Instructions:

• Please check that this question paper consists of 11 pages.
• Code number given on the right hand side of the question paper should be written on the title page  of the answer book by the candidate.
• Please check that this question paper consists of 30 questions.
• Please write down the serial number of the question before attempting it.
• 15 minutes times has been allotted to read this question paper. The question paper will be distributed at 10:15 am. From 10:15 am to 10:30 am, the students will read the question paper only and will not write any answer  on the answer book during this period.

SUMMATIVE ASSESSMENT – II

MATHEMATICS

Time allowed: 3 hours                                                             Maximum Marks: 80

General Instructions:

(i) All questions are compulsory

(ii) The question paper consists of 30 questions divided into four sections – A, B, C and D

(iii) Section A consists of 6 questions of 1 mark each. Section B consists of 6 questions of 2 marks each. Section C consists of 10 questions of 3 marks each. Section D consists of 8 questions of 4 marks each.

(iv) There is no overall choice. However, an internal choice has been provided in two questions of 1 mark, two questions of 2 marks, four questions of 3 marks each and three questions of 4 marks each. You have to attempt only one of the alternative in all such questions.

(v) Use of calculator is not permitted.

SECTION – A

Question number 1 to 6 carry 1 mark each.

Question 1: For what values of $k$ does the quadratic equation $4x^2 - 12x - k = 0$ have no real roots ?

Given equation: $4x^2 - 12x - k = 0$

Comparing it with $ax^2 + bx + c = 0, a = 4, b = -12$ and $c = -k$

When $b^2 - 4ac < 0$, there are no real roots.

$\therefore (-12)^2 - 4 (4)(-k) <0$

$\Rightarrow 144 + 16k < 0$

$\Rightarrow k < -9$

Therefore for $k < -9$, there will be no real roots.

$\\$

Question 2: Find the distance between the points $(a, b)$ and $(- a, - b)$.

Given points $(a, b)$ and $(-a, -b)$

Therefore the distance between the two points $= \sqrt{(-a-a)^2 + (-b-b)^2} = \sqrt{4a^2 + 4b^2} = 2 \sqrt{a^2 + b^2}$

$\\$

Question 3: Find a rational number between $\sqrt{2}$ and $\sqrt{7}$.

Or

Write the number of zeroes in the end of a number whose prime factorization is $2^2 \times 5^3 \times 3^2 \times 17$.

$\sqrt{2} = 1.412$ and $\sqrt{7} = 2.645$

Therefore $2.5$ is between $\sqrt{2}$ and $\sqrt{7}$

$\therefore 2.5 =$ $\frac{25}{10}$ $=$ $\frac{5}{2}$ is a rational number between $\sqrt{2}$ and $\sqrt{7}$

Or

The given expression:

$2^2 \times 5^3 \times 3^3 \times 17 = (2 \times 5) \times (2 \times 5) \times 5 \times 3^3 \times 17$

$= 10 \times 10 \times 5 \times 3^3 \times 17$

We know that zeros in an expression are a result of number of $10's$ in it.

Hence the expression we can see that there will be $2$ zeros in the given expression.

$\\$

Question 4: Let $\triangle ABC \sim \triangle DEF$ and their areas be respectively, $64 \ cm^2$ and $121 \ cm^2$. If $EF = 15.4$ cm, find $BC$.

Since $\triangle ABC \sim \triangle DEF$

$\frac{ar(\triangle ABC)}{ar(\triangle DEF)}$ $= \Big($ $\frac{BC}{EF}$ $\Big)^2$

$\Rightarrow \frac{64}{121}$ $= \Big($ $\frac{BC}{EF}$ $\Big)^2$

$\Rightarrow \frac{BC}{EF}$ $=$ $\frac{8}{11}$

$\Rightarrow BC =$ $\frac{8}{11}$ $\times 15.4 = 11.2$ cm

$\\$

Question 5: Evaluate: $\frac{\tan 65^o}{\cot 25^o}$

Or

Express $( \sin 67^o + \cos 75^o)$ in terms of trigonometric ratios of the angle between $0^o$ and $45^o$.

