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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: Write the discriminant of the quadratic equation (x + 5)^2 = 2 (5x - 3) .

Answer:

Given Equation: (x + 5)^2 = 2 (5x - 3)

\Rightarrow x^2 + 25 + 10x = 10x - 6

\Rightarrow x^2 + 31 = 0

Comparing it with ax^2 + bx + c= 0 .   a = 1, b = 0 and c = 31

Determinant = b^2 - 4ac = -4(1)(31) = -124

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Question 2: Find after how many places of decimal the decimal form of the number \frac{27}{2^3 . 5^4 . 3^2} will terminate.

Or

Express 429 as a product of its prime factors.

Answer:

Simplifying \frac{27}{2^3 . 5^4 . 3^2} = \frac{3^3}{2^3 5^4 3^2} = \frac{3}{2^3 5^4} = \frac{3}{5000} = 0.0006

So the decimal form will terminate are 4 digits.

Or

429 = 3 \times 11 \times 13

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Question 3: Find the sum of first 10 multiples of 6 .

Answer:

AP is 6, 12, 18, \cdots

a = 6, d = 12-6 = 6, n = 10

S_{10} = \frac{n}{2} (2a + (n-1)d)

= \frac{10}{2} (2 \times 6 + (10-1) \times 6)

= 5 (12 + 54) = 330

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Question 4: Find the value(s) of x , if the distance between the points A(0, 0) and B(x, - 4) is 5 units.

Answer:

Given: A(0,0) and B(x, -4)

AB = 5 units

\therefore 5 = \sqrt{(x-0)^2 + (-4-0)^2}

\Rightarrow 25 = x^2 + 16

\Rightarrow x^2 = 9

\Rightarrow x = \pm 3

Therefore B could be (3, -4) or (-3, -4)

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Question 5: Two concentric circles of radii a and b (a > b) are given. Find the length of the chord of the larger circle which touches the smaller circle.2019-07-02_22-52-42

Answer:

Length of the chord = 2 \Big( \sqrt{a^2 - b^2} \Big)

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Question 6: In Figure 1, PS = 3 cm, QS = 4 cm, \angle PRQ = \theta, \angle PSQ = 90^o, PQ \perp RQ and RQ = 9 cm. Evaluate \tan \theta .

2019-05-23_6-49-18
Figure 1

Or

If \tan \alpha = \frac{5}{12} , find the value of \sec \alpha .

Answer:2019-07-02_22-53-06

PQ = \sqrt{4^2 + 3^2} = 5 cm

\tan \theta = \frac{5}{9}

Or

\tan \alpha = \frac{5}{12} 2019-07-02_22-53-38

AB = \sqrt{12^2 + 5^2} = 13

\therefore \cos \alpha = \frac{12}{13} \Rightarrow \sec \alpha = \frac{13}{12}

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Section – B

Question number 7 to 12 carry 2 mark each.

Question 7: Points A(3, 1), B(5, 1), C(a, b) and D(4, 3) are vertices of a parallelogram ABCD . Find the values of a and b .

Or

Points P and Q trisect the line segment joining the points A(- 2, 0) and B(0, 8) such that P is near to A . Find the coordinates of points P and Q .

Answer:

Let midpoint of BD be O

\therefore O (x, y) = \Big( \frac{5+4}{2} , \frac{1+3}{2} \Big) = (4.5, 2)

O is also the midpoint of AC

O(x, y) = \Big( \frac{3+a}{2} , \frac{1+b}{2} \Big)

Therefore \frac{3+a}{2} = 4.5 \Rightarrow a = 9-3 = 6

\frac{1+b}{2} = 2 \Rightarrow b = 4-1 = 3

Or

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

Using section formula [P divides AB in the ratio of 1:2 ]

\Rightarrow P(x_1, y_1) = \Big( \frac{1 \times 0 + 2 \times (-2)}{1+2} , \frac{1 \times 8 + 2 \times 0}{1+2} \Big) = \Big( \frac{-4}{3} , \frac{8}{3} \Big)

Similarly,

Using section formula [Q divides AB in the ratio of 2:1 ]

