2019 CBSE (Class 10) Board Paper Solution Mathematics : 30-2-1

Instructions:

Please check that this question paper consists of 11 pages.
Code number given on the right hand side of the question paper should be written on the title page of the answer book by the candidate.
Please check that this question paper consists of 30 questions.
Please write down the serial number of the question before attempting it.
15 minutes times has been allotted to read this question paper. The question paper will be distributed at 10:15 am. From 10:15 am to 10:30 am, the students will read the question paper only and will not write any answer on the answer book during this period.

SUMMATIVE ASSESSMENT – II

MATHEMATICS

Time allowed: 3 hours Maximum Marks: 80

General Instructions:

(i) All questions are compulsory

(ii) The question paper consists of 30 questions divided into four sections – A, B, C and D

(iii) Section A consists of 6 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 8 questions of 4 marks each.

(iv) There is no overall choice. However, an internal choice has been provided in two questions of 1 mark, two questions of 2 marks, four questions of 3 marks each and three questions of 4 marks each. You have to attempt only one of the alternative in all such questions.

(v) Use of calculator is not permitted.

SECTION – A

Question number 1 to 6 carry 1 mark each.

Question 1: If HCF $\displaystyle (336, 54) = 6$ , find LCM $\displaystyle (336, 54)$ .

Answer:

We know $\displaystyle HCF \times LCM =$ product of the two numbers

$\displaystyle \therefore 6 \times LCM = 336 \times 54$

$\displaystyle \Rightarrow LCM = \frac{336 \times 54}{6} = 3024$

$\displaystyle \\$

Question 2: Find the nature of roots of the quadratic equation $\displaystyle 2x^2 - 4x + 3 = 0$ .

Answer:

Given equation $\displaystyle 2x^2 - 4x + 3 = 0$

$\displaystyle \text{Therefore Roots } = \frac{4 \pm \sqrt{16 - 4 \times 2 \times 3}}{4} = \frac{4 \pm \sqrt{-8}}{4}$

Therefore the nature of the roots is imaginary

$\displaystyle \\$

Question 3: Find the common difference of the Arithmetic Progression $\displaystyle \text{(A.P.) } \frac{1}{a}, \frac{3-a}{3a}, \frac{3-2a}{3a} , \cdots (a \neq 0)$

Answer:

$\displaystyle \text{Given AP: } \frac{1}{a}, \frac{3-a}{3a}, \frac{3-2a}{3a} , \cdots (a \neq 0)$

$\displaystyle \text{Common difference } d = a_2 - a_1, a_3 - a_2, \cdots$

$\displaystyle \therefore d = \frac{3-a}{3a} - \frac{1}{a} = \frac{3 -a - 3}{3a} = - \frac{1}{3}$

$\displaystyle \text{To confirm, we also do } a_3 - a_2 = \frac{3-2a}{3a} - \frac{3-a}{3a} = \frac{3 - 2a - 3 + a}{3a} = - \frac{1}{3}$

$\displaystyle \text{Hence } d = - \frac{1}{3}$

$\displaystyle \\$

Question 4: Evaluate : $\displaystyle \sin^2 60^o+ 2 \tan 45^o - \cos^2 30^o$

$\displaystyle \text{If } \sin A = \frac{3}{4} , \text{ calculate} \sec A$

Answer:

$\displaystyle \sin^2 60^o+ 2 \tan 45^o - \cos^2 30^o$

$\displaystyle = \Big( \frac{\sqrt{3}}{2} \Big)^2 + 2 (1) - \Big( \frac{\sqrt{3}}{2} \Big)^2$

$\displaystyle = 2$

$\displaystyle \text{If } \sin A = \frac{3}{4}$

$\displaystyle \therefore AB = \sqrt{4^2 - 3^2} = \sqrt{7}$

$\displaystyle \therefore \sec A = \frac{4}{\sqrt{7}}$

$\displaystyle \\$

Question 5: Write the coordinates of a point $\displaystyle P$ on $\displaystyle x-axis$ which is equidistant from the points $\displaystyle A(- 2, 0)$ and $\displaystyle B(6, 0)$ .