$\frac{\tan 65^o}{\cot 25^o}$ $=$ $\frac{\tan 65^o}{\cot (90^o-65^o)}$ $=$ $\frac{\tan 65^o}{\tan 65^o}$ $= 1$

Or

$\sin 67^o + \cos 75^o = \sin (90^o-23^o) + \cos (90^o-15^o) = \cos 23^o + \sin 15^o$

$\\$

Question 6: Find the number of terms in the A.P. : $18, 15$ $\frac{1}{2}$ $, 13, \cdots , - 47$

Given AP: $18, 15\frac{1}{2}, 13, \ldots , -47$

$a = 18, d = 15.5 - 18 = -2.5$

Let $-47$ be the $n^{th}$ term

$T_n = a + (n-1) d$

$\Rightarrow -47 = 18 + (n-1) (-2.5)$

$\Rightarrow -65 = -2.5 (n-1)$

$\Rightarrow 26 = n - 1$

$\Rightarrow n = 27$

Therefore $-47$ is the $27^{th}$ term in the given AP

$\\$

Section – B

Question number 7 to 12 carry 2 mark each.

Question 7: A bag contains $15$ balls, out of which some are white and the others are black. If the probability of drawing a black ball at random from the bag is $\frac{2}{3}$, then find how many white balls are there in the bag.

Total number of balls $= 15$

Let the number of black balls $= x$

$\therefore \frac{x}{15}$ $=$ $\frac{2}{3}$ $\Rightarrow x = 10$

Hence the number of white balls $= 15 - 10 = 5$

$\\$

Question 8: A card is drawn at random from a pack of $52$ playing cards. Find the probability of drawing a card which is neither a spade nor a king.

Total number of cards $= 52$

Number of space cards $= 13$ (includes the king of space)

No of kings other than king of spade $= 3$

Therefore Probability $=$ $\frac{52 - (13+3)}{52}$ $=$ $\frac{36}{52}$ $=$ $\frac{9}{13}$

$\\$

Question 9: Find the solution of the pair of equations :

$\frac{3}{x}$ $+$ $\frac{8}{y}$ $= - 1;$ $\frac{1}{x}$ $-$ $\frac{2}{y}$ $= 2 , x, y \neq 0$

Or

Find the value(s) of $k$ for which the pair of equations $kx+2y = 3, 3x+6y = 10$ has a unique solution.

Given equation:

$\frac{3}{x}$ $+$ $\frac{8}{y}$ $= - 1$

$\frac{1}{x}$ $-$ $\frac{2}{y}$ $= 2$

Let $\frac{1}{x}$ $= u$ and $\frac{1}{y}$ $= v$

Therefore

$3u + 8 v = -1$   … … … … … iii)

$u - 2v = 2$   … … … … … iv)

Multiplying iv) by iii) and subtracting it from iii)

$\hspace*{1cm} 3u + 8 v = -1 \\ (-1) \underline {\hspace*{0.2cm}3u - 6v = 6} \\ \hspace*{1.5cm} 14v = -7$

$v=$ $\frac{-1}{2}$

Therefore $u = 2v + 2 = 2 ($ $\frac{-1}{2}$ $) + 2 = -1 +2 = 1$

Hence $\frac{1}{x}$ $= 1 \Rightarrow x = 1$

Also $\frac{1}{y}$ $=$ $\frac{-1}{2}$ $\Rightarrow y = -2$

Or

Given equations:

$kx + 2y = 3$

$3x + 6y = 10$

If the system of equations are $ax_1 + b_1y + c_1 = 0$ and $ax_2 + b_2y + c_2 = 0$ and they have unique solution then it satisfy the following:

$\frac{a_1}{a_2}$ $\neq$ $\frac{b_1}{b_2}$ $\neq$ $\frac{c_1}{c_2}$

$\Rightarrow$ $\frac{k}{3}$ $\neq$ $\frac{2}{6}$ $\neq$ $\frac{-3}{-10}$

From first two terms

$k \neq 1$

From First and Third term

$k \neq$ $\frac{9}{10}$

$\therefore k \neq 1$ or $k \neq$ $\frac{9}{10}$ for the equations to have a unique solution.

$\\$

Question 10: How many multiples of $4$ lie between $10$ and $205$ ?

