2015 CBSE (Class 10) Board Paper Solution Mathematics : 30-1

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: 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 $p$ .

Answer:

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.

Answer:

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 $\displaystyle 6$ .

Answer:

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 $\displaystyle \angle PRQ$ .

Answer:

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 $\displaystyle OR = PR + RQ$ .

Answer:

$\displaystyle \angle OPR = \angle OQR = 90^{\circ}$ … … … … … 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 $\displaystyle AC$ .

Answer:

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

Answer:

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

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

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

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

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.

Answer:

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 $\displaystyle p$ .

Answer:

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

Answer:

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

Answer:

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

Answer:

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

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.

Answer:

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

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

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

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

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

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 $\displaystyle AB$ .

Answer:

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.

Answer:

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 $\displaystyle \pi = \frac{22}{7}$ )

Answer:

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 $\displaystyle \pi = \frac{22}{7}$ )

Answer:

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

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.

Answer:

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

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

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

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

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 $\displaystyle \pi = 3.14$ ]

Answer:

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 $\displaystyle \pi = \frac{22}{7}$ )

Answer:

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.

Answer:

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

Answer:

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 ?

Answer:

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)

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

$\displaystyle \\$

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

Answer:

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.

$\displaystyle \\$

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.

Answer:

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.

$\displaystyle \\$

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 $\displaystyle AB' = 8 \text{ cm }$ .

Answer:

$\displaystyle \\$

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 $\displaystyle A$ .

Answer:

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

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

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

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

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

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

$\displaystyle \\$

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.

Answer:

Total number of cards $\displaystyle = 52$

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

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

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

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

$\displaystyle \\$

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.

Answer:

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 $\displaystyle k = 3$

$\displaystyle \\$

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).

Answer:

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 $\displaystyle = 2 \times 252 = 504 \ m^2$

$\displaystyle \\$

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 $\displaystyle \pi = \frac{22}{7}$ )