Answer:

Let $\displaystyle P(x, 0)$ be equidistant from $\displaystyle A$ and $\displaystyle B$

$\displaystyle AP = BP \Rightarrow AP^2 = BP^2$

$\displaystyle \Rightarrow (-2-x)^2 +(0-0)^2 = (6-x)^2 + (0-0)^2$

$\displaystyle \Rightarrow 4 + x^2 + 4x = 36 + x^2 - 12x$

$\displaystyle \Rightarrow 16x = 32 \Rightarrow x = 2$

$\displaystyle \therefore P$ is $\displaystyle (2, 0)$

$\displaystyle \\$

Question 6: In Figure 1, $\displaystyle ABC$ is an isosceles triangle right angled at $\displaystyle C$ with $\displaystyle AC = 4$ cm. Find the length of $\displaystyle AB$ .

In Figure 2, $\displaystyle DE \parallel BC$ . Find the length of side $\displaystyle AD$ , given that $\displaystyle AE = 1.8$ cm, $\displaystyle BD = 7.2$ cm and $\displaystyle CE = 5.4$ cm.

Answer:

Since $\displaystyle \triangle ABC$ is isosceles

$\displaystyle \Rightarrow AC = BC$

(hypotenuse $\displaystyle AB$ has to be the largest side, hence the other two sides are equal)

$\displaystyle \therefore AB = \sqrt{4^2 + 4^2} = 4\sqrt{2}$ cm

Since $\displaystyle DE \parallel BC$ by Thales Theorem

$\displaystyle \frac{AD}{BD} = \frac{AE}{EC}$

$\displaystyle \Rightarrow \frac{AD}{7.2} = \frac{1.8}{5.4}$

$\displaystyle \Rightarrow AD = \frac{1.8}{5.4} \times 7.2 = 2.4 \ cm$

$\displaystyle \\$

Section – B

Question number 7 to 12 carry 2 mark each.

Question 7: Write the smallest number which is divisible by both $\displaystyle 306$ and $\displaystyle 657$ .

Answer:

$\displaystyle 306 = 2 \times 3^2 \times 17$

$\displaystyle 657 = 3^2 \times 73$

LCM of $\displaystyle 306$ and $\displaystyle 657 = 2 \times 3^2 \times 17 \times 73 = 22338$

Hence the smallest number which is divisible by $\displaystyle 306$ and $\displaystyle 657$ is $\displaystyle 22338$

$\displaystyle \\$

Question 8: Find a relation between $\displaystyle x$ and $\displaystyle y$ if the points $\displaystyle A(x, y), B( - 4, 6)$ and $\displaystyle C( - 2, 3)$ are collinear.

Find the area of a triangle whose vertices are given as $\displaystyle (1, - 1) (- 4, 6)$ and $\displaystyle (- 3, - 5)$ .

Answer:

If $\displaystyle A$ , $\displaystyle B$ and $\displaystyle C$ are collinear, then the area of the $\displaystyle \triangle ABC$ should be $\displaystyle 0$ .

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

$\displaystyle \Rightarrow 0 = \frac{1}{2} \times |x(6-3) + (-4)(3-y) + (-2)(y-6) |$

$\displaystyle \Rightarrow 0 = | 3x - 12 + 4y - 2y + 12 |$

$\displaystyle \Rightarrow 0 = |3x + 2y |$

$\displaystyle \Rightarrow 3x + 2y = 0$

Let the points be $\displaystyle A(1, -1), B(-4, 6)$ and $\displaystyle C(-3, -5)$

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

Here $\displaystyle x_1 = 1, y_1 = -1, x_2 = -4, y_2 = 6, x_3 = -3, y_3 = -5$

Substituting in the formula above

$\displaystyle \text{Area of } \triangle ABC = \frac{1}{2} [1(6-(-5)) + (-4) (-5 - (-1)) + (-3)(-1-6) ]$

$\displaystyle = \frac{1}{2} [ 11+ 16 + 21 ] = \frac{1}{2} \times 48 = 24 \text { sq. units.}$

$\displaystyle \\$

Question 9: The probability of selecting a blue marble at random from a jar that contains only blue, black and green marbles is $\displaystyle \frac{1}{5}$ . The probability of selecting a black marble at random from the same jar is $\displaystyle \frac{1}{4}$ . If the jar contains $\displaystyle 11$ green marbles, find the total number of marbles in the jar.