Or

Determine the A.P. whose third term is $16$ and $7^{th}$ term exceeds the $5^{th}$ term by $12$.

Since the multiples of $4$ lie between $10$ and $205$ are $12, 16, 20, \ldots , 204$

$\therefore a = 12, d = 16-12 = 4$

$T_n = a + (n-1)d$

$\Rightarrow 204 = 12 + (n-1) (4)$

$\Rightarrow 192 = (n-1)(4)$

$\Rightarrow 48 = n - 1$

$\Rightarrow n = 49$

Therefore there are $49$ multiples of $4$ between $10$ and $205$

Or

Let for the AP, first term $= a$ and common difference $= d$

Given $T_3 = 16$

$\therefore a + (3 - 1) d = 16$

$\Rightarrow a + 2d = 16$   … … … … … i)

$T_7 = a + 6d$

$T_5 = a + 4d$

Given $T_7 = T_5 + 12$

$\Rightarrow a + 6d = a + 4d + 12$

$\Rightarrow 2d = 12$

$\Rightarrow d = 6$

Therefore form i) we get $a = 16 - 2( 6) = 4$

Therefore the AP is $4, 10, 16, 22, \ldots$

$\\$

Question 11: Use Euclid’s division algorithm to find the HCF of $255$ and $867$.

$255 ) \overline{867} ( 3 \\ \hspace*{0.7cm} \underline{765} \\ \hspace*{0.8cm} 102 ) \overline{255} ( 2 \\ \hspace*{1.5cm} \underline{204} \\ \hspace*{1.7cm} 51 ) \overline{102} ( 2 \\ \hspace*{2.25cm} \underline{102} \\ \hspace*{2.5cm} \times$

According to Eculid’s division theorem, any positive number can be expressed as $a = bq+r$ where $q$ is the quotient, $b$ is the divisor and $r$ is the remainder and $0 \leq r < b$

$\therefore 867 = 51 \times 17 + 0$

So HCF of $867$ and $255$ is $51$

$\\$

Question 12: The point $R$ divides the line segment $AB$, where $A( 4, 0)$ and $B(0, 6)$ such that $AR =$ $\frac{3}{4}$ $AB$. Find the coordinates of $R$.

Given points $A(-4, 0)$ and $B(0, 6)$

$AR =$ $\frac{3}{4}$ $AB$

$\Rightarrow$ $\frac{AR}{AB}$ $=$ $\frac{3}{4}$

$\Rightarrow$ $\frac{AR}{AR + RB}$ $=$ $\frac{3}{4}$

$\Rightarrow 4AR = 3AR + 3RB$

$\Rightarrow$ $\frac{AR}{RB}$ $=$ $\frac{3}{1}$

Using sections formula, $R(x, y) = \Big($ $\frac{3 \times 0 + 1 \times (-4)}{3+1}$ $,$ $\frac{3 \times 6 + 1 \times 0}{3+1}$ $\Big) = \Big( -1,$ $\frac{9}{2}$ $\Big)$

$\\$

Section – C

Question number 13 to 22 carry 3 mark each.

Question 13: Prove that : $( \sin \theta + 1 + \cos \theta) ( \sin \theta - 1 + \cos \theta) . \sec \theta \ \mathrm{cosec} \theta = 2$

Or

$\sqrt{\frac{\sec \theta -1}{\sec \theta +1}}$ $+$ $\sqrt{\frac{\sec \theta +1}{\sec \theta -1}}$ $= 2 \ \mathrm{cosec} \theta$

LHS $= ( \sin \theta + 1 + \cos \theta) ( \sin \theta - 1 + \cos \theta) . \sec \theta \ \mathrm{cosec} \theta$

$= [ (\sin \theta + \cos \theta )^2 -1 ] \sec \theta \ \mathrm{cosec} \theta$

$= [ \sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta -1] \sec \theta \ \mathrm{cosec} \theta$

$= [ 1 + 2 \sin \theta \cos \theta -1 ] \sec \theta \ \mathrm{cosec} \theta$

$= 2 \sin \theta \cos \theta \sec \theta \ \mathrm{cosec} \theta$

$= 2\$ $\frac{\sin \theta \cos \theta }{\sin \theta \cos \theta }$

$= 2$. Hence proved.

