2018 CBSE (Class 10) Board Paper Solution Mathematics : 30(b)

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: Find the value of $\displaystyle k$ for which the roots of the quadratic equation

$\displaystyle (k - 5)x^2 + 2(k - 5) x + 2 = 0$ are equal.

Answer:

Given equation: $\displaystyle (k - 5)x^2 + 2(k - 5) x + 2 = 0$

For roots to be equal, $\displaystyle b^2 - 4ac = 0$

$\displaystyle \Rightarrow [ 2(k-5)]^2 - 4 (k-5)(2) = 0$

$\displaystyle \Rightarrow k^2 +25-10k-2k +10 = 0$

$\displaystyle \Rightarrow k^2 - 12k + 35 = 0$

$\displaystyle \Rightarrow (k-7)(k-5) = 0$

$\displaystyle \Rightarrow k = 7 \text{ or } k = 5$

$\displaystyle \\$

Question 2: Find the value of $\displaystyle y$ for which the distance between the points $\displaystyle (2, -3) \text{ and } (10, y)$ is $\displaystyle 10$ units.

Answer:

Given distance between the points $\displaystyle (2, -3) \text{ and } (10, y)$ is $\displaystyle 10$ units.

$\displaystyle \Rightarrow \sqrt{(10-2)^2 + (y+3)^2} = 10$

$\displaystyle \Rightarrow 8^2 + (y+3)^2 = 10^2$

$\displaystyle \Rightarrow y^2 + 9 + 6y = 36$

$\displaystyle \Rightarrow y^2 + 6y - 27 = 0$

$\displaystyle \Rightarrow y^2 + 9y - 3y - 27 = 0$

$\displaystyle \Rightarrow y(y+9) -3(y+9) = 0$

$\displaystyle \Rightarrow (y+9) (y-3) = 0$

$\displaystyle \Rightarrow y = -9 \text{ or } y = 3$

$\displaystyle \\$

$\displaystyle \text{Question 3: Write whether the rational number } \frac{13}{3125} \\ \\ \text{has a decimal expansion which is terminating or non-terminating repeating.}$

Answer:

$\displaystyle \frac{13}{3125} = \frac{13}{2^0 \times 5^5} = \frac{13}{2^0 \times 5^5} \times \frac{2^5}{2^5} = \frac{13 \times 2^5}{10^5} = \frac{416}{10^5} = 0.00416$

$\displaystyle \\$

$\displaystyle \text{Question 4: Write the } m^{th} \text{ term of the A.P. } \frac{1}{k}$ , $\displaystyle \frac{1+k}{k}$ , $\displaystyle \frac{1+2k}{k} , \cdots$

Answer:

From the given AP

$\displaystyle a = \frac{1}{k} \text{ and } d = \frac{1+k}{k} - \frac{1}{k} = \frac{k}{k} = 1$

$\displaystyle \therefore T_m = a+(m-1)d$

$\displaystyle \Rightarrow T_m = \frac{1}{k} + (m-1)$

$\displaystyle \Rightarrow T_m = \frac{k(m-1)+1}{k}$

$\displaystyle \\$

Question 5: If $\displaystyle \sin \theta + \cos \theta = 2 \cos (90^o - \theta)$ , find the value of $\displaystyle \cot \theta$

Answer:

$\displaystyle \sin \theta + \cos \theta = 2 \cos (90^o - \theta)$

$\displaystyle \sin \theta + \cos \theta = 2 \sin \theta$

$\displaystyle \Rightarrow 1 + \cot \theta = \sqrt{2}$

$\displaystyle \Rightarrow \cot \theta = \sqrt{2} -1$

$\displaystyle \\$

Question 6: $\displaystyle DE$ is drawn parallel to the base $\displaystyle BC$ of a $\displaystyle \triangle ABC$ , meeting

$\displaystyle AB \text{ at} D \text{ and } AC \text{ at } E . \text{ If } \frac{AB}{BD} = 4 \text{ and } CE = 2 \text{ cm, find } AE .$

Answer:

$\displaystyle \text{Given: } DE \parallel BC, \frac{AB}{BD} = 4$

$\displaystyle \Rightarrow \triangle ABC \sim \triangle ADE$ by AAA criterion

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

$\displaystyle \frac{AB}{AB-BD} = \frac{AC}{AC-CE}$

$\displaystyle \frac{1}{1-\frac{BD}{AB} } = \frac{1}{1- \frac{CE}{AC}}$

$\displaystyle 1- \frac{CE}{AC} = 1- \frac{BD}{AB}$

$\displaystyle \frac{CE}{AC} = \frac{BD}{AB}$

$\displaystyle AC = \frac{AB}{BD} \times CE = 4 \times 2 = 8 \text{ cm }$

$\displaystyle \therefore AE = AC = CE = 8 - 2 = 6 \text{ cm }$

$\displaystyle \\$

Section – B

Question number 7 to 12 carry 2 mark each.

Question 7: A bag contains $\displaystyle 5$ red balls and some blue balls. If the probability of drawing a blue ball from the bag is three times that of a red ball, find the number of blue balls in the bag.

Answer:

Number of Red balls $\displaystyle = 5$

Let the number of Blue balls $\displaystyle = x$

$\displaystyle \therefore \frac{x}{5+x} = 3 \times \frac{5}{5+x}$

$\displaystyle \Rightarrow x = 15$

No of Blue balls in the bag $\displaystyle = 15$

$\displaystyle \\$

Question 8: The $\displaystyle 5^{th} \text{ and } 15^{th}$ terms of an A.P. are $\displaystyle 13 \text{ and } -17$ respectively. Find the sum of first $\displaystyle 21$ terms of the A.P.

