NOTE:

(I)  Please check that this question paper consists of 23 pages.

(II)  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.

(III)  Please check that this question paper consists of 40 questions.

(IV)  Please write down the serial number of the question before attempting it.

(V)  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.

MATHEMATICS (STANDARD) – Theory

—————————————————————————————————————————————–Time allowed: 3 hours                                                             Maximum Marks: 80                      —————————————————————————————————————————————–

General Instructions:

Read the following instructions very carefully and strictly follow them:

(i) This question paper comprises four sections – A, B, C and D. This question paper carries 40 questions. All questions are compulsory.

(ii) Section A – Question numbers 1 to 20 comprises of 20 questions of one mark each.

(iii) Section B – Question numbers 21 to 26 comprises of 6 questions of two marks each.

(iv) Section C – Question numbers 27 to 34 comprises of 8 questions of three marks each.

(v) Section D – Question numbers 35 to 40 comprises of 6 questions of four marks each.

(vi) There is no overall choice in the question paper. However, an internal choice has been provided in 2 questions of one mark, 2 questions of two marks, 3 questions of three marks and 3 questions of four marks. You have to attempt only one of the choices in such questions.

(vii) In addition to this, separate instructions are given with each section and question, wherever necessary.

(viii) Use of calculators is not permitted.

Section – A

Question numbers 1 to 10 are multiple choice questions of 1 mark each. Select the correct option.

Question 1:  On dividing a polynomial p(x) by x^2 - 4 , quotient and remainder are found to be x and 3 respectively. The polynomial p(x) is:

(A) 3x^2 + x - 12       (B) x^3 - 4x + 3        (C) x^2 + 3x - 4        (D) x^3 - 4x - 3

Answer:

\fbox B

We know that p(x) = q(x). g(x) + r(x)

Where q(x) = x ,    g(x) = x^2 - 4   and r(x) = 3

Substituting all the above values we get

p(x) = x ( x^2 - 4) + 3

\Rightarrow p(x) = x^2 - 4x + 3

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Question 2: In Figure – 1, ABC is an isosceles triangle, right angled at C . Therefore:

Figure - 21
Figure – 1

(A) AB^2 = 2AC^2       (B) BC^2 = 2AB^2        (C) AC^2 = 2AB^2       (D) AB^2 = 4AC^2

Answer:

\fbox A

Given BC = AC

Using pythagoras theorem

AB^2 = AC^2 + BC^2

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

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Question 3:  The point on the x- axis which is equidistant from (-4, 0) and (10, 0) is

(A) (7, 0)             (B) (5, 0)             (C) (0, 0)         (D) (3, 0)

(OR)

The centre of a circle whose end points of a diameter are (-6, 3) and (6, 4) is

(A) (8, -1)                (B) (4, 7)                (C) \Big(0, \frac{7}{2} \Big)           (D) \Big(4, \frac{7}{2} \Big)

Answer:

\fbox D

Given points A(-4, 0) and B(10, 0)

Let the equidistant point be C(x, 0) as it is on x-axis.

Therefore AC = BC

\Rightarrow \sqrt{(x+4)^2 + (0-0)^2} = \sqrt{ (10-x)^2 + (0-0)^2}

\Rightarrow (x+4)^2 = (10-x)^2

\Rightarrow x^2 + 16 + 8x = 100 + x^2 - 20 x

\Rightarrow 28 x = 84

\Rightarrow x = 3

Therefore the required point is (3, 0)

OR

\fbox C

Given points A(-6, 3) and B(6, 4)

Let the center is O(x, y)

\therefore x = \frac{-6+6}{2} = 0

and y = \frac{3+4}{2} = \frac{7}{2}

\therefore O(x, y) = \Big( 0, \frac{7}{2} \Big)

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Question 4:  The value(s) of k for which the quadratic equation 2x^2 + kx + 2 = 0 has equal roots, is

(A) 4           (B) \pm 4           (C) - 4           (D) 0

Answer:

\fbox B

Given 2x^2 + kx + 2 = 0

\therefore a = 2, b = k and c = 2

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

\therefore k^2 - 4 ( 2) ( 2) = 0

\Rightarrow k^2 = 16

\Rightarrow k = \pm 4

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Question 5: Which of the following is not A.P.?

(A) - 1.2, 0.8, 2.8, \ldots         (B) 3,3+ \sqrt{2} , 3+ 2 \sqrt{2}, 3+ 3 \sqrt{2} \dots

(C) \frac{4}{3} , \frac{7}{3} , \frac{9}{3} , \frac{12}{3} , \ldots         (D) \frac{-1}{5} , \frac{-2}{5} , \frac{-3}{5} , \ldots

Answer:

\fbox C

For - 1.2, 0.8, 2.8, \ldots

d_1 = 0.8 - ( - 1.2) = 2 \hspace{1.0cm} d_2 = 2.8 - 0.8 = 2 \hspace{1.0cm} \Rightarrow d_1 = d_2 .

Therefore this is an A.P.

For 3,3+ \sqrt{2} , 3+ 2 \sqrt{2}, 3+ 3 \sqrt{2} \dots

d_1 = 3+\sqrt{2} - 3 = \sqrt{2} \hspace{1.0cm} d_2 = 3+2\sqrt{2} - 3 - \sqrt{2} = \sqrt{2} \hspace{1.0cm} \Rightarrow d_1 = d_2 .