Or

LHS $=$ $\sqrt{\frac{\sec \theta -1}{\sec \theta +1}}$ $+$ $\sqrt{\frac{\sec \theta +1}{\sec \theta -1}}$

$=$ $\frac{(\sqrt{\sec \theta - 1})^2 + (\sqrt{\sec \theta + 1})^2}{\sqrt{(\sec \theta + 1)(\sec \theta - 1)}}$

$=$ $\frac{\sec \theta - 1 + \sec \theta + 1}{\sqrt{\sec^2 \theta -1}}$

$=$ $\frac{2}{\cos \theta } \sqrt{\frac{1}{\frac{1}{\cos^2 \theta }-1}}$

$=$ $\frac{2}{\cos \theta } \sqrt{\frac{\cos^2 \theta}{1- \cos^2 \theta}}$

$= 2 \$ $\frac{\cos \theta}{\cos \theta } \sqrt{\frac{1}{\sin^2 \theta}}$

$=$ $\frac{2}{\sin \theta}$

$= 2 \ \mathrm{cosec} \theta$

hence proved.

$\\$

Question 14: In what ratio does the point $P(- 4, y)$ divide the line segment joining the points $A(- 6, 10)$ and $B(3, - 8)$ ? Hence find the value of $y$.

Or

Find the value of $p$ for which the points $(- 5, 1), (1, p)$ and $(4, - 2)$ are collinear.

Let $P$ divides $AB$ in the ratio of $k:1$

$\therefore P(-4, y) = \Big($ $\frac{3k-6}{k+1}$ $,$ $\frac{-8k + 10}{k + 1}$ $\Big)$

$\Rightarrow -4 =$ $\frac{3k-6}{k+1}$

$\Rightarrow -4k - 4 = 3k - 6$

$\Rightarrow 7k = 2$

$\Rightarrow k =$ $\frac{2}{7}$

Therefore ratio is $2:7$

$\therefore y =$ $\frac{-8k + 10}{k + 1}$

$=$ $\frac{-8(\frac{2}{7}) + 10}{(\frac{2}{7}) + 1}$

$=$ $\frac{-16+70}{2 + 7}$

$=$ $\frac{54}{9}$ $= 6$

Or

Given point $A(-5, 1), B(1, p)$ and $C(4, -2)$

If the points are collinear are of the $\triangle ABC = 0$

$\therefore 0 =$ $\frac{1}{2}$ $[ x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) ]$

$\Rightarrow -5(p+2) + 1(-2-1) + 4(1-p) = 0$

$\Rightarrow -5p - 10 - 3 + 4 - 4p = 0$

$\Rightarrow -9p - 9 = 0$

$\Rightarrow p = -1$

$\\$

Question 15: $ABC$ is a right triangle in which $\angle B = 90^o$. If $AB = 8$ cm and $BC = 6$ cm, find the diameter of the circle inscribed in the triangle.

$AC = \sqrt{AB^2 + BC^2} = \sqrt{8^2 + 6^2} = 10$ cm

Area of  $\triangle ABC =$ Area of $\triangle AOB +$ Area of  $\triangle BOC +$ Area of $\triangle AOC$

$\frac{1}{2}$ $\times BC \times AB =$ $\frac{1}{2}$ $\times AB \times r +$ $\frac{1}{2}$ $\times BC \times r =$ $\frac{1}{2}$ $\times AC \times r$

$\Rightarrow 6 \times 8 = 8r + 6r + 10r$

$\Rightarrow 48 = 24r$

$\Rightarrow r = 2 cm$

$\\$

Question 16: In Figure 1, $BL$ and $CM$ are medians of a $\triangle ABC$ right-angled at $A$. Prove that $4 (BL^2 + CM^2) = 5 BC^2$.

Or

Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.