Answer:

Let the first term $\displaystyle = a$ and the common difference $\displaystyle = d$

$\displaystyle T_5 = a + (5-1)d = 13$

$\displaystyle \Rightarrow a + 4d = 13$ … … … … … i)

$\displaystyle T_{15} = a + (15-1)d = -17$

$\displaystyle \Rightarrow a + 14d = -17$ … … … … … ii)

Subtracting ii) from i)

$\displaystyle 10d = - 30 \Rightarrow d = -3$

$\displaystyle \therefore a = 13 - 4(-3) = 25$

$\displaystyle \therefore a = 25 \text{ and } d = -3$

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

$\displaystyle \therefore S_{21} = \frac{21}{2} [2(25) + (21-1)(-3) ]$

$\displaystyle \Rightarrow S_{21} = \frac{21}{2} [50) + 20(-3) ]$

$\displaystyle \Rightarrow S_{21} = \frac{21}{2} [50 -60 ]$

$\displaystyle \Rightarrow S_{21} = 21 \times (-5) = -105$

Therefore the sum of the first $\displaystyle 21$ terms is $\displaystyle - 105$

$\displaystyle \\$

Question 9: Using Euclid’s Division Algorithm, find the HCF of $\displaystyle 255 \text{ and } 867$ .

Answer:

$\displaystyle 255 ) \overline{867} ( 3 \\ \hspace*{0.7cm} \underline{765} \\ \hspace*{0.8cm} 102 ) \overline{255} ( 2 \\ \hspace*{1.5cm} \underline{204} \\ \hspace*{1.7cm} 51 ) \overline{102} ( 2 \\ \hspace*{2.25cm} \underline{102} \\ \hspace*{2.5cm} \times$

According to Eculid’s division theorem, any positive number can be expressed as $\displaystyle a = bq+r$ where $\displaystyle q$ is the quotient, $\displaystyle b$ is the divisor and $\displaystyle r$ is the remainder and $\displaystyle 0 \leq r < b$

$\displaystyle \therefore 867 = 51 \times 17 + 0$

So HCF of $\displaystyle 867 \text{ and } 255$ is $\displaystyle 51$

$\displaystyle \\$

Question 10: If the point $\displaystyle (0, 2)$ is equidistant from the points $\displaystyle (3, k) \text{ and } (k, 5)$ , find the value of $\displaystyle k$ .

Answer:

Given point $\displaystyle (0, 2)$ is equidistant from the points $\displaystyle (3, k) \text{ and } (k, 5)$

Applying distance formula

$\displaystyle \sqrt{(3-0)^2+(k-2)^2} = \sqrt{(k-0)^2+(5-2)^2}$

$\displaystyle \Rightarrow 9 + (k-2)^2 = k^2 + 9$

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

$\displaystyle \Rightarrow k = 1$

$\displaystyle \\$

Question 11: Find the value of $\displaystyle 'a'$ for which the pair of linear equations $\displaystyle 2x + 3y = 7 \text{ and } 4x + ay = 14$ has infinitely many solutions.

Answer:

If the system of equations are $\displaystyle ax_1 + b_1y + c_1 = 0 \text{ and } ax_2 + b_2y + c_2 = 0$ and they have infinitely many solutions then it satisfy the following:

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

$\displaystyle \frac{2}{4} = \frac{3}{a} = \frac{7}{14}$

From first two terms $\displaystyle a = 6$

From Last two terms $\displaystyle a = 6$

Therefore for $\displaystyle a = 6$ equations $\displaystyle 2x + 3y = 7 \text{ and } 4x + ay = 14$ has infinitely many solutions.

$\displaystyle \\$

Question 12: A card is drawn at random from a well shuffled pack of $\displaystyle 52$ playing cards. Find the probability of getting (i) a red king (ii) a queen or a jack

Answer:

Total number of cards $\displaystyle = 52$

Number of Red kings $\displaystyle = 2$

Number of queens and jacks $\displaystyle = 8$

$\displaystyle \text{i) Probability (Red King) } = \frac{2}{52} = \frac{1}{26}$

$\displaystyle \text{ii) Probability (Queen or Jack) } = \frac{8}{52} = \frac{2}{13}$

$\displaystyle \\$

Section – C

Question number 13 to 22 carry 3 mark each.

Question 13: Show that any positive odd integer is of the form $\displaystyle 4q + 1 \text{ or } 4q + 3$ for some integer $\displaystyle q$ .

Answer:

Let $\displaystyle a$ be any positive integer.

Eculid’s division theorem, any positive number can be expressed as $\displaystyle a = bq+r$ where $\displaystyle q$ is the quotient, $\displaystyle b$ is the divisor and $\displaystyle r$ is the remainder and $\displaystyle 0 \leq r < b$

Take $\displaystyle b = 4 \Rightarrow a = 4q + r$

Since $\displaystyle 0 \leq r < 4$ , the possible remainders are $\displaystyle 0, 1, 2, 3$

That is $\displaystyle a$ can be $\displaystyle 4q, 4q+1, 4q+2 \text{ or } 4q+3$

Since $\displaystyle a$ is odd, $\displaystyle a$ cannot be $\displaystyle 4q \text{ or } 4q+2$

Therefore any odd integer is of the form $\displaystyle 4q+1 \text{ or } 4q+3$ .