\Rightarrow P(x_2, y_2) = \Big( \frac{2 \times 0 + 1 \times (-2)}{2+1} , \frac{2 \times 8 + 1 \times 0}{2+1} \Big) = \Big( \frac{-2}{3} , \frac{16}{3} \Big)

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Question 8: Solve the following pair of linear equations :
3x - 5y = 4
2y + 7 = 9x

Answer:

Given equations:

3x - 5y = 4    … … … … … i)
2y + 7 = 9x

\Rightarrow 9x - 2y = 7    … … … … … ii)

Multiplying i) by 3 and subtracting ii) from the resultant equation

\hspace*{1cm} 9x - 15y = 12 \\ \underline {(-) \hspace*{0.4cm} 9x - 2y = \hspace*{0.2cm} 7} \\ \hspace*{1.3cm} -13y = \hspace*{0.2cm} 5

\Rightarrow y = \frac{-5}{13} 

Substituting into i)

3x - 5 \Big( \frac{-5}{13} \Big) = 4

\Rightarrow x = \frac{9}{13} 

Hence x = \frac{9}{13}  and y = \frac{-5}{13} 

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Question 9: If HCF of 65 and 117 is expressible in the form 65n - 117 , then find the value of n .

Or

On a morning walk, three persons step out together and their steps measure 30 cm, 36 cm and 40 cm respectively. What is the minimum distance each should walk so that each can cover the same distance in complete steps ?

Answer:

117 = 3 \times 3 \times 13

65 = 5 \times 13

Therefore HCF of 117 and 65 is 13

\therefore 65n - 117 = 13

\Rightarrow 65n = 130

\Rightarrow n = 2

Or

Finding LCM of 30, 36 and 40

2 | \underline{30, 36, 40} \\ 5 | \underline{15, 18, 20} \\ 3 | \underline{3, 18, 4} \\ 2 | \underline{1, 6, 4} \\ \hspace*{0.2cm}|  1, 3, 2

Therefore LCM = 2 \times 5 \times 3 \times 2 \times 3 \times 2 = 360

Therefore the minimum distance each should walk should be 360 cm.

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Question 10: A die is thrown once. Find the probability of getting (i) a composite number, (ii) a prime number.

Answer:

Total possible outcomes are 1, 2, 3, 4, 5,6

Therefore total number of events = 6

i) Composite numbers are 4 and 6

Therefore Probability (Composite Numbers) = \frac{2}{6} = \frac{1}{3}

ii) Prime numbers are 2, 3, 5

Therefore Probability (Prime Numbers) = \frac{3}{6} = \frac{1}{2}

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Question 11: Using completing the square method, show that the equation x^2 - 8x + 18 = 0 has no solution.

Answer:

Given equation x^2 - 8x + 18 = 0

\Rightarrow x^2 - 8x = -18

\Rightarrow x^2 - 8x+16 = -18 + 16

\Rightarrow (x-4)^2 = -2

A square of a number is always positive. Hence the given equation has no solution.

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Question 12: Cards numbered 7 to 40 were put in a box. Poonam selects a card at random. What is the probability that Poonam selects a card which is a multiple of 7 ?

Answer:

Total number of cards = (40-7)+1 = 33 + 1 = 34

Favorable outcomes = 7, 14, 21, 28, 35 \Rightarrow 5

Therefore Probability = \frac{5}{34}

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Section – C

Question number 13 to 22 carry 3 mark each.

Question 13: The perpendicular from A on side BC of a \triangle ABC meets BC at D such that DB = 3CD . Prove that 2AB^2 = 2AC^2 + BC^2 .