Given $BL$ and $CM$ are medians

$\Rightarrow BM = MA \Rightarrow AM =$ $\frac{1}{2}$ $AB$

$\Rightarrow CL = AL \Rightarrow AL =$ $\frac{1}{2}$ $AC$

From

$\triangle ABC, BC^2 = AC^2 + AB^2$   … … … … … i)

$\triangle BAL, BL^2 = AL^2 + AB^2$   … … … … … ii)

$\triangle ACM, CM^2 = AC^2 + AM^2$   … … … … … iii)

$BL^2 + CM^2 = AL^2 + AB^2 + AC^2 + AM^2$

$BL^2 + CM^2 = \Big($ $\frac{1}{2}$ $AC \Big)^2 + AB^2 + AC^2 + \Big($ $\frac{1}{2}$ $AB \Big)^2$

$BL^2 + CM^2 =$ $\frac{1}{4}$ $AC^2 + AB^2 + AC^2 +$ $\frac{1}{4}$ $AB^2$

$BL^2 + CM^2 =$ $\frac{1}{4}$ $(AC^2 + AB^2) + AB^2 + AC^2$

$4 (BL^2 + CM^2) = 5 BC^2$

Hence Proved.

Or

Given: $ABCD$ is a rhombus

To prove: $AB^2 + BC^2 + CD^2 + DA^2 = AC^2 + BD^2$

We know that the diagonals of a rhombus bisect at right angles

Therefore from $\triangle AOB, AB^2 = AO^2 + OB^2$

$\Rightarrow AB^2 = \Big($ $\frac{AC}{2}$ $\Big)^2 + \Big($ $\frac{BD}{2}$ $\Big)^2$

$\Rightarrow 4 AB^2 = AC^2 + BD^2$

Since $AB = BC = CD = DA$

$\Rightarrow AB^2 + BC^2 + CD^2 + DA^2 = AC^2 + BD^2$

Hence proved.

$\\$

Question 17: In Figure 2, two concentric circles with centre $O$, have radii $21$ cm and $42$ cm. If $\angle AOB = 60^o$, find the area of the shaded region.

Area of shaded area $=$ $\frac{300}{360}$ $\pi \times (42)^2 -$ $\frac{300}{360}$ $\pi \times (21)^2$

$=$ $\frac{5}{6}$ $\pi (42^2 - 21^2)$

$=$ $\frac{5}{6}$ $\times$ $\frac{22}{7}$ $\times 1323$

$= 3465 \ cm^2$

$\\$

Question 18: Calculate the mode of the following distribution :

 Class: 10-15 15-20 20-25 25-30 30-35 Frequency: 4 7 20 8 1

$\\$

Question 19: A cone of height $24$ cm and radius of base $6$ cm is made up of modelling clay. A child reshapes it in the form of a sphere. Find the radius of the sphere and hence find the surface area of this sphere.

Or

A farmer connects a pipe of internal diameter $20$ cm from a canal into a cylindrical tank in his field which is $10$ m in diameter and $2$ m deep. If water flows through the pipe at the rate of $3$ km/hr, in how much time will the tank be filled ?

Volume of cone $=$ $\frac{1}{3}$ $\pi r^2 h =$ $\frac{1}{3}$ $\pi (6)^2 (24) = 288 \pi \ cm^3$

Let the radius of the sphere be $R$

$\therefore$ $\frac{4}{3}$ $\pi R^3 = 288 \pi$

$\Rightarrow R^3 = 3 \times 72 = 216$

$\Rightarrow R = 6$ cm

Surface are of the sphere $= 4 \pi R^2 = 4 \times$ $\frac{22}{7}$ $\times 36 = 452.57 \ cm^2$

Or

Rate of flow of water $=$ $\frac{3 \times 1000}{60}$ $\frac{m}{min}$ $= 50$ $\frac{m}{min}$

Volume of tank $= \pi r^2 h = \pi (5)^2 \times 2 = 50 \pi m^3$

Volume of water flowing per minute $= \pi (0.1)^2 \times 50$ $\frac{m^3}{min}$ $= 0.5 \pi$ $\frac{m^3}{min}$

Therefore the time taken to fill the tank $=$ $\frac{50 \pi m^3}{0.5 \pi \frac{m^3}{min} }$ $= 100$ minutes

$\\$

Question 20: Prove that $2 + 3 \sqrt{3}$ is an irrational number when it is given that $\sqrt{3}$ is an irrational number.

Let us assume that $2 +3 \sqrt{3}$ is a rational number.

