrt[3]{1728} \\[1em] \Rightarrow b = 12.$

Substituting value of b in equation (1), we get :

$\Rightarrow a = \dfrac{36}{12}$ = 3.

Hence, the numbers are 3 and 12.

Dividing the polynomial f(x) = 2x³ + 3x² - 3x - 2 by (x + 2), we get :

⇒ x + 2 = 0

⇒ x = -2

f(-2) = 2(-2)³ + 3(-2)² - 3(-2) - 2

= 2(-8) + 3(4) + 6 - 2

= -16 + 12 + 6 - 2

= -18 + 18

= 0.

Dividing the polynomial f(x) = 2x³ + 3x² - 3x - 2 by (x - 1), we get :

⇒ x - 1 = 0

⇒ x = 1

f(1) = 2(1)³ + 3(1)² - 3(1) - 2

= 2(1) + 3(1) - 3 - 2

= 2 + 3 - 3 - 2

= 0.

Dividing the polynomial f(x) = 2x³ + 3x² - 3x - 2 by (x - 2), we get :

⇒ x - 2 = 0

⇒ x = 2

f(2) = 2(2)³ + 3(2)² - 3(2) - 2

= 2(8) + 3(4) - 6 - 2

= 16 + 12 - 6 - 2

= 20.

Since, f(2) ≠ 0

∴ x - 2 does not divide the polynomial 2x³ + 3x² - 3x - 2.

Dividing the polynomial f(x) by (x + 2)(x - 1) or by (x² + x - 2), we get :

$\begin{array}{l} \phantom{x^2 + x - 2)}{\quad 2x + 1} \\ x^2 + x - 2\overline{\smash{\big)}\quad 2x^3 + 3x^2 - 3x - 2} \\ \phantom{x^2 + x - 2)}\phantom{)}\underline{\underset{-}{+}2x^3 \underset{-}{+}2x^2 \underset{+}{-}4x} \\ \phantom{{x^2 + x - 2}2x^3 + 20x^3}x^2 + x - 2 \\ \phantom{{x^2 + x - 2)}2x^3 + x^3}\underline{\underset{-}{+}x^2 \underset{-}{+} x \underset{+}{-} 2} \\ \phantom{{x^2 + x - 2)}{x^3-2x^{2}(31)}{x}}\times \end{array}$

∴ 2x³ + 3x² - 3x - 2 = (x² + x - 2)(2x + 1)

= (x + 2)(x - 1)(2x + 1).

Hence, 2x³ + 3x² - 3x - 2 = (x + 2)(x - 1)(2x + 1).

Given,

$\Rightarrow M = \dfrac{1}{2}A - 2B + 5l \\[1em] = \dfrac{1}{2}\begin{bmatrix*}[r] -6 & 0 \\ 4 & 2 \end{bmatrix*} - 2\begin{bmatrix*}[r] 1 & 0 \\ 1 & 3 \end{bmatrix*} + 5\begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -3 & 0 \\ 2 & 1 \end{bmatrix*} - \begin{bmatrix*}[r] 2 & 0 \\ 2 & 6 \end{bmatrix*} + \begin{bmatrix*}[r] 5 & 0 \\ 0 & 5 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] -3 - 2 + 5 & 0 - 0 + 0 \\ 2 - 2 + 0 & 1 - 6 + 5 \end{bmatrix*} \\[1em] = \begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*}.$

Hence, M = $\begin{bmatrix*}[r] 0 & 0 \\ 0 & 0 \end{bmatrix*}.$

(a) A.P. = 21, 42, 63, .........

The above sequence is an A.P. with first term (a) = 21 and common difference (d) = 21.

By formula,

T_n = a + (n - 1)d

= 21 + (n - 1)21

= 21 + 21n - 21

= 21n.

Hence, the n^th term = 21n.

(b) A.P. = 21, 42, 63, 84, 105, ........

Hence, there are four two digit numbers i.e. 21, 42, 63, 84 in the A.P.