Answer:

$\displaystyle \text{P(selecting Blue marble) } = \frac{1}{5}$

$\displaystyle \text{P(selecting Black marble) } = \frac{1}{4}$

$\displaystyle \text{P(selecting Green marble) } = 1 - \frac{1}{5} - \frac{1}{4} = \frac{20-9}{20} = \frac{11}{20}$

Therefore there are a total of $\displaystyle 20$ marbles in the Jar.

$\displaystyle \\$

Question 10: Find the value(s) of $\displaystyle k$ so that the pair of equations $\displaystyle x + 2y = 5$ and $\displaystyle 3x + ky + 15 = 0$ has a unique solution.

Answer:

Pair of equations $\displaystyle x + 2y = 5$ and $\displaystyle 3x + ky + 15 = 0$

If the intersecting lines have a unique solution, then

$\displaystyle \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$

$\displaystyle \Rightarrow \frac{1}{3} \neq \frac{2}{k} \neq \frac{-5}{15}$

From the first two terms, $\displaystyle k \neq 6$

From the second and the third terms, $\displaystyle k \neq -6$

Hence all values of $\displaystyle k$ except $\displaystyle 6$ and $\displaystyle -6$

$\displaystyle \\$

Question 11: The larger of two supplementary angles exceeds the smaller by $\displaystyle 18^o$ . Find the angles.

Sumit is $\displaystyle 3$ times as old as his son. Five years later, he shall be two and a half times as old as his son. How old is Sumit at present ?

Answer:

Let $\displaystyle x$ and $\displaystyle y$ be the two supplementary angles

$\displaystyle \therefore x + y = 180^o$

Also given $\displaystyle x = y + 18^o$ (given one angle is greater than the other by $\displaystyle 18^o$ )

Solving the two equations

$\displaystyle 2y + 18 = 180^o \Rightarrow 2y = 162^o \Rightarrow y = 81^o$

$\displaystyle \therefore x = 81^o + 18^o = 99^o$

Hence the two angles are $\displaystyle 81^o$ ad $\displaystyle 99^o$

Let the age of son be $\displaystyle x$ years

Therefore are of Sumit is $\displaystyle 3x$

Age of son $\displaystyle 5$ years later $\displaystyle = x + 5$

Age of Sumit $\displaystyle 5$ years later $\displaystyle = 3x + 5$

Give: $\displaystyle 3x+5 = 2.5 (x + 5)$

$\displaystyle \Rightarrow 3x + 5 = 2.5 x + 12.5$

$\displaystyle \Rightarrow 0.5 x = 7.5$

$\displaystyle \Rightarrow x = 15$ years

Therefore age of Sumit $\displaystyle = 3 \times 15 = 45$ years

$\displaystyle \\$

Question 12: Find the mode of the following frequency distribution :

Class Interval:	25-30	30-35	35-40	40-45	45-50	50-55
Frequency:	25	34	50	42	38	14

Answer:

$\\$

Section – C

Question number 13 to 22 carry 3 mark each.

Question 13: Prove that $2 + 5 \sqrt{3}$ is an irrational number, given that $\sqrt{3}$ is an irrational number.

Using Euclid’s Algorithm, find the HCF of $2048$ and $960$ .

Answer:

Let us assume that $5 +2 \sqrt{3}$ is a rational number.

$\displaystyle \text{The rational number is in the form of } \frac{p}{q}$

$\displaystyle \therefore 5 + 2\sqrt{3} = \frac{p}{q}$

$\displaystyle 2\sqrt{3} = \frac{p}{q} - 5$

$\displaystyle 2\sqrt{3} = \frac{p-5q}{q}$

$\displaystyle \sqrt{3} = \frac{p-5q}{2q}$

$\displaystyle \therefore \sqrt{3} \text{ is a rational no as } \frac{p-5q}{2q} \text{ is a rational number }$

But it is given that $\sqrt{3}$ is an irrational number which contradicts our initial assumption.

Hence $5 + 2 \sqrt{3}$ is an irrational number.