The rational number is in the form of $\frac{p}{q}$

$\therefore 2 + 3\sqrt{3} =$ $\frac{p}{q}$

$3\sqrt{3} =$ $\frac{p}{q}$ $- 2$

$3\sqrt{3} =$ $\frac{p-2q}{q}$

$\sqrt{3} =$ $\frac{p-2q}{3q}$

$\therefore \sqrt{3}$ is a rational no as $\frac{p-2q}{3q}$ is a rational number

But it is given that $\sqrt{3}$ is an irrational number which contradicts our initial assumption.

Hence $2 + 3 \sqrt{3}$ is an irrational number.

$\\$

Question 21: Sum of the areas of two squares is $157 \ m^2$. If the sum of their perimeters is $68$ m, find the sides of the two squares.

Let the sides of the two squares be $a$ and $b$ respectively

$\therefore a^2 + b^2 = 157$    … … … … … i)

Also $4a + 4 b = 68$

$\Rightarrow a + b = 17$    … … … … … ii)

From i) $(a+b)^2 - 2ab = 157$

$17^2 - 2ab = 157$

$289 - 157 = 2ab$

$ab = 66$    … … … … … iii)

From ii) $b = 17 - a$

$\therefore a (17-a) = 66$

$17a - a^2 -= 66$

$a^2 - 17a + 66 = 0$

$(a-11) (a-6) =0$

$a =11$ or $a = 6$

When $a = 11, b = 17 - 11 = 6$

When $a = 6, b = 17 - 6 = 11$

Hence the sides of the two squares are $6$ units and $11$ units

$\\$

Question 22: Find the quadratic polynomial, sum and product of whose zeroes are $-1$ and $- 20$ respectively. Also find the zeroes of the polynomial so obtained.

Sum of zeros: $\alpha + \beta = 1$

Product of zeros: $\alpha \beta = -12$

The quadratic polynomial is of the form:

$x^2 -$ (sum of zeros) $x +$ (product of zeros) $= 0$

$x^2 - x - 12 = 0$

$\Rightarrow \alpha + \beta = 1 \Rightarrow \beta = (1-\alpha)$

$\alpha \beta = -12$

$\Rightarrow \alpha (1 - \alpha ) = -12$

$\Rightarrow \alpha - \alpha^2 = -12$

$\Rightarrow \alpha ^2 - \alpha - 12 = 0$

$\Rightarrow (\alpha -4)(\alpha +3) = 0$

$\Rightarrow \alpha = 4$ or $-3$

When $\alpha = 4, \beta = 1-4 = -3$

When $\alpha = -3, \beta = 1 - (-3) = 4$

Therefore zeros are $(4, -3)$

$\\$

Section – D

Question number 23 to 30 carry 4 mark each.

Question 23: A plane left $30$ minutes later than the scheduled time and in order to reach its destination $1500$ km away on time, it has to increase its speed by $250$ km/hr from its usual speed. Find the usual speed of the plane.

Or

Find the dimensions of a rectangular park whose perimeter is $60$ m and area $200 \ m^2$.

Let the speed of the plane $= x$ $\frac{km}{hr}$

New speed of the plane $= (x + 250)$ $\frac{km}{hr}$

$\therefore$ $\frac{1500}{x}$ $-$ $\frac{1500}{x+250}$ $=$ $\frac{30}{60}$

$\frac{1500x + 1500 \times 250 - 1500 x}{x(x+250)}$ $=$ $\frac{1}{2}$

$1500 \times 250 \times 2 = x(x+250)$

$x^2 + 250 x = 750000$

$x^2 + 250x - 750000=0$

$x^2 +1000x - 750x - 750000 = 0$

$x(x+1000) - 750(x+1000) = 0$

$(x+1000)(x-750)=0$

$x = 750$ or $-1000$ ( not possible as speed cannot be negative)

$\therefore x = 750$ $\frac{km}{hr}$

Or

Let length $= l$ and breadth $= b$

$\therefore lb = 200$   … … … … … i)

$2(l+b) = 60$

$l+b = 30$   … … … … … ii)

From ii) $b = (30-l)$

Substituting in i)

$l(30-l) = 200$

$30l - l^2 = 200$

$l^2 - 30l + 200 = 0$

$l^2 - 20l - 10l +200 = 0$

$l(l-20) - 10(l-20) = 0$

$(l-10)(l-20) = 0$

When $l = 10$ m, $b = 30 - 10 = 20$ m

When $l = 20$ m, $b = 30 -20 = 10$ m

Hence the dimensions are $10$ m by $20$ m

$\\$

Question 24: Find the value of $x$, when in the A.P. given below $2 + 6 + 10 + \cdots + x = 1800$.