(c) By formula,

Sum of A.P. = $\dfrac{n}{2}(a + a_n)$

$\text{Sum of first 10 terms of A.P.} = \dfrac{10}{2}(a + a_{10}) \\[1em] = 5[a + a + (n - 1)d] \\[1em] = 5[2a + (n - 1)d] \\[1em] = 5[2 \times 21 + (10 - 1) \times 21] \\[1em] = 5[42 + 9 \times 21] \\[1em] = 5[42 + 189] \\[1em] = 5 \times 231 \\[1em] = 1155.$

Hence, sum of first 10 terms of the A.P. = 1155.

Terms of the sequence given by $(\sqrt{3})^n$ are :

⇒ $(\sqrt{3})^1, (\sqrt{3})^2, (\sqrt{3})^3, (\sqrt{3})^4, (\sqrt{3})^5$, .......

⇒ $\sqrt{3}, 3, 3\sqrt{3}, 9, 9\sqrt{3}$

(a) Ratio between terms = $\dfrac{3}{\sqrt{3}} = \sqrt{3}$.

Hence, the sequence is a G.P. with common ratio = $\sqrt{3}$.

(b) By formula,

Sum of G.P. (S) = $\dfrac{a(r^n - 1)}{(r - 1)}$

Substituting values we get :

$\Rightarrow S_{10} = \dfrac{\sqrt{3}[(\sqrt{3})^{10} - 1]}{\sqrt{3} - 1} \\[1em] \Rightarrow p(3 + \sqrt{3})= \dfrac{\sqrt{3}(243 - 1)}{\sqrt{3} - 1} \\[1em] \Rightarrow p = \dfrac{\sqrt{3}(243 - 1)}{(\sqrt{3} - 1)(3 + \sqrt{3})} \\[1em] \Rightarrow p = \dfrac{\sqrt{3} \times 242}{(\sqrt{3} - 1)\sqrt{3}(\sqrt{3} + 1)} \\[1em] \Rightarrow p = \dfrac{242}{3 - 1} \\[1em] \Rightarrow p = \dfrac{242}{2} = 121.$

Hence, p = 121.

(a) From figure,

On reflecting in origin A, B and C becomes A', B' and C'.

Coordinates of A' = (-4, -5), B' = (0, -3) and C' = (-3, 0).

Hence, on reflecting A, B and C in origin the coordinates are (-4, -5), (0, -3) and (-3, 0) respectively.

(b) Point C(3, 0) lies on x-axis.

∴ It remains invariant on reflecting in x-axis.

Point B(0, 3) lies on y-axis.

∴ It remains invariant on reflecting in y-axis.

Hence, the invariant point for x-axis is point C and for y-axis is point B.

By two point form,

Equation of line :

⇒ y - y₁ = $\dfrac{y_2 - y_1}{x_2 - x_1}$(x - x₁)

Equation of AB :

$\Rightarrow y - 9 = \dfrac{2 - 9}{4 - (-3)}[x - (-3)] \\[1em] \Rightarrow y - 9 = \dfrac{-7}{4 + 3}[x + 3] \\[1em] \Rightarrow y - 9 = \dfrac{-7}{7}[x + 3] \\[1em] \Rightarrow y - 9 = -1[x + 3] \\[1em] \Rightarrow y - 9 = -x - 3 \\[1em] \Rightarrow y + x = -3 + 9 \\[1em] \Rightarrow x + y = 6.$

Solving equation y = 2 + 3x and x + y = 6 simultaneously,

⇒ x + y = 6 .......(1)

⇒ y = 2 + 3x .......(2)

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

⇒ x + (2 + 3x) = 6

⇒ 4x + 2 = 6

⇒ 4x = 6 - 2

⇒ 4x = 4

⇒ x = $\dfrac{4}{4}$

⇒ x = 1.

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

⇒ y = 2 + 3(1) = 2 + 3 = 5.

Let (1, 5) divide the line AB in the ratio k : 1.

