Short Answer Questions 2

(a) Given, total GST on wheat flour and basmati rice is ₹ 45.

Total price of wheat flour = 35 × 5 = ₹ 175.

Total price of basmati rice = 180 × 5 = ₹ 900.

Given,

Total GST on wheat flour and basmati rice is ₹ 45.

$$\begin{aligned}\therefore \frac{x}{100} \times 175 + \frac{5}{100} \times 900 = 45 \\[1em] \Rightarrow 1.75x + 45 = 45 \\[1em] \Rightarrow 1.75x = 45 - 45 \\[1em] \Rightarrow 1.75x = 0 \\[1em] \Rightarrow x = 0.\end{aligned}$$

Hence, x = 0%.

(b) From table,

Total price of surf excel detergent = 220 × y = ₹ 220y.

GST on detergent powder = 18%

CGST = $\frac{\text{GST}}{2} = \frac{18}{2}$ = 9%.

Given,

CGST paid for detergent powder is ₹ 39.60

$$\begin{aligned}\therefore \frac{9}{100} \times 220y = 39.60 \\[1em] \Rightarrow 9 \times 220y = 39.60 \times 100 \\[1em] \Rightarrow y = \frac{39.60 \times 100}{9 \times 220} \\[1em] \Rightarrow y = \frac{3960}{1980} = 2.\end{aligned}$$

Hence, y = 2 kg.

(c) For wheat flour,

Total price = 35 × 5 = ₹ 175

GST = 0%

So, total price = ₹ 175.

For basmati rice,

Total price = 180 × 5 = ₹ 900

GST = 5%

So, total price = ₹ 900 + $\frac{5}{100} \times 900$ = ₹ 900 + ₹ 45 = ₹ 945.

For detergent powder,

Total price = 220 × 2 = ₹ 440

GST = 18%

So, total price = ₹ 440 + $\frac{18}{100} \times 440$ = ₹ 440 + ₹ 79.20 = ₹ 519.20

Total bill = ₹ 175 + ₹ 945 + ₹ 519.20 = ₹ 1639.20

Hence, total bill = ₹ 1639.20

(a) Let money be deposited for n months.

By formula,

M.V. = P × n + $\frac{P \times n(n + 1)}{2 \times 12} \times \frac{r}{100}$

Substituting values we get :

$$\begin{aligned}\Rightarrow 11826 = 600 \times n + \frac{600 \times n(n + 1)}{2 \times 12} \times \frac{12}{100} \\[1em] \Rightarrow 11826 = 600n + \frac{600(n^2 + n)}{200} \\[1em] \Rightarrow 11826 = 600n + 3(n^2 + n) \\[1em] \Rightarrow 11826 = 600n + 3n^2 + 3n \\[1em] \Rightarrow 3n^2 + 603n = 11826 \\[1em] \Rightarrow 3(n^2 + 201n) = 11826 \\[1em] \Rightarrow n^2 + 201n = \frac{11826}{3} \\[1em] \Rightarrow n^2 + 201n = 3942 \\[1em] \Rightarrow n^2 + 201n - 3942 = 0 \\[1em] \Rightarrow n^2 + 219n - 18n - 3942 = 0 \\[1em] \Rightarrow n(n + 219) - 18(n + 219) = 0 \\[1em] \Rightarrow (n - 18)(n + 219) = 0 \\[1em] \Rightarrow n - 18 = 0 \text{ or } n + 219 = 0 \\[1em] \Rightarrow n = 18 \text{ or } n = -219.\end{aligned}$$

Since, no. of months cannot be negative.

∴ n = 18.

Hence, number of monthly installments = 18.

(b) By formula,

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

Substituting values we get :

$$\begin{aligned}\text{Interest } = \frac{600 \times 18(18 + 1)}{2 \times 12} \times \frac{12}{100} \\[1em] = \frac{600 \times 18 \times 19}{200} \\[1em] = 3 \times 18 \times 19 \\[1em] = \text{₹ 1026}.\end{aligned}$$

Hence, total interest paid = ₹ 1026.

(c) Total amount deposited by Amit = P × n

= ₹ 600 × 18

= ₹ 10800.

Hence, total amount deposited by Amit = ₹ 10800.

(a) No. of shares Aman sells = 500

Aman sells the share when they rise to ₹ 200.

Sale proceeds = 500 × ₹ 200 = ₹ 1,00,000.

Hence, sale proceeds = ₹ 1,00,000.

(b) Given,

Aman invests half of the sale proceeds in ₹ 10, 12% shares at ₹ 25.

∴ Investment = $\frac{\text{Sale proceeds}}{2} = \frac{100000}{2}$ = ₹ 50,000.

Hence, investment in ₹ 10, 12% shares at ₹ 25 = ₹ 50,000.

(c) By formula,

Income = No. of shares × $\frac{\text{Rate of dividend}}{100}$ × Nominal value of share

= 500 × $\frac{10}{100}$ × 100

= ₹ 5,000.

Hence, original income = ₹ 5,000.

(d) Aman invests ₹ 50,000 in each of the new shares.

For 1st share :

N.V. = ₹ 10

Dividend = 12%

M.V. = ₹ 25

No. of shares bought = $\frac{50000}{25}$ = 2000.

Income = No. of shares × $\frac{\text{Rate of dividend}}{100}$ × Nominal value of share

= 2000 × $\frac{12}{100} \times 10$

= ₹ 2400.

For 2nd share :

N.V. = ₹ 400

Dividend = 9%

M.V. = ₹ 500

No. of shares bought = $\frac{50000}{500}$ = 100.

Income = No. of shares × $\frac{\text{Rate of dividend}}{100}$ × Nominal value of share

= 100 × $\frac{9}{100} \times 400$

= ₹ 3,600.

New income = ₹ 3,600 + ₹ 2,400 = ₹ 6000

Change in income = New income - Original income = ₹ 6,000 - ₹ 5,000 = ₹ 1,000.

Hence, change in income = ₹ 1,000 (increase).