$960 ) \overline{2048} ( 2 \\ \hspace*{1cm} \underline{1920} \\ \hspace*{1cm} 128 ) \overline{960} ( 7 \\ \hspace*{2cm} \underline{896} \\ \hspace*{2cm} 64 ) \overline{128} ( 2 \\ \hspace*{3cm} \underline{128} \\ \hspace*{3cm} \times$

Hence HCF $= 64$

$\\$

Question 14: Two right triangles $ABC$ and $DBC$ are drawn on the same hypotenuse $BC$ and on the same side of $BC$ . If $AC$ and $BD$ intersect at $P$ , prove that $AP \times PC = BP \times DP$ .

Diagonals of a trapezium $PQRS$ intersect each other at the point $O, PQ \parallel RS$ and $PQ = 3RS$ . Find the ratio of the areas of triangles $POQ$ and $ROS$ .

Answer:

To prove: $AP \times PC = BP \times PD$

Consider $\triangle ABP$ and $\triangle CDP$

$\angle BAP = \angle CDP = 90^o$ (given)

$\angle BPA = \angle CPD$ ( vertically opposite angles)

$\therefore \triangle ABP \sim \triangle CDP$ (By AA criterion)

$\displaystyle \therefore \frac{BP}{PC} = \frac{AP}{PD}$

$\Rightarrow AP \times PC = BP \times PD$

Hence proved.

Given $PQRS$ is a trapezium

$PQ \parallel RS$ and $PQ = 3 RS$

Consider $\triangle POQ$ and $\triangle SRO$

$\angle SOR = \angle POQ$ (vertically opposite angles)

$\angle RSO = \angle OQP$ (alternate angles)

$\angle SRO = \angle OPQ$ (alternate angles)

$\therefore \triangle POQ \sim \triangle SRO$ (By AAA criterion)

By property of similar triangles

$\displaystyle \frac{ar (\triangle POQ)}{ar(\triangle SOR)} = \frac{PQ^2}{RS^2} = \Big( \frac{PQ}{RS} \Big)^2 = \Big( \frac{3}{1} \Big)^2 = \frac{9}{1}$

$\\$

Question 15: 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 $\angle AOB = 90^o$ .

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

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

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

$\Rightarrow DA \parallel EB$

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 16: Find the ratio in which the line $\displaystyle x - 3y = 0$ divides the line segment joining the points $\displaystyle (- 2, - 5)$ and $\displaystyle (6, 3)$ . Find the coordinates of the point of intersection.

Answer:

Let $\displaystyle P(x, y)$ divides point $\displaystyle A(-2, -5) \text{ and } B (6, 3)$ in the ratio $\displaystyle k : 1$

$\displaystyle \text{Using section formula: } P(x, y) = \Big( \frac{mx_2+nx_1}{m+n} , \frac{my_2+ny_1}{m+n} \Big)$

$\displaystyle A(-2, -5) \text{ and } B(6, 3)$ in the ratio of $\displaystyle k:1$

$\displaystyle \Rightarrow x = \frac{6k-2}{k+1} \text{ and } \Rightarrow y = \frac{3k-5}{k+1}$

Since $\displaystyle P(x, y)$ lies on the line $\displaystyle x-3y = 0$

$\displaystyle \frac{6k-2}{k+1} - 3 \Big( \frac{3k-5}{k+1} \Big) = 0$

$\displaystyle \Rightarrow 6k - 2 - 9k + 15 = 0$

$\displaystyle \Rightarrow -3k +13 = 0$

$\displaystyle \Rightarrow k = \frac{13}{3}$

Therefore $\displaystyle P$ divided $\displaystyle A \text{ and } B$ in the ratio of $\displaystyle 13:3$

$\displaystyle \therefore x = \frac{6(\frac{13}{3})-2}{\frac{13}{3}+1} = \frac{(26-2)\times 3}{16} = \frac{9}{2}$

$\displaystyle y = \frac{3(\frac{13}{3})-5}{(\frac{13}{3})+1} = \frac{(13-5) \times 3}{16} = \frac{3}{2}$

$\displaystyle \text{Hence } P( \frac{9}{2} , \frac{3}{2} ) \text{which is the point of intersection.}$

$\displaystyle \\$

$\displaystyle \text{Question 17: Evaluate: } \Big( \frac{3 \sin 43^{\circ}}{\cos 47^{\circ}} \Big)^2 - \frac{\cos 37^{\circ} \mathrm{cosec} 53^{\circ}}{\tan 5^{\circ} \tan 25^{\circ} \tan 45^{\circ} \tan 65^{\circ} \tan 85^{\circ}}$

