2016 CBSE (Class 10) Board Paper Solution Mathematics : 30-1-1

Instructions:

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Please check that this question paper consists of 30 questions.
Please write down the serial number of the question before attempting it.
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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: In Fig. 1, $PQ$ is a tangent at point $C$ to a circle with center $O$ . If $AB$ is the diameter and $\angle CAB = 30^o$ , find $\angle PCA$ .

Answer:

Construction: Join $\displaystyle OC$

Now in $\displaystyle \triangle AOC$

$\displaystyle AO = OC$ (radius of the same circle)

Therefore , $\displaystyle \angle OAC= \angle OCA = 30^{\circ}$

Also, $\displaystyle \angle OCP= 90^{\circ}$

Therefore, $\displaystyle \angle PCA= 90^{\circ} - 30^{\circ} = 60^{\circ}$

$\\$

Question 2: For what value of $\displaystyle k$ will $\displaystyle k+9, 2k-1$ , and $\displaystyle 2k+7$ are the consecutive terms of an A.P.?

Answer:

If three terms $\displaystyle x$ , $\displaystyle y$ and $\displaystyle z$ are in A.P. then, $\displaystyle 2 y = x + z$

Since $\displaystyle k + 9, 2 k - 1$ and $\displaystyle 2 k + 7$ are in A.P.

$\displaystyle \therefore 2(2k-1) = (k+9) + (2k+7)$ $\displaystyle \Rightarrow 4k - 2 = 3k + 16$ $\displaystyle \Rightarrow k = 18$

$\displaystyle \\$

Question 3: A ladder, leaning against a wall, makes an angle of $\displaystyle 60^{\circ}$ with the horizontal. If the foot of the ladder is $\displaystyle 2.5 \text { m }$ away from the wall, find the length of the ladder.

Answer:

Let $\displaystyle AC$ be a ladder, $\displaystyle AB$ is the wall and $\displaystyle BC$ is the ground. The situation given in the question is shown in the figure.

$\displaystyle \text{Therefore } \frac{BC}{AC} = \cos 60^{\circ} = \frac{1}{2}$

$\displaystyle \Rightarrow \frac{1}{2} = \frac{2.5}{AC}$

$\displaystyle \Rightarrow AC = 5 \text { m }$ . Therefore, the length of ladder is 5 m.

$\displaystyle \\$

Question 4: A card is drawn at random from a well shuffled pack of $\displaystyle 52$ playing cards. Find the probability of getting neither a Red card or the queen.

Answer:

There are $\displaystyle 26$ red cards in a deck of $\displaystyle 52$ cards

Also there are $\displaystyle 4$ queens in a deck of $\displaystyle 52$ cards, $\displaystyle 2$ queens are black and $\displaystyle 2$ are red.

Now, numbers of card which are neither red nor queen are $\displaystyle = 52 - (26 + 2) = 24$

$\displaystyle \text{Therefore Probability of getting neither a red card nor a queen is } \\ \\ = \frac{24}{52} = \frac{6}{13}$

$\displaystyle \\$

Section – B

Question number 5 to 10 carry 2 mark each.

Question 5: If $\displaystyle -5$ is the root of the quadratic equation $\displaystyle 2x^2 +px - 15 = 0$ and the quadratic equation $\displaystyle p(x^2+x)+ k = 0$ has equal roots, find the value of $\displaystyle k$ .

Answer:

If $\displaystyle -5$ is root of $\displaystyle 2x^2 +px - 15 = 0$

$\displaystyle \Rightarrow 2(-5)^2 + p (-5) - 15 = 0$

$\displaystyle \Rightarrow 50 - 5p - 15 = 0$ $\displaystyle \Rightarrow 5p = 35$ $\displaystyle \Rightarrow p = 7$

For $\displaystyle p(x^2+x)+ k = 0$ to have equal roots,

$\displaystyle b^2 - 4ac = 0$ (Determinant should be zero)

$\displaystyle \Rightarrow p^2 - 4pk = 0 \Rightarrow k = \frac{p}{2} = \frac{7}{2}$

$\displaystyle \\$

Question 6: Let $P$ and $Q$ be points of trisection of the line segment joining the points $A(2, -2)$ and $B(-7, 4)$ such that $P$ is nearer to $A$ . Find the coordinates of $P$ and $Q$ .