$\displaystyle \\$

Question 14: The ten’s digit of a number is twice its unit’s digit. The number obtained by interchanging the digits is $\displaystyle 36$ less than the original number. Find the original number.

Answer:

Let tens digit be $\displaystyle x$ and unit digit be $\displaystyle y$

Therefore the number $\displaystyle =10x+y$

Interchanged number $\displaystyle =10 y+x$

And (tens digit is twice the unit ) That is, $\displaystyle x=2y$ … … … … … i)

Now according to the question

$\displaystyle 10 y+x=10x+y-36$

$\displaystyle \Rightarrow 9y =9x-36$

$\displaystyle \Rightarrow y=x-4 \text{from i)}$

$\displaystyle \Rightarrow y=2y-4$

$\displaystyle \Rightarrow y=4$

Then $\displaystyle \Rightarrow x=2y=2 (4)=8$

Therefore the number is $\displaystyle 84$

$\displaystyle \\$

Question 15: The line segment joining the points $\displaystyle A(2, 1) \text{ and } B(5, -8)$ is trisected at the points $\displaystyle P \text{ and } Q$ , where $\displaystyle P$ is nearer to $\displaystyle A$ . If $\displaystyle P$ lies on the line $\displaystyle 2x - y + k = 0$ , find the value of $\displaystyle k$ .

The $\displaystyle x-$ coordinate of a point $\displaystyle P$ is twice its $\displaystyle y-$ coordinate. If $\displaystyle P$ is equidistant from the points $\displaystyle Q(2, -5) \text{ and } R(-3, 6)$ , find the coordinates of $\displaystyle P$ .

Answer:

$\displaystyle AP : AB = 1:3$ (trisects)

$\displaystyle \frac{AP}{AB} = \frac{1}{3}$

$\displaystyle \Rightarrow \frac{AP}{AP+PB} = \frac{1}{3}$

$\displaystyle \Rightarrow 3AP = AP + PB$

$\displaystyle \Rightarrow 2AP = PB$

$\displaystyle \Rightarrow \frac{AP}{PB} = \frac{1}{2}$

$\displaystyle \Rightarrow AP : PB = 1:2$

Applying section formula

$\displaystyle \text{ Coordinates of } P = \Big( \frac{1 \times 5 + 2 \times 2}{1 + 2} , \frac{1 \times (-8) + 2 \times (1)}{1 + 2} \Big)$

$\displaystyle \Rightarrow P = \Big( \frac{5+4}{3} , \frac{-8+2}{3} \Big)$

$\displaystyle \Rightarrow P = (3, -2)$

$\displaystyle \text{Since } P(3, -2)$ lies on $\displaystyle 2x - y + k = 0$

$\displaystyle \therefore 2(3) - (-2) + k = 0$

$\displaystyle \Rightarrow 6 + 2 + k = 0$

$\displaystyle \Rightarrow k = -8$

Let $\displaystyle P(x, y)$ be the required point

Given: $\displaystyle P$ is equidistant from the points $\displaystyle Q(2, -5) \text{ and } R(-3, 6)$

$\displaystyle \therefore PQ = PR \Rightarrow PQ^2 = PR^2$

$\displaystyle \Rightarrow (x-2)^2 + (y + 5)^2 = (x+3)^2 + (y-6)^2$

$\displaystyle \Rightarrow x^2 + 4 - 4x + y^2 + 25 + 10y = x^2 + 9 + 6x + y^2 + 36 - 12y$

$\displaystyle \Rightarrow -10x + 22y - 16 = 0$

Now $\displaystyle x = 2y$

$\displaystyle \Rightarrow -10(2y) + 22y = 16$

$\displaystyle \Rightarrow -20y + 22y = 16$

$\displaystyle \Rightarrow 2y = 18$

$\displaystyle \Rightarrow y = 8$

$\displaystyle \therefore x = 2 \times 8 = 16$

Hence $\displaystyle P$ is $\displaystyle (16, 8)$

$\displaystyle \\$

$\displaystyle \text{Question 16: Show that } 1 , \frac{1}{2} \text{ and } -2 \text{ are the zeroes of the polynomial } \\ \\ 2x^3 + x^2 - 5x + 2 .$

Answer:

Comparing given cubic equation to $\displaystyle ax^3 + bx^2 + cx + d$ we get $\displaystyle a = 2, b = 1, c = -5 \text{ and } d = 2$

$\displaystyle \text{Take } \alpha = \frac{1}{2} , \ \ \beta = 1 \text{ and } \gamma = -2$

Now we verify the relations between zeros and their coefficients

$\displaystyle \alpha + \beta + \gamma = \frac{1}{2} + 1 - 2 = \frac{-1}{2} = \frac{-b}{a}$

$\displaystyle \alpha . \beta + \beta . \gamma + \gamma . \alpha = \frac{1}{2} \times 1 + 1 \times (-2) + (-2) \times \frac{1}{2} = \frac{1}{2} -2 - 1 = \frac{-5}{2} = \frac{c}{a}$

$\displaystyle \alpha . \beta . \gamma = \frac{1}{2} \times 1 \times (-2) = -1 = \frac{-2}{2} = \frac{-d}{a}$

Hence verified.

Question 17: Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.

Answer:

Given: Circle with center $\displaystyle O$ . Let $\displaystyle PA \text{ and } PB$ be tangents from external point P.