Given,

$\frac{11 + 3x}{5} \ge 3 - x \gt -\frac{3}{2}$

Solving L.H.S. of the equation,

$$\begin{aligned}\Rightarrow \frac{11 + 3x}{5} \ge 3 - x \\[1em] \Rightarrow 11 + 3x \ge 5(3 - x) \\[1em] \Rightarrow 11 + 3x \ge 15 - 5x \\[1em] \Rightarrow 3x + 5x \ge 15 - 11 \\[1em] \Rightarrow 8x \ge 4 \\[1em] \Rightarrow x \ge \frac{4}{8} \\[1em] \Rightarrow x \ge \frac{1}{2} \text{ ...........(1)}\end{aligned}$$

Solving R.H.S. of the equation,

$$\begin{aligned}\Rightarrow 3 - x \gt -\frac{3}{2} \\[1em] \Rightarrow x \lt 3 + \frac{3}{2} \\[1em] \Rightarrow x \lt \frac{6 + 3}{2} \\[1em] \Rightarrow x \lt \frac{9}{2} \text{ .......(2)}\end{aligned}$$

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

$\Rightarrow \frac{1}{2} \le x \lt \frac{9}{2}$

Hence, solution set = {$x : \frac{1}{2} \le x \lt \frac{9}{2}, x ∈ R$}

(a) Given, equation : 5𝑥² − 9𝑥 + 4 = 0

Comparing above equation with ax² + bx + c = 0, we get :

a = 5, b = -9 and c = 4.

Discriminant (D) = b² - 4ac = (-9)² - 4 × 5 × 4 = 81 - 80 = 1.

Since, D > 0 and a perfect square.

Hence, equation 5𝑥² − 9𝑥 + 4 = 0 has real roots.

(b) Solving,

⇒ 5𝑥² − 9𝑥 + 4 = 0

⇒ 5x² - 5x - 4x + 4 = 0

⇒ 5x(x - 1) - 4(x - 1) = 0

⇒ (5x - 4)(x - 1) = 0

⇒ 5x - 4 = 0 or x - 1 = 0

⇒ 5x = 4 or x = 1

⇒ x = $\frac{4}{5}$ or x = 1.

Hence, roots of the equation are 1 and $\frac{4}{5}$.

(a) Since,

$\frac{1400}{28000} = \frac{5600}{112000} = \frac{1}{20}$.

Hence, number of customers and profit per week are in proportion but not in continued proportion.

(b) As,

Number of customers and profit per week are in proportion.

$$\begin{aligned}\therefore \frac{x}{32140} = \frac{1}{20} \\[1em] \Rightarrow x = 32140 \times \frac{1}{20} = 1607 \\[1em]\end{aligned}$$

and

$$\begin{aligned}\therefore \frac{3212}{y} = \frac{1}{20} \\[1em] \Rightarrow y = 3212 \times 20 = 64240.\end{aligned}$$

Hence, x = 1607 and y = 64240.

(a) Simplifying 9𝑥² - 4, we get :

⇒ 9𝑥² - 4

⇒ (3𝑥)² - 2²

⇒ (3𝑥 + 2)(3𝑥 - 2).

We know that,

If (x - a) is a factor of f(x), then f(a) = 0.

Given,

9𝑥² - 4 is a factor of 9𝑥³ - m𝑥² - n𝑥 + 8.

∴ (3𝑥 + 2) and (3𝑥 - 2) are the factors of 9𝑥³ - m𝑥² - n𝑥 + 8.

⇒ 3𝑥 + 2 = 0

⇒ 3𝑥 = -2

⇒ 𝑥 = $-\frac{2}{3}$

Substituting x = $-\frac{2}{3}$ in 9𝑥³ - mx² - nx + 8, we get remainder = 0.

$$\begin{aligned}\Rightarrow 9 \times \Big(-\frac{2}{3}\Big)^3 - m \times \Big(-\frac{2}{3}\Big)^2 - n \times \Big(-\frac{2}{3}\Big) + 8 = 0 \\[1em] \Rightarrow 9 \times -\frac{8}{27} - m \times \frac{4}{9} + \frac{2n}{3} + 8 = 0 \\[1em] \Rightarrow -\frac{8}{3} - \frac{4m}{9} + \frac{2n}{3} + 8 = 0 \\[1em] \Rightarrow \frac{-24 - 4m + 6n + 72}{9} = 0 \\[1em] \Rightarrow \frac{6n - 4m + 48}{9} = 0 \\[1em] \Rightarrow 6n - 4m + 48 = 0 \\[1em] \Rightarrow 2(3n - 2m + 24) = 0 \\[1em] \Rightarrow 3n - 2m + 24 = 0 \\[1em] \Rightarrow 3n - 2m = -24 \text{ ..........(1)}\end{aligned}$$

⇒ 3𝑥 - 2 = 0

⇒ 3𝑥 = 2

⇒ 𝑥 = $\frac{2}{3}$

Substituting 𝑥 = $\frac{2}{3}$ in 9𝑥³ - m𝑥² - n𝑥 + 8, we get remainder = 0.

$$\begin{aligned}\Rightarrow 9 \times \Big(\frac{2}{3}\Big)^3 - m \times \Big(\frac{2}{3}\Big)^2 - n \times \Big(\frac{2}{3}\Big) + 8 = 0 \\[1em] \Rightarrow 9 \times \frac{8}{27} - m \times \frac{4}{9} - \frac{2n}{3} + 8 = 0 \\[1em] \Rightarrow \frac{8}{3} - \frac{4m}{9} - \frac{2n}{3} + 8 = 0 \\[1em] \Rightarrow \frac{24 - 4m - 6n + 72}{9} = 0 \\[1em] \Rightarrow \frac{96 - 6n - 4m}{9} = 0 \\[1em] \Rightarrow 96 - 6n - 4m = 0 \\[1em] \Rightarrow 4m + 6n = 96 \\[1em] \Rightarrow 2(2m + 3n) = 96 \\[1em] \Rightarrow 2m + 3n = \frac{96}{2} \\[1em] \Rightarrow 2m + 3n = 48 \text{ ..........(2)}\end{aligned}$$

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

⇒ 3n - 2m + 2m + 3n = -24 + 48

⇒ 6n = 24

⇒ n = $\frac{24}{6}$

⇒ n = 4.

