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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 AB be the tower and BC be it’s shadow.

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

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

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

\Rightarrow \tan \theta = \tan 30^o

\Rightarrow \theta = 30^o

Therefore sunset is at an altitude  of 30^o .

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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 6 .

Answer:

Two dice are tossed. Therefore

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 = 6 \times 6 = 36

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

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

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

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

Answer:

Given: PT is a tangent so \angle OPT=90^o

\angle QPT =60^o

\angle OPQ = \angle OPT- \angle QPT = 90^o-60^o = 30^o

\angle OPQ+\angle OQP+ \angle POQ=180^o       (sum of the angles in a triangle is 180^o )

\angle POQ=180^o-30^o-30^o

\angle POQ=120^o

\angle OPQ= \angle OQP = 30^o

\angle POQ= 120^o .  Also, \angle PRQ= \frac{1}{2} reflex \angle POQ

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

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Section – B

Question number 5 to 10 carry 2 mark each.

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

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

Answer:

\angle OPR = \angle OQR = 90^o    … … … … …  i)2019-11-16_9-34-48

And in \triangle OPR and \triangle OQR

\angle OPR = \angle OQR = 90^o (from equation i)

OP = OQ (Radii of same circle)

OR = OR (common side)

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

So, RP = RQ    … … … … …  ii)

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

\angle PRQ = \angle ORP + \angle ORQ

Substitute \angle PQR = 120^o (given) And from equation iii) we get

\angle ORP + \angle ORP = 120^o   \Rightarrow 2 \angle ORP = 120^o   \Rightarrow \angle ORP = 60^o

And we know \cos 0^o = \frac{Adjacent}{Hypotenuse}

So in \triangle OPR we get

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

\cos 60^o = \frac{PR}{OR}

\frac{1}{2} = \frac{PR}{OR}   (we know \cos 60^o = \frac{1}{2} )

OR = 2PR

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

OR = PR  + RQ

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

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

Answer:

Given OD = 3 cm

Construction: Join OA, OB and OC

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

Let AE = AF = x 2019-11-16_11-37-43

BD=6 cm, BE = 6 cm (equal tangents)

DC=9 cm, CF=9 cm (equal tangents)

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

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

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

Therefore

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

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

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

\Rightarrow 36 = 2x+ 30

\Rightarrow x = 3

Therefore sides are 15 cm, 9 cm, 12 cm

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

Answer:

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

4x^2 +4bx -a^2 + b^2 =0

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

(2x + b)^2 -a^2 = 0

Use formula:  a^2-b^2 = (a -b)(a + b)

(2x + b -a)(2x + b +a) =0

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

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

Answer:

Let the first term be a and the common difference be d

S_5 + S_7 = 167

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

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

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

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

S_{10} = 235

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

\Rightarrow 10a + 45 d = 235

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

Solving i) and ii) we get

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

\Rightarrow d = 5

Substituting in ii) we get

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

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

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

Answer:

A(4,7), B(p,3)  and C(7,3)

AB^2+BC^2=AC^2

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

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

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

AB^2 = (p-4)^2 + 16

BC^2 = (7-p)^2 + ( 3- 3)^2  = (7-p)^2

AC^2 = (4-7)^2 + (7-3)^2  = 25

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

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

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

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

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

\Rightarrow p = 7 or  p = 4

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

Answer:

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

Therefore the area of \triangle ABC = 0

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

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

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

\Rightarrow  2x + y + 3  = 0

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Section – C

Question number 11 to 20 carry 3 mark each.

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

Answer:

Let a be the first term and d be the common difference

\therefore T_{14} = a + 13 d

T_8= a + 7d

Given, T_{14} = 2 T_8

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

\Rightarrow a + d = 0    … … … … … i)

T_6 = a + 5d

Given T_6 = -8

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

Solving i) and ii) we get

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

\Rightarrow d = - 2

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

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

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

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

Answer:

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

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

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

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

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

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

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

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

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

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

or x = \frac{\sqrt{2} - 2\sqrt{2}}{\sqrt{3}} = \frac{-\sqrt{2}}{\sqrt{3}} = - \sqrt{\frac{2}{3}}

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

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

\tan 60^o = \frac{1500\sqrt{3}}{AC}

\Rightarrow AC = \frac{1500\sqrt{3}}{\sqrt{3}} = 1500 m

\tan 30^o = \frac{1500\sqrt{3}}{AE}

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

\therefore CE = 4500 - 1500 = 3000 m

Time taken to travel CE = 15 sec

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

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

Answer:

Let the coordinates of point P be P(x, y)

Given AP = \frac{3}{7} AB

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

\Rightarrow 7AP = 3AP + 3 PB

\Rightarrow 4AP = 3 P

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

Hence P divides AB in the ratio of 3:4

Now using section’s formula

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

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

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

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

Answer:

Let the number of red balls be x and number of Blue balls be y

No of Orange balls = 10

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

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

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

\Rightarrow 4x = x + y + 10

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

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

\Rightarrow 3y = x + y + 10

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

Solving i) and ii) we get

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

\Rightarrow y = 8

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

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

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

Answer:2019-07-20_8-30-20

Radius (r) = 14 cm \theta = 60^o

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

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

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

We draw OM \perp AB

\therefore \angle OMB = \angle OMA = 90^o

In \triangle OMA and \triangle OMB

\angle OMA = \triangle OMB = 90^o  (by construction)