Therefore this is an A.P.

For \frac{4}{3} , \frac{7}{3} , \frac{9}{3} , \frac{12}{3} , \ldots

d_1 = \frac{7}{3} - \frac{4}{3} = \frac{3}{3} = 1 \hspace{1.0cm} d_2 = \frac{9}{3} - \frac{7}{3} = \frac{2}{3} \hspace{1.0cm} \Rightarrow d_1 \neq d_2 .

Therefore this is NOT an A.P.

For \frac{-1}{5} , \frac{-2}{5} , \frac{-3}{5} , \ldots

d_1 = \frac{-2}{5} - \frac{-1}{5} = \frac{-1}{5} \hspace{1.0cm} d_2 = \frac{-3}{5} - \frac{-2}{5} = \frac{-1}{5} \hspace{1.0cm} \Rightarrow d_1 = d_2 .

Therefore this is an A.P.

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Question 6:  The pair of linear equations \frac{3x}{2} + \frac{5y}{3} = 7 and 9x+10y=14 is

(A) Consistent         (B) Inconsistent

(C) Consistent with one solution         (D) consistent with many solutions

Answer:

\fbox B

Given: \frac{3x}{2} + \frac{5y}{3} = 7

\Rightarrow 9x + 10y = 42

\Rightarrow a_1 = 9, \hspace{0.5cm} b_1 = 10 , \hspace{0.5cm} c_1 = 42

For 9x+10y=14

\Rightarrow a_2 = 9, \hspace{0.5cm} b_2 = 10 , \hspace{0.5cm} c_2 = 14

For consistent equations \frac{a_1}{a_2} =  \frac{b_1}{b_2} = \frac{c_1}{c_2}

In our case this is \frac{9}{9} =  \frac{10}{10} \neq \frac{42}{14} .

Hence the equations are inconsistent.

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Question 7: In Figure – 2, PQ is tangent to the circle with centre at O , at the point B . If \angle AOB =100^{\circ} , then \angle ABP is equal to:

Figure - 3
Figure – 2

(A) 50^{\circ}         (B) 40^{\circ}         (C) 60^{\circ}         (D) 80^{\circ}

Answer:

\fbox A

\triangle AOB is an isosceles triangle because OA = OB (radius of the same circle)

\therefore \angle OAB = \angle OBA

Now  \angle OAB + \angle OBA + 100^{\circ} = 180^{\circ}

\Rightarrow \angle ABO = 40^{\circ}

\therefore \angle ABP = 90^{\circ} - 40^{\circ} = 50^{\circ}    since \angle OBP = 90^{\circ}

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Question 8: The radius of a sphere (in cm) whose volume is 12 \pi \ cm^3 , is

(A) 3               (B) 3 \sqrt{3}               (C) 3^{\frac{2}{3}}               (D) 3^{\frac{1}{3}}

Answer:

\fbox C

Volume of sphere = 12 \pi \ cm^3

Let the radius = R cm

\therefore \frac{4}{3} \pi (R)^3 = 12 \pi

\Rightarrow R^3 = 9

\Rightarrow R = 3^{\frac{2}{3}}

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Question 9:  The distance between the points (m, -n) and (-m, n) is

(A) \sqrt{m^2+n^2}                 (B) m+n               (C) 2 \sqrt{m^2+n^2}         (D) \sqrt{2m^2+2n^2}

Answer:

\fbox C

Given points (m, -n) and (-m, n)

Distance between points = \sqrt{ (-m-m)^2 + ( n - (-n))^2}

= \sqrt{(-2m)^2 + (2n)^2} = \sqrt{4(m^2 + n^2)} = 2 \sqrt{m^2 + n^2}

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Question 10:  In Figure – 3, from an external point P , two tangents PQ and PR are drawn to a circle of radius 4 cm with centre O . If \angle QPR = 90^{\circ} , then length of PQ is:

Figure - 1
Figure – 3

(A) 3 cm                 (B) 4 cm                 (C) 2 cm               (D) 2\sqrt{2} cm

Answer:

\fbox B

PQ = PR

Given \angle QPR = 90^{\circ}

Also \angle OQP = \angle ORP = 90^{\circ}

\therefore \angle QOR = 360^{\circ} - ( 90^{\circ}+90^{\circ}+90^{\circ}) = 90^{\circ}

Therefore PQOR is a square

Hence all sides are equal.

Therefore PQ = 4 cm

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In Question Nos. 11 to 15,  Fill in the blanks. Each question carries 1 mark.

Question 11:  The probability of an event that is sure to happen, is \underline{ \hspace{2.0cm}}

Answer:

\underline{ \hspace{0.5cm} 1 \hspace{0.5cm} }

The probability of an event is a number describing the chance that the event will happen. An event that is certain to happen has a probability of 1.