To Prove: $\displaystyle \angle APB + \angle AOB = 180^o$

Proof: Since $\displaystyle PA$ is a tangent, $\displaystyle OA \perp PA \Rightarrow \angle OAP = 90^o$

Similarly, since $\displaystyle PB$ is a tangent, $\displaystyle OB \perp PB \Rightarrow \angle OBP = 90^o$

In Quadrilateral $\displaystyle OAPB$

$\displaystyle \angle OAP + \angle APB + \angle OBP + \angle AOB = 360^o$

$\displaystyle \Rightarrow 90^o + \angle APB + 90^o + \angle AOB = 360^o$

$\displaystyle \Rightarrow \angle APB + \angle AOB = 180^o$

Hence proved.

$\displaystyle \\$

Question 18: $\displaystyle S \text{ and } T$ are points on the sides $\displaystyle PR \text{ and } QR$ of $\displaystyle \triangle PQR$ such that $\displaystyle \angle P = \angle RTS$ . Show that $\displaystyle \triangle RPQ \sim \triangle RTS$ .

$\displaystyle \text{In an equilateral } \triangle ABC, D \text{ is a point on the side } BC \text{ such that } \\ \\ BD = \frac{1}{3} BC . \text{ Prove that } 9AD^2 = 7AB^2 .$

Answer:

Given: $\displaystyle \triangle PQR$ and points $\displaystyle S \text{ and } T$ on sides of $\displaystyle PR \text{ and } QR$

$\displaystyle \angle P = \angle RTS$

To Prove: $\displaystyle \triangle RPQ \sim \triangle RTS$

Proof: In $\displaystyle \triangle RPQ \text{ and } \triangle RTS$

$\displaystyle \angle P = \angle RTS$ (given)

$\displaystyle \angle PRQ = \angle TRS = \angle R$ (common angle)

$\displaystyle \therefore \triangle RPQ \cong \triangle RTS$ (by AA criterion)

Hence proved.

Given: $\displaystyle \triangle ABC$ is an equilateral triangle.

$\displaystyle \Rightarrow AB = BC = CA$

$\displaystyle \text{Also } BD = \frac{1}{3} BC$

To Prove: $\displaystyle 9 AD^2 = 7 AB^2$

Construction: Draw $\displaystyle AE \perp BC$

Proof: Consider $\displaystyle \triangle AEB \text{ and } \triangle AEC$

$\displaystyle AE$ is common

$\displaystyle \angle AEB = \angle AEC = 90^o$

$\displaystyle AB = AC$ (equilateral triangle)

$\displaystyle \triangle AEB \cong \triangle AEC$ (By RHS criterion)

$\displaystyle \therefore BE = EC$

$\displaystyle \Rightarrow BE = \frac{BC}{2}$

$\displaystyle \Rightarrow BD + DE = \frac{BC}{2}$

$\displaystyle \Rightarrow \frac{BC}{3} + DE = \frac{BC}{2}$

$\displaystyle \Rightarrow DE = \frac{BD}{6}$

Using Pythagoras theorem,

In $\displaystyle \triangle AEB$

$\displaystyle AB^2 = AE^2 + BE^2 \Rightarrow AE^2 = AB^2 - BE^2$ … … … … … i)

In $\displaystyle \triangle AED$

$\displaystyle AD^2 = AE^2 + DE^2 \Rightarrow AE^2 = AD^2 - DE^2$ … … … … … ii)

From i) and ii)

$\displaystyle AB^2 - BE^2 = AD^2 - DE^2$

$\displaystyle \Rightarrow AB^2 - \Big( \frac{BC}{2} \Big)^2 = AD^2 - \Big( \frac{BC}{6} \Big)^2$

$\displaystyle \Rightarrow AB^2 - \frac{BC^2}{4} = AD^2 - \frac{BC^2}{36}$

$\displaystyle \Rightarrow AB^2 = AD^2 + \frac{8}{36} BC^2$

$\displaystyle AB^2 = AD^2 + \frac{2}{9} BC^2$

But $\displaystyle BC = AB$

$\displaystyle \therefore AB^2 - \frac{2}{9} AB^2 = AD^2$

$\displaystyle \Rightarrow 7AB^2 = 9AD^2$

Hence proved.

$\displaystyle \\$

$\displaystyle \text{Question 19: Prove that: } \frac{1}{\mathrm{cosec} \theta + \cot \theta} - \frac{1}{\sin \theta} = \frac{1}{\sin \theta} - \frac{1}{\mathrm{cosec} \theta - \cot \theta}$

If $\displaystyle \tan \theta + \sin \theta = m, \tan \theta - \sin \theta = n$ , show that $\displaystyle m^2 - n^2 = 4 \sqrt{mn}$

Answer:

$\displaystyle \text{Prove } \frac{1}{\mathrm{cosec} \theta + \cot \theta} - \frac{1}{\sin \theta} = \frac{1}{\sin \theta} - \frac{1}{\mathrm{cosec} \theta - \cot \theta}$

$\displaystyle \text{or Prove } \frac{1}{\mathrm{cosec} \theta + \cot \theta} + \frac{1}{\mathrm{cosec} \theta - \cot \theta} = \frac{2}{\sin \theta}$

$\displaystyle \Rightarrow \frac{\mathrm{cosec} \theta - \cot \theta + \mathrm{cosec} \theta + \cot \theta}{\mathrm{cosec}^2 \theta - \cot^2 \theta} = \frac{2}{\sin \theta}$

$\displaystyle \Rightarrow \frac{2 \mathrm{cosec} \theta}{\mathrm{cosec}^2 \theta - \cot^2 \theta} = \frac{2}{\sin \theta}$

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

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

$\displaystyle \Rightarrow \frac{2}{ \sin \theta }$ $\displaystyle \frac{ \sin^2 \theta }{ \sin^2 \theta } = \frac{2}{ \sin \theta }$

$\displaystyle \Rightarrow \frac{2}{ \sin \theta } = \frac{2}{ \sin \theta }$

Hence proved.