Or

AD and PM are medians of triangles ABC and PQR respectively where \triangle ABC \sim \triangle PQR . Prove that \frac{AB}{PQ} = \frac{AD}{PM} 

Answer:

Given: \triangle ABC with AD \perp BC Also DB = 2 CD

To Prove: 2AB^2 = 2AC^2 + BC^2 .2019-07-02_22-32-49

Proof: Let BC = x

\Rightarrow CD + DB = x \Rightarrow CD + 3 CD = x \Rightarrow 4 CD = x \Rightarrow  CD = \frac{x}{4} 

Also DB = 3CD \Rightarrow DB = \frac{3x}{4} 

Using Pythagoras theorem, in right angled triangle \triangle ADC

AC^2 = AD^2 + DC^2

\Rightarrow AC^2 = AD^2 + \Big( \frac{x}{4} \Big)^2

\Rightarrow AC^2 = AD^2 + \Big( \frac{x^2}{16} \Big)  … … … … … i)

Using Pythagoras theorem, in right angled triangle \triangle ADB

AB^2 = AD^2 + DB^2

\Rightarrow AB^2 = AD^2 + \Big( \frac{3x}{4} \Big)^2

\Rightarrow AB^2 = AD^2 + \Big( \frac{9x^2}{16} \Big)  … … … … … ii)

Subtracting ii) from i)

AC^2 - AB^2 = AD^2 + \Big( \frac{x^2}{16} \Big) - AD^2 - \Big( \frac{9x^2}{16} \Big)

\Rightarrow AC^2 - AB^2 =  \Big( \frac{x^2}{16} \Big) - \Big( \frac{9x^2}{16} \Big)

\Rightarrow AC^2 - AB^2 = - \frac{8}{16} x^2

\Rightarrow AC^2 - AB^2 = - \frac{1}{2} x^2

\Rightarrow 2AC^2 - 2AB^2 = -x^2    (Putting x = BC back in the equation)

\Rightarrow 2AC^2 - 2AB^2 = -BC^2

\Rightarrow 2AB^2 = 2AC^2 + BC^2 .

Hence Proved

Or

Given: \triangle ABC and \triangle PQR 2019-07-02_22-30-18

AD is the median of \triangle ABC and PM is the median of \triangle PQR

Also \triangle ABC \sim \triangle PQR

To Prove: \frac{AB}{PQ} = \frac{AD}{PM} 

Proof: Since AD is a median \Rightarrow BD = DC = \frac{1}{2} BC

Similarly, PM is the median \Rightarrow QM = MR = \frac{1}{2} QR

Now, \triangle ABC \sim \triangle PQR

\Rightarrow \frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR}   (corresponding sides of similar triangles)

So,  \frac{AB}{PQ} = \frac{BC}{QR}

\Rightarrow \frac{AB}{PQ} = \frac{2BD}{2QM}  (Since AD and PM are medians)

\Rightarrow \frac{AB}{PQ} = \frac{BD}{QM}   … … … … … i)

Also since \triangle ABC \sim \triangle PQR

\angle B = \angle Q   (corresponding angles of similar triangles)

Now in \triangle ABD and \triangle PQM

\angle B = \angle Q

\frac{AB}{PQ} = \frac{BD}{QM}  from i)

Hence \triangle ABD \sim \triangle PQM (By SAS criterion)

Since corresponding sides of similar triangles are proportional

\frac{AB}{PQ} = \frac{AD}{PM} . Hence proved.

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Question 14: Check whether g(x) is a factor of p(x) by dividing polynomial p(x) by polynomial g(x) , where p(x) = x^5 - 4x^3 + x^2 + 3x + 1 , g(x) = x^3 - 3x + 1

Answer:

If p(x) divisible by g(x) \Rightarrow   the remainder of \frac{p(x)}{g(x)} = 0

x^3-3x+1 ) \overline{x^5 - 4x^3 + x^2 + 3x + 1} ( x^2-1 \\ \hspace*{1.5cm}(-) { x^5-3x^3+x^2} \\ \hspace*{2cm}  \overline{-x^3+3x+1} \\ \hspace*{1.5cm}(-) \underline{ -x^3+3x-1} \\ \hspace*{4.0cm}  {2} 

Therefore remainder is 2 .

Therefore g(x) is not a factor of p(x) .

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Question 15: Find the area of the triangle formed by joining the mid-points of the sides of the triangle ABC , whose vertices are A(0, - 1), B(2, 1) and C(0, 3) .