Answer:

Since $\displaystyle P$ and $\displaystyle Q$ be the points of trisection of the line segment joining the points $\displaystyle A (2, -2)$ and $\displaystyle B (-7, 4)$ such that $\displaystyle P$ is nearer to $\displaystyle A$ . Therefore $\displaystyle P$ divides line segment in the rail $\displaystyle 1:2$ and $\displaystyle Q$ divides in $\displaystyle 2:1$

By section formula Coordinates of $\displaystyle P$ are

$\displaystyle = \Bigg( \frac{1 \times (-7) + 2 \times (2) }{1 + 2} , \frac{1 \times (4) + 2 \times (-2) }{1 + 2} \Bigg)= \Big( \frac{-3}{3}, \frac{0}{3} \Big) = (-1, 0)$

By section formula Coordinates of Q are

$\displaystyle = \Bigg( \frac{2 \times (-7) + 1 \times (2) }{2 + 1} , \frac{2 \times (4) + 1 \times (-2) }{2 + 1} \Bigg) = \Big( \frac{-12}{3}, \frac{6}{3} \Big) = (-4, 2)$

$\displaystyle \\$

Question 7: In Fig. 2, a quadrilateral $ABCD$ is drawn to circumscribe a circle, with center $O$ , in such a way that the sides $AB, BC, CD$ , and $DA$ touch the circle at the points $P, Q, R$ and $S$ respectively. Prove that $AB + CD = BC + DA$ .

Answer:

Since tangents drawn from the exterior point to a circle are equal in length.

As, $\displaystyle DR$ and $\displaystyle DS$ are tangents from exterior point $\displaystyle D$ so, $\displaystyle DR = DS$ … … i)

Similarly,

$\displaystyle AP$ and $\displaystyle AS$ are tangents from exterior point $\displaystyle A$ so, $\displaystyle AP = AS$ … … … ii)

$\displaystyle BP$ and $\displaystyle BQ$ are tangents from exterior point $\displaystyle B$ so, $\displaystyle BP = BQ$ … … … iii)

$\displaystyle CR$ and $\displaystyle CQ$ are tangents from exterior point $\displaystyle C$ so, $\displaystyle CR = CQ$ … … … iv)

Adding i) , ii), iii) and iv) , we get

$\displaystyle DR + AP + BP + CR = DS + AS + BQ + CQ$

$\displaystyle (DR + CR) + (AP + BP) = (DS + AS) + (BQ + CQ)$

$\displaystyle CD + AB = DA + BC$

$\displaystyle AB + CD = BC + DA$

Hence proved.

$\displaystyle \\$

Question 8: Prove that the points $\displaystyle (3, 0), (6, 4)$ and $\displaystyle ( -1, 3)$ are the vertices of a right angled isosceles triangle.

Answer:

If $\displaystyle A (3, 0), B (6, 4)$ and $\displaystyle C (-1, 3)$ be the vertices of the triangle. From distance formula,

$\displaystyle AB = \sqrt{(6-3)^2 + (4-0)^2} = \sqrt{9+16} = \sqrt{25} = 5$

$\displaystyle BC = \sqrt{(-1--6)^2 + (3-4)^2} = \sqrt{49+1} = \sqrt{50} = 5\sqrt{2}$

$\displaystyle CA = \sqrt{(-1-3)^2 + (3-0)^2} = \sqrt{16+9} = \sqrt{25} = 5$

Therefore we see that $\displaystyle AB^2 + CA^2 = BC^2$

This implies that $\displaystyle \triangle ABC$ is a right angled isosceles triangle.

$\displaystyle \\$

Question 9: The $\displaystyle 4^{th}$ term of an AP is zero. Prove that the $\displaystyle 25^{th}$ term of the AP is three times its $\displaystyle 11^{th}$ term.

Answer:

Let $\displaystyle a$ be the first term and $\displaystyle d$ be the common difference of an A.P. having $\displaystyle n$ terms.

Its $\displaystyle n^{th}$ term is given by, $\displaystyle a_n = a+ (n-1)d$

Therefore $\displaystyle a_4 = a + 3d = 0 \Rightarrow a = - 3d$

$\displaystyle a_{25} = a + 24d \Rightarrow a_{25} = -3d + 24 d = 21d$ … … … … … i)

$\displaystyle a_{11} = a+ 10 d \Rightarrow a_{11} = -3d + 10d = 7 d$ … … … … … ii)

From equation i) and ii)

$\displaystyle a_{25} = 3 a_{11}$ Hence proved.

$\displaystyle \\$

Question 10: In Fig. 3, from an external point $P$ , two tangents $PT$ and $PS$ are drawn to a circle with center $O$ and radius $r$ . If $OP=2r$ , show that $\angle OTS = \angle OST = 30^o$ .