Given:

$\displaystyle \tan \theta + \sin \theta = m$

$\displaystyle \tan \theta - \sin \theta = n$

$\displaystyle \therefore m^2 - n^2 = ( \tan \theta + \sin \theta )^2 - ( \tan \theta - \sin \theta )^2$

$\displaystyle = \tan^2 \theta + \sin^2 \theta + 2 \tan \theta \sin \theta - \tan^2 \theta - \sin^2 \theta + 2 \tan \theta \sin \theta$

$\displaystyle = 4 \tan \theta \ \sin \theta$

$\displaystyle 4 \sqrt{mn} = 4 \sqrt{( \tan \theta + \sin \theta)( \tan \theta - \sin \theta)}$

$\displaystyle = 4 \sqrt{\tan^2 \theta - \sin^2 \theta }$

$\displaystyle = 4 \sqrt{\frac{ \sin^2 \theta }{ \cos^2 \theta } - \sin^2 \theta }$

$\displaystyle = 4 \sqrt{ \sin^2 \theta \Big( \frac{1}{ \cos^2 \theta} - 1 \Big) }$

$\displaystyle = 4 \sqrt{\sin^2 \theta \frac{1 - \cos^2 \theta }{\cos^2 \theta } }$

$\displaystyle = 4 \sqrt{ \frac{ \sin^2 \theta . \sin^2 \theta }{ \cos^2 \theta } }$

$\displaystyle = 4 \sqrt{ \tan^2 \theta . \sin^2 \theta }$

$\displaystyle = 4 \tan \theta \sin \theta$

$\displaystyle \therefore m^2 - n^2 = 4 \sqrt{ mn }$

Hence proved.

$\displaystyle \\$

Question 20: A chord of a circle, of radius $\displaystyle 15$ cm, subtends an angle of $\displaystyle 60^o$ at the center of the circle. Find the area of major and minor segments (Take $\displaystyle \pi = 3.14, \sqrt{3} = 1.73$ )

Answer:

Radius $\displaystyle (r) = 15$ cm $\displaystyle \theta = 60^o$

Therefore Area of $\displaystyle OAPB = \frac{60}{360} \times \pi r^2 = \frac{1}{6} \times 3.14 \times 15^2 = 117.75 \ cm^2$

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

We draw $\displaystyle OM \perp AB$

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

In $\displaystyle \triangle OMA \text{ and } \triangle OMB$

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

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

$\displaystyle OM$ is common

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

$\displaystyle \Rightarrow \angle AOM = \angle BOM$

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

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

Also, since $\displaystyle \triangle OMB \cong \triangle OMA$

$\displaystyle \therefore BM = AM$

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

$\displaystyle \Rightarrow AB = 2 BM$ … … … … … i)

In right $\displaystyle \triangle OMB$

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

$\displaystyle \Rightarrow \frac{AM}{15} = \frac{1}{2}$

$\displaystyle \Rightarrow AM = \frac{15}{2}$

$\displaystyle \Rightarrow AB = 2 \times \frac{15}{2} = 15$

Similarly, In right $\displaystyle \triangle OMA$

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

$\displaystyle \Rightarrow \frac{OM}{15} = \frac{\sqrt{3}}{2}$

$\displaystyle \Rightarrow OM = \frac{\sqrt{3}}{2} \times 15$

$\displaystyle \therefore \text{ Area of } \triangle AOB = \frac{1}{2} \times 15 \times \frac{\sqrt{3}}{2} \times 15 = 97.3125 \ cm^2$

Therefore area of segment $\displaystyle APB = 117.75 - 97.3125 = 20.4375 \ cm^2$

Area of major segment $\displaystyle = \pi r^2 - 20.4375 = 3.14 \times 15^2 - 20.4375 = 686.0625 \ cm^2$

$\displaystyle \\$

Question 21: A sphere of diameter $\displaystyle 12$ cm is dropped in a right circular cylindrical vessel, partly filled with water. If the sphere is completely submerged in water, the water level in the vessel rises by $\displaystyle 3 \frac{5}{9}$ cm. Find the diameter of the cylindrical vessel.

A cylinder whose height is two-third of its diameter, has the same volume as that of a sphere of radius $\displaystyle 4$ cm. Find the radius of base of the cylinder.

Answer:

Volume of sphere $\displaystyle =$ Volume of water displaced

$\displaystyle \frac{4}{3} \pi {r_1}^3 = \pi {r_2}^2 h$

$\displaystyle \Rightarrow \frac{4}{3} \times 6^3 = {r_2}^2 \times \frac{32}{9}$

$\displaystyle \Rightarrow {r_2}^2 = \frac{4}{3} \times \frac{6^3 \times 9}{32} = 81$

$\displaystyle \Rightarrow r_2 = 9 \text{ cm }$

Therefore diameter $\displaystyle = 2 \times 9 = 18\text{ cm }$

Let the diameter of the cylinder $\displaystyle = x$

$\displaystyle \text{Therefore Radius } (r_2) = \frac{x}{2} , \text{ Height } (h) = \frac{2x}{3}$