Answer:

$\displaystyle \Big( \frac{3 \sin 43^{\circ}}{\cos 47^{\circ}} \Big)^2 - \frac{\cos 37^{\circ} \mathrm{cosec} 53^{\circ}}{\tan 5^{\circ} \tan 25^{\circ} \tan 45^{\circ} \tan 65^{\circ} \tan 85^{\circ}}$

$\displaystyle = \Big( \frac{3 \sin 43^{\circ}}{\cos (90-43)^{\circ}} \Big)^2 - \frac{\cos 37^{\circ} \frac{1}{\sin (90-37)^{\circ} } }{\tan 5^{\circ} \tan 25^{\circ} .1. \tan (90-25)^{\circ} \tan (90-5)^{\circ}}$

$\displaystyle = \Big( \frac{3 \sin 43^{\circ}}{\sin 43^{\circ}} \Big)^2 - \frac{\cos 37^{\circ} \frac{1}{\cos 37^{\circ} } }{\tan 5^{\circ} \tan 25^{\circ} .1. \cot 25^{\circ} \cot 5^{\circ}}$

$\displaystyle = 3^2 -1 = 9-1 = 8$

$\displaystyle \\$

Question 18: In Figure 4, a square $OABC$ is inscribed in a quadrant $OPBQ$ . If $OA = 15$ cm, find the area of the shaded region. (Use $\pi = 3.14$ )

In Figure 5, $ABCD$ is a square with side $2 \sqrt{2}$ cm and inscribed in a circle. Find the area of the shaded region. (Use $\pi = 3.14$ )

Answer:

$OA = OB = 15$ cm (since $OABC$ is a square)

$\therefore OB = \sqrt{15^2 + 15^2} = 15 \sqrt{2} =$ Radius

Area of square $OABC = 15 \times 15 = 225 \ cm^2$

$\displaystyle \text {Area of quadrant } = \frac{1}{4} (\pi (15 \sqrt{2})^2 ) = 353.25 \ cm^2$

Shaded area $= 353.25 - 225 = 128.25 \ cm^2$

Area of square $= 2\sqrt{2} \times 2\sqrt{2} = 8 \ cm^2$

Diameter $AC = \sqrt{(2\sqrt{2})^2 + (2\sqrt{2})^2} = \sqrt{8+8} = 4 \ cm$

Therefore Radius of circle $= 2 \ cm$

Area of circle $= \pi (2)^2 = 3.14 \times 4 = 12.56 \ cm^2$

Therefore shaded area $= 12.56 - 8 = 4.56 \ cm^2$

$\\$

Question 19: A solid is in the form of a cylinder with hemispherical ends. The total height of the solid is $20$ cm and the diameter of the cylinder is $7$ cm. Find the total volume of the solid. (Use $\pi = \frac{22}{7}$ )

Answer:

Radius of hemisphere $= 3.5 \ cm$

Therefore height of cylinder $= 20-3.5 -3.5 = 13 \ cm$

$\displaystyle \text{Therefore total volume }= 2 \times ( \frac{1}{2} \times \frac{4}{3} \times \pi (3.5)^2 ) + \pi (3.5)^2 \times 13$

$= 179.67 + 500.5 = 680.17 \ cm^2$

$\\$

Question 20: The marks obtained by $100$ students in an examination are given below :

Marks:	30-35	35-40	40-45	45-50	50-55	55-60	60-65
Number of Students:	14	16	28	23	18	8	3

Answer:

Marks ( Class interval)	No. of Students (Frequency)	Mean Value $latex (x) $	$fx$
30-35	14	32.5	455
35-40	16	37.5	600
40-45	28	42.5	1190
45-50	23	47.5	1092.5
50-55	18	52.5	945
55-60	8	57.5	460
60-65	3	62.5	187.5
	$\Sigma f = 110$		$\Sigma fx = 4930$

$\displaystyle \text{Mean } = \frac{\Sigma fx}{\Sigma f} = \frac{4930}{110} = 44.818$

$\displaystyle \\$

Question 21: For what value of k, is the polynomial $\displaystyle f(x) = 3x^4 - 9x^3 + x^2 + 15x + k$ completely divisible by $\displaystyle 3x^2 - 5$ ?

