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SUMMATIVE ASSESSMENT – II

MATHEMATICS

Time allowed: 3 hours                                                             Maximum Marks: 90

General Instructions:

(i) All questions are compulsory

(ii) The question paper consists of 31 questions divided into four sections – A, B, C and D

(iii) Section A consists of 4 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 11 questions of 4 marks each.

(iv) Use of calculator is not permitted.

SECTION – A

Question number 1 to 6 carry 1 mark each.

Question 1: If the quadratic equation $px^2 - 2 \sqrt{5} px + 15 = 0$ has two equal roots, then find the value of $p$.

The given quadratic equation is $px^2 - 2 \sqrt{5} px + 15 = 0$

For real roots, Discriminant $= 0$

$\Rightarrow b^2 - 4ac = 0$

$\Rightarrow ( - 2 \sqrt{5})^2 - 4 (p) \times 15 = 0$

$\Rightarrow 20p^2 - 60 p = 0$

$\Rightarrow 20p(p - 3) = 0$

$\Rightarrow p = 0 or p = 3$

$p = 0$ is not possible as whole equation will become $0$ then. Hence $p = 3$.

$\\$

Question 2: In Figure 1, a tower $AB$ is $20$ m high and $BC$, its shadow on the ground, is $20 \sqrt{3}$ m long. Find the Sun’s altitude.

Let $\displaystyle AB$ be the tower and $\displaystyle BC$ be it’s shadow.

$\displaystyle \tan \theta = \frac{AB}{BC}$

$\displaystyle \Rightarrow \tan \theta = \frac{20}{20\sqrt{3}}$

$\displaystyle \Rightarrow \tan \theta = \frac{1}{\sqrt{3}}$

$\displaystyle \Rightarrow \tan \theta = \tan 30^{\circ}$

$\displaystyle \Rightarrow \theta = 30^{\circ}$

Therefore sunset is at an altitude of $\displaystyle 30^{\circ}$.

$\displaystyle \\$

Question 3: Two different dice are tossed together. Find the probability that the product of the two numbers on the top of the dice is $\displaystyle 6$.

Two dice are tossed. Therefore

$\displaystyle S = \begin{bmatrix} (1, 1), (1, 2), (1, 3), (1 4), (1, 5), (1, 6) \\ (2, 1), (2, 2), (2, 3), (2 4), (2, 5), (2, 6) \\ (3, 1), (3, 2), (3, 3), (3 4), (3, 5), (3, 6) \\ (4, 1), (4, 2), (4, 3), (4 4), (4, 5), (4, 6) \\ (5, 1), (5, 2), (5, 3), (5 4), (5, 5), (5, 6) \\ (6, 1), (6, 2), (6, 3), (6 4), (6, 5), (6, 6) \end{bmatrix}$

Therefore the total number of outcomes when two dices are tossed $\displaystyle = 6 \times 6 = 36$

Favorable events of getting the product as $\displaystyle 6$ are $\displaystyle ( 1, 6), (2, 3), (3, 2), (6, 1) = 4$

$\displaystyle P \text{( getting products as 6)} = \frac{4}{36} = \frac{1}{9}$

$\displaystyle \\$

Question 4: In Figure 2, $\displaystyle PQ$ is a chord of a circle with centre $\displaystyle O$ and $\displaystyle PT$ is a tangent. If $\displaystyle \angle QPT = 60^{\circ}$, find $\displaystyle \angle PRQ$.

Given: $\displaystyle PT$ is a tangent so $\displaystyle \angle OPT=90^{\circ}$

$\displaystyle \angle QPT =60^{\circ}$

$\displaystyle \angle OPQ = \angle OPT- \angle QPT = 90^{\circ} -60^{\circ} = 30^{\circ}$

$\displaystyle \angle OPQ+\angle OQP+ \angle POQ=180^{\circ}$ (sum of the angles in a triangle is $\displaystyle 180^{\circ}$)

$\displaystyle \angle POQ=180^{\circ} -30^{\circ} -30^{\circ}$

$\displaystyle \angle POQ=120^{\circ}$

$\displaystyle \angle OPQ= \angle OQP = 30^{\circ}$

$\displaystyle \angle POQ= 120^{\circ} \text{ also } \angle PRQ= \frac{1}{2} \text{ reflex } \angle POQ$

$\displaystyle \angle PRQ= \frac{1}{2} (360-120) = \frac{240}{2} = 120^{\circ}$

$\displaystyle \\$

Section – B

Question number 5 to 10 carry 2 mark each.

Question 5: In Figure 3, two tangents $\displaystyle RQ$ and $\displaystyle RP$ are drawn from an external point $\displaystyle R$ to the circle with centre $\displaystyle O$. If $\displaystyle \angle PRQ = 120^{\circ}$, then prove that $\displaystyle OR = PR + RQ$.