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Question 12:  Simplest form of  \frac{1 + \tan^2 A}{1 + \cot^2 A} is \underline{ \hspace{2.0cm}}

Answer:

\underline{ \hspace{0.5cm} \tan^2 A \hspace{0.5cm} }

\frac{1 + \tan^2 A}{1 + \cot^2 A} = \Bigg[ \frac{(\cos^2 A + \sin^2 A)}{\cos^2 A} \Bigg]  \Bigg[ \frac{\sin^2A}{(\sin^2 A + \cos^2 A)} \Bigg] = \frac{\sin^2 A}{ \cos^2 A} = \tan^2 A

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Question 13:  AOBC is a rectangle whose three vertices are A(0, -3), O(0, 0) and B(4, 0) . The length of its diagonals is \underline{ \hspace{2.0cm}}

Answer:

\underline{ \hspace{0.5cm} 5 \ units \hspace{0.5cm} }

AB = \sqrt{(4-0)^2 + ( 0+3)^2} = \sqrt{16+9} = \sqrt{25} = 5

\therefore AB = 5 units

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Question 14:  In the formula \overline{x} = \frac{\Sigma f_i u_i}{\Sigma f_i} \times h  ,  u_i =  \underline{ \hspace{2.0cm}}

Answer:

\underline{ \hspace{0.5cm} u_i = \frac{x_i - a}{h} \hspace{0.5cm} }

Given \overline{x} = \frac{\Sigma f_i u_i}{\Sigma f_i} \times h

\therefore u_i = \frac{x_i - a}{h}   where h = class size, a = assumed mean and x_i = class mark

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Question 15:  All concentric circles are \underline{ \hspace{2.0cm}} to each other.

Answer:

\underline{ \hspace{0.5cm} Similar \hspace{0.5cm} }

All concentric circles are similar to each other.

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Question Nos. 16 to 20 are short answer type questions of 1 mark each.

Question 16:  Find the sum of the first 100 natural numbers.

Answer:

S_{100} = 1 + 2 + 3 + \ldots + 100

First term (a) = 1       Common difference (d) = 2 -1 = 1      Last term (l) = 100

S_n = \frac{n}{2} (a+l)

\therefore S_{100} = \frac{100}{2} ( 1 + 100 ) = 50 \times 101 = 5050

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Question 17:  In Figure  – 4, the angle of elevation of the top of a tower from a point C on the ground, which is 30 m away from the foot of the tower, is 30^{\circ} . Find the height of the tower.

Figure - 4
Figure – 4

Answer:

\tan 30^{\circ} = \frac{AB}{30}

\Rightarrow \frac{1}{\sqrt{3}} = \frac{AB}{30}

AB = \frac{30}{\sqrt{3}} = 10\sqrt{3} m

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Question 18:  The LCM of two numbers is 182 and their HCF is 13 . If one of the numbers is 26 , find the other.

Answer:

We know

LCM \times HCF = Product of two numbers

182 \times 13 = 26 \times x

\Rightarrow x = \frac{182 \times 13}{26} = 91

Therefore the other number = 91

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Question 19:  Form a quadratic polynomial, the sum and product of whose zeroes are (-3) and 2 respectively.

OR

Can (x^2 - 1) be a remainder while dividing x^4 - 3x^2 + 5x - 6 by (x^2 + 3) ?

Answer:

Let \alpha and \beta are two roots  of equation

Therefore \alpha + \beta = - 3

and \alpha \beta = 2

A quadratic equation can be written in the form

p(x) = x^2 - (\ sum \ of \ roots\ ) x + product \ of \ roots

\Rightarrow p(x) =x^2 - ( \alpha + \beta) x + \alpha \beta

\Rightarrow p(x) = x^2 + 3x + 2

OR

(x^2 -1) cannot be the remainder because the degree of the remainder should be less than the degree of the divisor

Long division

\begin{array}{r l l} x^2+3 ) & \overline{ x^4 - 3x^2 + 5x - 6 } & ( x^2 - 6  \\  (-) & x^4 + 3x^2 & \\  \hline& \hspace{0.5cm} -6x^2 + 5x - 6 & \\  (-) & \hspace{0.5cm} -6x^2 - 18 & \\ \hline & \hspace{2.0cm} 5x + 12 & \end{array}

The remainder is 5x+12

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Question 20:  Evaluate:  \frac{2 \tan 45^{\circ} \times \cos 60^{\circ}}{\sin 30^{\circ}}

Answer:

\frac{2 \tan 45^{\circ} \times \cos 60^{\circ}}{\sin 30^{\circ}} = \frac{2 \times 1 \times ( \frac{1}{2} )}{( \frac{1}{2} )} = 2

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Section – B

Question Nos. 21 to 26 carry 2 marks each.

Question 21:  In the Figure – 5, DE \parallel AC and DF \parallel AE Prove that: \frac{BF}{BE} = \frac{FE}{EC}

Figure - 71
Figure – 5

Answer:

Given DE \parallel AC and DF \parallel AE

To Prove: \frac{BF}{FE} = \frac{BE}{EC}

Proof: In \triangle ABC, DE \parallel AC

Note: line drawn parallel to one side of the triangle, intersects the other two sides in distinct points, then it divides the other two sides in the same ratio.

\therefore \frac{BE}{EC} = \frac{BD}{DA}    … … … … … i)

Similarly, in \triangle AEB, DF \parallel AE

\therefore \frac{BF}{FE} = \frac{BD}{DA}      … … … … … ii)

From i) and ii)

\frac{BF}{FE} = \frac{BE}{EC}

Hence proved.