$\displaystyle \text{Find the zeroes of the quadratic polynomial } 7y^2 - \frac{11}{3} y - \frac{2}{3} \text{and verify the}$ relationship between the zeroes and the coefficients.

Answer:

$\displaystyle f(x) \text{ divisible by } P(x) \Rightarrow \text{ the remainder of } \frac{f(x)}{P(x)} = 0$

$\displaystyle 3x^2-5 ) \overline{3x^4 - 9x^3 + x^2 + 15x + k} ( x^2-3x+2 \\ \hspace*{1cm}(-) { 3x^4-5x^2} \\ \hspace*{2cm} \overline{-9x^3 + 6x^2 + 15x+ k} \\ \hspace*{2cm}(-) \underline{ -9x^3+15x} \\ \hspace*{3.5cm} {6x^2+k} \\ \hspace*{3cm} (-) \underline{6x^2-10} \\ \hspace*{4cm} k+10$

Remainder should be $\displaystyle 0 \Rightarrow k+10 = 0 \Rightarrow k = -10$

$\displaystyle (3x^2-5) \text{ and } (x^2 - 3x+2)$ are factors of $\displaystyle f(x)$

$\displaystyle \Rightarrow (\sqrt{3} x - \sqrt{5}) \text{ and } (\sqrt{3} x + \sqrt{5})$ are factors

$\displaystyle \Rightarrow x = \frac{\sqrt{5}}{\sqrt{3}} \text{ or } - \frac{\sqrt{5}}{\sqrt{3}}$

Also $\displaystyle x^2 - 3x + 2 = (x-2)(x-1)$ are factors $\displaystyle \Rightarrow x = 2, 1$

$\displaystyle \text{Hence } \frac{\sqrt{5}}{\sqrt{3}} , - \frac{\sqrt{5}}{\sqrt{3}} , 2, 1 \text{ are zeros of } f(x)$

$\displaystyle \text{Given polynomial } P(y) = y^2 - \frac{11}{3} y - \frac{2}{3} = 0$

$\displaystyle \Rightarrow 21y^2 - 11y - 2 = 0$

$\displaystyle \Rightarrow 7y(3y-2) + (3y-2) = 0$

$\displaystyle \Rightarrow (3y-2)(7y+1) = 0$

$\displaystyle \Rightarrow y = \frac{2}{3} \text{ or } y = - \frac{1}{7}$

$\displaystyle \text{Comparing } P(y) \text{ with } ax^2 + bx + c =0 \text{ we get } a = 7, b = - \frac{11}{3} \text{ and } c = - \frac{2}{3}$

$\displaystyle \text{Sum of zeros } = - \frac{b}{a} = - (- \frac{11}{3} ) ( \frac{1}{7} ) = \frac{11}{21}$ … … … … … i)

$\displaystyle \text{Sum of roots } = \frac{2}{3} + \Big( \frac{-1}{7} \Big) = \frac{11}{21}$ … … … … … ii)

Therefore i) = ii)

$\displaystyle \text{Products of zeros } = \frac{c}{a} = \frac{-\frac{2}{3}}{7} = \frac{-2}{21}$ … … … … … iii)

$\displaystyle \text{Product of roots } = \frac{2}{3} \times \Big( \frac{-1}{7} \Big) = \frac{-2}{21}$ … … … … … iv)

Therefore iii) = iv)

$\displaystyle \\$

Question 22: Write all the values of $p$ for which the quadratic equation $x^2 + px + 16 = 0$ has equal roots. Find the roots of the equation so obtained.

Answer:

Given quadratic equation is: $x^2 + px + 16 = 0$

If the roots are equal then the discriminant is $0$

i.e. $b^2 - 4ac = 0$

$\Rightarrow p^2 - 4 (1) (16) = 0$

$\Rightarrow p^2 = 64$

$\Rightarrow p = \pm 8$

When $p = 8, x^2 + 8x + 16 = 0$

$\Rightarrow (x+4)(x+4) = 0$

$\Rightarrow x = -4$

When $p = -8, x^2 - 8x + 16 = 0$

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

$\Rightarrow x = 4$

Therefore the roots of equation are $4, -4$

$\\$

Section – D

Question number 23 to 30 carry 4 mark each.