First term of AP $= a = 2$

Common difference $= d = 6-2 = 4$

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

$\frac{n}{2}$ $[ 2 \times 2 + (n-1)(4)] = 1800$

$n [ 2 + (n-1) 2 ] = 1800$

$n [ 2n + 2 - 2] = 1800$

$2n^2 = 1800$

$n^2 = 900$

$\Rightarrow n = 30$ ( total number of terms)

$x$ is the $n^{th}$ term

$\therefore x = a + (n-1) d$

$x = 2 + (30-1) (4)$

$x = 118$

$\\$

Question 25: If $\sec \theta + \tan \theta = m$ , show that $\frac{m^2 -1 }{m^2 + 1}$ $= \sin \theta$

$\sec \theta + \tan \theta = m$

$\Rightarrow \sec^2 \theta + \tan^2 \theta + 2 \sec \theta \ \tan \theta = m^2$

$\Rightarrow \sec^2 \theta + \tan^2 \theta + 2 \sec \theta \ \tan \theta - 1 = m^2 - 1$

$\because \tan^2 \theta = \sec^2 \theta - 1$

$\Rightarrow 2 \tan^2 \theta + 2 \sec \theta \ \tan \theta = m^2 - 1$

$\Rightarrow 2 \tan \theta (\tan \theta + \sec \theta ) = m^2 -1$   … … … … … i)

Similarly,

$\sec^2 \theta + \tan^2 \theta + 2 \sec \theta \ \tan \theta + 1 = m^2 + 1$

$\because \sec^2 \theta = \tan^2 \theta + 1$

$\Rightarrow 2 \sec^2 \theta + 2 \sec \theta \ \tan \theta = m^2 + 1$

$\Rightarrow 2 \sec \theta (\tan \theta + \sec \theta ) = m^2 +1$  … … … … … ii)

Dividing i) by ii) we get

$\frac{m^2-1}{m^2+1}$ $=$ $\frac{2 \tan \theta (\tan \theta + \sec \theta) }{2 \sec \theta (\tan \theta + \sec \theta)}$

$\Rightarrow$ $\frac{m^2-1}{m^2+1}$ $= \sin \theta$

$\\$

Question 26: In $\triangle ABC$ (Figure 3), $AD \perp BC$. Prove that $AC^2 = AB^2 + BC^2 - 2BC \times BD$

Given $AD \perp BC$

In $\triangle ABD \ \ \ \ \ AB^2 = AD^2 + BD^2 \Rightarrow AD^2 = AB^2 -BD^2$   … … … … … i)

In $\triangle ADC \ \ \ \ \ AC^2 = AD^2 + DC^2 \Rightarrow AD^2 = AC^2 -DC^2$   … … … … … ii)

Frim i) and ii)

$AB^2 - BD^2 = AC^2 -DC^2$

$\Rightarrow AC^2 = AB^2 + DC^2 - BD^2$

$\Rightarrow AC^2 = AB^2 + (DC - BD)(DC + BD)$

$\Rightarrow AC^2 = AB^2 + BC(DC-BD)$

$\Rightarrow AC^2 = AB^2 + BC [(BC-BD)-BD]$

$\Rightarrow AC^2 = AB^2 + BC[BC-2BD]$

$\Rightarrow AC^2=AB^2 + BC^2 - 2 BC \times BD$

Hence Proved.

$\\$

Question 27: A moving boat is observed from the top of a $150$ m high cliff moving away from the cliff. The angle of depression of the boat changes from $60^o$ to $45^o$ in $2$ minutes. Find the speed of the boat in m/min.

Or

There are two poles, one each on either bank of a river just opposite to each other. One pole is $60$ m high. From the top of this pole, the angle of depression of the top and foot of the other pole are $30^o$ and $60^o$ respectively. Find the width of the river and height of the other pole.