$\displaystyle \angle OPR = \angle OQR = 90^{\circ}$ … … … … … i)

And in $\displaystyle \triangle OPR$ and $\displaystyle \triangle OQR$

$\displaystyle \angle OPR = \angle OQR = 90^{\circ}$ (from equation i)

$\displaystyle OP = OQ$ (Radii of same circle)

$\displaystyle OR = OR$ (common side)

$\displaystyle \triangle OPR \cong \triangle OQR$ ( by RHS criterion)

So, $\displaystyle RP = RQ$ … … … … … ii)

And $\displaystyle \angle ORP = \angle ORQ$ … … … … … iii)

$\displaystyle \angle PRQ = \angle ORP + \angle ORQ$

Substitute $\displaystyle \angle PQR = 120^{\circ}$ (given) And from equation iii) we get

$\displaystyle \angle ORP + \angle ORP = 120^{\circ} \Rightarrow 2 \angle ORP = 120^{\circ} \Rightarrow \angle ORP = 60^{\circ}$

$\displaystyle \text{And we know } \cos 0^{\circ} = \frac{Adjacent}{Hypotenuse}$

So in $\displaystyle \triangle OPR$ we get

$\displaystyle \cos \angle ORP = \frac{PR}{OR}$

$\displaystyle \cos 60^{\circ} = \frac{PR}{OR}$

$\displaystyle \frac{1}{2} = \frac{PR}{OR} \text{ we know } \cos 60^{\circ} = \frac{1}{2}$

$\displaystyle OR = 2PR$

$\displaystyle OR = PR + PR$ (substitute value from equation ii) we get)

$\displaystyle OR = PR + RQ$

$\displaystyle \\$

Question 6: In Figure 4, a triangle $\displaystyle ABC$ is drawn to circumscribe a circle of radius $\displaystyle 3 \text{ cm }$, such that the segments $\displaystyle BD$ and $\displaystyle DC$ are respectively of lengths $\displaystyle 6 \text{ cm }$ and $\displaystyle 9 \text{ cm }$. If the area of $\displaystyle \triangle ABC$ is $\displaystyle 54 \ cm^2$, then find the lengths of sides $\displaystyle AB$ and $\displaystyle AC$.

Given $\displaystyle OD = 3 \text{ cm }$

Construction: Join $\displaystyle OA, OB$ and $\displaystyle OC$

Proof: Area of the $\displaystyle \triangle ABC =$ area of $\displaystyle \triangle OBC +$ area of $\displaystyle \triangle OAC +$ area of $\displaystyle \triangle OBA$

Let $\displaystyle AE = AF = x$

$\displaystyle BD=6 \text{ cm }$, $\displaystyle BE = 6 \text{ cm }$ (equal tangents)

$\displaystyle DC=9 \text{ cm }$, $\displaystyle CF=9 \text{ cm }$ (equal tangents)

$\displaystyle \text{ Area of } \triangle OBC = \frac{1}{2} \times 15 \times 3 = \frac{45}{2} cm^2$

$\displaystyle \text{ Area of } \triangle OAC = \frac{1}{2} \times (x+9) \times 3 = \frac{3(x+9)}{2} cm^2$

$\displaystyle \text{ Area of } \triangle OAB = \frac{1}{2} \times (x+6) \times 3 = \frac{3(x+6)}{2} cm^2$

Therefore

$\displaystyle 54 = \frac{45}{2} + \frac{3(x+9)}{2} + \frac{3(x+6)}{2}$

$\displaystyle \Rightarrow 54 = \frac{3}{2} \Big( (x+9) + (x+6) + 15 \Big)$

$\displaystyle \Rightarrow 54 = \frac{3}{2} (2x+30)$

$\displaystyle \Rightarrow 36 = 2x+ 30$

$\displaystyle \Rightarrow x = 3$

Therefore sides are $\displaystyle 15 \text{ cm }$, $\displaystyle 9 \text{ cm }$, $\displaystyle 12 \text{ cm }$

$\displaystyle \\$

Question 7: Solve the following quadratic equation for $\displaystyle x : 4x^2 + 4bx - (a^2 - b^2) = 0$

$\displaystyle 4x^2 +4bx -(a^2 -b^2) = 0$

$\displaystyle 4x^2 +4bx -a^2 + b^2 =0$

$\displaystyle (2x)^2 + 2.(2x).b + b^2 -a^2 =0$

$\displaystyle (2x + b)^2 -a^2 = 0$

Use formula: $\displaystyle a^2-b^2 = (a -b)(a + b)$

$\displaystyle (2x + b -a)(2x + b +a) =0$

$\displaystyle x = \frac{a - b}{2} , - \frac{a + b}{2}$

$\displaystyle \\$

Question 8: In an AP, if $\displaystyle S_5 + S_7 = 167$ and $\displaystyle S_10 = 235$, then find the AP, where $\displaystyle S_n$ denotes the sum of its first $\displaystyle n$ terms.