Question 23: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then prove that the other two sides are divided in the same ratio.

Answer:

Given: In $\triangle ABC, DE$ intersects $AB$ and $AC$ at $D$ and $E$ respectively.

$\displaystyle \text{To Prove: } \frac{AD}{DB} = \frac{AE}{EC}$

Construct: $DM \perp AC$ and $EN \perp AB$

Proof:

$\displaystyle \text{Area of } \triangle ADE = \frac{1}{2} \times AD \times EN$

$\displaystyle \text{Area of } \triangle BDE = \frac{1}{2} \times BD \times EN$

$\displaystyle \therefore \frac{ar(\triangle ADE)}{ar(\triangle BDE)} = \frac{AD}{BD}$ … … … … … i)

Now $ar(\triangle BDE) = ar(\triangle CDE)$ … … … … … ii)

(Because they have the same base $DE$ and are between the same parallel)

$\displaystyle \text{Similarly, } \frac{ar(\triangle ADE)}{ar(\triangle CDE)} = \frac{AE}{EC}$ … … … … … iii)

$\displaystyle \text{From i) , ii) and iii) we get } \frac{AD}{DB} \frac{AE}{EC} \text{. Hence proved.}$

$\\$

Question 24: Amit, standing on a horizontal plane, finds a bird flying at a distance of $200$ m from him at an elevation of $30^o$ . Deepak standing on the roof of a $50$ m high building, finds the angle of elevation of the same bird to be $45^o$ . Amit and Deepak are on opposite sides of the bird. Find the distance of the bird from Deepak.

Answer:

$\displaystyle \frac{h}{200} = \sin 30^o = \frac{1}{2}$

$\Rightarrow h = 100$ m

$\displaystyle \therefore \frac{100-50}{BD} = \sin 45^o = \frac{1}{\sqrt{2}}$

$\therefore BD = 50\sqrt{2}$

$\\$

Question 25: A solid iron pole consists of a cylinder of height $220$ cm and base diameter $24$ cm, which is surmounted by another cylinder of height $60$ cm and radius $8$ cm. Find the mass of the pole, given that $1 \ cm^3$ of iron has approximately $8$ gm mass. (Use $\pi = 3.14$ )

Answer:

Volume of pole $= \pi (12)^2 \times 220 + \pi (8)^2 \times 60$

$= 31680 \pi + 3840 \pi$

$= 35520 \times 3.14$

$= 111532.8 \ cm^3$

$\displaystyle \text{Weight of the pole } = 111532.8 \ cm^3 \times 8 \frac{gm}{cm^3} = 892262.4 \ gm = 892.26 \ kg$

$\\$

Question 26: Construct an equilateral $\triangle ABC$ with each side $5$ cm. Then construct another triangle whose sides are $\frac{2}{3}$ times the corresponding sides of $\triangle ABC$ .

Draw two concentric circles of radii $2$ cm and $5$ cm. Take a point $P$ on the outer circle and construct a pair of tangents $PA$ and $PB$ to the smaller circle. Measure $PA$ .

Answer:

$\\$

Question 27: Change the following data into ‘less than type’ distribution and draw its ogive :

Class Interval:	30-40	40-50	50-60	60-70	70-80	80-90	90-100
Frequency:	7	5	8	10	6	6	8

Answer:

$\\$

$\displaystyle \text{Question 28: Prove that: } \frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \sec \theta \ \mathrm{cosec} \theta$

$\displaystyle \text{Prove that: } \frac{\sin \theta}{\cot \theta + \mathrm{cosec} \theta } = 2 + \frac{\sin \theta}{\cot \theta - \mathrm{cosec} \theta}$

Answer:

$\displaystyle \text{LHS } = \frac{\tan \theta}{1- \cot \theta} + \frac{\cot \theta}{1 - \tan \theta}$

$\displaystyle = \frac{\tan \theta}{1- \frac{1}{\tan \theta}} + \frac{\frac{1}{\tan \theta}}{1 - \tan \theta}$

$\displaystyle = \frac{\tan^2 \theta}{\tan \theta -1 } + \frac{1}{\tan \theta (1 - \tan \theta)}$

