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SUMMATIVE ASSESSMENT – II

MATHEMATICS

Time allowed: 3 hours                                                             Maximum Marks: 80

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 4 carry 1 mark each.

2019-08-06_8-44-16

Question 1: What is the common difference of an A.P. in which a_{21} - a_7 = 84 ?

Answer:

Given: a_{21} -a_7 = 84

a-{21} = a + (21-1) d = a + 20d

a_7 = a + (7-1) d = a + 6 d

\therefore (a + 20 d ) - (a + 6d) = 84

14 d = 84

\Rightarrow d = 6

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Question 2: If the angle between two tangents drawn from an external point P to a circle of radius a and center O , is 60^o , then find the length of OP .

Answer:2019-08-04_8-19-23

Radius = a

\therefore \frac{OA}{PA} = \tan 30^o

\Rightarrow \frac{OA}{PA} = \frac{1}{\sqrt{3}}

\Rightarrow PA = \sqrt{3} a

Also, \frac{OA}{OP} = \sin 30^o = \frac{1}{2}

\Rightarrow OP = 2a

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Question 3: If a tower 30 m high, casts a shadow 10 \sqrt{3} m long on the ground, then what is the angle of elevation of the sun?

Answer:

\frac{30}{10\sqrt{3}} = \tan \alpha

\Rightarrow \tan \alpha = \sqrt{3} = \tan 60^o

\Rightarrow \alpha = 60^o

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Question 4: The probability of selecting a rotten apple randomly from a heap of 900 apples is 0.18 . What is the number of rotten apples in the heap?

Answer:

Let the number of rotten apples = x

\therefore \frac{x}{900} = 0.18

\Rightarrow x = 0.18 \times 900 = 162

Hence the number of rotten apples = 162

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Section – B

Question number 5 to 10 carry 2 mark each.

Question 5: Find the value of p , for which one root of the quadratic equation px^2 - 14x + 8 = 0 is 6 times the other.

Answer:

Given equation: px^2 - 14x + 8 = 0

\Rightarrow a = p, b = -14, c = 8

Let \alpha and \beta be the roots

Given \beta = 6 \alpha

\alpha . \beta = \frac{c}{a}

\Rightarrow \alpha (6 \alpha) = \frac{8}{p}

\Rightarrow 6 {\alpha}^2 = \frac{8}{p}  … … … … … i)

\alpha + \beta = - \frac{b}{a}

6 \alpha + \alpha = \frac{14}{p}

7 \alpha = \frac{14}{p}

\Rightarrow \alpha = \frac{2}{p}  … … … … … ii)

Solving i) and ii)

6 \Big(  \frac{2}{p} \Big)^2 = \frac{8}{p}

\frac{6 \times 4}{p^2} = \frac{8}{p}

\Rightarrow p = \frac{6 \times 4}{8^2} = 3

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Question 6: Which term of the progression 20, 19 \frac{1}{4} , 18 \frac{1}{2} , 17 \frac{3}{4} , \ldots  is the first negative term?

Answer:

Given AP is 20, 19 \frac{1}{4} , 18 \frac{1}{2} , 17 \frac{3}{4} , \ldots

\therefore a = 20

d = 19 \frac{1}{4} - 20 = - \frac{3}{4}

Confirming it once again: d = 18 \frac{1}{2} - 19 \frac{1}{4} = - \frac{3}{4}

\therefore a = 20 and d= - \frac{3}{4}

Let n^{th}   term be zero

\therefore a _n = a + (n-1)d

\Rightarrow 0 = 20 + (n-1) (- \frac{3}{4} )

\Rightarrow 3(n-1) = 80

3n = 83

n = \frac{83}{3} = 27.67

\therefore n^{th} term where the value is very close is zero

\therefore (n+1)^{th} term or 28^{th} term will be negative

Also 28^{th} term

a_{28} = 20 + (28 - 1)(- \frac{3}{4} ) = - \frac{1}{4}

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Question 7: Prove that the tangents drawn at the end points of a chord of a circle make equal angles with the chord.