Answer:

In $\displaystyle \triangle OTP, \angle OTP = 90^{\circ}$ (tangent to a circle perpendicular to the radius through the point of contact)

$\displaystyle \sin \angle OTP = \frac{OT}{OP}$

$\displaystyle \Rightarrow \sin \angle OTP = \frac{r}{2r}$ $\displaystyle \Rightarrow \sin \angle OTP = \frac{1}{2}$ $\displaystyle \Rightarrow \angle OTP = 30^{\circ}$

We know $\displaystyle \angle OTP + \angle OPT + \angle TOP = 180^{\circ}$

$\displaystyle \Rightarrow 90^{\circ} + 30^{\circ} + \angle TOP = 180^{\circ}$ $\displaystyle \Rightarrow \angle TOP = 60^{\circ}$

Consider $\displaystyle \triangle OTP$ and $\displaystyle \triangle OSP$

$\displaystyle OT = OS$ (radius of the same circle)

$\displaystyle OP = OS$ (tangents from the same external point)

$\displaystyle OP$ is common

$\displaystyle \therefore \triangle OTP \cong \triangle OSP$ (by SSS criterion)

$\displaystyle \Rightarrow \angle TPO = \angle SPO$

$\displaystyle OP$ is bisector of $\displaystyle \angle TPS$

Also $\displaystyle \triangle TQP \cong \triangle SQP$ (by SAS criterion)

Therefore $\displaystyle \angle OQT = OQS = 90^{\circ}$

In $\displaystyle \triangle OTQ$ , $\displaystyle \angle OQT + \angle OTQ + \angle TOQ = 180^{\circ}$ (angles of a triangle)

$\displaystyle \Rightarrow 90^{\circ} + \angle OTQ + \angle 60^{\circ} = 180^{\circ}$ $\displaystyle \Rightarrow \angle OTQ = 30^{\circ}$

Similarly, $\displaystyle \angle OST = 30^{\circ}$

Section – C

Question number 11 to 20 carry 3 mark each.

Question 11: In Fig. 4, $O$ is the center of the circle such that the diameter $AB = 13$ cm and $AC = 12$ cm. $BC$ is joined. Find the area of shaded region. (Take $\pi = 3.14$ )

Answer:

In $\displaystyle \triangle ABC \angle C = 90^{\circ}$ (angle in a semicircle)

Therefore $\displaystyle BC^2 + AC^2 = AB^2$

$\displaystyle \Rightarrow BC^2 + 12^2 = 13^2$

$\displaystyle \Rightarrow BC^2 = 25 \Rightarrow BC = 5 \text { cm }$

Area of shaded region $\displaystyle =$ Area of semicircle $\displaystyle -$ Area of triangle

$\displaystyle = \frac{1}{2} \times 3.14 \times \Big( \frac{13}{2} \Big)^2 - \frac{1}{2} \times 5 \times 12$

$\displaystyle = 66.3325 - 30 = 36.3325 \ cm^2$

$\displaystyle \\$

$\\$

Question 12: In Fig. 5, a tent is an a shape of cylinder surmounted by a conical top of same diameter. If the height of the diameter of the cylindrical part are $2.$ 1 m and $3$ m respectively and the slant height of the conical part is $2.8$ m, find the cost of the canvas needed to make the tent if the canvas is available at the rate of $500$ Rs sq. meter. (Use $\pi =$ $\frac{22}{7}$ )

Answer:

Canvas needed to make the tent $\displaystyle =$ Curved surface area of the conical part $\displaystyle +$ curved surface area of cylindrical part.

$\displaystyle \text{Radius of the conical part = radius of cylindrical part } = r = \frac{3}{2} \text { m }$

Slant height of the conical part is $\displaystyle = l = 2.$ 8 m.

Height of cylindrical part is $\displaystyle h = 2.1 \text { m }$

$\displaystyle \text{Curved surface area of conical part } = \pi r l = \frac{22}{7} \times \frac{3}{2} \times 2.8 = 13.2 m^2$

$\displaystyle \text{Curved surface area of cylindrical part } = 2 \pi r h = 2 \times \frac{22}{7} \times \frac{3}{2} \times 2.1 = 19.8 \ m^2$

Total surface area $\displaystyle = 13.2 + 19.8 = 33 \ m^2$

Therefore cost of canvas used $\displaystyle = 33 \times 500 = Rs. \ 16500$

$\displaystyle \\$

Question 13: If the point $\displaystyle P(x, y)$ is equidistant from the points $\displaystyle A(a+b, b-a)$ and $\displaystyle B(a-b, a+b)$ . Prove that $\displaystyle bx = ay$ .