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Question 22: Show that 5+ 2 \sqrt{7} is an irrational number, where \sqrt{7} is given to be an irrational number

OR

Check whether 12^n can end with the digit 0 for any natural number n .

Answer:

Let 5 + 2\sqrt{7} be a rational number

\therefore 5 + 2 \sqrt{7} = \frac{p}{q} ,  where q \neq 0 and p, q \in Z

\Rightarrow 2 \sqrt{7} = \frac{p}{q} - 5

\Rightarrow 2 \sqrt{7} = \frac{p-5q}{q}

\Rightarrow \sqrt{7} = \frac{p-5q}{2q} = \frac{Integer}{Integer}

\Rightarrow \sqrt{7} is a rational number. But it is given that \sqrt{7} is an irrational number.

Hence our assumption that 5 + 2\sqrt{7} is a rational number is wrong.

Hence 5 + 2 \sqrt{7} is an irrational number.

OR

For any number to end with a 0 or 5 , must be divisible by 5 .

12 = 2 \times 2 \times 3 = 2^2 \times 3

\Rightarrow 12^n = (2^2 \times 3)^n = 2^{2n} \times 3^n

This factorization does not contain any term of 5 .

Hence there is no value of n \in N for which 12^n ends with digit 0 or 5

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Question 23:  If A, B and C are interior angles of a \triangle ABC , then show that \cos \Big( \frac{B+C}{2} \Big) = \sin \Big( \frac{A}{2} \Big)

Answer:

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

Given A + B + C = 180^{\circ}

\Rightarrow B+C = 180^{\circ}-A

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

Therefore LHS = \cos \Big( \frac{B+C}{2} \Big) = \cos \Big( 90^{\circ} - \frac{A}{2} \Big) = \sin \frac{A}{2} = RHS.

Hence proved.

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Question 24:  In Figure – 6, a quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB + CD = BC + AD .

Figure - 5
Figure – 6

OR

In Figure – 7, find the perimeter of \triangle ABC , if AP = 12 cm.

Figure - 6
Figure – 7

Answer:

2020-07-13_8-22-50Given ABCD is a quadrilateral

The quadrilateral touches the circle at P, Q, R and S

To prove: AB + CD = AD + BC

Proof:

We know that the lengths of the tangents drawn from an external point to a circle are equal. Therefore

AP = AS    … … … … … i)                      BP = BQ    … … … … … ii)

CR = CQ    … … … … … iii)                    DR = DS    … … … … … iv)

Adding i), ii), iii) and iv) we get

AP + BP + CR + DR = AS + BQ + CQ + DS

\Rightarrow ( AP + BP) + ( CR + DR) = ( AS + DS) + ( BQ + CQ)

\Rightarrow AB + CD = AD + BC

Hence proved.

OR

Figure - 6

AP = AB + BP    … … … … … i)

AQ = AC + CQ      … … … … … ii)

We know that the lengths of the tangents drawn from an external point to a circle are equal. Therefore

BP = BD and CD = CQ . Also AP = AQ

Therefore we can write i) as

AP = AB + BD    … … … … … iii)

and similarly ii) as AQ = AC + DC    … … … … … iv)

Adding iii) and iv) we get

AP + AQ = AB + ( BD + DC) + AC = AB + BC + AC = Perimeter of the \triangle ABC

Therefore the perimeter of \triangle ABC = 12 + 12 = 24 cm

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Question 25:  Find the mode of the following distribution.

Marks 0-10 10-20 20-30 30-40 40-50 50-60
Number of Students 4 6 7 12 5 6

Answer:

Marks Number of Students
0-10 4
10-20 6
20-30 7 f_0
30-40 12 f_1
40-50 5 f_2
50-60 6

Mode = l + \Big(  \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \Big)  \times h

= 30 + \Big( \frac{12-7}{24-7-5}   \Big)  \times 10

= 30 + \frac{5}{12} \times 10

= 30 + 4.167 = 34.17

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Question 26:  2 cubes, each of volume 125 \ cm^3 , are joined end to end. Find the surface area of the resulting cuboid.

Answer:

Volume of cube = 125 \ cm^3

Therefore the side of  the cube = 5 \ cm^3

Dimension of the new cuboid

l = 10 cm          h = 5 cm          b = 5 cm

Therefore surface area of cuboid = 2 ( lh + hb + bh)

= 2 ( 10 \times 5 + 5 \times 5 + 5 \times 10) = 2 ( 50 + 25 + 50) = 250 \ cm^2

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Section – C

Question Nos. 27 to 34 carry 3 marks each.

Question 27:  A fraction becomes \frac{1}{3} when 1 is subtracted from the numerator and it becomes \frac{1}{4} when 8 is added to its denominator. Find the fraction.

OR

The present age of a father is three years more than three times the age of his son. Three years hence the father’s age will be 10 years more than twice the age of the son. Determine their present ages.