$\displaystyle \Rightarrow \frac{4}{3} \pi {r_1}^3 = \pi {r_2}^2 h$

$\displaystyle \Rightarrow \frac{4}{3} \times 4^3 = \Big( \frac{x}{2} \Big)^2 \times \Big( \frac{2x}{2} \Big)$

$\displaystyle \Rightarrow x^3 = \frac{4 \times 4^3 \times 4}{2} = 512$

$\displaystyle \Rightarrow x = 8 \text{ cm }$

$\displaystyle \text{Hence Radius of the base cylinder } (r_2) = \frac{8}{2} = 4 \text{ cm }$

$\displaystyle \\$

Question 22: The following table gives the daily income of $50$ laborer:

Daily Income (Rs.):	100-120	120-140	140-160	160-180	180-200
Number of Laborer:	12	14	8	6	10

Find the mean and mode of the above data.

Answer:

We have

Daily Income	Mid Value $(x_i)$	Frequency $(f_i)$	Cumulative Frequency $(cf)$	$f_ix_i$
100-120	110	12	12	1320
120-140	130	14	26	1820
140-160	150	8	34	1200
160-180	170	6	40	1020
180-200	190	10	50	1900
		$\Sigma f_i = 50$		$\Sigma f_ix_i = 7260$

$\displaystyle \text{Mean } ( \overline{x}) = \frac{\Sigma f_ix_i }{\Sigma f_i } = \frac{7260}{50} = 145.2$

$\displaystyle N = 50 \Rightarrow \frac{N}{2} = 25$

Cumulative frequency just greater than $\displaystyle 25$ is $\displaystyle 26$ and corresponding class is $\displaystyle 120-140$

Thus, the Median class is $\displaystyle 120-140$

$\displaystyle \therefore L = 120, h = 20, f=14, C = 12, \frac{N}{2} = 25$

$\displaystyle \text{Median } (M) = L + \frac{\frac{N}{2} - C}{f} \times h$

$\displaystyle = 120 + \frac{25-12}{14} \times 20$

$\displaystyle = 138.57$

Mode $\displaystyle = 3 M - 2 (\overline{x}) = 3 \times 138.57 - 2 \times 145.2 = 125.31$

$\displaystyle \\$

Section – D

Question number 23 to 30 carry 4 mark each.

Question 23: Two taps together can fill a tank in $\displaystyle 6$ hours. The tap of larger diameter takes $\displaystyle 9$ hours less than the smaller one to fill the tank separately. Find the time in which each tap can fill the tank separately.

$\displaystyle \text{Solve for } x: \frac{x+1}{x-1} - \frac{x-1}{x+1} = \frac{5}{6} , x \neq 1, -1$

Answer:

Let the time taken by the smaller tap $\displaystyle = x$ hours

Therefore the time taken by the larger tap $\displaystyle = (x-9)$ hours

$\displaystyle \text{In 1 hour, smaller tap fills } \frac{1}{x} \text{ of the tank. }$

$\displaystyle \text{In 1 hour, larger tap fills } \frac{1}{x-9} \text{ of the tank.}$

$\displaystyle \text{In 1 hour, both taps fills } \frac{1}{6} \text{of the tank. }$

$\displaystyle \therefore \frac{1}{x} - \frac{1}{x-9} = \frac{1}{6}$

$\displaystyle \Rightarrow 6(x - 9 + x) = x(x-9)$

$\displaystyle \Rightarrow 12x - 54 = x^2 - 9x$

$\displaystyle \Rightarrow x^2 - 21x + 54 = 0$

$\displaystyle \Rightarrow x^2 - 18x - 3x +54 = 0$

$\displaystyle \Rightarrow x(x-18) - 3(x-18) = 0$

$\displaystyle \Rightarrow (x-18)(x-3) = 0$

$\displaystyle \Rightarrow x = 18 \ or \ x = 3$

When $\displaystyle x = 18$ , time taken by the larger tap to fill the tank $\displaystyle = 18 -9 = 9$ hours. $\displaystyle x = 3$ is not possible because then the time taken by the larger tap would be negative which is not possible.

$\displaystyle \frac{x+1}{x-1} - \frac{x-1}{x+1} = \frac{5}{6}$

$\displaystyle \Rightarrow \frac{(x+1)^2 - (x-1)^2}{x^2 - 1} = \frac{5}{6}$

$\displaystyle \Rightarrow \frac{x^2 + 1 + 2x - x^2 -1 +2x}{x^2 - 1} = \frac{5}{6}$

$\displaystyle \Rightarrow 4x(6) = 5(x^2 - 1)$

$\displaystyle \Rightarrow 5x^2 - 24x - 5 = 0$

$\displaystyle \Rightarrow (5x+1)(x-5) = 0$

$\displaystyle \Rightarrow x = 5 \ or \ x = - \frac{1}{5}$

$\displaystyle \\$

Question 24: Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Prove that in a triangle, if the square of one side is equal to sum of the squares of the other two sides, the angle opposite the first side is a right angle.