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

⇒ 3(4) - 2m = -24

⇒ 12 - 2m = -24

⇒ -2m = -24 - 12

⇒ -2m = -36

⇒ m = $\frac{-36}{-2}$ = 18.

Hence, m = 18 and n = 4.

(b) Substituting value of m and n in 9𝑥³ - m𝑥² - n𝑥 + 8, we get :

9𝑥³ - 18𝑥² - 4𝑥 + 8

Dividing 9𝑥³ - 18𝑥² - 4𝑥 + 8 by 9𝑥² - 4, we get :

$$\begin{array}{l} \phantom{9x^2 - 4)}{\quad x - 2} \\ 9x^2 - 4\overline{\smash{\big)}\quad 9x^3 - 18x^2 - 4x + 8} \\ \phantom{9x^2 - 4)}\phantom{)}\underline{\underset{-}{+}9x^3 \phantom{- 18x^2} \underset{+}{-}4x} \\ \phantom{{9x^2 - 4}{+9x^3 - }}-18x^2 \phantom{- 4x} + 8 \\ \phantom{{9x^2 - 4)}{+9x^3 - }}\underline{\underset{+}{-}18x^2 \phantom{- 4x} \underset{-}{+} 8} \\ \phantom{{9x^2 - 4)}{x^3-2x^{2}(31)}{x}}\times \end{array}$$

∴ 9𝑥³ - 18𝑥² - 4𝑥 + 8 = (9𝑥² - 4)(𝑥 - 2)

= (3𝑥 + 2)(3𝑥 - 2)(𝑥 - 2).

Hence, factors are (3𝑥 + 2), (3𝑥 - 2) and (𝑥 - 2).

Marks scored	No. of students	Cumulative frequency
0-10	4	4
10-20	5	9
20-30	9	18
30-40	7	25
40-50	13	38
50-60	12	50
60-70	15	65
70-80	11	76
80-90	14	90
90-100	10	100

(a) By formula,

Probability that student has scored less than 20

= $\frac{\text{No. of students who scored less than 20}}{\text{Total no. of students}} = \frac{9}{100}$.

Hence, probability that the student has scored less than 20 = $\frac{9}{100}$.

(b) From table,

No. of students who scored less than 60 = 50

No. of students who scored less than 30 = 18

∴ No. of students who score below 60 but 30 or more = 50 - 18 = 32.

Probability that student has scored below 60 but 30 or more = $\frac{\text{No. of students scoring between 60 and 30}}{\text{Total no. of students}} = \frac{32}{100} = \frac{8}{25}$.

Hence, probability that the student has scored below 60 but 30 or more = $\frac{8}{25}$.

(c) From table,

No. of students who scored less than 70 = 65

Total no. of students = 100

∴ No. of students who score more than or equal to 70 = 100 - 65 = 35.

Probability that student has scored more than or equal to 70 = $\frac{\text{No. of students scoring 70 or more}}{\text{Total no. of students}} = \frac{35}{100} = \frac{7}{20}$.

Hence, probability that the student has scored more than or equal to 70 = $\frac{7}{20}$.

(d) No. of students those who have scored more than 89 = No. of students who has scored between 90-100 = 10.

Probability that student has scored more than 89 = $\frac{\text{No. of students scoring \textgreater 89}}{\text{Total no. of students}} = \frac{10}{100} = \frac{1}{10}$.

Hence, probability that the student has scored more than 89 = $\frac{1}{10}$.

Given,

AB is a null matrix.

Let AB = X, where X is a null matrix of order a × b.

⇒ AB = X

⇒ A_{2 × 2} × B_{2 × 1} = X_{a × b}

We know that,

The resultant matrix has no. of rows equal to the rows in the first matrix and no. of columns equal to the no. of columns in the second matrix.

∴ a = 2 and b = 1.

$$\therefore \begin{bmatrix*}[r] x & 1 \\ y & 2 \end{bmatrix*}_{2 \times 2} \begin{bmatrix*}[r] x \\ x - 2 \end{bmatrix*}_{2 \times 1} = \begin{bmatrix*}[r] 0 \\ 0 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] x^2 + 1(x - 2) \\ xy + 2(x - 2) \end{bmatrix*} = \begin{bmatrix*}[r] 0 \\ 0 \end{bmatrix*} \\[1em] \Rightarrow \begin{bmatrix*}[r] x^2 + x - 2 \\ xy + 2x - 4 \end{bmatrix*} = \begin{bmatrix*}[r] 0 \\ 0 \end{bmatrix*} \text{ .............(1)}$$

From equation (1) :

⇒ x² + x - 2 = 0

⇒ x² + 2x - x - 2 = 0

⇒ x(x + 2) - 1(x + 2) = 0

⇒ (x - 1)(x + 2) = 0

⇒ x - 1 = 0 or x + 2 = 0

⇒ x = 1 or x = -2.

From equation (1) :

⇒ xy + 2x - 4 = 0

Substituting x = 1, we get :

⇒ 1.y + 2.1 - 4 = 0

⇒ y + 2 - 4 = 0

⇒ y - 2 = 0

⇒ y = 2.

Substituting x = -2, we get :

⇒ (-2).y + 2.(-2) - 4 = 0

⇒ -2y - 4 - 4 = 0

⇒ -2y - 8 = 0

⇒ -2y = 8

⇒ y = $\frac{8}{-2}$ = -4.

Hence, x = 1, y = 2 or x = -2, y = -4.

(a) Let no. of terms be n.