Answer:

Let the fraction be \frac{x}{y}

Therefore \frac{x-1}{y} = \frac{1}{3}

\Rightarrow 3x - 3 = y

\Rightarrow 3x - y = 3      … … … … … i)

Also \frac{x}{y+8} = \frac{1}{4}

\Rightarrow 4x = y + 8

\Rightarrow 4x - y = 8      … … … … … ii)

Subtracting i) from ii) we get

\begin{array}{r r} & 4x-y = 8 \\ (-) & 3x - y = 3 \\ \hline & x = 5 \end{array}

Substituting in i)

y = 3x - 3 = 3(5) - 3 = 12

Therefore the fraction is \frac{5}{12}

OR

Let the present age of the son  = x years

Therefore the present age of father = 3x + 3

After 3 years

(3x+3) + 3 = 2 ( x+3) + 10

\Rightarrow 3x + 6 = 2x + 16

\Rightarrow x = 10 years

Therefore present age of father = 3(10) + 3 = 33 years and the present age of son is 10 years.

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Question 28:  Use Euclid Division Lemma to show that the square of any positive integer is either of the form 3q \ or \ 3q + 1 for some integer q .

Answer:

As per Euclid’s Division Lemma, if a and b are two positive integers, than

Let a be the positive integer and b = 3

Therefore a = 3m + r   where 0 \leq r < b and m is any positive integer.

Therefore r can be either 0, 1 or 2 .

If   r = 0 , then a = 3m

\therefore a^2 = (3m)^2 = 3 \times 3m^2 = 3q where q = 3m^2

If   r = 1 , then a = 3m + 1

\therefore a^2 = ( 3m+1)^2 = 9m^2 + 6m + 1

\Rightarrow  a^2 = 3(3m^2 + 2m) + 1 = 3q+1 where q = 3m^2 + 2m

If   r = 2 , then a = 3m + 2

\therefore a^2 = ( 3m+2)^2 = 9m^2 + 12m + 4

\Rightarrow  a^2 = 3(3m^2 + 4m+1) + 1 = 3q+1 where q = 3m^2 + 4m + 1

Therefore the square of any positive number is of the form 3q or 3q+1 for some integer q

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Question 29:  Find the ratio in which the y- axis divides the line segment joining the points (6, -4) and (-2, -7) . Also find the point of intersection.

OR

Show that the points (7, 10), (-2, 5) and (3, -4) are vertices of an isosceles right triangle.

Answer:

Given y-axis divides the line segment.

Therefore the point = ( 0, y)

2020-07-13_8-14-57

Applying section formula

p(x,y) = \Bigg( \frac{kx_2+x_1}{k+1} , \frac{ky_2+y_1}{k+1}    \Bigg)

\Rightarrow (0, y) = \Bigg( \frac{k(-2)+6}{k+1} , \frac{k(-7)+(-4)}{k+1}    \Bigg)

\Rightarrow (0, y) = \Bigg( \frac{-2k+6}{k+1} , \frac{-7k-4}{k+1}    \Bigg)

Comparing

0 = \frac{-2k+6}{k+1}

\Rightarrow -2k + 6 = 0

\Rightarrow k = 3

Therefore the ratio is 3:1

\therefore y = \frac{-7(3)-4}{3+1} = \frac{-25}{4}

Hence the point of intersection is \Big( 0, \frac{-25}{4} \Big)

OR

Given three vertices of a triangle

2020-07-13_8-10-39

\therefore AB = \sqrt{ (-2-7)^2 + ( 5-10)^2} = \sqrt{81+25} = \sqrt{106}

Similarly, BC = \sqrt{ (3-(-2))^2 + ( -4-5)^2} = \sqrt{25+ 81} = \sqrt{106}

And CA = \sqrt{ (7-3)^2 + ( 10-(-4))^2} = \sqrt{16+ 196} = \sqrt{212}

Since AB = BC, \triangle ABC is an isosceles triangle.     … … … … … i)

Applying Pythagoras theorem

AB^2 + BC^2 = AC^2

\Rightarrow (\sqrt{106})^2 + (\sqrt{106})^2 = (\sqrt{212})^2

\Rightarrow 106 + 106 = 212

Therefore \triangle ABC is a right angled triangle.     … … … … … ii)

Hence from i) and ii), \triangle ABC is an isosceles right angled triangle.

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Question 30:  Prove that \sqrt{ \frac{1+\sin A}{1 - \sin A} } = \sec A + \tan A

Answer:

LHS = \sqrt{ \frac{1+\sin A}{1 - \sin A} }

= \sqrt{ \frac{1+\sin A}{1 - \sin A} \times \frac{1+\sin A}{1 + \sin A} }

= \sqrt{ \frac{(1+\sin A)^2}{1 - \sin^2 A} }

= \sqrt{ \frac{(1+\sin A)^2}{\cos^2 A} }

= \frac{1+\sin A}{\cos A}

= \sec A + \tan A = RHS. Hence proved.

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Question 31:  For an A.P., it is given that the first term a = 5 , common difference d = 3 and the n^{th} term a_n = 50 . Find n and sum of first n terms S_n of the A.P.

