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• 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 $k$ for which the roots of the quadratic equation
$(k - 5)x^2 + 2(k - 5) x + 2 = 0$ are equal.

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

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

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

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

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

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

$\Rightarrow k = 7$ or $k = 5$

$\\$

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

Given distance between the points $(2, -3)$ and $(10, y)$ is $10$ units.

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

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

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

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

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

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

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

$\Rightarrow y = -9$ or $y = 3$

$\\$

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

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

$\\$

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

From the given AP

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

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

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

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

$\\$

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

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

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

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

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

$\\$

Question 6: $DE$ is drawn parallel to the base $BC$ of a $\triangle ABC$, meeting $AB$ at $D$ and $AC$ at $E$. If  $\frac{AB}{BD}$ $= 4$ and $CE = 2$ cm, find $AE$.

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

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

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

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

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

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

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

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

$\therefore AE = AC = CE = 8 - 2 = 6$ cm

$\\$

Section – B

Question number 7 to 12 carry 2 mark each.

Question 7: A bag contains $5$ 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 $= 5$

Let the number of Blue balls $= x$

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

$\Rightarrow x = 15$

No of Blue balls in the bag $= 15$

$\\$

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

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

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

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

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

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

Subtracting ii) from i)

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

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

$\therefore a = 25$ and $d = -3$

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

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

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

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

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

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

$\\$

Question 9: Using Euclid’s Division Algorithm, find the HCF of $255$ and $867$.

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

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

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

So HCF of $867$ and $255$ is $51$

$\\$

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

Given point $(0, 2)$ is equidistant from the points $(3, k)$ and $(k, 5)$

Applying distance formula

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

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

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

$\Rightarrow k = 1$

$\\$

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

If the system of equations are $ax_1 + b_1y + c_1 = 0$ and $ax_2 + b_2y + c_2 = 0$ and they have infinitely many solutions then it satisfy the following:
$\frac{a_1}{a_2}$ $=$ $\frac{b_1}{b_2}$ $=$ $\frac{c_1}{c_2}$

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

From first two terms $a = 6$

From Last two terms $a = 6$

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

$\\$

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

Total number of cards $= 52$

Number of Red kings $= 2$

Number of queens and jacks $= 8$

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

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

$\\$

Section – C

Question number 13 to 22 carry 3 mark each.

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

Let $a$ be any positive integer.

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

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

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

That is $a$ can be $4q, 4q+1, 4q+2$ or $4q+3$

Since $a$ is odd, $a$ cannot be $4q$ or $4q+2$

Therefore any odd integer is of the form $4q+1$ or $4q+3$.

$\\$

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

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

Therefore the number $=10x+y$

Interchanged number $=10 y+x$

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

Now according to the question

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

$\Rightarrow 9y =9x-36$

$\Rightarrow y=x-4$  from i)

$\Rightarrow y=2y-4$

$\Rightarrow y=4$

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

Therefore the number is $84$

$\\$

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

Or

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

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

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

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

$\Rightarrow 3AP = AP + PB$

$\Rightarrow 2AP = PB$

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

$\Rightarrow AP : PB = 1:2$

Applying section formula

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

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

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

Since $P(3, -2)$ lies on $2x - y + k = 0$

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

$\Rightarrow 6 + 2 + k = 0$

$\Rightarrow k = -8$

Or

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

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

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

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

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

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

Now $x = 2y$

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

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

$\Rightarrow 2y = 18$

$\Rightarrow y = 8$

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

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

$\\$

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

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

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

Now we verify the relations between zeros and their coefficients

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

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

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

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 $O$. Let $PA$ and $PB$ be tangents from external point P.

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

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

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

In Quadrilateral $OAPB$

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

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

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

Hence proved.

$\\$

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

Or

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

Given: $\triangle PQR$  and points $S$ and $T$ on sides of $PR$ and $QR$

$\angle P = \angle RTS$

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

Proof: In $\triangle RPQ$ and $\triangle RTS$

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

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

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

Hence proved.