Let the first term be $\displaystyle a$ and the common difference be $\displaystyle d$

$\displaystyle S_5 + S_7 = 167$

$\displaystyle \Rightarrow \frac{5}{2} [ 2a + 4d ] + \frac{7}{2} [ 2a + 6d ] = 167$

$\displaystyle \Rightarrow 5(a+ 2d) + 7 ( a + 3d) = 167$

$\displaystyle \Rightarrow 5a + 10 d + 7 a + 21 d = 167$

$\displaystyle \Rightarrow 12 a + 31 d = 167$ … … … … … i)

$\displaystyle S_{10} = 235$

$\displaystyle \Rightarrow \frac{10}{2} [ 2a + 9d ] = 235$

$\displaystyle \Rightarrow 10a + 45 d = 235$

$\displaystyle \Rightarrow 2a + 9 d = 47$ … … … … … ii)

Solving i) and ii) we get

$\displaystyle \hspace*{1.3cm} 12 a + 31 d = 167 \\ (-) \underline {6 \times [ 2a + 9 d = 47 ] \hspace*{1.3cm}} \\ \hspace*{1.8cm} -23d = - 115$

$\displaystyle \Rightarrow d = 5$

Substituting in ii) we get

$\displaystyle 2a = 47 - 9(5) = 2 \Rightarrow a = 1$

Hence the AP is $\displaystyle 1, 6, 11, 16, \cdots$

$\displaystyle \\$

Question 9: The points $\displaystyle A(4, 7), B(p, 3)$ and $\displaystyle C(7, 3)$ are the vertices of a right triangle, right-angled at $\displaystyle B$. Find the value of $\displaystyle p$.

$\displaystyle A(4,7), B(p,3)$ and $\displaystyle C(7,3)$

$\displaystyle AB^2+BC^2=AC^2$

$\displaystyle AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$\displaystyle \Rightarrow AB=\sqrt{(p-4)^2+(3-7)^2}$

$\displaystyle \Rightarrow AB=\sqrt{(p-4)^2+16}$

$\displaystyle AB^2 = (p-4)^2 + 16$

$\displaystyle BC^2 = (7-p)^2 + ( 3- 3)^2 = (7-p)^2$

$\displaystyle AC^2 = (4-7)^2 + (7-3)^2 = 25$

$\displaystyle \therefore (p-4)^2 + 16 + (7-p)^2 = 25$

$\displaystyle \Rightarrow p^2 + 16 - 8p + 16 + 49 + p^2 - 14 p = 25$

$\displaystyle \Rightarrow 2p^2 - 22 p + 56 = 0$

$\displaystyle \Rightarrow p^2 - 11 p + 28 = 0$

$\displaystyle \Rightarrow (p-7)(p-4) = 0$

$\displaystyle \Rightarrow p = 7$ or $\displaystyle p = 4$

$\displaystyle \\$

Question 10: Find the relation between $\displaystyle x$ and $\displaystyle y$ if the points $\displaystyle A(x, y), B(- 5, 7)$ and $\displaystyle C(- 4, 5)$ are collinear.

It is given that $\displaystyle A(x, y), B(- 5, 7)$ and $\displaystyle C(- 4, 5)$ are collinear.

Therefore the area of $\displaystyle \triangle ABC = 0$

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

$\displaystyle \Rightarrow \frac{1}{2} [ x(7-5) -5(5-y) -4 (y - 7) ] = 0$

$\displaystyle \Rightarrow 2x -25 + 5y - 4 y + 28 = 0$

$\displaystyle \Rightarrow 2x + y + 3 = 0$

$\displaystyle \\$

Section – C

Question number 11 to 20 carry 3 mark each.

Question 11: The $\displaystyle 14^{th}$ term of an AP is twice its $\displaystyle 8^{th}$ term. If its $\displaystyle 6^{th}$ term is $\displaystyle - 8$, then find the sum of its first $\displaystyle 20$ terms

Let $\displaystyle a$ be the first term and $\displaystyle d$ be the common difference

$\displaystyle \therefore T_{14} = a + 13 d$

$\displaystyle T_8= a + 7d$

Given, $\displaystyle T_{14} = 2 T_8$

$\displaystyle \Rightarrow a + 13 d = 2a + 14 d$

$\displaystyle \Rightarrow a + d = 0$ … … … … … i)

$\displaystyle T_6 = a + 5d$

Given $\displaystyle T_6 = -8$

$\displaystyle \therefore a + 5d = -8$ … … … … … ii)

Solving i) and ii) we get

$\displaystyle \hspace*{1.0cm} a + d = 0 \\ \underline{(+) - a - 5d = 8} \\ \hspace*{1.0cm} -4d = 8$

$\displaystyle \Rightarrow d = - 2$

$\displaystyle \text{Now } S_n = \frac{n}{2} [2a + (n-1) d ]$

$\displaystyle \Rightarrow S_{20} = \frac{20}{2} [2(2) + (20-1) (-2) ]$

$\displaystyle \Rightarrow S_{20} = 10(4-38) = - 340$

$\displaystyle \\$

Question 12: Solve for $\displaystyle x : 3 x^2 - 2 \sqrt{2} x - 2 \sqrt{3} = 0$

$\displaystyle \sqrt{3} x^2 - 2 \sqrt{2} x - 2 \sqrt{3} = 0$

On comparing it with $\displaystyle ax^2 + bx + c = 0$ we get

$\displaystyle a = \sqrt{3}, b = - 2 \sqrt{2}$ and $\displaystyle c = - 2 \sqrt{3}$

$\displaystyle \text{We know } x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$\displaystyle = \frac{-(- 2 \sqrt{2}) \pm \sqrt{(- 2 \sqrt{2})^2 - 4(\sqrt{3})( -2 \sqrt{3})}}{2(\sqrt{3})}$

