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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 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.

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.

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

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

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

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

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

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

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

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

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

$\displaystyle \\$

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

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

$\displaystyle \\$

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

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 $\displaystyle \cot \theta$

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

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

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

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

$\displaystyle \\$

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

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

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

$\displaystyle \Rightarrow \triangle ABC \sim \triangle ADE$ 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.

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.

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 $\displaystyle 255 \text{ and } 867$.

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

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 $\displaystyle k$.

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.

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

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 $\displaystyle q$.

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.

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 $\displaystyle k$.

Or

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 $\displaystyle P$.

$\displaystyle AP : AB = 1:3$ (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 $\displaystyle 2x - y + k = 0$

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

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

$\displaystyle \Rightarrow k = -8$

Or

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 $\displaystyle (16, 8)$

$\displaystyle \\$

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

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.

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 $\displaystyle \triangle RPQ \sim \triangle RTS$.

Or

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

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.

Or

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.

$\displaystyle \\$

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

Or

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

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

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

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

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

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

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

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

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

Hence proved.

Or

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 $\displaystyle \pi = 3.14, \sqrt{3} = 1.73$ )

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.

Or

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.

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

Or

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.

We have

 Daily Income Mid Value $(x_i)$$(x_i)$ Frequency $(f_i)$$(f_i)$ Cumulative Frequency $(cf)$$(cf)$ $f_ix_i$$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$$\Sigma f_i = 50$ $\Sigma f_ix_i = 7260$$\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.

Or

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

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.

Or

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

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

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

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

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

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

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

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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.

Or

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.

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

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

Or

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.

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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 $\displaystyle \triangle ABC$.

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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.

Or

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.

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

Or

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

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

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

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

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

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

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

$\displaystyle = 2 \times 2$

$\displaystyle = 4 =$ RHS

Hence Proved.

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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 $\displaystyle \sqrt{3} = 1.73$)

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 $\displaystyle = 2 \text{ m }$

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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 $\displaystyle \pi = \frac{22}{7}$ )

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

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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$$x$ 30-40 12 40-50 17 50-60 20 60-70 7 70-80 9 80-90 7 90-100 4

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

$\displaystyle 76+x+y = 100 \Rightarrow x + y = 24$ … … … … … i)

Median class $\displaystyle = 50 -60$

$\displaystyle L = 50, \frac{N}{2} = 25, cf = 36+x, C = 10, f = 20, Median = 52.5$

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

$\displaystyle \Rightarrow 52.5 = 50 + \Big( \frac{50-36-x}{20} \Big) \times 10$

$\displaystyle \Rightarrow 52.5 = 50 + \frac{14-x}{2}$

$\displaystyle \Rightarrow 52.5 = 50 + 7 - \frac{x}{2}$

$\displaystyle \Rightarrow \frac{x}{2} = 57 - 52.5 = 4.5$

$\displaystyle \Rightarrow x = 2 \times 4.5 = 9$

$\displaystyle \therefore y = 24 - 9 = 15$

Hence $\displaystyle x = 9 \text{ and } y = 15$