Or

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

$\Rightarrow AB = BC = CA$

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

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

Construction: Draw $AE \perp BC$

Proof: Consider $\triangle AEB$ and $\triangle AEC$

$AE$ is common

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

$AB = AC$ (equilateral triangle)

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

$\therefore BE = EC$

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

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

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

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

Using Pythagoras theorem,

In $\triangle AEB$

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

In $\triangle AED$

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

From i) and ii)

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

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

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

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

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

But $BC = AB$

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

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

Hence proved.

$\\$

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

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

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

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

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

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

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

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

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

Hence proved.

Or

Given:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Hence proved.

$\\$

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

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

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

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

We draw $OM \perp AB$

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

In $\triangle OMA$ and $\triangle OMB$

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

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

$OM$ is common

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

$\Rightarrow \angle AOM = \angle BOM$

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

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

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

$\therefore BM = AM$

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

$\Rightarrow AB = 2 BM$  … … … … … i)

In right $\triangle OMB$

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

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

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

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

Similarly, In right $\triangle OMA$

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

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

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

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

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

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

$\\$

Question 21: A sphere of diameter $12$ 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 $3$ $\frac{5}{9}$ 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 $4$ cm. Find the radius of base of the cylinder.

Volume of sphere $=$ Volume of water displaced

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

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

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

$\Rightarrow r_2 = 9$ cm

Therefore diameter  $= 2 \times 9 = 18$ cm

Or

Let the diameter of the cylinder $= x$

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

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

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

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

$\Rightarrow x = 8$ cm

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

$\\$

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

 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$

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

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

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

Thus, the Median class is $120-140$

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

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

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

$= 138.57$

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

$\\$

Section – D

Question number 23 to 30 carry 4 mark each.

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

Or

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 $= x$ hours

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

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

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

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

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

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

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

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

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

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

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

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

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

Or

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

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

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

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

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

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

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

$\\$

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: $\triangle ABC \sim \triangle DEF$

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 $AG \perp BC$ and $DH \perp EF$

Proof:

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

Now in $\triangle ABG$ and $\triangle DEH$

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

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

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

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

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

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

From i) and ii)

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

Similarly, we can prove that

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

Hence proved

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

Or

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

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

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

Proof: From $\triangle PQR$

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

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

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

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

$\therefore AC = PR$   … … … … … iii)

Now in $\triangle ABC$ and $\triangle PQR$

$AB = PQ$ (by construction)

$BC = QR$ (by construction)

$AC = PR$ (from iii)

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

$\therefore \angle B = \angle Q$

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

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

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Question 25: Write the steps of construction for drawing a $\triangle ABC$ in which
$BC = 8 \ cm, \angle B = 45^o$ and $\angle C = 30^o$. Now write the steps of construction for drawing a triangle whose sides are $\frac{3}{4}$ of the corresponding sides of $\triangle ABC$.

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

Or

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

Sum of the first n terms $= S_n$

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

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

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

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

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

$\therefore a = 8, d = 10$

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

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

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

$\Rightarrow 16 = m - 1$

$\Rightarrow m = 17$

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

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

$\Rightarrow t_{20} = 198$

Hence $m = 17$ and $t_{20} = 198$

Or

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

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

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

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

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

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

Subtracting i) from ii) we get

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

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

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

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

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

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

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

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

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

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

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

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

$= 2 \times 2$

$= 4 =$ RHS

Hence Proved.

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

In $\triangle ABC$

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

From $\triangle ABD$

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

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

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

Substituting i) and ii)  we get

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

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

Therefore the height of the pedestal $= 2$ m

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

Radius $(r)$ of drum $= 0.7$ m

Height $(h)$ of the drum $= 2$ m

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

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

Therefore amount spent $= 3 \times 3.08 \times 350 = 3234$ 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$

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

Median class $= 50 -60$

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

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

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

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

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

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

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

$\therefore y = 24 - 9 = 15$

Hence $x = 9$ and $y = 15$

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