Answer:

Since the point $\displaystyle P (x, y)$ is equidistant from the points $\displaystyle A (a + b, b - a)$ and $\displaystyle B (a - b, a + b)$

Therefore $\displaystyle PA = PB$

Using distance formula

$\displaystyle \sqrt{ (x - (a+b))^2 + (y - (b-a))^2 } = \sqrt{ (x - (a-b))^2 + (y - (a+b))^2 }$

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

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

$\displaystyle \Rightarrow - 2x(a+b) - 2y(b-a) = -2x(a-b) - 2y (a+b)$

$\displaystyle \Rightarrow -2xa -2xb-2yb+2ya = - 2xa+2xb-2ya-2yb$

$\displaystyle \Rightarrow -4bx= - 4ay$

$\displaystyle \Rightarrow bx = ay$

$\displaystyle \\$

Question 14: In Fig. 6, find the area of the shaded region, enclosed between two concentric circles of radii $7$ cm and $14$ cm where $\angle AOC = 40^o$ . (Use $\pi =$ $\frac{22}{7}$ )

Answer:

Area of shaded region $\displaystyle =$ Area of circular ring $\displaystyle -$ Area of region ABDC

$\displaystyle = \Big( \pi (14^2 - 7^2) - \frac{40}{360} \pi (14^2 - 7^2) \Big)$

$\displaystyle = \pi (14^2 - 7^2) \times \Big( \frac{360-40}{360} \Big)$

$\displaystyle = \frac{22}{7} \times 147 \times \frac{8}{9}$

$\displaystyle = 22 \times 21 \times \frac{8}{9} = 410.67 \ cm^2$

$\displaystyle \\$

Question 15: If the ratio of the sum first $\displaystyle n$ terms of two APs is $\displaystyle (7n+1): (4n+27)$ , find the ratio of their $\displaystyle m^{th}$ term.

Answer:

For first AP, let the first term be $\displaystyle a_1$ and the common difference be $\displaystyle d_1$ and Sum of the first $\displaystyle n$ terms be $\displaystyle S_1$

Similarly, for Second AP, let the first term be $\displaystyle a_2$ and the common difference be $\displaystyle d_2$ and Sum of the first $\displaystyle n$ terms be $\displaystyle S_2$

$\displaystyle \text{Therefore } \frac{S_1}{S_2} = \frac{\frac{n}{2} (2a_1+ (n-1)d_1)}{\frac{n}{2} (2a_2+ (n-1)d_2)} = \frac{2a_1+ (n-1)d_1}{ 2a_2+ (n-1)d_2}$

$\displaystyle \Rightarrow \frac{2a_1+ (n-1)d_1}{ 2a_2+ (n-1)d_2} = \frac{7n+1}{4n+27}$

For $\displaystyle m^{th}$ term, replace $\displaystyle n$ by $\displaystyle 2m-1$

$\displaystyle \Rightarrow \frac{2a_1+ (2m-1-1)d_1}{ 2a_2+ (2m-1-1)d_2} = \frac{7(2m-1)+1}{4(2m-1)+27}$

$\displaystyle \Rightarrow \frac{a_1+ (m-1)d_1}{ a_2+ (m-1)d_2} = \frac{14m-6}{8m+23}$

Hence the ratio of the $\displaystyle m^{th}$ term is $\displaystyle 14m-6 : 8m + 23$

$\displaystyle \\$

$\displaystyle \text{Question 16: Solve for } x: \frac{1}{(x-1)(x-2)} + \frac{1}{(x-2)(x-3)} = \frac{2}{3} , x \neq 1, 2, 3$

Answer:

$\displaystyle \text{Given } \frac{1}{(x-1)(x-2)} + \frac{1}{(x-2)(x-3)} = \frac{2}{3} , x \neq 1, 2, 3$

$\displaystyle \Rightarrow \frac{(x-3) + (x-1)}{(x-1)(x-2)(x-3)} = \frac{2}{3}$

$\displaystyle \Rightarrow \frac{2x-4}{(x-1)(x-2)(x-3)} = \frac{2}{3}$

$\displaystyle \Rightarrow \frac{x-2}{(x-1)(x-2)(x-3)} = \frac{1}{3}$

$\displaystyle \Rightarrow \frac{1}{(x-1)(x-3)} = \frac{1}{3}$

$\displaystyle \Rightarrow 3 = (x-1)(x-3)$

$\displaystyle \Rightarrow 3 = x^2 - 4x + 3$

$\displaystyle \Rightarrow x(x-4) = 0$

$\displaystyle \Rightarrow x = 0 \ or \ x = 4$

$\displaystyle \\$

Question 17: A conical vessel with a base of radius $\displaystyle 5 \text { cm }$ and height $\displaystyle 24 \text { cm }$ , is full of water. This water is emptied into a cylindrical vessel of base radius $\displaystyle 10 \text { cm }$ . Find the height to which the water will rise in the cylindrical vessel. (Use $\displaystyle \pi = \frac{22}{7}$ )

Answer:

Let $\displaystyle r, h, l$ be the radius, height and slant height of the cone.

Let $\displaystyle R$ and $\displaystyle H$ be the radius and height of the cylinder.