Answer:

Mid-point of AB (x_1, y_1) = \Big( \frac{0+2}{2} , \frac{-1+1}{2} \Big) = (1, 0)

Mid-point of BC (x_2, y_2) = \Big( \frac{2+0}{2} , \frac{1+3}{2} \Big) = (1, 2)

Mid-point of AC (x_3, y_3) = \Big( \frac{0+0}{2} , \frac{-1+3}{2} \Big) = (0, 1)

Area of \triangle PQR = \frac{1}{2} \Big[ x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \Big]

= \frac{1}{2} \Big[ 1(2-1) + 1(1-0) + 0(0-2) \Big]

= \frac{1}{2} [ 1+1] = 1 sq. units.

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Question 16: Draw the graph of the equations x - y + 1 = 0 and 3x + 2y - 12 = 0 . Using this graph, find the values of x and y which satisfy both the equations.

Answer:

For x - y + 1 = 0 : When x =0, y = 1 and when y = 0, x = -1

For 3x + 2y - 12 = 0 : When x = 0, y = 6 and when y = 0, x = 4

From the graph the two lines intersect at (2, 3) . This is the point that satisfy both the equations.

2019-06-30_17-39-13

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Question 17: Prove that \sqrt{3} is an irrational number.

Or

Find the largest number which on dividing 1251, 9377 and 15628 leaves remainders 1, 2 and 3 respectively.

Answer:

Assume \sqrt{3} is a rational number

i.e. it can be expressed as a rational fraction of the form \frac{a}{b}  where a, b are relatively prime numbers.

Since \sqrt{3} = \frac{a}{b} 

We have 3 = \frac{a^2}{b^2} or a^2 = 3 b^2

If b is even, then a is also even in which case a/b is not in simplest form. If b is odd then a is also odd. Therefore:

a = 2n+1

b = 2m+1

(2n+1)^2 = 3 (2m+1)^2

4n^2 + 1 + 4n = 12m^2 + 3 + 12m

4n^2 + 4n = 12m^2 + 2 + 12m

2n^2 + 2n = 6m^2 + 1 + 6m

2(n^2+n) = 2(3m^2+3) + 1

Since (n^2+n) is an integer, the left hand side is even. Since (3m^2+3m) is an integer, the right hand side is odd and we have found a contradiction, therefore our hypothesis is false.

Therefore \sqrt{3} is an irrational number

Or

Since, 1, 2 and 3 are the remainders of 1251, 9377 and 15628 respectively. Thus, after subtracting these remainders from the numbers.

We have the numbers, 1251 - 1 = 1250, 9377 - 2 = 9375 and 15628 - 3 = 15625

Which is divisible by the required number.

Now, required number = HCF of 1250, 9375 and 15625

By Euclid’s division algorithm a = bq + r , 0 \leq r < b    … … … … … (i)

[ \because dividend = divisor \times quotient + remainder ]

For largest number, put a = 15625 and b = 9375

15625 = 9375 \times 1 + 6250 \ \ \ [ \because r \neq 0 ]

\Rightarrow 9375 = 6250 \times 1 + 3125 \ \ \ [ \because r \neq 0 ]

\Rightarrow 6250 = 3125 \times 2 + 0

Now r = 0

\therefore HCF (15625 and 9375) = 3125

Now, we take c = 1250   and d = 3125  , then again using Euclid’s division algorithm, d = cq + r, 0 \leq r < c

\Rightarrow 3125 = 1250 \times 2 + 625     [ \because r \neq 0 ]

\Rightarrow 1250 = 625 \times 2 + 0     [ Now r = 0 ]

\therefore HCF (1250, 9375 and 15625) = 625

Hence, 625 is the largest number which divides 1251, 9377 and 15628 leaving remainder 1, 2 and 3 respectively

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Question 18: A, B and C are interior angles of a triangle ABC . Show that

(i) \sin \Big( \frac{B+C}{2} \Big) = \cos \frac{A}{2}

(ii) If \angle A = 90^o , then find the value of \tan \Big( \frac{B+C}{2} \Big)

Or

If \tan (A + B) = 1 and \tan (A - B) = \frac{1}{\sqrt{3}} , 0^o < A + B < 90^o , A > B , then find the values of A and B .