Answer:

Given: First term (a) = 5 , Common difference (d) = 3

n^{th} term = a_n = 50

We know, a_n = a + ( n-1) d

Therefore 50 = 5 + ( n-1)(3)

\Rightarrow 45 = 3(n-1)

\Rightarrow 15 = n - 1

\Rightarrow n = 16

We know S_n = \frac{n}{2} \Big[ a+ a_n  \Big]

\therefore S_{16} = \frac{16}{2} \Big[ 5+ 50  \Big] = 8 \times 55 = 440

Hence n = 16, S_{16} = 440

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Question 32:  Construct a \triangle ABC with sides BC = 6 cm, AB = 5 cm and \angle ABC = 60^{\circ} . Then construct a triangle whose sides are \frac{3}{4} of the corresponding sides of \triangle ABC .

OR

Draw a circle of radius 3.5 cm. Take a point P outside the circle at a distance of 7 cm from the center of the circle and construct a pair of tangents to the circle from that point.

Answer:

Step 1: Construct \triangle ABC 2020-07-18_10-09-09

  • Draw base BC of side 6 cm
  • Draw \angle B = 60^{\circ}
  • Taking B as a center, 5 cm as radius, draw an arc. Let the point where arc intersect the ray be point A
  • Join AC
  • This completes the construction of \triangle ABC

Now we need to make a triangle which is \frac{3}{4} times its size. Therefore the scale factor = \frac{3}{4} < 1

Step 2:

  • Draw any ray BX making an acute angle with BC on the side opposite to vertex A
  • Mark 4 points – B_1, B_2, B_3 and B_4 so that BB_1 = B_1B_2 = B_2B_3 = B_3B_4
  • Join B_4 C and draw a line through B_3 parallel to B_4C to intersect BC at C'
  • Draw a line through C' parallel to the line AC to intersect BA at A'
  • Therefore \triangle A'BC' is the required triangle

Now consider \triangle ABC and \triangle A'BC'

\angle B is common

Since A'C' \parallel AC

\angle A'C'B = \angle ACB

\therefore \triangle ABC \sim \triangle A'BC' ( AA similarity)

Therefore \frac{A'B}{AB} = \frac{A'C'}{AC} = \frac{BC'}{BC} = \frac{3}{4}

By construction \frac{BC'}{BC} = \frac{3}{4}

Therefore our construction is justified.

OR

2020-07-18_9-56-54Draw a circle of radius 3.5 cm

Mark any point P at a distance of 7 cm from the center

Join OP and locate the midpoint M

Taking M as a center draw a circle of radius 3.5 cm

The two circles intersect at A and B

Join AP and BP and they would be the tangents to the circle from the point P .

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Question 33:  Read the following passage and answer the questions given at the end :

Diwali Fair

A game in a booth at a Diwali Fair involves using a spinner first. Then, if the spinner stops on an even number, the player is allowed to pick a marble from a bag. The spinner and the marbles in the bag are represented in Figure – 8.

Prizes are given, when a black marbles is picked. Shweta plays the same once.

Figure - 9
Figure – 8

(i) What is the probability that she will be allowed to pick a marble from the bag?

(ii) Suppose she is allowed to pick a marble from the bag, what is the probability of getting a prize, when it is given that the bag contains 20 balls out of which 6 are black?

Answer:

i) Even numbers in the spinner = 5 = n(f)

Total numbers in the spinner = 6 = n(s)

Therefore Probability to pick marble = \frac{n(f)}{n(s)} = \frac{5}{6}

ii) Number of black marbles = 6

Number of white marbles = 14

Therefore total number of marbles = 20

Therefore the probability of picking a black marble = \frac{6}{20} = \frac{3}{10}

Therefore the probability of winning = \frac{5}{6} \times \frac{3}{10} = \frac{1}{4}   or 25\%

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Question 34:  In Figure-9, a square OPQR is inscribed in a quadrant OAQB of a circle. If the radius of circle is 6 \sqrt{2} cm, find the area of the shaded region.

2020-07-18_10-17-15
Figure – 9

Answer:

2020-07-18_10-16-54Area of the quadrant of the circle =

\frac{1}{4} (\pi r^2) = \frac{1}{4} \pi (6\sqrt{2}) = 18 \pi = 18 \times 3.14 = 56.52 cm

Now a^2 + a^2 = (6\sqrt{2})^2

\Rightarrow 2a^2 = 36 \times 2 \Rightarrow a = 6 cm

Therefore area of square OPQR = 36 \ cm^2

Hence the shaded area = 56.52 - 36 = 20.52 \ cm^2

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Section – D

Question Nos. 35 to 40 carry 3 marks each.

Question 35:  Obtain other zeroes of the polynomial p(x) = 2x^4 - x^3 - 11x^2 + 5x + 5 if two of its zeores are \sqrt{5} and - \sqrt{5} .

OR

What minimum must be added to 2x^3 - 3x^2 + 6x + 7 so that the resulting polynomial will be divisible by x^2 - 4x + 8 ?

Answer:

Given p(x) = 2x^4 - x^3 - 11x^2 + 5x + 5

Given that \sqrt{5} and -\sqrt{5} are zeros of  p(x)

This means that ( x - \sqrt{5}) and (x+\sqrt{5}) are factors of p(x) .