$\displaystyle = \frac{2\sqrt{2} \pm \sqrt{8 + 24}}{2\sqrt{3}}$

$\displaystyle = \frac{2\sqrt{2} \pm \sqrt{32}}{2\sqrt{3}}$

$\displaystyle = \frac{\sqrt{2} \pm \sqrt{8}}{\sqrt{3}}$

$\displaystyle = \frac{\sqrt{2} \pm 2\sqrt{2}}{\sqrt{3}}$

$\displaystyle \text{Hence } x = \frac{\sqrt{2} + 2\sqrt{2}}{\sqrt{3}} = \frac{3\sqrt{2}}{\sqrt{3}} = \sqrt{6}$

$\displaystyle \text{or } x = \frac{\sqrt{2} - 2\sqrt{2}}{\sqrt{3}} = \frac{-\sqrt{2}}{\sqrt{3}} = - \sqrt{\frac{2}{3}}$

$\displaystyle \\$

Question 13: The angle of elevation of an aeroplane from a point A on the ground is $\displaystyle 60^{\circ}$. After a flight of $\displaystyle 15$ seconds, the angle of elevation changes to $\displaystyle 30^{\circ}$. If the aeroplane is flying at a constant height of $\displaystyle 1500 \sqrt{3} \text{ m }$, find the speed of the plane in km/hr.

$\displaystyle \tan 60^{\circ} = \frac{1500\sqrt{3}}{AC}$

$\displaystyle \Rightarrow AC = \frac{1500\sqrt{3}}{\sqrt{3}} = 1500 \text{ m }$

$\displaystyle \tan 30^{\circ} = \frac{1500\sqrt{3}}{AE}$

$\displaystyle \Rightarrow AE = 1500\sqrt{3} \times \sqrt{3} = 4500 \text{ m }$

$\displaystyle \therefore CE = 4500 - 1500 = 3000 \text{ m }$

Time taken to travel $\displaystyle CE = 15$ sec

$\displaystyle \therefore Speed = \frac{3000 \ m}{15 \ sec} = 200 \frac{m}{sec} = 200 \times \frac{3600}{1000} \frac{km}{hr} = 720 \text{ km/hr }$

$\displaystyle \\$

Question 14: If the coordinates of points $\displaystyle A$ and $\displaystyle B$ are $\displaystyle (-2, - 2)$ and $\displaystyle (2, - 4)$ respectively, find the coordinates of $\displaystyle P$ such that $\displaystyle AP = \frac{3}{7} AB$, where $\displaystyle P$ lies on the line segment $\displaystyle AB$.

Let the coordinates of point $\displaystyle P \text{ be } P(x, y)$

$\displaystyle \text{Given } AP = \frac{3}{7} AB$

$\displaystyle \Rightarrow AP = \frac{3}{7} (AP + PB)$

$\displaystyle \Rightarrow 7AP = 3AP + 3 PB$

$\displaystyle \Rightarrow 4AP = 3 P$

$\displaystyle \Rightarrow \frac{AP}{PB} = \frac{3}{4}$

Hence $\displaystyle P$ divides $\displaystyle AB$ in the ratio of $\displaystyle 3:4$

Now using section’s formula

$\displaystyle x = \frac{3(2) + 4(-2)}{3+4} = \frac{6-8}{7} = - \frac{2}{7}$

$\displaystyle y = \frac{3(-4) + 4(-2)}{3+4} = \frac{-12-8}{7} = - \frac{20}{7}$

$\displaystyle \therefore \text{Coordinates of } P = \Big( - \frac{2}{7} , - \frac{20}{7} \Big)$

$\displaystyle \\$

Question 15: The probability of selecting a red ball at random from a jar that contains only red, blue and orange balls is $\displaystyle \frac{1}{4}$ . The probability of selecting a blue ball at random from the same jar is $\displaystyle \frac{1}{3}$ . If the jar contains $\displaystyle 10$ orange balls, find the total number of balls in the jar.

Let the number of red balls be $\displaystyle x$ and number of Blue balls be $\displaystyle y$

No of Orange balls $\displaystyle = 10$

$\displaystyle \therefore$ Total number of balls $\displaystyle = x + y + 10$

$\displaystyle P(Red \ ball) = \frac{1}{4} \text{ and } P(Blue \ ball) = \frac{1}{3}$

$\displaystyle \therefore \frac{x}{x+y+10} = \frac{1}{4}$

$\displaystyle \Rightarrow 4x = x + y + 10$

$\displaystyle \Rightarrow 3x - y = 10$ … … … … … i)

$\displaystyle \text{Similarly, } \frac{y}{x+y+10} = \frac{1}{3}$

$\displaystyle \Rightarrow 3y = x + y + 10$

$\displaystyle \Rightarrow -x + 2y = 10$ … … … … … ii)

Solving i) and ii) we get

$\displaystyle \hspace*{1.8cm} 3x - y = 10 \\ (+) \underline{ 3 \times [ -x + 2y = 10 ] } \\ \hspace*{2.7cm} 5y = 40$