Answer:

Given: A circle with center O, PA and PB are tangents drawn at the end A and B on chord AB 2019-08-04_8-27-22.jpg

To prove: \angle PAB = \angle PBA

Construction: Join OA and OB

Proof: In \triangle OAB , we have

OA = OB   (radius of the same circle)

\Rightarrow \angle 2 = \angle  1    (angles opposite to equal sides of a triangle)

\angle OBP = \angle OAP = 90^o (radius is \perp tangents)

\Rightarrow \angle 2 + \angle 3 = \angle  1 + \angle  4

\Rightarrow \angle  3 = \angle  4

\Rightarrow \angle PAB = \angle  PBA

Hence proved.

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Question 8: A circle touches all the four sides of a quadrilateral ABCD . Prove that AB + CD = BC + DA

Answer:

Given: A circle touches quadrilateral ABCD on all four sides.

To prove: AB + CD = AD + BC 2019-08-04_8-31-54.jpg

Proof: Since tangents to a circle from external points are equal.

\therefore AP = AS, DQ = DR, CS = CR, BS = BP

\therefore AP + DQ + CS + BS = AP + DR + RC + PB

\Rightarrow AD + BC = AB + CD

Hence proved.

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Question 9: A line intersects the y-axis and x-axis at the points P and Q respectively. If (2, - 5) is the mid-point of PQ , then find the coordinates of P and Q .2019-08-04_8-36-30.jpg

Answer:

Let P be (0, y) and Q be (x, 0)

\therefore \frac{x+0}{2} = 2 \Rightarrow x = 4

Also \therefore \frac{0+y}{2} = -5 \Rightarrow y = -10

\therefore P(0, -10) and Q(4, 0)

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Question 10: If the distances of P(x, y) from A(5, 1) and B(- 1, 5) are equal, then prove that 3x = 2y .2019-08-04_8-38-24

Answer:

Given: AP = BP \Rightarrow AP^2 = BP^2

\therefore (5-x)^2 + (1-y)^2 = (-1-x)^2 + (5-y)^2

\Rightarrow 25 + x^2 - 10x + 1 + y^2 -2y = 1 + x^2 + 2x + 25 + y^2 - 10y

\Rightarrow 25 - 10x + 1 - 2y = 1 + 2x + 25 - 10y

\Rightarrow -10x - 2y = 2x - 10y

\Rightarrow 8y = 12x

\Rightarrow 2y = 3x \ or \ 3x = 2y

Hence proved.

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Section – C

Question number 13 to 22 carry 3 mark each.

Question 11: If ad \neq bc , then prove that the equation (a^2 + b^2) x^2 + 2 (ac + bd) x + (c^2 + d^2) = 0 has no real roots.

Answer:

The equation is of the form ax^2 + bx+ c =0

\therefore  D = b^2 - 4ac

\Rightarrow D = [ 2(ac+bd) ]^2 - 4 [ (a^2+b^2)(c^2+d^2) ]

\Rightarrow D = 4 [ a^2c^2+b^2d^2+2abcd - a^2c^2-a^2d^2-b^2c^2-b^2d^2 ]

\Rightarrow D = 4 [ 2abcd - a^2d^2 - b^2c^2 ]

\Rightarrow D = - (ad-bc)^2

Since ad \neq bc , the equation has no real roots.

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Question 12: The first term of an A.P. is 5 , the last term is 45 and the sum of all its terms is 400 . Find the number of terms and the common difference of the A.P.

Answer:

We know a = 5

Let n^{th} term be the last

\therefore 45 = 5 + (n-1) d

\Rightarrow (n-1)d = 40    … … … … … i)

Also the sum of the n terms is 400

\therefore 400 = \frac{n}{2} [ 2 \times 5 + (n-1)d ]    … … … … … ii)

Substituting i) in ii) we get

400 = \frac{n}{2} [10 + 40]

\Rightarrow n = \frac{400}{25} = 16

\therefore d = \frac{40}{n-1} = \frac{40}{16-1} = \frac{8}{3}

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Question 13: On a straight line passing through the foot of a tower, two points C and D are at distances of 4 m and 16 m from the foot respectively. If the angles of elevation from C and D of the top of the tower are complementary, then find the height of the tower.