Answer:

i) To prove: \sin \Big( \frac{B+C}{2} \Big) = \cos \Big( \frac{A}{2} \Big)

In \triangle ABC , sum of angles = 180^o

A + B + C = 180^o

\Rightarrow B+ C = 180^o - A

\Rightarrow \frac{B+C}{2} = 90^o - \frac{A}{2}

\Rightarrow \sin \Big( \frac{B+C}{2} \Big) = \sin \Big( 90 - \frac{A}{2} \Big)

\Rightarrow \sin \Big( \frac{B+C}{2} \Big) = \cos \Big( \frac{A}{2} \Big)    … … … … … i)

Hence Proved.

ii) Similarly, \cos \Big( \frac{B+C}{2} \Big) = \cos \Big( 90 - \frac{A}{2} \Big)

\Rightarrow \cos \Big( \frac{B+C}{2} \Big) = \sin \Big( \frac{A}{2} \Big)    … … … … … ii)

From i) and ii) and \angle A = 90^o

\tan \Big( \frac{B+C}{2} \Big) = \frac{\cos 45^o}{\sin 45^o} = \frac{1}{\tan 45^o} = 1

\therefore \tan \Big( \frac{B+C}{2} \Big) = 1

Or

Given \tan (A+B) = 1 \Rightarrow \tan (A+B) = \tan 45^o

\therefore A+B = 45^o    … … … … … i)

Similarly, \tan (A-B) = \frac{1}{\sqrt{3}} \Rightarrow \tan (A-B) = \tan 30^o

\therefore A-B = 30^o    … … … … … ii)

Solving i) and ii)

2A = 75^o  \Rightarrow A = 37.5^o

From i) B = 45^o - A = 45^o - 37.5^o = 7.5^o

Hence A = 37.5^o and B = 7.5^o

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Question 19: In Figure 2, PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at P and Q intersect at a point T . Find the length TP .

2019-05-23_6-51-25
Figure 2

 

Or

Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the center of the circle.

Answer:

TP = TQ (Length of the tangents from an external point to a circle are equal)

In \triangle TPQ TP = TQ 2019-06-24_7-12-03

OT is the bisector of \angle PTQ

\therefore OT \perp PQ

Since OT \perp PQ

PR = RQ ( perpendicular from center to a chord bisects the chord)

\therefore PR = QR = \frac{1}{2} PQ = 4 cm

In Right triangle ORP

OP^2 = OR^2+ PR^2

\Rightarrow 5^2 = OR^2 + 4^2

\Rightarrow OR^2 = 25 - 16 = 9

\Rightarrow OR = 3 cm

Let TP = x

In Right triangle PRT

TP^2 = PR^2 + RT^2

\Rightarrow x^2 = 4^2 + RT^2    … … … … … i)

Since TP is a tangent, OP \perp TP

\angle OPT = 90^o

In Right triangle OPT

OT^2 = OP^2 + TP^2

\Rightarrow OT^2 = 5^2 + x^2

\Rightarrow (OR+RT)^2 = 5^2 + x^2

\Rightarrow (3+RT)^2 = 5^2 + x^2

\Rightarrow 9 + RT^2 + 6 RT = 25 + x^2

From i) x^2 = 16+RT^2

9 + RT^2 + 6 RT = 25 + 16+RT^2

\Rightarrow 6RT = 32

\Rightarrow RT = \frac{32}{6} = \frac{16}{3} 

\therefore x^2 = 16 + \Big( \frac{16}{3} \Big)^2 = \frac{400}{9} 

\Rightarrow x = \frac{20}{3} 

Hence TP is \frac{20}{3}  cm

Or

Given: A circle with center O touches the side AB, BC, CD and DA of a quadrilateral ABCD at points P, Q, R and S .