Since volume will remain same

Volume of water in conical vessel $\displaystyle =$ Volume in cylindrical vessel

$\displaystyle \text{Therefore } \frac{1}{3} \pi r^2 h = \pi R^2 H$

$\displaystyle \Rightarrow \frac{1}{3} \pi 5^2 (24) = \pi 10^2 H$

$\displaystyle \Rightarrow H = 2 \text { cm }$

$\displaystyle \\$

Question 18: A sphere of diameter $\displaystyle 12 \text { cm }$ , is dropped into a right circular cylindrical vessel, partly filled with water. If the sphere is completely submerged in water, the water level in the cylindrical vessel rises by $\displaystyle 3\frac{5}{9} \text { cm }$ . Find the diameter of the cylindrical vessel.

Answer:

Diameter of sphere $\displaystyle = 12 \ cm \Rightarrow$ radius $\displaystyle (r) = 6 \text { cm }$

$\displaystyle \text{Rise in water level in cylinder } = 3 \frac{5}{9} = \frac{32}{9} \text { cm }$

Let the radius of cylinder is $\displaystyle R$

$\displaystyle\text{Therefore } \frac{4}{3} \pi (6)^3 = \pi R^2 \frac{32}{9}$

$\displaystyle \Rightarrow R^2 = \frac{4}{3} \times 6^3 \times \frac{9}{32} = 3\sqrt{3} \text { cm }$

$\displaystyle \\$

Question 19: A man standing on the deck of the ship, which is $\displaystyle 10 \text { m }$ eter above the water level, observes the angle of elevation of the top of the hill as $\displaystyle 60^{\circ}$ and the angle of depression of the base of the hill as $\displaystyle 30^{\circ}$ . Find the distance of the hill from the ship and the height of the ship.

Answer:

The man is standing on the deck of the ship at point $\displaystyle A$ .

Let $\displaystyle CE$ be the hill with base at $\displaystyle C$ .

It is given that the angle of elevation of point $\displaystyle E$ from $\displaystyle A$ is $\displaystyle 60^{\circ}$ and the angle of depression of point $\displaystyle C$ from $\displaystyle A$ is $\displaystyle 30^{\circ}$

So, $\displaystyle \angle DAE=60^{\circ}$

$\displaystyle \angle CAD=30^{\circ}$

$\displaystyle \angle CAD= \angle ACB=30^{\circ}$ (alternate angles)

$\displaystyle AB =10 \text { m }$

Suppose $\displaystyle ED = h \text { m }$ and $\displaystyle BC = x \text { m }$

In $\displaystyle \triangle EAD$ , we have $\displaystyle \tan 60^{\circ} = \frac{ED}{AD}$

$\displaystyle \Rightarrow \sqrt{3} = \frac{h}{x} (BC = AD = x)$

$\displaystyle \Rightarrow h = \sqrt{3} x$

In $\displaystyle \triangle ABC$ , we have $\displaystyle \tan 30^{\circ} = \frac{AB}{BC}$

$\displaystyle \Rightarrow \frac{1}{\sqrt{3}} = \frac{10}{x}$ $\displaystyle \Rightarrow x = 10\sqrt{3} \text { m }$

Therefore distance of the hill from the ship is $\displaystyle 10 \sqrt{3} \text { m }$

And height of the hill is $\displaystyle h = x\sqrt{3}+ 10 = 10 \sqrt{3} \times \sqrt{3} + 10 = 40 \text { m }$

$\displaystyle \\$

Question 20: Three different coins are tossed together. Find the probability of getting (i) exactly two heads (ii) at least two heads (iii) at least two tails

Answer:

If three different coins tossed together,

Total possible outcomes are $\displaystyle \{$ TTT, TTH , THT, THH, HTT, HTH, HHT, HHH $\displaystyle \}$

Number of total possible outcomes $\displaystyle = 8$

i ) Exactly two heads (favorable outcomes) $\displaystyle = \{$ TTH , THT, HTT $\displaystyle \}$

Number of favorable outcomes =3.

$\displaystyle \text{Therefore } P( E ) = \frac{3}{8}$

ii ) At least two heads is (favorable outcomes) $\displaystyle = \{$ THH, HTH, HHT, HHH $\displaystyle \}$

Number of favorable outcomes =4.

$\displaystyle P( E ) \text{Therefore } = \frac{4}{8} = \frac{1}{2}$

iii ) At least two tails is (favorable outcomes) $\displaystyle =\{$ TTT, TTH , THT, HTT $\displaystyle \}$

Number of favorable outcomes =4.

$\displaystyle P( E ) \text{Therefore } = \frac{4}{8} = \frac{1}{2}$

$\displaystyle \\$

Section – D

Question number 21 to 31 carry 4 mark each.