Answer:2019-08-04_8-45-29.jpg

{\alpha}_1 + {\alpha}_2  = 90^o

From \triangle ABC:  \frac{h}{4} = \tan {\alpha}_1    … … … … … i)

From \triangle ABD:  \frac{h}{16} = \tan {\alpha}_2 = \tan (90^o - {\alpha}_1) = \cot {\alpha}_1

\therefore \frac{h}{16} = \frac{1}{\tan {\alpha}_1}

\Rightarrow \tan {\alpha}_1 = \frac{16}{h}     … … … … … ii)

Substituting ii) in i) we get

\frac{h}{4} = \frac{16}{h}

\Rightarrow h^2 = 4 \times 16

\Rightarrow h = 2 \times 4 = 8 m

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Question 14: A bag contains 15 white and some black balls. If the probability of drawing a black ball from the bag is thrice that of drawing a white ball, find the number of black balls in the bag.

Answer:

Let the number of black balls = x

\therefore P(Black \ Balls) = \frac{x}{x+15}

and P(White \ Balls) = \frac{15}{x+15}

Given \frac{x}{x+15} = 3 \times \frac{15}{x+15}

\Rightarrow x = 45

Therefore number of Black balls = 45

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Question 15: In what ratio does the point \Big( \frac{24}{11} , y \Big) divide the line segment joining the points divide the line segment joining the points P(2, - 2) and Q(3, 7) ? Also find the value of y .

Answer:

Given P(2, -2), Q(3, 7) and R( \frac{24}{11} , y)

Let R divides P and Q in the ratio k:1 . Using sections formula,

\frac{24}{11} = \frac{3 \times k + 1 \times 2}{k+1}

\Rightarrow 24(k+1) = 33k + 22

\Rightarrow 24k + 24 = 33k + 22

\Rightarrow k = \frac{2}{9}

Hence  R divides P and Q in the ratio of 2:9

Similarly, y = \frac{k \times 7 + 1 \times (-2)}{k+1}

\Rightarrow y = \frac{\frac{2}{9} \times 7 + 1 \times (-2)}{\frac{2}{9}+1}

\Rightarrow y = \frac{14-18}{2+9} = - \frac{4}{11}

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Question 16: Three semicircles each of diameter 3 cm, a circle of diameter 4·5 cm and a semicircle of radius 4·5 cm are drawn in the given figure. Find the area of the shaded region.

2019-05-25_20-02-03

Answer:

Area of large semi circle = \frac{1}{2} \times \pi  (4.5)^2 = 10.125 \pi \ cm^2

Area of large circle = \pi ( \frac{4.5}{2} )^2 = 5.0625 \pi \ cm^2

Area of two small semi circles = 2 \times [ \frac{1}{2} \pi  \times (1.5)^2 ] = 2.25 \pi \ cm^2

Area of shaded small semi circle = \frac{1}{2} \pi (1.5)^2 = 1.125 \pi \ cm^2

Therefore shaded area = 10.125 \pi - 5.0625 \pi - 2.25 \pi + 1.125 \pi = 3.9375 \pi = 12.375 \ cm^2

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Question 17: In the given figure, two concentric circles with centre O have radii 21 cm and 42 cm. If \angle AOB = 60^o , find the area of the shaded region. [ Use \pi = \frac{22}{7} ]

2019-05-25_19-51-26

Answer:

Area of OAB = \frac{60}{360} \times \pi (42)^2 = 294 \pi \ cm^2 

Area of OCD = \frac{60}{360} \times \pi (21)^2 = 73.5 \pi \ cm^2

Therefore area of CDBA = 294 \pi - 73.3 \pi = 220.5 \pi \ cm^2

Shaded Area = \frac{300}{360} \times \pi (42)^2 - \frac{300}{360} \times \pi (21)^2

= 1470 \pi - 367.5 \pi = 1102.5 \times \frac{22}{7} = 3465 \ cm^2

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Question 18: Water in a canal, 5.4 m wide and 1.8 m deep, is flowing with a speed of 25 km/hour. How much area can it irrigate in 40 minutes, if 10 cm of standing water is required for irrigation ?

Answer:

Cross Section of canal = 5.4 \times 1.8 = 9.72 \ m^2

Speed of water = 25 \frac{km}{hr} = 25 \times \frac{1000 \ m}{60 \ min} = \frac{2500}{6}\frac{m}{min}

Therefore volume of water flow  in 1 minute = 9.72 \times \frac{2500}{6} = 4050 \frac{m^3}{min}

Let the area irrigated in 40 \ min = x \ m^2

\therefore x \times \frac{10}{100} = 40 \times 4050

\Rightarrow x = 1620000 \ m^2

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Question 19: The slant height of a frustum of a cone is 4 cm and the perimeters of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum.