By section formula,

(x, y) = $\Big(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big)$

Substituting values we get :

$\Rightarrow (1, 5) = \Big(\dfrac{k \times 4 + 1 \times -3}{k + 1}, \dfrac{k \times 2 + 1 \times 9}{k + 1}\Big) \\[1em] \Rightarrow (1, 5) = \Big(\dfrac{4k - 3}{k + 1}, \dfrac{2k + 9}{k + 1}\Big) \\[1em] \Rightarrow 1 = \dfrac{4k - 3}{k + 1} \text{ or } 5 = \dfrac{2k + 9}{k + 1} \\[1em] \Rightarrow 1(k + 1) = 4k - 3 \text{ or } 5(k + 1) = 2k + 9 \\[1em] \Rightarrow k + 1 = 4k - 3 \text{ or } 5k + 5 = 2k + 9 \\[1em] \Rightarrow 4k - k = 1 + 3 \text{ or } 5k - 2k = 9 - 5 \\[1em] \Rightarrow 3k = 4 \text{ or } 3k = 4 \\[1em] \Rightarrow k = \dfrac{4}{3} \\[1em] \Rightarrow k : 1 = \dfrac{4}{3} : 1 = 4 : 3.$

Hence, the line y = 2 + 3x divides the line segment AB in the ratio 4 : 3.

(a) Given,

BD is equally inclined to both axes.

∴ Slope of BD = tan 45° = 1.

Hence, slope of BD = 1.

(b) Since, diagonals of a square are perpendicular and product of slope of perpendicular lines equals to -1.

∴ Slope of BD × Slope of AC = -1

⇒ 1 × Slope of AC = -1

⇒ Slope of AC = -1.

Since, diagonals meet at point (-2, -2).

By point-slope form,

⇒ y - y₁ = m(x - x₁)

⇒ y - (-2) = -1[x - (-2)]

⇒ y + 2 = -1[x + 2]

⇒ y + 2 = -x - 2

⇒ x + y + 2 + 2 = 0

⇒ x + y + 4 = 0.

Hence, equation of AC is x + y + 4 = 0.

Given,

ABCD is a rectangle where side BC is twice side AB.

⇒ BC = 2AB

In right angled triangle ABC,

By pythagoras theorem,

⇒ AC² = AB² + BC²

⇒ AC² = AB² + (2AB)²

⇒ AC² = AB² + 4AB²

⇒ AC² = 5AB²

⇒ AC = $\sqrt{5}$ AB.

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.

$\therefore \dfrac{\text{area of △BAP}}{\text{area of △ACQ}} = \dfrac{BA^2}{AC^2} \\[1em] \Rightarrow \dfrac{\text{area of △BAP}}{\text{area of △ACQ}} = \dfrac{BA^2}{(\sqrt{5}BA)^2} \\[1em] \Rightarrow \dfrac{\text{area of △BAP}}{\text{area of △ACQ}} = \dfrac{BA^2}{5BA^2} \\[1em] \Rightarrow \dfrac{\text{area of △BAP}}{\text{area of △ACQ}} = \dfrac{1}{5} \\[1em]$

Area of △BAP : Area of △ACQ = 1 : 5.

Hence, Area of △BAP : Area of △ACQ = 1 : 5.

(a) In △ ABD and △ CBA,

⇒ ∠ABD = ∠CBA (Common angle)

⇒ ∠BAD = ∠ACB (Given)

∴ △ ABD ~ △ CBA (By A.A. axiom)

Hence, proved that △ ABD ~ △ CBA.

(b) We know that,

Corresponding sides of similar triangle are proportional.

$\therefore \dfrac{AB}{BC} = \dfrac{BD}{BA} \\[1em] \Rightarrow \dfrac{8}{x + 4} = \dfrac{4}{8} \\[1em] \Rightarrow 4(x + 4) = 8 \times 8 \\[1em] \Rightarrow 4x + 16 = 64 \\[1em] \Rightarrow 4x = 64 - 16 \\[1em] \Rightarrow 4x = 48 \\[1em] \Rightarrow x = \dfrac{48}{4} = \text{ 12 cm}.$

Hence, x = 12 cm.