Question 21: Due to heavy flood in a state, thousands were rendered homeless. $\displaystyle 50$ schools collectively offered to the state government to provide place and canvas for $\displaystyle 1500$ tents to be fixed by government and decided to share the whole expenditure equally. The lower part of the tent is cylindrical of base radius $\displaystyle 2.8 \text { m }$ and height $\displaystyle 3.5 \text { m }$ , with conical upper part of the same base radius but of height $\displaystyle 2.1 \text { m }$ . If the canvas used to make the tent costs Rs. $\displaystyle 120$ sq. m, find the amount shared by each school to set up the tents. What value is generated by the above problem.(Use $\displaystyle \pi = \frac{22}{7}$ )

Answer:

Let the radius of base of both cylinder and cone be $\displaystyle r$

Let the height of cone be $\displaystyle H$ and that of cylinder be $\displaystyle h$ .

Let the slant height of cone be $\displaystyle l$ , Then,

Area of tent $\displaystyle =$ CSA of cylindrical base $\displaystyle +$ CSA of cone

$\displaystyle = 2 \pi r h + \pi r l$

$\displaystyle = 2 \pi r h + \pi r \sqrt{H^2 + r^2}$

$\displaystyle = 2 \pi r h + \pi r \sqrt{2.1^2 + 2.8^2}$

$\displaystyle = 2 \pi r h + \pi r \sqrt{12.25}$

$\displaystyle = 2 \pi r h + \pi r \times 3.5$

$\displaystyle = \frac{22}{7} \times 2.8 ( 2 \times 3.5 + 3.5)$

$\displaystyle = \frac{22}{7} \times 2.8 \times 10.5$

$\displaystyle = 92.4 \ m^2$

Cost of canvas used in making one tent $\displaystyle = Rs. \ 92.4 \times 120 = Rs.11088$

Cost of canvas used in making $\displaystyle 1500$ tents $\displaystyle = Rs. \ 11088 \times 1500= Rs.16632000$

$\displaystyle \text{Share of one school } = Rs. \ \frac{16632000}{50} = Rs. 332640$

$\displaystyle \\$

Question 22: Prove that the lengths of the tangents drawn from an external point to the circle are equal.

Answer:

Let us draw a circle with center $\displaystyle O$ and two tangents $\displaystyle PQ$ and $\displaystyle PR$ are drawn from a external point $\displaystyle P$ to the circle as shown in the figure given below,

Consider $\displaystyle \triangle PQO$ and $\displaystyle \triangle PRO$

$\displaystyle OQ = OR$ (Radius)

$\displaystyle OP$ is common

$\displaystyle \angle OQP = \angle ORP = 90^{\circ}$

Therefore $\displaystyle \triangle PQO \cong \triangle PRO$ (By SAS criterion)

Hence $\displaystyle PQ = PR$

$\displaystyle \\$

Question 23: Draw a circle of radius $4$ cm. Draw two tangents to the circle inclined at an angle of $60^o$ to each other.

Answer:

i) Draw a circle of radius $\displaystyle 4 \text { cm }$ with center $\displaystyle O$ .

ii) Take a point $\displaystyle A$ on the circumference of circle and join $\displaystyle OA$ .

iii) Draw a perpendicular to $\displaystyle OA$ at $\displaystyle A$

iv) Draw a radius $\displaystyle OB$ , making an angle of $\displaystyle 120^{\circ} (=180^{\circ} - 60^{\circ} )$ with $\displaystyle OA$

v) Draw a perpendicular to $\displaystyle OB$ at $\displaystyle B$ . Let these perpendiculars intersect at $\displaystyle P$ .

$\displaystyle \\$

Question 24: In Fig. 7, two equal circles with center $O$ and $O'$ , touch each other at $X$ . $OO'$ produced meets the circle with center $O'$ at $A$ . $AC$ is a tangent to the circle with center $O$ , at the point $C$ . $O'D$ is perpendicular to $AC$ . Find the value of $\frac{DO'}{CO}$ .

Answer:

As, circles are equal, so their radius are also equal.