$\displaystyle \Rightarrow y = 8$

$\displaystyle \therefore x = \frac{10+8}{3} = 6$

Hence the total number of balls $\displaystyle = 6 + 8 + 10 = 24$

$\displaystyle \\$

Question 16: Find the area of the minor segment of a circle of radius $\displaystyle 14 \text{ cm }$, when its central angle is $\displaystyle 60^{\circ}$. Also find the area of the corresponding major segment. (Use $\displaystyle \pi = \frac{22}{7}$ )

Radius $\displaystyle (r) = 14 \text{ cm }$ $\displaystyle \theta = 60^{\circ}$

$\displaystyle \text{Therefore Area of } OAPB = \frac{60}{360} \times \pi r^2 =$

$\displaystyle \frac{1}{6} \times \frac{22}{7} \times 14^2 = 102.67 \ cm^2$

Area of $\displaystyle \triangle AOB = \frac{1}{2} \times base \times height$

We draw $\displaystyle OM \perp AB$

$\displaystyle \therefore \angle OMB = \angle OMA = 90^{\circ}$

In $\displaystyle \triangle OMA$ and $\displaystyle \triangle OMB$

$\displaystyle \angle OMA = \triangle OMB = 90^{\circ}$ (by construction)

$\displaystyle OA = OB$ (both are radius of the same circle)

$\displaystyle OM$ is common

$\displaystyle \therefore \triangle OMA \cong \triangle OMB$ (by RHS criterion)

$\displaystyle \Rightarrow \angle AOM = \angle BOM$

$\displaystyle \therefore \angle AOM = \angle BOM = \frac{1}{2} \angle BOA$

$\displaystyle \Rightarrow \angle AOM = \angle BOM = \frac{1}{2} \times 60 = 30^{\circ}$

Also, since $\displaystyle \triangle OMB \cong \triangle OMA$

$\displaystyle \therefore BM = AM$

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

$\displaystyle \Rightarrow AB = 2 BM$ … … … … … i)

In right $\displaystyle \triangle OMB$

$\displaystyle \frac{AM}{AO} = \sin 30^{\circ}$

$\displaystyle \Rightarrow \frac{AM}{14} = \frac{1}{2}$

$\displaystyle \Rightarrow AM = \frac{14}{2}$

$\displaystyle \Rightarrow AB = 2 \times \frac{14}{2} = 14$

Similarly, In right $\displaystyle \triangle OMA$

$\displaystyle \frac{OM}{AO} = \cos 30^{\circ}$

$\displaystyle \Rightarrow \frac{OM}{14} = \frac{\sqrt{3}}{2}$

$\displaystyle \Rightarrow OM = \frac{\sqrt{3}}{2} \times 14$

$\displaystyle \therefore \text{ Area of } \triangle AOB = \frac{1}{2} \times 14 \times \frac{\sqrt{3}}{2} \times 14 = 84.87 \ cm^2$

Therefore area of segment $\displaystyle APB = 102.67-84.87 = 17.80 \ cm^2$

Area of major segment $\displaystyle = \pi r^2 - 17.80 = 3.14 \times 14^2 - 17.80 = 598.2 \ cm^2$

$\displaystyle \\$

Question 17: Due to sudden floods, some welfare associations jointly requested the government to get $\displaystyle 100$ tents fixed immediately and offered to contribute $\displaystyle 50\%$ of the cost. If the lower part of each tent is of the form of a cylinder of diameter $\displaystyle 4.2 \text{ m }$ and height $\displaystyle 4 \text{ m }$ with the conical upper part of same diameter but of height $\displaystyle 2.8 \text{ m }$, and the canvas to be used costs Rs. $\displaystyle 100$ per sq. m, find the amount, the associations will have to pay. What values are shown by these associations ? (Use $\displaystyle \pi = \frac{22}{7}$ )

$\displaystyle l = \sqrt{r^2 + h^2} = \sqrt{(2.1)^2 + (2.8)^2} = \sqrt{12.25} = 3.5 \text{ m }$

Surface Area of canvas $\displaystyle =$ CSA of Cone $\displaystyle +$ CSA of cylinder

$\displaystyle = \pi r l + 2 \pi r H$

$\displaystyle = \frac{22}{7} (2.1) [ 3.5 + 2 \times 4 ]$

$\displaystyle = 22 \times 0.3 \times 11.5 = 75.9 \ m^2$

$\displaystyle \text{Total cost of canvas } = 100 \frac{Rs}{m^2} \times 75.9 \ m^2 = 7590$ Rs.

$\displaystyle \text{Association will pay } = \frac{50}{100} \times 7590 = 3795$ Rs. per tent

Hence for $\displaystyle 100$ tents, Association will pay $\displaystyle = 3795 \times 100 = 379500$ Rs.

This is an example of showing humanity and helpful nature of the association.

$\displaystyle \\$

Question 18: A hemispherical bowl of internal diameter $\displaystyle 36 \text{ cm }$ contains liquid. This liquid is filled into $\displaystyle 72$ cylindrical bottles of diameter $\displaystyle 6 \text{ cm }$. Find the height of the each bottle, if $\displaystyle 10\%$ liquid is wasted in this transfer.