Scale factor (k) = 2 cm to 1 km or 2 cm to 100000 cm = $\dfrac{2}{100000} = \dfrac{1}{50000}$

(a) Length of cart track on map = k × Length of cart track on ground

$7.6 = \dfrac{1}{50000}$ × Length of cart track on ground

Length of cart track on ground = 7.6 × 50000 = 380000 cm = $\dfrac{380000}{100000}$ = 3.8 km

Hence, length of actual cart track on ground = 3.8 km

(b) Area of model grid square = k² × Area of actual grid square

⇒ 4 = $\Big(\dfrac{1}{50000}\Big)^2$ × Area of actual grid square

⇒ 4 = $\Big(\dfrac{1}{25 \times 10^8}\Big)$ × Area of actual grid square

⇒ Area of actual grid square = $4 \times 25 \times 10^8$

⇒ Area of actual grid square = 100 × 10⁸ = 10² × 10⁸ = 10¹⁰ cm²

= $\dfrac{10^{10}}{10^{10}}$ km² = 1 km².

Hence, actual area of a grid square = 1 km².

We know that,

Locus of the points equidistant from two sides is the angle bisector between two sides.

Locus of the points equidistant from two points is the perpendicular bisector of the line joining two points.

Steps of construction :

1. Draw a line segment AB = 7 cm.

2. With A and B as center and radii 5 cm and 6 cm respectively draw arcs intersecting at C.

3. Join AC and BC.

4. Draw BD, angle bisector of B.

5. Draw XY perpendicular bisector of AB.

6. Mark point P, the intersection point of XY and BD.

7. Measure AP.

Hence, AP = 3.8 cm.

We know that,

The length of tangents drawn from an external point to the circle are equal.

⇒ AE = AH = x (let)

⇒ BE = BF = y (let)

⇒ CF = CG = z (let)

⇒ DG = DH = k (let)

Given,

⇒ AB = 15

⇒ AE + EB = 15

⇒ x + y = 15 .......(1)

⇒ BC = 18

⇒ BF + CF = 18

⇒ y + z = 18 ........(2)

⇒ AD = 24

⇒ AH + DH = 24

⇒ x + k = 24 ..........(3)

Adding equations (2) and (3), we get :

⇒ y + z + x + k = 18 + 24

⇒ x + y + z + k = 42

⇒ 15 + z + k = 42 [From equation (1)]

⇒ z + k = 42 - 15

⇒ z + k = 27.

⇒ CD = CG + DG = k + z = 27 cm.

Hence, length of CD = 27 cm.

(a) By formula,

Each interior angle of a n sided polygon = $\dfrac{(n - 2) \times 180°}{n}$

$= \dfrac{(6 - 2) \times 180°}{6} \\[1em] = \dfrac{4 \times 180°}{6} \\[1em] = \dfrac{2}{3} \times 180° \\[1em] = 120°.$

∴ ∠FAB = 120°

From figure,

∠FAG and ∠FAB form a linear pair.

∴ ∠FAB + ∠FAG = 180°

⇒ 120° + ∠FAG = 180°

⇒ ∠FAG = 180° - 120° = 60°.

Hence, ∠FAG = 60°.

(ii) Each interior angle of regular hexagon = 120°.

Hence, ∠BCD = 120°.

(iii) Join FC.

We know that,

Angle in a semicircle is a right angle.

∴ ∠FBC = 90°

In triangle BCD,

⇒ BC = CD (Sides of regular hexagon)

⇒ ∠CBD = ∠CDB = x (Angles opposite to equal sides are equal)

By angle sum property of triangle,

⇒ ∠BCD + ∠CDB + ∠CBD = 180°

⇒ 120° + x + x = 180°

⇒ 2x = 180° - 120°

⇒ 2x = 60°

⇒ x = $\dfrac{60°}{2}$ = 30°.

From figure,

⇒ ∠FBD = ∠FBC - ∠CBD = 90° - x = 90° - 30° = 60°.

We know that,

The angle between a tangent and a chord through the point of contact is equal to an angle in the alternate segment.

⇒ ∠PDE = ∠FBD = 60°.

Hence, ∠PDE = 60°.

(a) In △ BAE,

⇒ ∠AEB = 90° (Angle in a semicircle is a right angle)

By angle sum property of triangle,

⇒ ∠AEB + ∠EBA + ∠BAE = 180°

⇒ 90° + 33° + ∠BAE = 180°

⇒ ∠BAE + 123° = 180°

⇒ ∠BAE = 180° - 123° = 57°.

Hence, ∠BAE = 57°.