Answer:

We know that the curved surface area of frustum = \pi (r_1+r_2) l

Here l = 4 \  cm

For r_1: 2 \pi r_1 = 18 \Rightarrow r_1 = \frac{9}{\pi}

For r_2: 2 \pi r_2 = 6 \Rightarrow r_1 = \frac{3}{\pi}

Now curved surface area = \pi (r_1+r_2) l = \pi \Big( \frac{9}{\pi} + \frac{3}{\pi} \Big) l = 12 \times 4 = 48 \ cm^2

Hence the curved surface area  = 48 \ cm^2

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Question 20: The dimensions of a solid iron cuboid are 4.4 m \times 2.6 m \times 1.0 m. It is melted and recast into a hollow cylindrical pipe of 30 cm inner radius and thickness 5 cm. Find the length of the pipe.

Answer:

Dimensions of cuboid: l = 4.4 \ m, b = 2.6 \ m, h = 1 \ m

Therefore volume of cuboid = lbh = 4.4 \times 2.6 \times 1 = 11.44 \ m^3

Dimension of the pipe: Inner radius (r) = 30 \ cm = 0.3 \ m

Thickness = 5 \ cm = 0.05 \ m

Therefore outer radius (R) = 0.30 + 0.05 = 0.35 \ m

Let l be the length of the pipe

\therefore \pi (R^2 - r^2)  \times l = 11.44

\Rightarrow \frac{22}{7} \Big( (0.35)^2 - (0.30)^2 \Big) \times l = 11.44

\Rightarrow \frac{22}{7} \times 0.0325 \times l = 11.44

\Rightarrow l = \frac{7 \times 11.44}{22\times 0.0325} = 112 \ m

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Section – D

Question number 21 to 30 carry 4 mark each.

Question 21: Solve for x: \frac{1}{x+1} + \frac{3}{5x+1} = \frac{5}{x+4} , x \neq -1, - \frac{1}{5} , -4

Answer:

\frac{1}{x+1} + \frac{3}{5x+1} = \frac{5}{x+4}

\Rightarrow \frac{5x+1 + 3(x+1)}{(x+1)(5x+1)} = \frac{5}{x+4}

\Rightarrow (8x+4)(x+4) = 5(x+1)(5x+1)

\Rightarrow 8x^2 + 4x + 32x + 16 = 25x^2 + 25x + 5x + 5

\Rightarrow 6x + 11 = 17x^2

\Rightarrow 17x^2 - 6x - 11 = 0

\Rightarrow 17x^2 - 17x + 11x - 11 = 0

\Rightarrow 17x(x-1) + 11 (x-1) = 0

\Rightarrow (x-1)(17x + 11) = 0

\Rightarrow x = 1 \ or \ x = - \frac{11}{17}

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Question 22: Two taps running together can fill a tank in 3 \frac{1}{3} hours. If one tap takes 3 hours more than the other to fill the tank, then how much time will each tap take to fill the tank ?

Answer:

Time taken by larger tap to fill the tank = x hours

Therefore Time taken by smaller tap to fill the tank = (x+3) hours

Portion of the tank filled by larger tap in 1 hour = \frac{1}{x}

Portion of the tank filled by Smaller tap in 1 hour = \frac{1}{x+3}

Given: Time taken when both taps are running to fill the tank = 3 \frac{1}{13} = \frac{40}{13} hours

Portion of tank filled in 1 hour when both taps are running = \frac{1}{\frac{40}{13}} = \frac{13}{40}

\therefore \frac{1}{x} + \frac{1}{x+3} = \frac{13}{40}

\Rightarrow 40(x + 3 + x) = 13x(x + 3)

\Rightarrow 40 (2x + 3) = 13x(x+3)

\Rightarrow 80x + 120 = 13x^2 + 39x

\Rightarrow 13x^2 - 41x- 120 = 0

\Rightarrow 13x^2 - 65x + 24x - 120 = 0

\Rightarrow 13x( x - 5) + 24( x - 5) = 0

\Rightarrow (x-5) ( 13x + 24) = 0

\Rightarrow x = 5 \ or \ x = - \frac{24}{13} (not possible as time cannot be negative)

Therefore the time taken by the larger tap to fill the tank = 5 hours

and the time taken by the smaller tap to fill the tank = 8 hours

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Question 23: If the ratio of the sum of the first n terms of two A.Ps is (7n + 1) : (4n + 27) , then find the ratio of their 9^{th} terms.