To Prove: \angle AOB + \angle COD = 180^o and  \angle AOD + \angle BOC = 180^o

Construction: Join OP, OQ, OR and OS

Proof: Since the two tangents from an external point to a circle subtend equal angles at the center;

\angle 1 = \angle 2, \angle 3 = \angle 4, \angle 5 = \angle 6 and \angle 7 = \angle 8

Since the sum of all the angles subtended at the point is 360^o

\angle 1 + \angle 2 + \angle 3 + \angle 4 + \angle 5 + \angle 6 + \angle 7 + \angle 8 = 360^o

\Rightarrow  2 (\angle 2 + \angle 4 + \angle 6 + \angle 7) = 360^o

\Rightarrow \angle 2 + \angle 3 + \angle 6 + \angle 7 = 180^o

\Rightarrow (\angle 2+\angle 3) + ( \angle 6+\angle 7) = 180^o

\Rightarrow \angle COD + \angle AOB = 180^o

Similarly

\angle 1 + \angle 2 + \angle 3 + \angle 4 + \angle 5 + \angle 6 + \angle 7 + \angle 8 = 360^o

\Rightarrow  2 (\angle 1 + \angle 4 + \angle 5 + \angle 8) = 360^o

\Rightarrow \angle 1 + \angle 4 + \angle 5 + \angle 8 = 180^o

\Rightarrow (\angle 1+\angle 8) + ( \angle 4+\angle 5) = 180^o

\Rightarrow \angle AOD + \angle BOC = 180^o

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Question 20: Water in a canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes if 8 cm of standing water is needed ?

Answer:

Speed of water = \frac{10 \ km}{hr} = \frac{10 \times 1000 \ m}{60 \ min} = \frac{1000 \ m}{6 \ min} 

Rate of flow = 6 \times 1.5 \ m^2 \times \frac{1000 \ m}{6 \ min} = \frac{1500 \ m^3}{min} 

Therefore volume of water in 30 mins = 1500 \frac{m^3}{min} \times 30 \ min = 45000 \ m^3

Let Area of irrigation = x

\therefore x \times \frac{8}{100} \ m = 45000 m^3

x = \frac{45000 \times 100}{8} \ m^2 = 562500 \ m^2

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Question 21: A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.

Number of Days: 0-6 6-12 12-18 18-24 24-30 30-36 35-42
Number of Students: 10 11 7 4 4 3 1

Answer:

Class Interval f_i x_i f_ix_i
0-6 10 3 30
6-12 11 9 99
12-18 7 15 105
18-24 4 21 84
24-30 4 27 108
30-36 3 33 99
36-42 1 39 39
\Sigma f_i = 40 \Sigma f_ix_i = 564

Mean  value = \frac{\Sigma f_ix_i}{\Sigma f_i} = \frac{564}{40} = 14.1

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Question 22: A car has two wipers which do not overlap. Each wiper has a blade of
length 21 cm sweeping through an angle 120^o . Find the total area cleaned at each sweep of the blades. (Take \pi = \frac{22}{7} )

Answer:

Total area = 2 \Big[ \frac{120}{360} \times \pi (21)^2 \Big]

= 2 \Big[ \frac{1}{3} \times \frac{22}{7} \times 21 \times 21\Big]

= 2 \times 22 \times 21

= 924 \ cm^2

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Section – D

Question number 23 to 30 carry 4 mark each.

Question 23: A pole has to be erected at a point on the boundary of a circular park of diameter 13 m in such a way that the difference of its distances from two diametrically opposite fixed gates A and B on the boundary is 7 m. Is it possible to do so ? If yes, at what distances from the two gates should the pole be erected ?

Answer:

Let AC = a and BC = b

\therefore a - b = 7

\Rightarrow a = b+ 7

We know that diameter will subtend a right angle on circumference.

\therefore AB^2 = a^2 + b^2

\Rightarrow 13^2 = (b+7)^2 + b^2

\Rightarrow 169 = 2b^2 + 14 b + 49

\Rightarrow 2b^2 + 14b - 120 =0

\Rightarrow b^2 + 7b-60 = 0

\Rightarrow (b+12)(b-5) = 0

\Rightarrow b = 5 \ or \ -12 (this is not possible as b cannot be negative)

\therefore b = 5 m

\Rightarrow 5+7 = 12 m

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Question 24: If m times the m^{th} term of an Arithmetic Progression is equal to n times its n^{th} term and m \neq n , show that the (m + n)^{th} term of the A.P. is zero.

Or

The sum of the first three numbers in an Arithmetic Progression is 18 . If the product of the first and the third term is 5 times the common difference, find the three numbers.