By formula,

Sum of A.P. = $\frac{n}{2}[2a + (n - 1)d]$

Substituting values we get :

$$\begin{aligned}\Rightarrow 65 = \frac{n}{2}[2 \times 20 + (n - 1) \times (-3)] \\[1em] \Rightarrow 65 = \frac{n}{2}[40 - 3n + 3] \\[1em] \Rightarrow 65 = \frac{n}{2}[43 - 3n] \\[1em] \Rightarrow 65 \times 2 = n(43 - 3n) \\[1em] \Rightarrow 130 = 43n - 3n^2 \\[1em] \Rightarrow 3n^2 - 43n + 130 = 0 \\[1em] \Rightarrow 3n^2 - 30n - 13n + 130 = 0 \\[1em] \Rightarrow 3n(n - 10) - 13(n - 10) = 0 \\[1em] \Rightarrow (3n - 13)(n - 10) = 0 \\[1em] \Rightarrow 3n - 13 = 0 \text{ or } n - 10 = 0 \\[1em] \Rightarrow 3n = 13 \text{ or } n = 10 \\[1em] \Rightarrow n = \frac{13}{3} \text{ or } n = 10.\end{aligned}$$

Since, no. of terms cannot be in fraction.

∴ n = 10.

Hence, no. of terms = 10.

(b) By formula,

Last term (l) = a + (n - 1)d

= 20 + (10 - 1) × (-3)

= 20 + 9 × (-3)

= 20 - 27

= -7.

Hence, last term = -7.

(a) P(2, -3) ⇒ P'(2, 3)

Since, sign of y-coordinate changes.

∴ L₁ = x-axis

Hence, point P becomes P' on reflection in x-axis.

(b) Origin remains invariant on reflection along both the axes.

L₂ is perpendicular to L₁.

It means L₂ is perpendicular to x-axis and passes through (0, 0).

∴ L₂ is y-axis.

P'(2, 3) on reflection in y-axis becomes P''(-2, 3).

Hence, coordinates of P'' = (-2, 3).

(c) x-axis and y-axis intersect at origin.

Hence, coordinates of intersection of lines L₁ and L₂ is (0, 0).

(d) P(2, -3) on reflection in origin becomes P'''(-2, 3).

Since, P'' and P''' have similar co-ordinates.

Hence, P'' and P''' are coincident points.

Let coordinates of C be (0, b) and D be (a, 0).

Given,

AC : AD = 1 : 4

Let AC = x and AD = 4x.

From figure,

⇒ AD = AC + CD

⇒ 4x = x + CD

⇒ CD = 4x - x = 3x.

AC : CD = 1 : 3.

By section formula,

$$\begin{aligned}\Rightarrow (x, y) = \Big(\frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \frac{m_1y_2 + m_2y_1}{m_1 + m_2}\Big) \\[1em] \Rightarrow (0, b) = \Big(\frac{1 \times a + 3 \times -2}{1 + 3}, \frac{1 \times 0 + 3 \times 6}{1 + 3}\Big) \\[1em] \Rightarrow (0, b) = \Big(\frac{a - 6}{4}, \frac{0 + 18}{4}\Big) \\[1em] \Rightarrow (0, b) = \Big(\frac{a - 6}{4}, \frac{18}{4}\Big) \\[1em] \Rightarrow \frac{a - 6}{4} = 0 \text{ and } b = \frac{18}{4} \\[1em] \Rightarrow a - 6 = 0 \text{ and } b = \frac{9}{2} \\[1em] \Rightarrow a = 6 \text{ and } b = \frac{9}{2}.\end{aligned}$$

C = (0 , b) = $\Big(0, \frac{9}{2}\Big)$ and D = (6, 0).

Given, D is the mid-point of CB.

$$\begin{aligned}\therefore (6, 0) = \Big(\frac{0 + p}{2}, \frac{\frac{9}{2} + q}{2}\Big) \\[1em] \Rightarrow (6, 0) = \Big(\frac{p}{2}, \frac{9 + 2q}{2 \times 2}\Big) \\[1em] \Rightarrow (6, 0) = \Big(\frac{p}{2}, \frac{9 + 2q}{4}\Big) \\[1em] \Rightarrow \frac{p}{2} = 6 \text{ and } \frac{9 + 2q}{4} = 0 \\[1em] \Rightarrow p = 12 \text{ and } 9 + 2q = 0 \\[1em] \Rightarrow p = 12 \text{ and } 2q = -9 \\[1em] \Rightarrow p = 12 \text{ and } q = -\frac{9}{2} \\[1em] \Rightarrow B = (p, q) = \Big(12, -\frac{9}{2}\Big).\end{aligned}$$

Hence, coordinates of B = $\Big(12, -\frac{9}{2}\Big), C = \Big(0, \frac{9}{2}\Big)$ and D = (6, 0).

Given,

Equation : x + 2y = 4

⇒ 2y = -x + 4

⇒ y = $-\frac{1}{2}x + 4$

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

⇒ m = $-\frac{1}{2}$

Let slope of line perpendicular to line x + 2y = 4 be m₁.

We know that,

Product of slope of perpendicular lines = -1.

$$\begin{aligned}\Rightarrow m_1 \times -\frac{1}{2} = -1 \\[1em] \Rightarrow m_1 = \frac{-1 \times 2}{-1} = 2.\end{aligned}$$

Substituting values we get :

⇒ y = mx + c

⇒ y = 2x + 2.

Simultaneously solving equation :

⇒ x + 2y = 4 .........(1)

⇒ y = 2x + 2 .......(2)

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

⇒ x + 2(2x + 2) = 4

⇒ x + 4x + 4 = 4

⇒ 5x = 4 - 4

⇒ 5x = 0

⇒ x = $\frac{0}{5}$ = 0.

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

⇒ y = 2(0) + 2 = 2.

Hence, equation of required line is y = 2x + 2 and point of intersection = (0, 2).

Since, ∆ ABC is enlarged along the side BC to ∆ AB'C'.

∴ ∆ ABC and ∆ AB'C' are similar triangles.

(a) We know that,

Ratio of corresponding sides of similar triangles are proportional.

$$\begin{aligned}\therefore \frac{AB}{AB'} = \frac{BC}{B'C'} \\[1em] \Rightarrow \frac{AB}{AB'} = \frac{3}{5}\end{aligned}$$

Let AB = 3x and AB' = 5x

From figure,

⇒ AB' = AB + BB'

⇒ 5x = 3x + BB'

⇒ BB' = 5x - 3x = 2x.