Answer:

Given: \frac{S_n}{ {S'}_n} = \frac{7n+1}{4n+27} 

\Rightarrow \frac{ \frac{n}{2} [ 2a + (n-1)d ] }{\frac{n}{2} [ 2a' + (n-1)d' ]} = \frac{7n+1}{4n+27} 

\frac{2a + (n-1)d}{2a' + (n-1)d'} = \frac{7n+1}{4n+27} 

\frac{a + \frac{(n-1)}{2} d}{a' + \frac{(n-1)}{2} d'} = \frac{7n+1}{4n+27}     … … … … … i)

We need to find:

\frac{T_9}{ {T'}_9} = \frac{a + (9-1)d}{a' + (9-1)d'} = \frac{a + 8d}{a' + 8d'}    … … … … … ii)

If we put \frac{n-1}{2} = 8 \Rightarrow n = 17    … … … … … iii)

Substituting n = 17 in i) we get

\frac{a + 8d}{a' + 8 d'} = \frac{7 \times 17 + 1}{4 \times 17 + 27} = \frac{120}{146} = \frac{24}{19} 

Hence \frac{T_9}{ {T'}_9} = \frac{24}{19} 

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

Answer:

2019-07-13_12-19-35.pngGiven: Let the circle with center O

Let P be an external point from which tangents PQ and PR are drawn as shown in the diagram

To prove: PQ = PR

Construction: Join OQ, OR and OP

Proof: As PQ is tangent OQ \perp PQ . Therefore \angle OQP = 90^o

Similarly, As PR is tangent OR \perp PR . Therefore \angle ORP = 90^o

(Note: Tangents at any point on a circle is perpendicular to the radius through the point of contact)

In \triangle OQP and \triangle ORP

\angle OQP = \angle ORP = 90^o

OP is common

OQ = OR   (radius of the same circle)

\therefore \triangle OQP \cong \triangle ORP

\Rightarrow PQ = PR

Therefore both tangents are equal in length.

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Question 25: In Figure 3, PQ and RS are two parallel tangents to a circle with center O  and another tangent AB with point of contact C intersecting PQ at A and RS at B . Prove that \angle AOB = 90^o .

2019-05-22_21-21-04
Figure 3

Answer:

Given: PQ \parallel RS , AB is a tangent

To prove: \angle AOB = 90^o

Join DE through O . DOE is diameter of the circle

\therefore \angle ADO = \angle BEO = 90^o 2019-06-25_7-49-32.png

\Rightarrow \angle ADO + \angle BEO = 180^o

\Rightarrow DA \parallel EB

We know that tangents to a circle from an external point are equally inclined to the line segment joining this point to the center

\therefore \angle 1 = \angle 2 and \angle 3 = \angle 4

Now DA \parallel RB and AB is transversal

\therefore (\angle 1 + \angle 2) + (\angle 3 + \angle 4) = 180^o

\Rightarrow 2 \angle 1 + 2 \angle 3 = 180^o

\Rightarrow \angle 1 + \angle 3 = 90^o

From \triangle AOB

\angle AOB + \angle 1 + \angle 3 = 180^o

\Rightarrow \angle AOB = 180^o - 90^0 = 90^o

Hence \angle AOB  = 90^o

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Question 26: Construct a triangle ABC with side BC = 7 cm, \angle B = 45^o, \angle A = 105^o . Then construct another triangle whose sides are \frac{3}{4} times the corresponding sides of the \triangle ABC .