(b) From figure,

⇒ ∠EAC = ∠EAB + ∠BAC

⇒ 82° = 57° + ∠BAC

⇒ ∠BAC = 82° - 57° = 25°.

We know that,

The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

⇒ ∠BOC = 2∠BAC = 2 × 25° = 50°.

Hence, ∠BOC = 50°.

(c) From figure,

∠BOC and ∠BOD form a linear pair.

∴ ∠BOC + ∠BOD = 180°

⇒ 50° + ∠BOD = 180°

⇒ ∠BOD = 180° - 50° = 130°.

In △ BOD,

⇒ ∠BOD + ∠ODB + ∠DBO = 180°

⇒ 130° + ∠ODB + 33° = 180°

⇒ ∠ODB + 163° = 180°

⇒ ∠ODB = 180° - 163° = 17°.

Hence, ∠ODB = 17°.

Given,

Diameter of container = 14 cm

Radius (R) = $\dfrac{\text{Diameter}}{2} = \dfrac{14}{2}$ = 7 cm

Height of container (H) = 16 cm

Diameter of rasgulla = 6 cm

Radius of rasgulla (r) = $\dfrac{\text{Diameter}}{2} = \dfrac{6}{2}$ = 3 cm

Volume of container = Volume of rasgullas + Volume of liquid

⇒ πR²H = $20 \times \dfrac{4}{3}$ πr³ + Volume of liquid

⇒ Volume of liquid = πR²H - $\dfrac{80}{3}$ πr³

⇒ Volume of liquid = π(R²H - $\dfrac{80}{3}r^3$)

⇒ Volume of liquid = 3.14 × (7² × 16 - $\dfrac{80}{3} \times 3^3$)

⇒ Volume of liquid = 3.14 × (784 - 720)

⇒ Volume of liquid = 3.14 × 64

⇒ Volume of liquid = 200.96 cm³.

Hence, volume of sweetened liquid in the container = 200.96 cm³.

Given,

Diameter of cone = 14 cm

Radius of cone (r) = $\dfrac{\text{Diameter}}{2} = \dfrac{14}{2}$ = 7 cm

Slant height (l) = 25 cm

Ratio of the radius and the height of a solid metallic right circular cylinder is 7 : 27.

Radius of cylinder (R) = 7x

Height of cylinder (H) = 27x

(a) Let height of cone be h cm.

By formula,

⇒ l² = r² + h²

⇒ 25² = 7² + h²

⇒ 625 = 49 + h²

⇒ h² = 625 - 49

⇒ h² = 576

⇒ h = $\sqrt{576}$ = 24 cm.

Hence, height of cone = 24 cm.

(b) Given,

A solid metallic right circular cylinder is melted and made into a cone.

∴ Volume of cylinder = Volume of cone

⇒ πR²H = $\dfrac{1}{3}πr^2h$

⇒ R²H = $\dfrac{1}{3}r^2h$

⇒ (7x)² × 27x = $\dfrac{1}{3} \times 7^2 \times 24$

⇒ 1323x³ = $\dfrac{1176}{3}$

⇒ 1323x³ = 392

⇒ x³ = $\dfrac{392}{1323}$

⇒ x³ = $\dfrac{8}{27}$

⇒ x³ = $\Big(\dfrac{2}{3}\Big)^3$

⇒ x = $\dfrac{2}{3}$

Height of cylinder (H) = 27x = $27 \times \dfrac{2}{3}$ = 18 cm.

Hence, height of cylinder = 18 cm.

(a) According to question,

⇒ AB = $\sqrt{3} \times BC$

⇒ AB = $\sqrt{3}BC$

⇒ BC = $\dfrac{AB}{\sqrt{3}}$

From figure,

⇒ tan θ = $\dfrac{BC}{AB}$

⇒ tan θ = $\dfrac{BC}{\sqrt{3}BC}$

⇒ tan θ = $\dfrac{1}{\sqrt{3}}$

⇒ tan θ = tan 30°

⇒ θ = 30°.

Hence, θ = 30°.