Answer:

Given: $\displaystyle \triangle ABC \sim \triangle DEF$

$\displaystyle \text{To Prove: } \frac{ar(\triangle ABC)}{ar(\triangle DEF)} = \frac{AB^2}{DE^2} = \frac{BC^2}{EF^2} = \frac{AC^2}{DF^2}$

Construction: Draw $\displaystyle AG \perp BC \text{ and } DH \perp EF$

Proof:

$\displaystyle \frac{ar(\triangle ABC)}{ar(\triangle DEF)} = \frac{\frac{1}{2} \times BC \times AG}{\frac{1}{2} \times EF \times DH} = \frac{BC}{EF} \times \frac{AG}{DH}$ … … … … … i)

Now in $\displaystyle \triangle ABG \text{ and } \triangle DEH$

$\displaystyle \angle B = \angle E$ (given by similarity)

$\displaystyle \angle AGB = \angle DHE = 90^o$ (by construction)

$\displaystyle \therefore \triangle ABG \cong \triangle DEH$ (by AA criterion)

$\displaystyle \therefore \frac{AB}{DE} = \frac{BG}{EH} = \frac{AG}{DH}$

$\displaystyle \text{But } \frac{AB}{DE} = \frac{BC}{EF} \text{ since } \triangle ABC \sim \triangle DEF$

$\displaystyle \therefore \frac{AG}{DH} = \frac{BC}{EF}$ … … … … … ii)

From i) and ii)

$\displaystyle \frac{ar(\triangle ABC)}{ar(\triangle DEF)} = \frac{BC}{EF} \times \frac{BC}{EF} = \frac{BC^2}{EF^2}$

Similarly, we can prove that

$\displaystyle \frac{ar(\triangle ABC)}{ar(\triangle DEF)} = \frac{AB^2}{DE^2} = \frac{AC^2}{DF^2}$

Hence proved

$\displaystyle \frac{ar(\triangle ABC)}{ar(\triangle DEF)} = \frac{AB^2}{DE^2} = \frac{BC^2}{EF^2} = \frac{AC^2}{DF^2}$

Given: $\displaystyle AC^2 = AB^2 + BC^2$

To Prove: $\displaystyle \angle B = 90^o$

Construction: $\displaystyle \triangle PQR$ is a right angled at $\displaystyle Q$ such that $\displaystyle PQ = AB \text{ and } QR = BC$

Proof: From $\displaystyle \triangle PQR$

$\displaystyle PR^2 = PQ^2 + QR^2$ (Pythagoras theorem)

$\displaystyle PR^2 = AB^2 + BC^2$ (by construction) … … … … … i)

But $\displaystyle AC^2 = AB^2 + BC^2$ (given) … … … … … ii)

$\displaystyle AC^2 = PR^2$ from i) and ii)

$\displaystyle \therefore AC = PR$ … … … … … iii)

Now in $\displaystyle \triangle ABC \text{ and } \triangle PQR$

$\displaystyle AB = PQ$ (by construction)

$\displaystyle BC = QR$ (by construction)

$\displaystyle AC = PR$ (from iii)

$\displaystyle \therefore \triangle ABC \cong \triangle PQR$ (by SSS criterion)

$\displaystyle \therefore \angle B = \angle Q$

But $\displaystyle \angle Q = 90^o$ by construction

$\displaystyle \therefore \angle B = 90^o$ . Hence proved.

$\displaystyle \\$

Question 25: Write the steps of construction for drawing a $\displaystyle \triangle ABC$ in which

$\displaystyle BC = 8 \ cm, \angle B = 45^o \text{ and } \angle C = 30^o$ . Now write the steps of construction for drawing a triangle whose sides are $\displaystyle \frac{3}{4}$ of the corresponding sides of $\displaystyle \triangle ABC$ .

Answer:

$\displaystyle \\$

Question 26: The sum of the first n terms of an A.P. is $\displaystyle 5n2 + 3n$ . If its $\displaystyle m^{th}$ term is $\displaystyle 168$ , find the value of $\displaystyle m$ . Also find the $\displaystyle 20^{th}$ term of the A.P.

The $\displaystyle 4^{th}$ and the last terms of an A.P. are $\displaystyle 11 \text{ and } 89$ respectively. If there are $\displaystyle 30$ terms in the A.P., find the A.P. and its $\displaystyle 23^{rd}$ term.

Answer:

Sum of the first n terms $\displaystyle = S_n$

$\displaystyle \therefore S_n = 5n^2 + 3n$

For $\displaystyle n=1, S_1 = 5(1)^2 + 3(1) = 8 = t_1$

For $\displaystyle n_2, S_2 = 5(2)^2 + 3(2) = 26$

$\displaystyle \therefore t_2 = S_2 - S_1 = 26-8 = 18$

$\displaystyle \therefore$ common difference $\displaystyle = t_2 - t_1 = 18-8 = 10$

$\displaystyle \therefore a = 8, d = 10$

Therefore $\displaystyle t_m = a + (m-1) d$

$\displaystyle \Rightarrow 168 = 8 + (m-1) (10)$

$\displaystyle \Rightarrow 160 = (m-1) (10)$

$\displaystyle \Rightarrow 16 = m - 1$

$\displaystyle \Rightarrow m = 17$

$\displaystyle t_{20} = a + (20-1) d$

$\displaystyle \Rightarrow t_{20} = 8 + (19)(10)$

$\displaystyle \Rightarrow t_{20} = 198$

Hence $\displaystyle m = 17 \text{ and } t_{20} = 198$

The $\displaystyle n^{th}$ of an AP where the first term is $\displaystyle a$ and the common difference is $\displaystyle d$

$\displaystyle a_n = a + (n-1)d$

$\displaystyle \Rightarrow a_4 = a + (4-1)d$

$\displaystyle \Rightarrow 11 = a + 3d$ … … … … … i)

Also $\displaystyle a_{30} = a + (30-1) d$

$\displaystyle \Rightarrow 89 = a + 29 d$ … … … … … ii)