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

OM is common

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

\Rightarrow \angle AOM = \angle BOM

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

\Rightarrow \angle AOM = \angle BOM = \frac{1}{2} \times 60 = 30^o

Also, since \triangle OMB \cong \triangle OMA

\therefore BM = AM

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

\Rightarrow AB = 2 BM  … … … … … i)

In right \triangle OMB

\frac{AM}{AO} = \sin 30^o

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

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

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

Similarly, In right \triangle OMA

\frac{OM}{AO} = \cos 30^o

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

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

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

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

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

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

Answer:2019-11-25_7-41-56

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

Surface Area of canvas = CSA of Cone + CSA of cylinder

= \pi r l + 2 \pi r H

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

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

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

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

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

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

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

Answer:

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

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

Volume of liquid to be transferred = 0.9 \times 3888 \pi = 3499.2 \pi \ cm^3

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

Let the height of Bottle = h

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

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

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

Therefore height of bottle is 5.4 cm

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Question 19: A cubical block of side 10 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. 5 per 100 sq. cm. [ Use \pi = 3.14 ]

Answer:

The greatest diameter of the hemisphere can be 10 cm

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

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

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

= 687.57 \ cm^2

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

\therefore Cost of painting = 678.57 \times 0.05 = 33.93 Rs.

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

Answer:

Diameter of cone (d) = 3.5 cm

Height of cone (h)= 3.0 cm

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

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

Let the radius of the sphere = r

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

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

\Rightarrow r = 10.5 cm

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

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Section – D

Question number 21 to 31 carry 4 mark each.

Question 21: The diagonal of a rectangular field is 16 metres more than the shorter side. If the longer side is 14 metres more than the shorter side, then find the lengths of the sides of the field.

Answer:

Let the shorter side = x

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

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

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

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

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

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

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

Hence the shorter side = 10 m, diagonal = 26 m and Longer side = 24 m

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

Answer:

Given AP: 8, 10, 12, ...

\therefore a = 8 and d = 2

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

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

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

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

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

= 5[ 252 - 18] = 1170

Therefore sum of the last 10 terms = 1170

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

Answer:

Total time taken = 3 hr

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

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

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

\Rightarrow  39x + 108 = x^2 + 6x

\Rightarrow  x^2 - 33x - 108 = x

\Rightarrow  x^2 - 36x + 3x - 108 = 0

\Rightarrow  x(x-36) + 3 (x - 36) = 0

\Rightarrow  (x-36)(x+ 3) = 0

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

\therefore x = 36 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.

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

Given: Circle with center O

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

To Prove: PQ = PR

Construction: Join OQ, OR and OP

Proof: Since PQ is a tangent, \angle PQO = 90^o

Similarly, \angle PRO = 90^o

Consider \triangle POQ and \triangle POR

OQ = OR      (radius of the circle)

\angle PQO = \angle POR = 90^o

PO is common

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

\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.

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

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

To Prove: AB \parallel ST

\angle OPT = 90^o (tangent)

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

\widehat{AP} = \widehat{PB}

\Rightarrow \angle AOP = \angle BOP

\Rightarrow \angle AOM = \angle BOM

In \triangle AOM and \triangle BOM

AO = OB     (radius)

OM is common

\angle AOM = \angle BOM

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

\Rightarrow \angle AMO = \angle BMO

\angle AMO + \angle BMO = 180^o

\Rightarrow \angle AMO = \angle BMO = 90^o

\therefore \angle AMO = \angle OPS = 90^o

\angle BMO = \angle OPT = 90^o

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

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

Answer:

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Question 27: At a point A, 20 metres above the level of water in a lake, the angle of elevation of a cloud is 30^o . The angle of depression of the reflection of the cloud in the lake, at A is 60^o . Find the distance of the cloud from A .

Answer:2019-11-25_7-35-59

\tan 30^o = \frac{h}{AD}    \Rightarrow AD = \sqrt{3} h

\tan 60^o = \frac{h+40}{AD}

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

\Rightarrow 3h = h + 40

\Rightarrow 2h = 40    \Rightarrow h = 20 m

\therefore EC = 20 + h = 20 + 20 = 40 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 = 52

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

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

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

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 k so that the area of the triangle with vertices (1, - 1), (- 4, 2k) and (- k, - 5) is 24 sq. units.

Answer:

Vertices: (1, -1), (-4, 2k), (-k, -5)

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

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

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

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

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

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

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

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Question 30: In Figure 5, PQRS is a square lawn with side PQ = 42 metres. Two circular flower beds are there on the sides PS and QR with centre at 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 = 42 \times 42 = 1764 \ m^2

Let OP = OS = x

\angle POS = 90^o (diagonals of square are perpendicular to each other)

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

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

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

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

\therefore Total area of flower bed = 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 10 cm and its base is of radius 4.2 cm. The rest of the cylinder is melted and converted into a cylindrical wire of 1.4 cm thickness. Find the length of the wire. (Use \pi = \frac{22}{7} )

Answer:

Volume of cylinder = \pi (4.2)^2 \times 10 = 176.4 \pi \ cm^3

Volume of 2 hemisphere = 2 \times \frac{2}{3} \times \pi (4.2)^3 = 98.784  \pi \ cm^3

\therefore Volume melted = 176.4 \pi - 98.78 \pi = 77.616 \pi \ cm^3

Diameter of cylindrical wire = 1.4 cm

\therefore 77.616 \pi = \pi \Big( \frac{1.4}{2} \Big)^2 \times l

l = \frac{77.616 \times 2 \times 2}{1.4 \times 1.4} = 158.4 m