⇒ AB : BB' = 3x : 2x = 3 : 2.

Hence, AB : BB' = 3 : 2.

(b) As,

$$\begin{aligned}\Rightarrow \frac{AB}{BB'} = \frac{3}{2} \\[1em] \Rightarrow \frac{AB}{4} = \frac{3}{2} \\[1em] \Rightarrow AB = \frac{3}{2} \times 4 \\[1em] \Rightarrow AB = 6 \text{ cm}.\end{aligned}$$

Hence, AB = 6 cm.

(c) Since, ∆ ABC is enlarged along the side BC to ∆ AB'C'.

∴ BC || B'C'

In ∆ ABC and ∆ AB'C',

⇒ ∠BAC = ∠B'AC' (Common angle)

⇒ ∠ABC = ∠AB'C' (Corresponding angle are equal)

∴ ∆ ABC ~ ∆ AB'C' (By A.A. axiom).

Hence, proved that ∆ ABC ~ ∆ AB'C'.

(d) We know that,

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

$$\begin{aligned}\therefore \frac{\text{Area of ∆ ABC}}{\text{Area of ∆ AB'C'}} = \Big(\frac{AB}{AB'}\Big)^2 \\[1em] \Rightarrow \frac{\text{Area of ∆ ABC}}{\text{Area of ∆ AB'C'}} = \Big(\frac{3}{5}\Big)^2 \\[1em] \Rightarrow \frac{\text{Area of ∆ ABC}}{\text{Area of ∆ AB'C'}} = \frac{9}{25}.\end{aligned}$$

Let area of ∆ ABC = 9a and area of ∆ AB'C' = 25a.

Area of quadrilateral BB'C'C = Area of ∆ AB'C' - Area of ∆ ABC = 25a - 9a = 16a.

∴ Area of ∆ ABC : Area of quadrilateral BB'C'C = 9a : 16a = 9 : 16.

Hence, area of ∆ ABC : area of quadrilateral BB'C'C = 9 : 16.

(a) Scale factor is greater than 1 as volume of model is greater than the original eye.

(b)

(i) By formula,

$$\begin{aligned}\Rightarrow k^3 = \frac{\text{Volume of model}}{\text{Volume of original human eye}} \\[1em] \Rightarrow k^3 = \frac{1404}{6.5} \\[1em] \Rightarrow k^3 = \frac{216}{1} \\[1em] \Rightarrow k^3 = 6^3 \\[1em] \Rightarrow k = 6.\end{aligned}$$

Hence, scale factor (k) = 6.

(ii) By formula,

$$\begin{aligned}\Rightarrow k = \frac{\text{Radius of model}}{\text{Radius of human eye}} \\[1em] \Rightarrow 6 = \frac{7.2}{\text{Radius of human eye}} \\[1em] \Rightarrow \text{Radius of human eye} = \frac{7.2}{6} = 1.2 \text{ cm}\end{aligned}$$

Diameter of human eye = 1.2 × 2 = 2.4 cm

Hence, diameter of human eye = 2.4 cm.

(iii) By formula,

$$\begin{aligned}\Rightarrow k^2 = \frac{\text{Surface area of model}}{\text{Surface area of human eye}} \\[1em] \Rightarrow 6^2 = \frac{651.6}{\text{Surface area of human eye}} \\[1em] \Rightarrow \text{Surface area of human eye} = \frac{651.6}{6^2} \\[1em] \Rightarrow \text{Surface area of human eye} = \frac{651.6}{36} = 18.1 \text{ cm}^2.\end{aligned}$$

Hence, surface area of human eye = 18.1 cm².

(a) We know that,

The radius of a circle and tangent are perpendicular at the point of contact.

∴ ∠PQO = 90°

In right-angled triangle PQO,

⇒ PO² = PQ² + OQ²

⇒ 25² = PQ² + 7²

⇒ 625 = PQ² + 49

⇒ PQ² = 625 - 49

⇒ PQ² = 576

⇒ PQ = $\sqrt{576}$ = 24 cm.

In △ POQ,

⇒ tan θ = $\frac{OQ}{PQ}$

⇒ tan θ = $\frac{7}{24}$ ...............(1)

From figure,

⇒ ∠PAS = 90°

⇒ PA = OP - OA = 25 - 7 = 18 cm.

⇒ tan θ = $\frac{AS}{PA}$

⇒ tan θ = $\frac{AS}{18}$ ...........(2)

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

$$\begin{aligned}\Rightarrow \frac{7}{24} = \frac{AS}{18} \\[1em] \Rightarrow AS = \frac{7}{24} \times 18 \\[1em] \Rightarrow AS = \frac{7 \times 3}{4} = \frac{21}{4} \text{ cm}.\end{aligned}$$

From figure,

⇒ ST = 2 × AS = $2 \times \frac{21}{4} = \frac{21}{2}$ = 10.5 cm

Hence, ST = 10.5 cm.

(b) From equation (1),

⇒ tan θ = $\frac{7}{24}$

⇒ tan θ = 0.292

⇒ tan θ = tan 16° 16'

⇒ θ = 16° 16'.

Hence, θ = 16° 16'.

(c) In △ POQ,

⇒ ∠POQ + ∠PQO + ∠QPO = 180°

⇒ ∠POQ + 90° + 16° 16' = 180°

⇒ ∠POQ + 106° 16' = 180°

⇒ ∠POQ = 180° - 106° 16' = 73° 44' = 74°.

We know that,

Tangent from an external point to the circle are equal in length.

∴ PQ = PR.

Also,

OR = OQ (Both equal to radius of circle)

In △ POQ and △ POR,

⇒ PQ = PR (Proved above)

⇒ OQ = OR (Radius of same circle)

⇒ PO = PO (Common side)

∴ △ POQ ≅ △ POR (By S.S.S. axiom)

We know that,

Corresponding parts of congruent triangles are equal.