$\displaystyle = - \frac{\tan^2 \theta}{1- \tan \theta } + \frac{1}{\tan \theta (1 - \tan \theta)}$

$\displaystyle = \frac{1 - \tan^3 \theta}{\tan \theta (1 - \tan \theta)}$

$\displaystyle = \frac{1 + \tan^2 \theta + \tan \theta}{\tan \theta }$

$\displaystyle = \frac{\sec^2 \theta + \tan \theta}{\tan \theta }$

$\displaystyle = \sec^2 \theta \cot \theta + 1$

$\displaystyle = \frac{1}{\cos^2 \theta} \times \frac{\cos \theta}{\sin \theta} + 1$

$\displaystyle = \frac{1}{\cos \theta \sin \theta} + 1$

$\displaystyle = 1 + \sec \theta \ \mathrm{cosec} \theta$

$\displaystyle \text{LHS } = \frac{\sin \theta}{\cot \theta+ \mathrm{cosec} \theta}$

$\displaystyle = \frac{\sin \theta}{\frac{\cos \theta}{\sin \theta} + \frac{1}{\sin \theta}}$

$\displaystyle = \frac{\sin^2 \theta}{\cos \theta+1}$

$\displaystyle = \frac{\sin^2 \theta}{\cos \theta+1} \times \frac{1 - \cos \theta}{1 - \cos \theta}$

$\displaystyle = \frac{\sin^2 \theta (1 - \cos \theta)}{1- \cos^2 \theta}$

$\displaystyle = \frac{\sin^2 \theta (1 - \cos \theta)}{\sin^2 \theta}$

$\displaystyle = 1- \cos \theta$

$\displaystyle \text{RHS } = 2 + \frac{\sin \theta}{\cot \theta - \mathrm{cosec} \theta}$

$\displaystyle = 2 + \frac{\sin \theta}{\frac{\cos \theta}{\sin \theta} - \frac{1}{\sin \theta}}$

$\displaystyle = 2 + \frac{\sin^2 \theta}{\cos \theta-1}$

$\displaystyle = 2 - \frac{\sin^2 \theta}{1-\cos \theta}\frac{(1 + \cos \theta)}{(1 + \cos \theta)}$

$\displaystyle = 2 - \frac{\sin^2 \theta}{1- \cos^2 \theta} (1 + \cos \theta)$

$\displaystyle = 2 - ( 1 + \cos \theta)$

$\displaystyle = 1 - \cos \theta$

$\displaystyle =$ LHS

Hence Proved.

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Question 29: Which term of the Arithmetic Progression $-7, -12, -17, -22, \cdots$ will be $- 82$ ? Is $-100$ any term of the A.P. ? Give reason for your answer.

How many terms of the Arithmetic Progression $45, 39, 33, \cdots$ must be taken so that their sum is $180$ ? Explain the double answer.

Answer:

Given AP: $-7, -12, -17, -22, \cdots$

$a_1 = -7$

Common difference $= -12 - (-7) = -5$

The $n^{th}$ term of AP

$T_n = a + (n-1)d$

$\Rightarrow -82 = -7 + (n-1)(-5)$

$\Rightarrow -82 = -7 - 5n + 5$

$\Rightarrow -82 = -2 - 5n$

$\Rightarrow -80 = -5n$

$\Rightarrow n = 16$

Hence $-82$ is the $16^{th}$ term

Since the numbers in the given $AP$ are not factors of $5, - 100$ will not be part of the AP.

Given AP is $45, 39, 33, \cdots$

$a_1 = 45$

Common difference $d = 39 - 45 = - 6$

$S_n = 180$

$\frac{n}{2}$ $(2a+(n-1)d) = 180$

$\frac{n}{2}$ $( 2 \times 45 + (n-1)(-6)) = 180$

$n(90-6n + 6) = 360$

$n(96-6n) = 360$

$n(16-n) = 60$

$n^2 - 16n + 60 = 0$

$(n-10)(n-6) = 0$

$n = 10$ or $n = 6$

We get double answer because the sum of the $7^{th}$ to $10^{th}$ term is $0$ .

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Question 30: In a class test, the sum of Arun’s marks in Hindi and English is $30$ . Had he got $2$ marks more in Hindi and $3$ marks less in English, the product of the marks would have been $210$ . Find his marks in the two subjects.