Answer:

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Question 27: An aeroplane is flying at a height of 300 m above the ground. Flying at this height, the angles of depression from the aeroplane of two points on both banks of a river in opposite directions are 45^o and 60^o respectively. Find the width of the river. [Use \sqrt{3} = 1.732 ]2019-08-04_8-56-46.jpg

Answer:

Let A be the aeroplane & BD is the width of river

In \triangle ABC, \frac{300}{BC} = \tan 45^o = 1

\Rightarrow BC = 300 m

In \triangle ABD, \frac{300}{CD} = \tan 60^o = \sqrt{3}

\Rightarrow CD = \frac{300}{\sqrt{3}} = 100\sqrt{3} = 173.2 m

Therefore the width of the river = BC + CD = 300 + 173.2 = 473.2 m

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Question 28: If the points A(k + 1, 2k), B(3k, 2k + 3) and C(5k - 1, 5k) are collinear, then find the value of k .

Answer:

Given: A (k+1, 2k), B( 3k, 2k+3), C(5k-1, 5k) are collinear

Therefore the area of \triangle ABC = 0

\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} [ (k+1)(2k+3-5k) + 3k(5k-2k) + (5k-1)(2k - 2k-3)  ] = 0

\Rightarrow \frac{1}{2} [ (k+1)(-3k+3) + 3k(3k) + (5k-1)(-3)  ] = 0

\Rightarrow \frac{1}{2} [ -3k^2+3k-3k+3+9k^2-15k+3 ] = 0

\Rightarrow \frac{1}{2} [ 6k^2 - 15k+6 ] = 0

\Rightarrow 6k^2 - 15k+6 = 0

\Rightarrow 2k^2 - 5k + 2 = 0

\Rightarrow 2k^2 - 4k - k + 2 = 0

\Rightarrow 2k(k-2)-1(k-2)=0

\Rightarrow (k-2)(2k-1) = 0

\Rightarrow k = 2 \ or \ k = \frac{1}{2} 

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Question 29: Two different dice are thrown together. Find the probability that the numbers obtained have
(i) even sum, and
(ii) even product.

Answer:

Number of possible outcomes:

(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)

Therefore total number of possible outcomes = 36

i) Even sum outcome

(1, 1),  (1, 3),  (1, 5)

(2, 2),  (2,4 ),  (2, 6)

(3, 1),  (3, 3),  (3, 5)

(4, 2),  (4,4 ),  (4, 6)

(5, 1),  (5, 3),  (5, 5)

(6, 2),  (6,4 ),  (6, 6)

Possible outcomes = 18

Therefore Probability (Even sum outcome) = \frac{18}{36} = \frac{1}{2}

ii) Even products

(1, 2),  (1,4 ),  (1, 6)

(2, 1), (2, 2), (2, 3), (2,4 ), (2, 5), (2, 6)

 (3, 2),  (3,4 ),  (3, 6)

(4, 1), (4, 2), (4, 3), (4,4 ), (4, 5), (4, 6)

 (5, 2),  (5,4 ),  (5, 6)

(6, 1), (6, 2), (6, 3), (6,4 ), (6, 5), (6, 6)

Possible outcomes = 27

Therefore Probability (Even product) = \frac{27}{36} = \frac{3}{4}

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Question 30: In the given figure, ABCD is a rectangle of dimensions 21 cm \times 14 cm. A semicircle is drawn with BC as diameter. Find the area and the perimeter of the shaded region in the figure.

2019-05-25_19-46-29

Answer:

Area of ABCD = 21 \times 14 = 294 \ cm^2

Area of semi circle = \frac{1}{2} [ \pi (7)^2 ] = 77 \ cm^2

Therefore area of shaded region = 294 - 77 = 217 \ cm^2

Perimeter of shaded region = 21 + 14 + 21 + \frac{1}{2} (2 \pi \times 7) = 56 + 22 = 78 \ cm

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Question 31: In a rain-water harvesting system, the rain-water from a roof of 22 m \times 20 m drains into a cylindrical tank having diameter of base 2 m and height 3.5 m. If the tank is full, find the rainfall in cm. Write your views on water conservation.

Answer:

Let the rainfall be x m

Therefore  volume of water from roof = volume of tank

\Rightarrow 22 \times 20 \times x = \pi (1)^2 \times 3.5

\Rightarrow x = \frac{22}{7} \times \frac{3.5}{22 \times 20}

\Rightarrow x = 0.025 \ m = 25 \ mm \ or \ 2.5 \ cm