That means $\displaystyle AO' = O'X = XO = r$

Consider $\displaystyle \triangle ADO'$ and $\displaystyle \triangle ACO$

$\displaystyle \angle ADO' = \angle ACO$

$\displaystyle \angle A$ is common

$\displaystyle \triangle ADO' \sim \triangle ACO$ (By AA criterion)

$\displaystyle \therefore \frac{AO'}{AO} = \frac{DO'}{CO}$

$\displaystyle \Rightarrow \frac{r}{3r} = \frac{DO'}{CO}$

$\displaystyle \Rightarrow \frac{DO'}{CO} = \frac{1}{3}$

$\displaystyle \\$

$\displaystyle \text{Question 25: Solve for } x: \frac{1}{x+1} + \frac{2}{x+2} = \frac{4}{x+4} , x \neq -1, -2, -4$

Answer:

$\displaystyle \text{Given } \frac{1}{x+1} + \frac{2}{x+2} = \frac{4}{x+4} , x \neq -1, -2, -4$

$\displaystyle \Rightarrow \frac{x+2 + 2 (x+1)}{(x+1)(x+2)} = \frac{4}{x+4}$

$\displaystyle \Rightarrow (3x+3)(x+4) = 4 (x+1)(x+2)$

$\displaystyle \Rightarrow 3x^2 + 3x + 12x + 12 = 4x^2 + 4x + 8x + 8$

$\displaystyle \Rightarrow 3x^2 + 15x + 12 = 4x^2 + 12x+8$

$\displaystyle \Rightarrow x^2 - 3x -4 = 0$

$\displaystyle \Rightarrow (x+1)(x-4) = 0$

$\displaystyle x = -1$ or $\displaystyle x = 4$

Since $\displaystyle x \neq -1, x = 4$

$\displaystyle \\$

Question 26: The angle of elevation of the top $\displaystyle Q$ of a vertical tower $\displaystyle PQ$ from a point $\displaystyle X$ on the ground is $\displaystyle 60^{\circ}$ . From a point $\displaystyle Y, 40 \text { m }$ vertically above $\displaystyle X$ , the angle of elevation of the top $\displaystyle Q$ of tower is $\displaystyle 45^{\circ}$ . Find the height of the tower $\displaystyle PQ$ and the distance $\displaystyle PX$ .

Answer:

Let $\displaystyle QM = y$ , therefore in $\displaystyle \triangle QMY$

$\displaystyle \tan 45^{\circ} = \frac{y}{MY} \Rightarrow \frac{y}{MY} = 1 \Rightarrow MY = y$

In $\displaystyle \triangle QPY$

$\displaystyle \tan 60^{\circ} = \frac{40+y}{PX}$

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

$\displaystyle \Rightarrow \sqrt{3}y = 40+y$

$\displaystyle \Rightarrow (\sqrt{3} - 1)y = 40$

$\displaystyle \Rightarrow y = \frac{40}{\sqrt{3} - 1}$

$\displaystyle \Rightarrow y = \frac{40}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1}$

$\displaystyle \Rightarrow y = 20 (\sqrt{3} + 1)$

Therefore the height of the tower $\displaystyle = 40 + 20 (\sqrt{3} + 1) = 94.64 \text { m }$

$\displaystyle \text{Now } \tan 60^{\circ} = \frac{h}{x}$

$\displaystyle \Rightarrow x = \frac{94.64}{\sqrt{3}} = 54.64 \text { m }$

Therefore horizontal distance is $\displaystyle 54.64 \text { m }$

$\displaystyle \\$

Question 27: The houses in a row are numbered consecutively from $\displaystyle 1$ to $\displaystyle 49$ . Show that there exists a value of $\displaystyle X$ such that the sum of the numbers of houses preceding the house numbered $\displaystyle X$ is equal to sum of the numbers of houses following $\displaystyle X$ .

Answer:

The number of houses preceding $\displaystyle X$ is $\displaystyle X - 1$

Sum of the number of houses preceding the house numbered, $\displaystyle X = 1 + 2 + 3+.... X - 1.$

Here, $\displaystyle a = 1, d =1$

Given

$\displaystyle S_{X-1} = S_{49} - S_X$

$\displaystyle \Rightarrow \frac{X-1}{2} \Big( 2a + (X-1-1)d \Big) = \frac{49}{2} (2a+48d) - \frac{X}{2} (2a + (X-1)d)$

Substituting $\displaystyle a = 1, d =1$

$\displaystyle \Rightarrow \frac{X-1}{2} \Big( 2 + (X-1-1) \Big) = \frac{49}{2} (2+48) - \frac{X}{2} (2a + (X-1))$

$\displaystyle \Rightarrow ({X-1}) \Big( 2 + (X-2) \Big) = \frac{49}{2} (2 +48 ) - ({X}) (2a + (X-1) )$

$\displaystyle \Rightarrow (X-1) X = 45 \times 50 - X ( 1+X)$

$\displaystyle \Rightarrow 2X^2 = 49 \times 50$

$\displaystyle \Rightarrow X^2 = 49 \times 25$

$\displaystyle \Rightarrow X = 7 \times 5 = 35$

$\displaystyle \\$

Question 28: In Fig. 8, the vertices of $\displaystyle \triangle ABC$ are $\displaystyle A( 4, 6), B(1, 5)$ and $\displaystyle C(7, 2)$ . A line segment $\displaystyle DE$ is drawn to intersect the sides $\displaystyle AB$ and $\displaystyle AC$ at $\displaystyle D$ and $\displaystyle E$ respectively such that $\displaystyle \frac{AD}{AB} = \frac{AE}{AC} = \frac{1}{3}$ . Calculate the area of $\displaystyle \triangle ADE$ and compare it with the area of $\displaystyle \triangle ABC$ .