$\displaystyle \text{Radius of hemispherical bowl } = \frac{36}{2} = 18 \text{ cm }$

$\displaystyle \text{Volume of hemispherical bowl } = \frac{2}{3} \pi (18)^3 = 3888 \pi \ cm^3$

$\displaystyle \text{Volume of liquid to be transferred } = 0.9 \times 3888 \pi = 3499.2 \pi \ cm^3$

$\displaystyle \text{Radius of cylindrical bottle } = \frac{6}{2} = 3 \text{ cm }$

Let the height of Bottle $\displaystyle = h$

$\displaystyle \therefore$ Volume of Bottle $\displaystyle = \pi r^2 h = \pi (3)^2 h$

Total volume of $\displaystyle 72$ bottles $\displaystyle = \pi (3)^2 h \times 72 = 648 \pi h \ cm^3$

$\displaystyle \text{Therefore } 648 \pi h = 3499.2 \pi \Rightarrow h = \frac{3499.2}{648} = 5.4 \text{ cm }$

Therefore height of bottle is $\displaystyle 5.4 \text{ cm }$

$\displaystyle \\$

Question 19: A cubical block of side $\displaystyle 10 \text{ cm }$ is surmounted by a hemisphere. What is the largest diameter that the hemisphere can have ? Find the cost of painting the total surface area of the solid so formed, at the rate of Rs. $\displaystyle 5$ per $\displaystyle 100$ sq. cm. [ Use $\displaystyle \pi = 3.14$ ]

The greatest diameter of the hemisphere can be $\displaystyle 10 \text{ cm }$

TSA of solid $\displaystyle =$ SA of cube $\displaystyle +$ CSA of hemisphere $\displaystyle -$ Area of base of hemisphere

$\displaystyle = 6 \times 10^2 + 2 \times \frac{22}{7} \times 5^2 - \frac{22}{7} \times 5^2$

$\displaystyle = 600 + \frac{22}{7} \times 25$

$\displaystyle = 687.57 \ cm^2$

$\displaystyle \text{Cost of painting } = \frac{5 \ Rs}{100 \ cm^2} = 0.05 \frac{Rs}{cm^2}$

$\displaystyle \therefore \text{Cost of painting } = 678.57 \times 0.05 = 33.93$ Rs.

$\displaystyle \\$

Question 20: $\displaystyle 504$ cones, each of diameter $\displaystyle 3.5 \text{ cm }$ and height $\displaystyle 3 \text{ cm }$, are melted and recast into a metallic sphere. Find the diameter of the sphere and hence find its surface area. (Use $\displaystyle \pi = \frac{22}{7}$ )

Diameter of cone $\displaystyle (d) = 3.5 \text{ cm }$

Height of cone $\displaystyle (h)= 3.0 \text{ cm }$

$\displaystyle \text{Volume of come } = \frac{1}{3} \pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times \Big( \frac{3.5}{2} \Big)^2 \times 3 = \frac{77}{8} \ cm^3$

$\displaystyle \therefore$ Volume of $\displaystyle 504$ cones $\displaystyle = 504 \times \frac{77}{8} = 4851 \ cm^3$

Let the radius of the sphere $\displaystyle = r$

$\displaystyle \therefore \frac{4}{3} \pi r^3 = 4851$

$\displaystyle \Rightarrow r^3 = \frac{3}{4} \times \frac{7}{22} \times 4851 = 1157.625 \ cm^3$

$\displaystyle \Rightarrow r = 10.5 \text{ cm }$

$\displaystyle \therefore \text{ Surface area of sphere } = 4 \pi r^2 = 4 \times \frac{22}{7} \times 10.5^2 = 1386 \ cm^2$

$\displaystyle \\$

Section – D

Question number 21 to 31 carry 4 mark each.

Question 21: The diagonal of a rectangular field is $\displaystyle 16 \text{ m }$etres more than the shorter side. If the longer side is $\displaystyle 14 \text{ m }$etres more than the shorter side, then find the lengths of the sides of the field.

Let the shorter side $\displaystyle = x$

$\displaystyle \therefore (x+14)^2 + x^2 = (x+16)^2$

$\displaystyle \Rightarrow x^2 + 196 + 28x + x^2 = x^2 + 256 + 32x$

$\displaystyle \Rightarrow x^2 - 4x - 60 = 0$

$\displaystyle \Rightarrow x^2 - 10x + 6 x - 60 = 0$

$\displaystyle \Rightarrow x(x-10) + 6(x-10) = 0$

$\displaystyle \Rightarrow (x-10)(x+6) = 0$

$\displaystyle \Rightarrow x = 10$ or $\displaystyle x = -6$ (not possible)

Hence the shorter side $\displaystyle = 10 \text{ m }$, diagonal $\displaystyle = 26 \text{ m }$ and Longer side $\displaystyle = 24 \text{ m }$

$\displaystyle \\$

Question 22: Find the $\displaystyle 60^{th}$ term of the AP $\displaystyle 8, 10, 12, \cdots$, if it has a total of $\displaystyle 60$ terms and hence find the sum of its last $\displaystyle 10$ terms