∴ ∠POR = ∠POQ = 74°

From figure,

⇒ ∠QOR = ∠POR + ∠POQ = 74° + 74° = 148°.

We know that,

The angle subtended by an arc of a circle at its center is twice the angle it subtends anywhere on the circle's circumference.

⇒ ∠QOR = 2∠QUR

⇒ ∠QUR = $\frac{∠QOR}{2} = \frac{148°}{2}$ = 74°.

Hence, ∠QUR = 74°.

Steps of construction :

1. Draw a line segment BC = 4.5 cm

2. Construct XB such that ∠XBC = 135°.

3. Cut AB = 5.5 cm from XB.

4. Join and measure AC.

5. Draw PQ and RS the perpendicular bisectors of BC and AB.

6. Mark point O the intersection of PQ and RS.

7. With O as center and radius as OA, draw a circle touching the vertices A, B and C.

On measuring AC = 9.1 cm and radius = 6.5 cm.

(a) Let radius of smaller and larger cone be r and R, respectively.

Let height of smaller and larger cone be h and H, respectively.

Let slant height of smaller and larger cone be l and L respectively.

Given,

Ratio of their slant heights is 2 : 1.

∴ l : L = 2 : 1.

Given,

Curved surface area of a right circular cone is half of the other right circular cone.

∴ πrl = $\frac{1}{2}πRL$

⇒ rl = $\frac{1}{2}RL$

⇒ $\frac{r}{R} = \frac{L}{2l} = \frac{1}{2} \times \frac{1}{2}$

⇒ $\frac{r}{R} = \frac{1}{4}$

⇒ r : R = 1 : 4

Hence, ratio of the radii = 1 : 4.

(b) Let volume of cone with smaller curved surface area be v and that with larger curved surface area be V.

Given,

Volumes are in the ratio 3 : 1.

∴ v : V = 3 : 1

$$\begin{aligned}\Rightarrow \frac{\text{Volume of larger CSA cone}}{\text{Volume of smaller CSA cone}} = \frac{V}{v} \\[1em] \Rightarrow \frac{\frac{1}{3}πR^2H}{\frac{1}{3}πr^2h} = \frac{1}{3} \\[1em] \Rightarrow \frac{4^2}{1^2} \times \frac{H}{h} = \frac{1}{3} \\[1em] \Rightarrow \frac{16}{1} \times \frac{H}{h} = \frac{1}{3} \\[1em] \Rightarrow \frac{H}{h} = \frac{1}{3 \times 16} \\[1em] \Rightarrow \frac{H}{h} = \frac{1}{48} \\[1em] \Rightarrow \frac{h}{H} = \frac{48}{1}.\end{aligned}$$

Hence, ratio of heights = 48 : 1.

Let AC be the plank and BC be the height from where the drum is rolled down.

⇒ sin θ = $\frac{\text{Perpendicular}}{\text{Hypotenuse}}$

⇒ sin 10° = $\frac{BC}{AC}$

⇒ 0.1736 = $\frac{BC}{10}$

⇒ BC = 1.736 ~ 1.74 m

Hence, height from which the cylindrical drum was rolled down = 1.74 m.

By formula,

Median = $\frac{\frac{n}{2}\text{ th term} + \Big(\frac{n}{2} + 1\Big)\text{ th term}}{2}$

Substituting values we get :

$$\begin{aligned}\Rightarrow 6 = \frac{\frac{12}{2}\text{ th term} + \Big(\frac{12}{2} + 1\Big)\text{ th term}}{2} \\[1em] \Rightarrow 6 \times 2 = \text{6th term + 7th term} \\[1em] \Rightarrow 12 = x + x + 2 \\[1em] \Rightarrow 12 = 2x + 2 \\[1em] \Rightarrow 12 - 2 = 2x \\[1em] \Rightarrow 2x = 10 \\[1em] \Rightarrow x = \frac{10}{2} = 5.\end{aligned}$$

Numbers :

2, 3, 3, 3, 4, 5, 7, 8, p, q, 8, 9.

Since, mode = 8, it means 8 occurs for the most times in the series.

Since, 3 occurs 3 times in the series,

∴ 8 must occur for atleast 4 times in order to be the mode.

∴ p = q = 8.

Hence, x = 5, p = 8 and q = 8.

Given,

5(2 - 4x) > 18 - 16x > 22 - 20x

Solving L.H.S. of the above equation :

⇒ 5(2 - 4x) > 18 - 16x

⇒ 10 - 20x > 18 - 16x

⇒ -16x + 20x < 10 - 18

⇒ 4x < -8

⇒ x < $-\frac{8}{4}$

⇒ x < -2 ...........(1)

Solving R.H.S. of the above equation :

⇒ 18 - 16x > 22 - 20x

⇒ 20x - 16x > 22 - 18

⇒ 4x > 4

⇒ x > $\frac{4}{4}$

⇒ x > 1 ............(2)

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

Solution set : {x : x < -2 or x > 1, x ∈ R}

Hence, solution set = {x : x < -2 or x > 1, x ∈ R}.

Given,

x³ + 2x² – ax + b leaves a remainder -6 when divided by x + 1.

⇒ x + 1 = 0

⇒ x = -1.

Let f(x) = x³ + 2x² – ax + b, then f(-1) = -6.

⇒ f(-1) = -6

⇒ (-1)³ + 2(-1)² - a(-1) + b = -6

⇒ -1 + 2(1) + a + b = -6

⇒ -1 + 2 + a + b = -6

⇒ a + b + 1 = -6

⇒ a + b = -6 - 1

⇒ a + b = -7 ........(1)

Given ,

x - 2 is a factor of f(x).

⇒ x - 2 = 0

⇒ x = 2.

f(2) = 0

⇒ 2³ + 2(2)² – 2a + b = 0

⇒ 8 + 2(4) - 2a + b = 0

⇒ 8 + 8 - 2a + b = 0

⇒ 16 - 2a + b = 0

⇒ 2a - b = 16 ......(2)

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

⇒ a + b + 2a - b = -7 + 16

⇒ 3a = 9

⇒ a = $\frac{9}{3}$ = 3.