Answer:

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

$\displaystyle = \frac{1}{2} | 4(5-2) + 1(2-6) + 7(6-5) |$

$\displaystyle = \frac{15}{2} = 7.5$ sq. units

Now in $\displaystyle \triangle ADE$ and $\displaystyle \triangle ABC$

$\displaystyle \frac{AD}{AB} = \frac{AE}{AC} = \frac{1}{3}$

$\displaystyle \angle A$ is common

$\displaystyle \therefore \triangle ADE \sim \triangle ABC$

$\displaystyle \text{Therefore } \frac{ar (\triangle ADE)}{ar (\triangle ABC)} = \Big( \frac{AD}{AB} \Big)^2= \Big( \frac{1}{3} \Big)^2$

$\displaystyle \Rightarrow ar (\triangle ADE) = \frac{ar (\triangle ABC)}{9} = \frac{7.5}{9} = \frac{5}{6}$ sq. units

$\displaystyle \\$

Question 29: A number $\displaystyle x$ is selected randomly from the numbers $\displaystyle 1, 2, 3$ and $\displaystyle 4$ . Another number $\displaystyle y$ is selected at random from the numbers $\displaystyle 1, 4, 9$ and $\displaystyle 16$ . Find the probability that the product of $\displaystyle x$ and $\displaystyle y$ is less than $\displaystyle 16$ .

Answer:

Number $\displaystyle x$ can be selected in four ways. Corresponding to each such way, there are four ways of selecting $\displaystyle y$ .

Therefore two numbers can be selected in one of the following ways,

$\displaystyle (1, 1), (1, 4), (1, 9), (1, 16)$

$\displaystyle (2, 1), (2, 4), (2, 9), (2, 16)$

$\displaystyle (3, 1), (3, 4), (3, 9), (3, 16)$

$\displaystyle (4, 1), (4, 4), (4, 9), (4, 16)$

Therefore, the two numbers can be selected in $\displaystyle 16$ ways

The number of combinations where $\displaystyle xy < 16$ are

$\displaystyle (1, 1), (1, 4), (1, 9), (2, 1), (2, 4), (3, 1), (3, 4), (4, 1)$

Therefore the number of possible combinations are $\displaystyle 8$

Therefore probability that the product $\displaystyle xy$ is less than $\displaystyle 16$ is $\displaystyle = \frac{8}{16} = \frac{1}{2}$

$\displaystyle \\$

Question 30: In Fig. 9, is shown a sector $\displaystyle OAP$ of a circle with center $\displaystyle O$ containing $\displaystyle \angle \theta$ . $\displaystyle AB$ is perpendicular to the radius $\displaystyle OA$ and meets $\displaystyle OP$ produced at $\displaystyle B$ . Prove that the perimeter of shaded region is $\displaystyle r [ \tan \theta + \sec \theta + \frac{\pi \theta}{180} - 1 ]$ .

Answer:

$\displaystyle \text{In } \triangle AOB, \cos \theta = \frac{OA}{OB}$

$\displaystyle \Rightarrow OB = \frac{OA}{\cos \theta}$ $\displaystyle \Rightarrow OB = r \sec \theta$

$\displaystyle PB = OP - OB = r \sec \theta - r$

$\displaystyle \text{Also } \tan \theta = \frac{AB}{OA}$

$\displaystyle AB = OA \tan \theta$ $\displaystyle \Rightarrow AB = r \tan \theta$

$\displaystyle \text{Length of arc } AP = \frac{\theta}{360^{\circ} } (2 \pi r) = \frac{\pi \theta}{180^{\circ} }$

Perimeter of shaded region $\displaystyle = PB + AB +$ Length of arc $\displaystyle AP$

$\displaystyle = \Big( r \sec \theta - r + r \tan \theta + \frac{\pi r \theta}{180^{\circ} } \Big)$

$\displaystyle = r\Big( \sec \theta + \tan \theta + \frac{\pi \theta}{180^{\circ} } - 1 \Big)$

$\displaystyle \\$

Question 31: A motor boat whose speed is $\displaystyle 24 \text { km }$ /h in still water takes $\displaystyle 1$ hour more to go $\displaystyle 32 \text { km }$ upstream than to return down stream to the same spot. Find the speed of the stream.