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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.

2019-08-06_8-44-16

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.

Answer:

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

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

Answer:

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

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

Answer:

\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

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

Answer:

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}

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

Answer:

\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

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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 .

Answer:2019-07-20_8-42-55

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

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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.

Answer:

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

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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.

Answer:

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

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

Answer:

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

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

Answer:

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

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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.

Answer:

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.

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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

Answer:

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}

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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 .

Answer:

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 .

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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.

Answer:

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

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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 .

Answer:

2019-06-24_7-10-31

 

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)

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

Answer:

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.

Answer:2019-07-20_8-39-33

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.

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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 .

Answer:

Given: \triangle PQR   and points S and T on sides of PR and QR 2019-07-20_8-34-04

\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.2019-07-14_10-35-07.png

\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.

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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}

Answer:

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.

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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 )2019-07-20_8-30-20

Answer:

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

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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.

Answer:

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

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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.

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

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

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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

Answer:

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}

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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.

Answer:

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} 2019-07-20_8-26-22

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 .

Answer:

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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.

Answer:

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

Answer:

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 )

Answer:2019-07-20_8-22-17

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} )

Answer:

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
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

 

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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