(b) In right angle triangle ABC,

By pythagoras theorem,

⇒ AC² = BC² + AB²

⇒ 15² = $\Big(\dfrac{AB}{\sqrt{3}}\Big)^2$ + AB²

⇒ 225 = $\dfrac{AB^2}{3}$ + AB²

⇒ 225 = $\dfrac{AB^2 + 3AB^2}{3}$

⇒ 225 = $\dfrac{4AB^2}{3}$

⇒ AB² = $\dfrac{225 \times 3}{4}$

⇒ AB² = $\dfrac{675}{4}$

⇒ AB² = 168.75

⇒ AB = $\sqrt{168.75}$

⇒ AB = 12.99 m

Rounding off,

AB = 13 m.

Hence, AB = 13 m.

Solving L.H.S. of the above equation, we get :

⇒ tan² θ + cos² θ - 1

⇒ $\dfrac{\text{sin}^2 θ}{\text{cos}^2 θ}$ - (1 - cos² θ)

⇒ $\dfrac{\text{sin}^2 θ}{\text{cos}^2 θ}$ - sin² θ

⇒ sin² θ $\Big(\dfrac{1}{\text{cos}^2 θ} - 1\Big)$

⇒ $\dfrac{\text{sin}^2 θ(1 - \text{cos}^2 θ)}{\text{cos}^2 θ}$

⇒ $\dfrac{\text{sin}^2 θ}{\text{cos}^2 θ}$. sin² θ

⇒ tan² θ. sin² θ.

Since. L.H.S. = R.H.S.

Hence, proved that tan² θ + cos² θ - 1 = tan² θ. sin² θ

(a) Class marks : 13, 15, 17, 19, 21, 23.

Difference between two consecutive class marks = 15 - 13 = 2.

Lower limit of class = Class mark - 1

Upper limit of class = Class mark + 1

(b)

Mean = $\dfrac{Σfx}{Σf} = \dfrac{504}{28}$ = 18.

Hence, mean = 18.

Given,

The mean of 5, 7, 8, 4 and m is n.

$\therefore \dfrac{5 + 7 + 8 + 4 + m}{5} = n \\[1em] \Rightarrow m + 24 = 5n \\[1em] \Rightarrow m = 5n - 24 \text{ .........(1)}$

The mean of 5, 7, 8, 4, m and n is m.

$\therefore \dfrac{5 + 7 + 8 + 4 + m + n}{6} = m \\[1em] \Rightarrow m + n + 24 = 6m$

Substituting value of m from equation (1) in above equation, we get :

⇒ 5n - 24 + n + 24 = 6m

⇒ 6n = 6m

⇒ n = m .........(2)

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

⇒ m = 5m - 24

⇒ 5m - m = 24

⇒ 4m = 24

⇒ m = $\dfrac{24}{4}$ = 6

⇒ n = 6.

Hence, m = 6 and n = 6.

As the bag contains red, blue and green marbles.

∴ Probability of selecting a red marble + Probability of selecting a blue marble + Probability of selecting a green marble = 1

⇒ $\dfrac{1}{3} + \dfrac{1}{5}$ + Probability of selecting a green marble = 1

⇒ $\dfrac{5 + 3}{15}$ + Probability of selecting a green marble = 1

⇒ Probability of selecting a green marble = $1 - \dfrac{8}{15}$

⇒ Probability of selecting a green marble = $\dfrac{15 - 8}{15}$

⇒ Probability of selecting a green marble = $\dfrac{7}{15}$.

$\therefore \dfrac{\text{No. of green marbles}}{\text{Total no. of marbles}} = \dfrac{7}{15} \\[1em] \Rightarrow \dfrac{14}{\text{Total no. of marbles}} = \dfrac{7}{15} \\[1em] \Rightarrow \text{Total no. of marbles} = \dfrac{15 \times 14}{7} = 30.$

(a) Given,

⇒ Probability of selecting a red marble = $\dfrac{1}{5}$.

$\therefore \dfrac{\text{No. of red marbles}}{\text{Total no. of marbles}} = \dfrac{1}{5} \\[1em] \Rightarrow \dfrac{\text{No. of red marbles}}{30} = \dfrac{1}{5} \\[1em] \Rightarrow \text{No. of red marbles} = \dfrac{30}{5} = 6.$

Hence, no. of red marbles = 6.

(b) Hence, total no. of marbles in the bag = 30.