Therefore ( x - \sqrt{5}) \times (x+\sqrt{5}) = (x^2 - 5) is a factor of p(x)

Long division

\begin{array}{r l l} x^2 - 5) & \overline{2x^4 - x^3 - 11x^2 + 5x + 5} & (2x^2 - x - 1 \\ (-) & 2x^4 - 10x^2 &   \\ \hline & \hspace{1.0cm} -x^3 -x^2 + 5x + 5  &  \\ (-)  & \hspace{1.0cm}-x^3 + 5x  &  \\ \hline & \hspace{2.0cm} -x^2 + 5  &  \\ (-)  & \hspace{2.0cm}-x^2 + 5 & \\ \hline & \hspace{3.0cm} 0 &    \end{array}

\therefore p(x) = ( 2x^2 - x - 1) ( x^2 - 5)

= (2x^2 - 2x + x - 1) ( x^2 - 5)

= [ 2x( x-1) + (x-1)] (x^2 - 1)

= ( 2x+1)(x-1) (x^2 - 5)

Therefore the other two zeros are \frac{-1}{2} , 1

OR

Long division

\begin{array}{r l l} x^2 - 4x + 8) & \overline{2x^3 - 3x^2 + 6x + 7} & (2x+5 \\ (-) & 2x^3 - 8x^2 + 16x &   \\ \hline & \hspace{1.0cm} 5x^2-10x+7  &  \\ (-)  & \hspace{1.0cm}5x^2 - 20x + 40  &  \\ \hline & \hspace{2.0cm} 10x - 33  &      \end{array}

So add ( -10x+33) to the polynomial 2x^3 - 3x^2 + 6x + 7 for it to be completely divisible by x^2 - 4x + 8

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Question 36:  Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Answer:

2020-07-18_10-28-00Given \triangle ABC \sim \triangle PQR

To prove:

\frac{ar(\triangle ABC)}{ar(\triangle PQR)} = \frac{AB^2}{PQ^2} = \frac{BC^2}{QR^2} = \frac{AC^2}{PR^2}

Construction: Draw AM \perp BC and PN \perp QR

Area of \triangle ABC = \frac{1}{2} \times BC \times AM      … … … … … i)

Area of \triangle PQR = \frac{1}{2} \times QR \times PN      … … … … … ii)

Dividing i) by ii)

\frac{ar(\triangle ABC)}{ar(\triangle PQR)} = \frac{BC \times AM}{QR \times PN}      … … … … … iii)

Consider \triangle ABM and \triangle PQN

\angle B = \angle Q as \triangle ABC \sim \triangle PQR

\angle M = \angle N = 90^{\circ}

\therefore \triangle ABM \sim \triangle PQN   (By AA similarity)

\therefore   \frac{AB}{PQ} = \frac{AM}{PN}

From i)

\frac{ar(\triangle ABC)}{ar(\triangle PQR)} = \frac{BC \times AM}{QR \times PN}

\Rightarrow \frac{ar(\triangle ABC)}{ar(\triangle PQR)} = \frac{BC }{QR } \times \frac{AB}{PQ}      … … … … … iv)

Since \triangle ABC \sim \triangle PQR

\Rightarrow  \frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR}

Substituting in iv)

\Rightarrow  \frac{ar(\triangle ABC)}{ar(\triangle PQR)} = \frac{AB }{PQ } \times \frac{AB}{PQ}   = \frac{AB^2}{PQ^2}

Similarly, we can prove

\frac{ar(\triangle ABC)}{ar(\triangle PQR)} = \frac{AB^2}{PQ^2} = \frac{BC^2}{QR^2} = \frac{AC^2}{PR^2}

Hence proved.

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Question 37:  Sum of the areas of two squares is 544 \ m^2 . If the difference of their perimeter is 32 m, find the sides of the two squares.

OR

A motor boat whose speed is 18 km/h in still water takes 1 hour more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream.

Answer:

Let the sides of the two square be x and y

\therefore x^2 + y^2 = 544      … … … … … i)

Perimeter of the two square will be 4x and 4y

\therefore 4x - 4y = 32

\Rightarrow x - y = 8      … … … … … ii)

Substituting ii) in i) we get

x^2 + ( x-8)^2 = 544

\Rightarrow x^2 + x^2 + 64 - 16x = 544

\Rightarrow 2x^2 - 16x - 480 = 0

\Rightarrow x^2 - 8x - 240 = 0

\Rightarrow x^2 - 20 x + 12 x - 240 = 0

\Rightarrow x( x - 20) + 12 ( x- 20) = 0

\Rightarrow (x-20)(x+ 12) =0

\Rightarrow x = 20 or x = - 12 (this is not possible)

\therefore x = 20 m

Hence y = 20-8 = 12 m

OR

Speed of the boat in still water = 18 km/hr

Let the speed of the stream = s

Therefore speed of the boat upstream = ( 18-s) km/hr

Similarly, the speed of the boat down stream = (18+s) km/hr

Time taken upstream = Time taken down stream + 1

\Rightarrow \frac{24}{18-s} = \frac{24}{18+s} + 1

\Rightarrow 24( 18+s) = 24(18-s) + (18-s)(18+s)

