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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.

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

Given: $a_{21} -a_7 = 84$

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

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

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

$14 d = 84$

$\Rightarrow d = 6$

$\\$

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 $OP$.

Radius $= a$

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

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

$\Rightarrow PA = \sqrt{3} a$

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

$\Rightarrow OP = 2a$

$\\$

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?

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

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

$\Rightarrow \alpha = 60^o$

$\\$

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?

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 $= 162$

$\\$

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.

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)

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

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

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

$\\$

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?

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

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

$\\$

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

Given: A circle with center $O, PA$ and $PB$ are tangents drawn at the end $A$ and $B$ on chord $AB$

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.

$\\$

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

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

To prove: $AB + CD = AD + BC$

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.

$\\$

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 $Q$.

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 $Q(4, 0)$

$\\$

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

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.

$\\$

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.

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.

$\\$

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.

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

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

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

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

$\\$

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.

${\alpha}_1 + {\alpha}_2 = 90^o$

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

$\\$

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.

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 $= 45$

$\\$

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 $y$.

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

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

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

$\\$

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.

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 $= 10.125 \pi - 5.0625 \pi - 2.25 \pi + 1.125 \pi = 3.9375 \pi = 12.375 \ cm^2$

$\\$

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 $\pi =$ $\frac{22}{7}$ ]

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

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

$\\$

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 ?

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 $40 \ min = x \ m^2$

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

$\Rightarrow x = 1620000 \ m^2$

$\\$

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.

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  $= 48 \ cm^2$

$\\$

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.

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

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

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

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

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

$\\$

Section – D

Question number 21 to 30 carry 4 mark each.

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

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

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

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

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

$\Rightarrow 6x + 11 = 17x^2$

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

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

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

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

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

$\\$

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 ?

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

$\\$

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.

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

$\\$

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

Given: 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.

$\\$

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 $\angle AOB = 90^o$.

Given: $PQ \parallel RS$, $AB$ is a tangent

To prove: $\angle AOB = 90^o$

Join $DE$ through $O$. $DOE$ is diameter of the circle

$\therefore \angle ADO = \angle BEO = 90^o$

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

$\Rightarrow DA \parallel EB$

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 $\angle AOB = 90^o$

$\\$

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 $\triangle ABC$.

$\\$

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 $\sqrt{3} = 1.732$]

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

$\\$

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 $k$.

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

Therefore the area of $\triangle ABC = 0$

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

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

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

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

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

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

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

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

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

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

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

$\\$

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

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) $=$ $\frac{27}{36}$ $=$ $\frac{3}{4}$

$\\$

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.

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 $= 21 + 14 + 21 +$ $\frac{1}{2}$ $(2 \pi \times 7) = 56 + 22 = 78 \ cm$

$\\$

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.

Let the rainfall be $x$ m
Therefore  volume of water from roof $=$ volume of tank
$\Rightarrow 22 \times 20 \times x = \pi (1)^2 \times 3.5$
$\Rightarrow x =$ $\frac{22}{7}$ $\times$ $\frac{3.5}{22 \times 20}$
$\Rightarrow x = 0.025 \ m = 25 \ mm \ or \ 2.5 \ cm$