$\frac{AB}{DB} = \tan 45^o = 1 \Rightarrow DB = AB = 150$ m

$\frac{AB}{BC}$ $= \tan 60^o = \sqrt{3} \Rightarrow BC =$ $\frac{150}{\sqrt{3}}$ $= 50 \sqrt{3}$

$\therefore DC = DB - CB = (150 - 50\sqrt{3} )$ m

Therefore speed $=$ $\frac{distance}{time}$ $=$ $\frac{150 - 50\sqrt{3}}{2}$ $\frac{m}{min}$

$= 75 - 25(1.732)$ $\frac{m}{min}$

$= 31.7$ $\frac{m}{min}$

Or

$\frac{AB}{CB}$ $= \tan 60^o = \sqrt{3}$

$\Rightarrow CB =$ $\frac{60}{\sqrt{3}}$ $= 20\sqrt{3}$ m (width of the river)

$\frac{AE}{20\sqrt{3}}$ $= \tan 30^o \Rightarrow AE =$ $\frac{20\sqrt{3}}{\sqrt{3}}$ $= 20$ m

Hence the height of the pole is $60-20 = 40$ m

$\\$

Question 28: Construct a triangle with sides $5$ cm, $6$ cm and $7$ cm and then another triangle whose sides are $\frac{3}{5}$ of the corresponding sides of the first triangle.

$\\$

Question 29: Calculate the mean of the following frequency distribution :

 Class: 10-30 30-50 50-70 70-90 90-110 110-130 Frequency: 5 8 12 20 3 2

Or

The following table gives production yield in kg per hectare of wheat of 100 farms of a village :

 Production Yield (kg/hectare): 40-45 45-50 50-55 55-60 60-65 65-70 Number of Farms: 4 6 16 20 30 24

Change the distribution to a ‘more than type’ distribution, and draw its ogive.

 Class Interval Frequency ( $f_i$$f_i$) $x_i$$x_i$ $f_ix_i$$f_ix_i$ 10-30 5 20 100 30-50 8 40 320 50-70 12 60 720 70-90 20 80 1600 90-110 3 100 300 110-130 2 120 2400 $\Sigma f_i = 50$$\Sigma f_i = 50$ $\Sigma f_ix_i = 3280$$\Sigma f_ix_i = 3280$

Mean $=$ $\frac{\Sigma f_ix_i}{\Sigma f_i}$ $=$ $\frac{3280}{50}$ $= 65.6$

Or

 Production Yield Cumulative Frequency More than 40 100 More than 45 96 More than 50 90 More than 55 74 More than 60 54 More than 65 24

$\\$

Question 30: A container opened at the top and made up of a metal sheet, is in the form of a frustum of a cone of height $16$  cm with radii of its lower and upper ends as $8$ cm and $20$ cm respectively. Find the cost of milk which can completely fill the container, at the rate of Rs. $50$ per liter. Also find the cost of metal sheet used to make the container, if it costs Rs. $10$ per $100 \ cm^2$. (Take $\pi = 3.14$ )

$h = 16$ cm, $r_1 = 8$ cm, $r_2 = 20$ cm

$l = \sqrt{h^2 + (r_2-r_1)^2} = \sqrt{254 + 144} = 20$ cm

Volume of frustum $=$ $\frac{1}{3} \pi h ({r_1}^2 + {r_2}^2 + r_1 r_2 )$

$=$ $\frac{1}{3}$ $\times 3.14 \times 16 (64+400+160)$

$= 10449.94 \ cm^3$

$= 10.4492$ liters

Cost of milk $= 50 \times 10.45 = 522.5$ Rs

Surface Area of the container $= \pi l (r_1+r_2) + \pi r_1^2$

$= 3.14 \times 20 (8 + 20+ 3.14 (8)^2$

$= 1758.4 + 200.96$

$= 1959.36 \ cm^2$

Cost of metal sheet $=$ $\frac{10 Rs}{100 cm^2}$ $= 0.1$ $\frac{Rs}{cm^2}$

Cost of sheet metal $= 1959.36 \ cm^2 \times 0.1$ $\frac{Rs}{cm^2}$ $= 195.936$ Rs

$\\$