Given AP: $\displaystyle 8, 10, 12, ...$

$\displaystyle \therefore a = 8$ and $\displaystyle d = 2$

$\displaystyle \therefore T_{60} = a + (60-1) d = 8 + 59 \times 2 = 126$

For sum of the last $\displaystyle 10$ terms, lets look at the series backwards

$\displaystyle \therefore a = 126, d = -2 n = 10$

$\displaystyle \text{We know } S_n = \frac{n}{2} [ 2a + (n-1)d ]$

$\displaystyle \Rightarrow S_{10} = \frac{10}{2} [ 2 \times 126 + (10-1)(-2) ]$

$\displaystyle = 5[ 252 - 18] = 1170$

Therefore sum of the last $\displaystyle 10$ terms $\displaystyle = 1170$

$\displaystyle \\$

Question 23: A train travels at a certain average speed for a distance of $\displaystyle 54$ km and then travels a distance of $\displaystyle 63$ km at an average speed of $\displaystyle 6$ km/h more than the first speed. If it takes $\displaystyle 3$ hours to complete the total journey, what is its first speed ?

Total time taken $\displaystyle = 3$ hr

$\displaystyle \therefore \frac{54}{x} + \frac{63}{x+6} = 3$

$\displaystyle \Rightarrow \frac{18}{x} + \frac{21}{x+6} = 1$

$\displaystyle \Rightarrow 18x + 108 + 21 x = x^2 + 6x$

$\displaystyle \Rightarrow 39x + 108 = x^2 + 6x$

$\displaystyle \Rightarrow x^2 - 33x - 108 = x$

$\displaystyle \Rightarrow x^2 - 36x + 3x - 108 = 0$

$\displaystyle \Rightarrow x(x-36) + 3 (x - 36) = 0$

$\displaystyle \Rightarrow (x-36)(x+ 3) = 0$

$\displaystyle \Rightarrow x = 36$ or $\displaystyle x = -3$ (not possible)

$\displaystyle \therefore x = 36 \text{ km/hr }$

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Question 24: Prove that the lengths of the tangents drawn from an external point to a circle are equal.

Given: Circle with center $\displaystyle O$

$\displaystyle PQ$ and $\displaystyle PR$ are tangents touching the circle at $\displaystyle Q$ and $\displaystyle R$ respectively.

To Prove: $\displaystyle PQ = PR$

Construction: Join $\displaystyle OQ, OR$ and $\displaystyle OP$

Proof: Since $\displaystyle PQ$ is a tangent, $\displaystyle \angle PQO = 90^{\circ}$

Similarly, $\displaystyle \angle PRO = 90^{\circ}$

Consider $\displaystyle \triangle POQ$ and $\displaystyle \triangle POR$

$\displaystyle OQ = OR$ (radius of the circle)

$\displaystyle \angle PQO = \angle POR = 90^{\circ}$

$\displaystyle PO$ is common

$\displaystyle \therefore \triangle POQ \cong \triangle POR$ (By RHS criterion)

$\displaystyle \therefore PQ = PR$. Hence proved.

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Question 25: Prove that the tangent drawn at the mid-point of an arc of a circle is parallel to the chord joining the end points of the arc.

Given: $\displaystyle P$ is the mid point of $\displaystyle \widehat{PQ}$, $\displaystyle O$ is the center, $\displaystyle AB$ is the chord

To Prove: $\displaystyle AB \parallel ST$

$\displaystyle \angle OPT = 90^{\circ}$ (tangent)

Since $\displaystyle P$ is the mid point of $\displaystyle \widehat{PQ}$,

$\displaystyle \widehat{AP} = \widehat{PB}$

$\displaystyle \Rightarrow \angle AOP = \angle BOP$

$\displaystyle \Rightarrow \angle AOM = \angle BOM$

In $\displaystyle \triangle AOM$ and $\displaystyle \triangle BOM$

$\displaystyle AO = OB$ (radius)

$\displaystyle OM$ is common

$\displaystyle \angle AOM = \angle BOM$

$\displaystyle \therefore \triangle AOM \cong \triangle BOM$ (By SAS criterion)

$\displaystyle \Rightarrow \angle AMO = \angle BMO$

$\displaystyle \angle AMO + \angle BMO = 180^{\circ}$

$\displaystyle \Rightarrow \angle AMO = \angle BMO = 90^{\circ}$

$\displaystyle \therefore \angle AMO = \angle OPS = 90^{\circ}$

$\displaystyle \angle BMO = \angle OPT = 90^{\circ}$

Corresponding angles are equal, hence $\displaystyle AB \parallel ST$. Hence proved.

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Question 26: Construct a $\displaystyle \triangle ABC$ in which $\displaystyle AB = 6 \text{ cm }$, $\displaystyle \angle A = 30^{\circ}$ and $\displaystyle \angle B = 60^{\circ}$. Construct another $\displaystyle \triangle AB'C'$ similar to $\displaystyle \triangle ABC$ with base $\displaystyle AB' = 8 \text{ cm }$.

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Question 27: At a point $\displaystyle A, 20 \text{ m }$etres above the level of water in a lake, the angle of elevation of a cloud is $\displaystyle 30^{\circ}$. The angle of depression of the reflection of the cloud in the lake, at $\displaystyle A$ is $\displaystyle 60^{\circ}$. Find the distance of the cloud from $\displaystyle A$.