Subtracting i) from ii) we get

$\displaystyle (89-11) = (29-3)d \Rightarrow d = \frac{78}{16} = 3$

$\displaystyle \therefore a = 11 - 3(3) = 2$

Therefore AP is $\displaystyle 2, 5, 8, 11, \ldots$

$\displaystyle a_{23} = a + (23-1)d$

$\displaystyle \Rightarrow a_{23} = 2 + 22 \times 3 = 68$

Hence the AP is $\displaystyle 2, 5, 8, 11, \ldots \text{ and } a_{23} = 68$

$\displaystyle \\$

Question 27: Prove that: $\displaystyle \Big( \frac{\sin A}{1 - \cos A} - \frac{1 - \cos A}{\sin A} \Big) . \Big( \frac{\cos A}{1 - \sin A} - \frac{1 - \sin A}{\cos A} \Big) = 4$

Answer:

$\displaystyle \text{LHS } = \Big( \frac{\sin A}{1 - \cos A} - \frac{1 - \cos A}{\sin A} \Big) . \Big( \frac{\cos A}{1 - \sin A} - \frac{1 - \sin A}{\cos A} \Big)$

$\displaystyle = \Big[ \frac{\sin^2 A - (1+ \cos^2 A - 2 \cos A)}{(1- \cos A) \sin A} \Big] \Big[ \frac{\cos^2 A - (1 + \sin^2 A - 2 \sin A)}{(1- \sin A) \cos A} \Big]$

$\displaystyle = \Big[ \frac{\sin^2 A - 1- \cos^2 A + 2 \cos A)}{(1- \cos A) \sin A} \Big] \Big[ \frac{\cos^2 A - 1 - \sin^2 A + 2 \sin A}{(1- \sin A) \cos A} \Big]$

$\displaystyle = \Big[ \frac{ 2 \cos A - 2 \cos^2 A}{(1- \cos A) \sin A} \Big] \Big[ \frac{2 \sin A - 2 \sin^2 A}{(1- \sin A) \cos A} \Big]$

$\displaystyle = \Big[ \frac{ 2 \cos A (1 - \cos A)}{(1- \cos A) \sin A} \Big] \Big[ \frac{2 \sin A( 1 - \sin A)}{(1- \sin A) \cos A} \Big]$

$\displaystyle = 2 \times 2$

$\displaystyle = 4 =$ RHS

Hence Proved.

$\displaystyle \\$

Question 28: A statue, $\displaystyle 1.46$ m tall, stands on a pedestal. From a point on the ground the angle of elevation of the top of the statue is $\displaystyle 60^o$ and from the same point angle of elevation of the top of the pedestal is $\displaystyle 45^o$ . Find the height of the pedestal. (use $\displaystyle \sqrt{3} = 1.73$ )

Answer:

In $\displaystyle \triangle ABC$

$\displaystyle \frac{BC}{AB} = \tan 45^o = 1 \Rightarrow AB = BC$ … … i)

From $\displaystyle \triangle ABD$

$\displaystyle \frac{DB}{AB} = \tan 60^o = \sqrt{3}$

$\displaystyle \Rightarrow \frac{1.46 + BC}{AB} = \tan 60^o = \sqrt{3}$

$\displaystyle \Rightarrow 1.46 + BC = \sqrt{3} AB$ … … … … … ii)

Substituting i) and ii) we get

$\displaystyle 1.46 + BC = \sqrt{3} BC$

$\displaystyle \Rightarrow BC = \frac{1.46}{\sqrt{3} - 1} = \frac{1.46}{1.73 - 1} = 2 \text{ m }$

Therefore the height of the pedestal $\displaystyle = 2 \text{ m }$

$\displaystyle \\$

Question 29: Sudhakar donated $\displaystyle 3$ cylindrical drums to store cereals to an orphanage. If radius of each drum is $\displaystyle 0.7$ m and height $\displaystyle 2$ m, find the volume of each drum. If each drum costs Rs. $\displaystyle 350 \ per \ m^3$ , find the amount spent by Sudhakar for orphanage. What value is exhibited in the question. (Use $\displaystyle \pi = \frac{22}{7}$ )

Answer:

Radius $\displaystyle (r)$ of drum $\displaystyle = 0.7 \text{ m }$

Height $\displaystyle (h)$ of the drum $\displaystyle = 2 \text{ m }$

$\displaystyle \text{Volume of each drum } = \pi r^2 h = \frac{22}{7} \times {0.7}^2 \times 2 = 3.08 \ m^3$

$\displaystyle \text{Cost of drum } = 350 \ \frac{Rs}{m^3}$

Therefore amount spent $\displaystyle = 3 \times 3.08 \times 350 = 3234 \text{ Rs. }$

$\displaystyle \\$

Question 30: The median of the following data is $52.5$ . If the total frequency is $100$ , find the values of $x$ and $y$ .

Classes	Frequency
0-10	2
10-20	5
20-30	$x$
30-40	12
40-50	17
50-60	20
60-70	7
70-80	9
80-90	7
90-100	4

Answer:

Classes	Frequency $(f_i)$	Cumulative Frequency $(cf)$
0-10	2	2
10-20	5	7
20-30	$x$	7 $+x$
30-40	12	19 $+x$
40-50	17	36 $+x$
50-60	20	56 $+x$
60-70	$y$	56 $+x+y$
70-80	9	65 $+x+y$
80-90	7	72 $+x+y$
90-100	4	76 $+x+y$