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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 .

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 .

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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.

2019-05-26_20-55-13
Figure 1

Answer:

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

\displaystyle \tan \theta = \frac{AB}{BC} 2019-11-16_9-00-32.png

\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 .

2019-05-26_20-53-09
Figure 2

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 .

2019-05-26_16-34-33
Figure 3

Answer:
2019-11-16_9-34-48

\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 .

2019-11-16_11-39-19
Figure 4

Answer:

Given \displaystyle OD = 3 \text{ cm } 2019-11-16_11-37-43

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:
2019-11-25_7-53-05.png

\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:
2019-07-20_8-30-20

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 2019-11-25_7-41-56

\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:
2019-11-25_7-45-03

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:
2019-11-25_7-49-41.png

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:
2019-11-25_7-35-59

\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.

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}  

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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.

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

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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).

2019-05-26_16-23-11
Figure 5

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} )

Answer:

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 }