$\displaystyle \tan 30^{\circ} = \frac{h}{AD} \Rightarrow AD = \sqrt{3} h$

$\displaystyle \tan 60^{\circ} = \frac{h+40}{AD}$

$\displaystyle \Rightarrow \sqrt{3} = \frac{h+40}{AD}$

$\displaystyle \Rightarrow 3h = h + 40$

$\displaystyle \Rightarrow 2h = 40 \Rightarrow h = 20 \text{ m }$

$\displaystyle \therefore EC = 20 + h = 20 + 20 = 40 \text{ m }$

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Question 28: A card is drawn at random from a well-shuffled deck of playing cards.

Find the probability that the card drawn is

(i) a card of spade or an ace.

(ii) a black king.

(iii) neither a jack nor a king.

(iv) either a king or a queen.

Total number of cards $\displaystyle = 52$

$\displaystyle \text{i) P( a card of spades or an ace) }= \frac{13 + 3}{52} = \frac{16}{52} = \frac{4}{13}$

$\displaystyle \text{ii) P(a black king) } = \frac{2}{52} = \frac{1}{26}$

$\displaystyle \text{iii) P( neither o jack nor a king) } = \frac{52 - (4 + 4)}{52} = \frac{44}{52} = \frac{11}{13}$

$\displaystyle \text{iv) P (either a king or a queen) } = \frac{4+4}{52} = \frac{8}{52} = \frac{2}{13}$

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Question 29: Find the values of $\displaystyle k$ so that the area of the triangle with vertices $\displaystyle (1, - 1), (- 4, 2k)$ and $\displaystyle (- k, - 5)$ is $\displaystyle 24$ sq. units.

Vertices: $\displaystyle (1, -1), (-4, 2k), (-k, -5)$

$\displaystyle \text{Area of triangle } = \frac{1}{2} [ x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) ]$

$\displaystyle \Rightarrow 24 = \frac{1}{2} [ 1 (2k + 5) + (-4)(-5+1) + (-k)(-1-2k) ]$

$\displaystyle \Rightarrow 48 = 2k + 5 + 16 + k + 2k^2$

$\displaystyle \Rightarrow 2k^2 + 3k - 27 =0$

$\displaystyle \Rightarrow 2k^2 + 9k - 6k - 27 = 0$

$\displaystyle \Rightarrow (2k+9)(k-3) = 0$

$\displaystyle \Rightarrow k = - \frac{9}{2}$ or $\displaystyle k = 3$

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Question 30: In Figure 5, $\displaystyle PQRS$ is a square lawn with side $\displaystyle PQ = 42 \text{ m }$etres. Two circular flower beds are there on the sides $\displaystyle PS$ and $\displaystyle QR$ with center at $\displaystyle O$, the intersection of its diagonals. Find the total area of the two flower beds (shaded parts).

Area of square lawn $\displaystyle = 42 \times 42 = 1764 \ m^2$

Let $\displaystyle OP = OS = x$

$\displaystyle \angle POS = 90^{\circ}$ (diagonals of square are perpendicular to each other)

$\displaystyle \therefore x^2 + x^2 = 42^2 \Rightarrow 2x^2 = 1764 \Rightarrow x^2 = 882$

$\displaystyle \text{Area of sector } POS = \frac{90}{360} \times \pi (x)^2 = \frac{1}{4} \pi x^2 = \frac{1}{4} \times \frac{22}{7} \times 882 = 693 \ m^2$

$\displaystyle \text{Area of } \triangle POS = \frac{1}{2} \times x \times x = \frac{1}{2} x^2 = \frac{1}{2} \times 882 = 441 \ m^2$

$\displaystyle \therefore$ Area of one flower bed $\displaystyle = 693 - 441 = 252 \ m^2$

$\displaystyle \therefore$ Total area of flower bed $\displaystyle = 2 \times 252 = 504 \ m^2$

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Question 31: From each end of a solid metal cylinder, metal was scooped out in hemispherical form of same diameter. The height of the cylinder is $\displaystyle 10 \text{ cm }$ and its base is of radius $\displaystyle 4.2 \text{ cm }$. The rest of the cylinder is melted and converted into a cylindrical wire of $\displaystyle 1.4 \text{ cm }$ thickness. Find the length of the wire. (Use $\displaystyle \pi = \frac{22}{7}$ )

Volume of cylinder $\displaystyle = \pi (4.2)^2 \times 10 = 176.4 \pi \ cm^3$
$\displaystyle \text{Volume of 2 hemisphere } = 2 \times \frac{2}{3} \times \pi (4.2)^3 = 98.784 \pi \ cm^3$
$\displaystyle \therefore$ Volume melted $\displaystyle = 176.4 \pi - 98.78 \pi = 77.616 \pi \ cm^3$
Diameter of cylindrical wire $\displaystyle = 1.4 \text{ cm }$
$\displaystyle \therefore 77.616 \pi = \pi \Big( \frac{1.4}{2} \Big)^2 \times l$
$\displaystyle l = \frac{77.616 \times 2 \times 2}{1.4 \times 1.4} = 158.4 \text{ m }$