Answer:

Let the first term of AP = a

common difference = d

We have to show that (m+n)^{th} term is zero or a + (m+n-1)d = 0

m^{th}  term = a + (m-1)d

n^{th} term = a + (n-1) d

Given that m{a +(m-1)d} = n{a + (n -1)d}

\Rightarrow am+m^2d-md=an+n^2d-nd

\Rightarrow am-an+m^2d-n^2d-md+nd=0

\Rightarrow a(m-n)+(m^2-n^2)d-(m-n)d=0

\Rightarrow a(m-n)+{(m-n)(m+n)}d-(m-n)d=0

\Rightarrow a(m-n) + {(m-n)(m+n) - (m-n)} d = 0

\Rightarrow a(m-n)+(m-n)(m+n-1)d=0

\Rightarrow (m-n){a + (m+n-1)d} = 0

\Rightarrow a+(m+n-1)d= \frac{0}{(m-n)}

\Rightarrow a+(m+n-1)d=0

Hence Proved

Or

Let first three numbers of A.P are (a-d) , a and (a+d)

Sum of the first three numbers = (a-d) + a +(a+d) = 3a = 18   or  a = 6

Product of first and third term = (a-d)(a+d) = a^2 -d^2 = 36 -d^2 = 5d   or  d^2 +5d -36 = 0   or  (d+9)(d-4) = 0

Hence d = 4   and three numbers are  2, 6, 10

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Question 25: Construct a triangle ABC with side BC = 6 cm, AB = 5 cm and \angle ABC = 60^o . Then construct another triangle whose sides are \frac{3}{4} of the corresponding sides of the triangle ABC .

Answer:

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Question 26: In Figure 3, a decorative block is shown which is made of two solids, a cube and a hemisphere. The base of the block is a cube with edge 6 cm and the hemisphere fixed on the top has a diameter of 4.2 cm. Find
(a) the total surface area of the block.
(b) the volume of the block formed. (Take \pi = \frac{22}{7} )

2019-05-23_7-00-19
Figure 3

Or

A bucket open at the top is in the form of a frustum of a cone with a capacity of 12308.8 \ cm^3 . The radii of the top and bottom circular ends are 20 cm and 12 cm respectively. Find the height of the bucket and the area of metal sheet used in making the bucket. (Use \pi = 3.14 )

Answer:

a) Total surface area

= 6 \times (6 \times 6) - \pi (2.1)^2 + \frac{1}{2} \Big( 4 \pi (2.1)^2 \Big)

= 216 - \frac{22}{7} \times 2.1 \times 2.1 + 2 \times \frac{22}{7} \times 2.1 \times 2.1

= 216 - 13.86 + 27.72

= 229.86 \ cm^2

b) Volume of the block

= 6^3 + \frac{1}{2} \Big( \frac{4}{3} \times \frac{22}{7} \times (2.1)^3 \Big)

= 216 + 19.404

= 235.404 \ cm^3

Or

2019-06-24_7-29-49.png

Volume = 12308.80 \ cm^3

r_1 = 20 cm  r_2 = 12 cm  h = ?

Volume of the bucket = \frac{\pi}{3} h ( {r_1}^2 + {r_2}^2 +r_1. r_2 )

\Rightarrow 12308.80 = 3.14 \times \frac{h}{3} ( 20^2 + 12^2 + 20 \times 12)

\Rightarrow h = 12308.80 \times \frac{3}{3.14 \times 784} = 15 cm

Therefore height of bucket = 15 cm

l = \sqrt{h^2 + (r_1 - r_2)^2} = \sqrt{15^2 + 8^2} = 17 cm

Therefore surface are of the metal sheet used

= curved surface area + base area

= \pi ( r_1 + r_2) l + \pi r^2

= 3.14 \times 32 \times 17 + 3.14 \times 12^2 = 2160.32 \ cm^2

Hence the height of the bucket is 15 cm and surface are of the metal sheet is 2160.32 \ cm^2

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Question 27: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, prove that the other two sides are divided in the same ratio.