Substituting value of a = 3 in equation (1), we get :

⇒ 3 + b = -7

⇒ b = -7 - 3 = -10.

Hence, a = 3 and b = -10.

(a) Order of matrix A = 2 × 2, Order of matrix C = 1 × 2

The product AC is not possible as the no. of columns in A is not equal to the no. of rows in C.

Hence, product AC is not possible.

(b) Given,

$$\Rightarrow X = AB + B^2 - DC \\[1em] \Rightarrow X = \begin{bmatrix*}[r] -1 & 3 \\ 2 & 0 \end{bmatrix*}\begin{bmatrix*}[r] 1 & -2 \\ 0 & 3 \end{bmatrix*} + \begin{bmatrix*}[r] 1 & -2 \\ 0 & 3 \end{bmatrix*}\begin{bmatrix*}[r] 1 & -2 \\ 0 & 3 \end{bmatrix*} - \begin{bmatrix*}[r] 4 \\ 1 \end{bmatrix*}\begin{bmatrix*}[r] 1 & -4 \end{bmatrix*} \\[1em] \Rightarrow X = \begin{bmatrix*}[r] -1 \times 1 + 3 \times 0 & -1 \times -2 + 3 \times 3 \\ 2 \times 1 + 0 \times 0 & 2 \times -2 + 0 \times 3 \end{bmatrix*} + \begin{bmatrix*}[r] 1 \times 1 + -2 \times 0 & 1 \times -2 + -2 \times 3 \\ 0 \times 1 + 3 \times 0 & 0 \times -2 + 3 \times 3 \end{bmatrix*} - \begin{bmatrix*}[r] 4 \times 1 & 4 \times -4 \\ 1 \times 1 & 1 \times -4 \end{bmatrix*} \\[1em] \Rightarrow X = \begin{bmatrix*}[r] -1 + 0 & 2 + 9 \\ 2 + 0 & -4 + 0 \end{bmatrix*} + \begin{bmatrix*}[r] 1 + 0 & -2 - 6 \\ 0 + 0 & 0 + 9 \end{bmatrix*} - \begin{bmatrix*}[r] 4 & -16 \\ 1 & -4 \end{bmatrix*} \\[1em] \Rightarrow X = \begin{bmatrix*}[r] -1 & 11 \\ 2 & -4 \end{bmatrix*} + \begin{bmatrix*}[r] 1 & -8 \\ 0 & 9 \end{bmatrix*} - \begin{bmatrix*}[r] 4 & -16 \\ 1 & -4 \end{bmatrix*} \\[1em] \Rightarrow X = \begin{bmatrix*}[r] -1 + 1 - 4 & 11 + (-8) - (-16) \\ 2 + 0 - 1 & (-4) + 9 - (-4) \end{bmatrix*} \\[1em] \Rightarrow X = \begin{bmatrix*}[r] -4 & 11 - 8 + 16 \\ 1 & -4 + 9 + 4 \end{bmatrix*} \\[1em] \Rightarrow X = \begin{bmatrix*}[r] -4 & 19 \\ 1 & 9 \end{bmatrix*}$$

Hence, X = $$\begin{bmatrix*}[r] -4 & 19 \\ 1 & 9 \end{bmatrix*}.$$

(a) In Δ AFE,

⇒ ∠AFE = ∠ACB = 105° (Corresponding angle are equal)

⇒ ∠EAF = $\frac{∠BAD}{2} = \frac{70°}{2}$ = 35° (AC is the bisector of ∠BAD)

By angle sum property of triangle,

⇒ ∠AFE + ∠EAF + ∠AEF = 180°

⇒ 105° + 35° + ∠AEF = 180°

⇒ 140° + ∠AEF = 180°

⇒ ∠AEF = 180° - 140° = 40°.

∠AGF = ∠ADC = 40° (Corresponding angles are equal)

In Δ AGF,

⇒ ∠GAF = $\frac{∠BAD}{2} = \frac{70°}{2}$ = 35° (AC is the bisector of ∠BAD)

In Δ AEF and Δ AGF,

⇒ ∠EAF = ∠GAF = 35° (AC being the bisector)

⇒ ∠AEF = ∠AGF = 40° (Proved above)

∴ Δ AEF ~ Δ AGF (By A.A. axiom)

Hence, proved that Δ AEF ~ Δ AGF.

(b) In Δ ACD,

⇒ ∠CAD = 35°

⇒ ∠ADC = 40°

By angle sum property of triangle,

⇒ ∠CAD + ∠ADC + ∠DCA = 180°

⇒ 35° + 40° + ∠DCA = 180°

⇒ ∠DCA + 75° = 180°

⇒ ∠DCA = 180° - 75° = 105°.

In Δ ACD and Δ ACB,

⇒ ∠ACB = ∠ACD = 105° (Proved above)

⇒ ∠BAC = ∠DAC = 35° (Proved above)

∴ Δ ACD ~ Δ ACB (By A.A. axiom)

(i) Given,

⇒ AE : EB = 2 : 3

Let AE = 2x and EB = 3x.

∴ AB = AE + EB = 2x + 3x = 5x.

We know that,

Ratio of corresponding sides are proportional.

$$\begin{aligned}\therefore \frac{AE}{AG} = \frac{AB}{AD} \\[1em] \Rightarrow \frac{AG}{AD} = \frac{AE}{AB} \\[1em] \Rightarrow \frac{AG}{AD} = \frac{2x}{5x} = \frac{2}{5}.\end{aligned}$$

Hence, AG : AD = 2 : 5.

(ii) In △ ABC and △ ADC,

⇒ ∠BAC = ∠DAC (Both equal to 35°)

⇒ ∠ACB = ∠ACD (Both equal to 105°)

⇒ ∠ABC = ∠ADC (Both equal to 40°)

∴ △ ABC and △ ADC are congruent.