\Rightarrow 192 + 24 s = 192 - 24s + 324 - s^2

\Rightarrow s^2 + 48s - 324 = 0

\Rightarrow s^2 + 54s - 6 s - 324 = 0

\Rightarrow s( s+ 54) - 6 (s+ 54) = 0

\Rightarrow (s+54)(s-6) = 0

\Rightarrow s = -54 (not possible) or s = 6 km/hr

Therefore the speed of the stream = 6 km/hr

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Question 38:  A solid toy is in the form of a hemisphere surmounted by a right circular cone of same radius. The height of the cone is 10 cm and the radius of the base is 7 cm. Determine the volume of the toy. Also find the area of the coloured sheet required to cover the toy.  \Big( \ Use \  \pi = \frac{22}{7} \ and  \ \sqrt{149} = 12.2 \Big)

Answer:

Volume of toy = \frac{1}{2} \Big( \frac{4}{3} \pi r^3 \Big) + \frac{1}{3} \pi r^2 h

= \frac{2}{3} \pi ( 7)^3 + \frac{1}{3} \pi (7^2) (10)

= \frac{686}{3} \pi + \frac{490}{3} \pi

= \frac{1176}{3} \times \frac{22}{7}

= 1232 \ cm^3

Area of colored sheet required to cover the toy

= CSA of cone + CSA of hemisphere

= \pi r l + \frac{1}{2} (4 \pi r^2)

= \pi ( 7) ( \sqrt{10^2 + 7^2} ) + 2\pi (7)^2

= ( 7\sqrt{149} + 98) \pi

= ( 7 \times 12.2 + 98) \times \frac{22}{7}

= 576.4 \ cm^2

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Question 39:  A statue 1.6 m tall, stands on the top of a pedestal.From a point on the ground, the angle of elevation of the top of the statue is 60^{\circ} and from the same point the angle of elevation of the top of the pedestal is 45^{\circ} . Find the height of the pedestal. ( Use \sqrt{3} = 1.73)

Answer:

2020-07-18_10-42-36Please refer to the adjoining diagram

In \triangle ABC

\frac{1.6+h}{PB} = \tan 60^{\circ}

\Rightarrow \frac{1.6+h}{PB} = \sqrt{3}

\Rightarrow PB = \frac{1.6+h}{\sqrt{3}}      … … … … … i)

Similarly, in \triangle ABP

\frac{h}{PB} = \tan 45^{\circ} = 1

\Rightarrow h = PB      … … … … … ii)

Therefore from i) and ii) we get

h = \frac{1.6+h}{\sqrt{3}}

\Rightarrow (\sqrt{3} -1) h = 1.6

\Rightarrow h = \frac{1.6}{\sqrt{3}-1} = \frac{1.6 (\sqrt{3}+1)}{3 - 1} = 0.8 \times (1.73+1) = 2.184 m

Therefore the height of the pedestal = 2.184 m

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Question 40:  For the following data, draw a ‘less than’ ogive and hence find the median of the distribution.

Age in years 0-10 10-20 20-30 30-40 40-50 50-60 60-70
Number of persons 5 15 20 25 15 11 9

OR

The distribution given below shows the number of wickets taken by bowlers in one-day cricket matches. Find the mean and the median of the numbers of wickets taken.

Number of wickets 20-60 60-100 100-140 140-180 180-220 220-260
Number of bowlers 7 5 16 12 2 3

Answer:

Less than frequency distribution

Age No of persons Class Cumulative Frequency
0-10 5 Less than 10 5
10-20 15 Less than 20 20
20-30 20 Less than 30 40
30-40 25 Less than 40 65
40-50 15 Less than 50 80
50-60 11 Less than 60 91
60-70 9 Less than 70 100

Now plotting less than ogive (10, 5), (20, 20), (30, 40), (40, 65), (50, 80), (60, 91), (70, 100)

2020-07-18_11-00-20

N = 100 \Rightarrow \frac{N}{2} = 50

Therefore Median = 34 (from graph)

OR

Number of wickets Number

of bowlers (f)

x_i u_i = \frac{x_i-a}{h} u_if_i
20-60 7 40 -3 -21
60-100 5 80 -2 -10
100-140 16 120 -1 16
140-180 12 160 0 0
180-220 2 200 1 2
220-260 3 240 2 6
\Sigma f = 45 \Sigma u_if_i = - 39

Assumed mean = 160

Class size = 40

\therefore Mean \overline{x} = a + \frac{\Sigma f_iu_i}{\Sigma f_i} \times h

\Rightarrow  \overline{x} = 160 + \frac{-39}{45} \times 40

\Rightarrow  \overline{x} = 160 - 34.67

\Rightarrow  \overline{x} = 125.33

To find median

Number of wickets ( C_i ) Number of bowlers (f) C_i f_i
20-60 7 7
60-100 5 12
100-140 16 28
140-180 12 40
180-220 2 42
220-260 3 45

N = 45 \Rightarrow \frac{N}{2} > 22.5

Median class = 100-140

f=16, \hspace{1.0cm} h = 40, \hspace{1.0cm} Cf = 12 , \hspace{1.0cm} l = 100

Median = l + \Bigg[  \frac{\Big( \frac{N}{2} - Cf \Big) }{f}   \times h \Bigg]

= 100 + \Bigg[  \frac{\Big( \frac{45}{2} - 12 \Big) }{f}   \times 40 \Bigg]

= 100 + \frac{105}{4}

= 100 + 26.25

= 126.25