Or

Prove that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Answer:

Given: In \triangle ABC, DE intersects AB and AC at D and E respectively.2019-06-30_8-07-41.png

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

Construct: DM \perp AC and EN \perp AB

Proof:

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

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

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

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

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

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

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

Or

Given: A right angled \triangle ABC , right angled at B

To prove: AC^2 = AB^2 + BC^2 2019-06-24_7-33-29.png

Draw: BD \perp AC

Proof: In \triangle ADB \& \angle ABC

\angle ADB = \angle ABC = 90^o

\angle BAD = \angle CAB (common angle)

\therefore \triangle ADB \sim \triangle ABC ( by AA criterion)

Therefore \frac{AD}{AB} = \frac{AB}{AC}  (corresponding sides are proportional)

AB^2 = AD \times AC    … … … … … i)

Similarly, \triangle BDC \sim \triangle ABC

and BC^2 = CD \times AC    … … … … … ii)

Adding i) and ii)

AB^2 + BC^2 = AD \times AC + CD \times AC

\Rightarrow AB^2 + BC^2 = AC ( AD + CD)

\Rightarrow AB^2 + BC^2 = AC^2 . Hence proved.

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Question 28: If 1 + \sin^2 \theta = 3 \sin \theta \cos \theta , then prove that \tan \theta = 1 or \tan \theta = \frac{1}{2}

Answer:

We have 1 + \sin^2 \theta = 3 \sin \theta \cos \theta

Dividing both sides with \cos^2 \theta

\Rightarrow \frac{1}{\cos^2 \theta} + \frac{\sin^2 \theta}{\cos^2 \theta} = \frac{3 \sin \theta \cos \theta}{\cos^2 \theta} 

\Rightarrow \sec^2 \theta + \tan^2 \theta = 3 \tan \theta

\Rightarrow 1 + \tan^2 \theta + \tan^2 \theta = 3 \tan \theta  (\because \sec^2 \theta = 1 + \tan^2 \theta)

\Rightarrow 2 \tan^2 \theta - 3 \tan \theta + 1 = 0

Let \tan \theta = y

\therefore 2y^2 - 3y + 1= 0

\Rightarrow 2y^2 - 2y - y + 1 = 0

\Rightarrow 2y(y -1) -1(y -1) = 0

\Rightarrow (2y -1)(y-1) = 0

\Rightarrow y = 1 or y = \frac{1}{2} 

Hence \tan \theta  = 1 or \tan \theta = \frac{1}{2} 

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Question 29: Change the following distribution to a ‘more than type’ distribution. Hence draw the ‘more than type’ ogive for this distribution.

Class Interval: 20-30 30-40 40-50 50-60 60-70 70-80 80-90
Frequency 10 8 12 24 6 25 15

Answer:

Less Than table

Less Than Cumulative Frequency
Less than 30 10
Less than 40 18
Less than 50 30
Less than 60 54
Less than 70 60
Less than 80 85
Less than 90 100

More than table

More Than Cumulative Frequency
More than 30 10
More than 40 18
More than 50 30
More than 60 54
More than 70 60
More than 80 85
More than 90 100

Draw the chart. The point of intersection is :

2019-07-02_21-59-37

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Question 30: The shadow of a tower standing on a level ground is found to be 40 m longer when the Sun’s altitude is 30^o than when it was 60^o . Find the height of the tower. (Given \sqrt{3} = 1.732 )

Answer:

Let h be the height of the tower.2019-07-02_21-45-14.png

\frac{h}{CB} = \tan 60^o = \sqrt{3}

\Rightarrow CB = \frac{h}{\sqrt{3}}    … … … … … i)

Also \frac{h}{40 + CB} = \tan 30^o = \frac{1}{\sqrt{3}} 

\Rightarrow \sqrt{3} h = 40 + CB

\Rightarrow CB = \sqrt{3}h - 40  … … … … … ii)

From i) and ii)

\frac{h}{\sqrt{3}} = \sqrt{3}h - 40

\Rightarrow h = 3h - 40\sqrt{3}

\Rightarrow 2h = 40 \sqrt{3}

\Rightarrow h = 20 \sqrt{3} = 20 \times 1.732 = 34.64 m

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