We know that,

Area of congruent triangles are equal.

Let area of △ ABC and area of △ ADC = x.

∴ Area of △ ABC : Area of △ ADC = x : x = 1 : 1.

Hence, area of △ ABC : area of △ ADC = 1 : 1.

(iii) From figure,

Area of quadrilateral ABCD = Area of △ ABC + Area of △ ADC = x + x = 2x.

∴ Area of quadrilateral ABCD : Area of △ ACB = 2x : x = 2 : 1.

Hence, area of quadrilateral ABCD : area of △ ACB = 2 : 1.

Join AT and BT.

(a) We know that,

Angle in a semicircle is a right angle.

∴ ∠CBT = 90°.

Hence, ∠CBT = 90°.

(b) In cyclic quadrilateral ATBC,

⇒ ∠CBT + ∠CAT = 180° (∵ Sum of opposite angles of a cyclic quadrilateral = 180°)

⇒ 90° + ∠CAT = 180°

⇒ ∠CAT = 180° - 90° = 90°.

In △ABC,

⇒ ∠CBA + ∠CAB + ∠ACB = 180° [By angle sum property of triangle]

⇒ 70° + ∠CAB + 50° = 180°

⇒ ∠CAB + 120° = 180°

⇒ ∠CAB = 180° - 120°

⇒ ∠CAB = 60°.

From figure,

∠BAT = ∠CAT - ∠CAB = 90° - 60° = 30°.

Hence, ∠BAT = 30°.

(c) From figure,

∠BTX = ∠BAT = 30° [Angle in same segment are equal]

∠PBT = ∠CBT - ∠CBA = 90° - 70° = 20°.

Hence, ∠PBT = 20°.

(d) Since, ∠PTB and ∠BTX are linear pairs.

⇒ ∠PTB = 180° - ∠BTX = 180° - 30° = 150°.

In △PBT,

⇒ ∠PBT + ∠PTB + ∠APT = 180° [By angle sum property of triangle]

⇒ 20° + 150° + ∠APT = 180°

⇒ ∠APT + 170° = 180°

⇒ ∠APT = 180° - 170°

⇒ ∠APT = 10°.

Hence, ∠APT = 10°.

Steps of construction :

1. Draw a line segment BC = 8 cm.

2. Draw BX perpendicular to BC.

3. With C as center and radius = 10 cm, draw an arc cutting BX at A.

4. Join AB and AC.

5. Draw AW, BY and CZ the angle bisectors of A, B and C respectively.

6. Mark the point of intersection as I.

7. Draw IR perpendicular to side AC.

8. With I as center and radius IR draw a circle, which is the required incircle.

9. Mark points P, Q and R where the incircle touches the sides AB, BC, and AC of the triangle respectively.

10. Measure CQ and CR.

From figure,

⇒ ∠IRC = ∠IQC = 90° (The radius from the center of the circle to the point of tangency is perpendicular to the tangent line.)

⇒ ∠RCI = ∠QCI = $\frac{∠C}{2}$ (As CZ is angle bisector)

In △ IRC,

⇒ ∠RIC = 180° - ∠RCI - ∠IRC [∵ Sum of ∠'s in a Δ = 180°]

⇒ ∠RIC = 180° - $\frac{∠C}{2}$ - 90°

⇒ ∠RIC = 90° - $\frac{∠C}{2}$ ............(1)

In △ IQC,

⇒ ∠QIC = 180° - ∠IQC - ∠ICQ [∵ Sum of ∠'s in a Δ = 180°]

⇒ ∠QIC = 180° - 90° - $\frac{∠C}{2}$

⇒ ∠QIC = 90° - $\frac{∠C}{2}$ ............(2)

Adding equations (1) and (2), we get :

⇒ ∠RIC + ∠QIC = 90° - $\frac{∠C}{2}$ + 90° - $\frac{∠C}{2}$

⇒ ∠RIQ = 180° - ∠C

⇒ ∠RIQ = 180° - ∠RCQ

⇒ ∠RIQ + ∠RCQ = 180°.

Hence, ∠RIQ + ∠QCR = 180°.

Difference between two consecutive class marks = 475 - 425 = 50.

Adjustment factor = $\frac{\text{Class width}}{2} = \frac{50}{2}$ = 25.

Lower class limit = Class mark - Adjustment factor

Upper class limit = Class mark + Adjustment factor

Class	Frequency
400-450	6
450-500	12
500-550	15
550-600	17
600-650	7
650-700	3

Steps of construction :

1. Take 2 cm along x-axis = ₹50 and 2 cm along y-axis = 5 workers.

2. Construct rectangles corresponding to the given data.

3. In highest rectangle, draw two st. lines AD and BC from corners of the rectangles on either side of the highest rectangle to the opposite corners of the highest rectangle.

4. Let K be the point of intersection of AD and BC. Through K, draw a vertical line to meet the x-axis at L. The abscissa of the point L represents 557.50.

Hence, mode = ₹ 557.50

Total no. of cards = 13 + 13 + 13 = 39 cards.

(a) P(that card drawn is a green card) = $\frac{\text{No. of green cards}}{\text{Total no. of cards}} = \frac{13}{39} = \frac{1}{3}$.

Hence, probability that card drawn is a green card = $\frac{1}{3}$.

(b) Even number cards are : 2, 4, 6, 8, 10, 12.

So, there are 6 even cards of each set.

∴ 18 cards.

P(that card drawn is a card with even number)

= $\frac{\text{No. of even number cards}}{\text{Total no. of cards}} = \frac{18}{39} = \frac{6}{13}$.

Hence, probability that card drawn is a card with even number = $\frac{6}{13}$.

(c) Multiples of three : 3, 6, 9, 12.

So, there are 4 cards of each red and black colour.

∴ 8 cards.

P(that card drawn is a red or black card with multiple of three)

= $\frac{\text{No. of red or black card with multiple of 3}}{\text{Total no. of cards}} = \frac{8}{39}$.

Hence, probability that card drawn is a red or black card with multiple of three = $\frac{8}{39}$.