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:  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

\\

Question 2: 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}

(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.

\\

Question 3: 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}}

\\

Question 4:  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}

\\

Question 5:  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

\\

Question 6:  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

\\

Question 7: In Figure – 2, DE \parallel BC . If \frac{AD}{DB}= \frac{3}{2} and AE=2.7 cm, then EC is equal to

Figure - 2
Figure – 2

(A) 2.0  cm        (B) 1.8  cm       (C) 4.0  cm       (D) 2.7  cm

Answer:

\fbox B

Since DE \parallel BC

\frac{AD}{DB} = \frac{AE}{EC}

\Rightarrow \frac{2.7}{EC} = \frac{3}{2}

\Rightarrow EC = \frac{2 \times 2.7}{3} = 1.8 cm

\\

Question 8:  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)

\\

Question 9:  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.

\\

Question 10: In Figure – 3, 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 – 3

(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}

\\

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

Question 11:  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

\\

Question 12:  If the probability of an event E happening is 0.023 , then P(\overline{E}) =  \underline{ \hspace{2.0cm}}

Answer:

\underline{ \hspace{0.5cm} 0.977 \hspace{0.5cm} }

We know P(E) = 0.023

Therefore P(\overline{E}) = 1 - P(E) =  1 - 0.023 = 0.977

\\

Question 13:  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.

\\

Question 14:  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.

\\

Question 15:  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

\\

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

Question 16:  Evaluate:  2 \sec 30^{\circ} \times \tan 60^{\circ}

Answer:

2 \sec 30^{\circ} \times \tan 60^{\circ} = 2 \times \frac{2}{\sqrt{3}} \times \sqrt{3} = 4

\\

Question 17:  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

\\

Question 18:  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

\\

Question 19:  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

\\

Question 20:  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

\\

Section – B

Question Nos. 21 to 26 carry 2 marks each.

Question 21:  A cone and a cylinder have the same radii but the height of the cone is 3 times that of the cylinder. Find the ratio of their volumes.

Answer:

Let the radius of cone and cylinder = r

Let the height of cylinder = h

Therefore height of cone = 3h

Volume of cone = \frac{1}{3} \pi r^2 (3h) = \pi r^2 h

Volume of  cylinder = \pi r^2 h

Therefore ratio of their volume = 1:1

\\

Question 22:  In Figure – 5, a quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB + CD = BC + AD .

Figure - 5
Figure – 5

OR

In Figure – 6, find the perimeter of \triangle ABC , if AP = 12 cm.

Figure - 6
Figure – 6

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

\\

Question 23:  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

\\

Question 24:   In Figure-7, If  PQ \parallel BC and PR \parallel CD , Prove that \frac{QB}{AQ} = \frac{DR}{AR}

Figure - 7
Figure – 7

Answer:

Given:  PQ \parallel BC and PR \parallel CD

In \triangle APQ and \triangle ACB

Since PQ \parallel CB

\angle APQ = \angle ACB

\angle AQP = \angle ABC

\therefore    \triangle APQ \sim \triangle ACB ( By AA similarity)

\Rightarrow \frac{AQ}{AB} = \frac{AP}{AC}      … … … … … i)

Now consider \triangle APR and \triangle AD C

\angle APR = \angle ACD

\angle ARP = \angle ADC

\therefore \triangle APR \sim \triangle ADC   ( By AA similarity)

\Rightarrow \frac{AR}{AD} = \frac{AP}{AC}      … … … … … ii)

From i) and ii)

\frac{AQ}{AB} = \frac{AR}{AD}

\Rightarrow \frac{AQ}{AQ+QB} = \frac{AR}{AR+DR}

\Rightarrow \frac{1}{1+\frac{QB}{AQ}} = \frac{1}{1+\frac{DR}{AR}}

\Rightarrow 1+ \frac{DR}{AR} = 1+ \frac{QB}{AQ}

\Rightarrow \frac{DR}{AR} = \frac{QB}{AQ}

Hence proved.

\\

Question 25: 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

\\

Question 26:  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.

\\

Section – C

Question Nos. 27 to 34 carry 3 marks each.

Question 27: Prove that:  (\sin^4 \theta - \cos^4 \theta+1) \mathrm{cosec}^2 \theta = 2

Answer:

LHS = (\sin^4 \theta - \cos^4 \theta+1) \mathrm{cosec}^2 \theta

= [ (\sin^2 \theta - \cos^2 \theta)(\sin^2 \theta + \cos^2 \theta) + 1 ] \mathrm{cosec}^2 \theta

= [ (\sin^2 \theta - \cos^2 \theta) + 1 ] \mathrm{cosec}^2 \theta

= [ (\sin^2 \theta + (1 - \cos^2 \theta)) + 1 ] \mathrm{cosec}^2 \theta

= ( 2 \sin^2 \theta)  \mathrm{cosec}^2 \theta

= 2 = RHS.

Hence proved.

\\

Question 28:  Find the sum: (-5) + ( -8) + (-11) + \ldots + (-230)

Answer:

First term (a) = - 5

Common difference (d) = -8 - ( - 5) = - 3

a_n = l = - 230

Therefore Number of terms n = \frac{l-a}{d} + 1 = \frac{-230 - ( - 5)}{(-3)} + 1 = 75+1 = 76

S_n = \frac{n}{2} [a+l] = \frac{76}{2} [ - 5 - 230 ] = 38 \times ( - 235) = -8930

\\

Question 29:  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 centre 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 .

\\

Question 30:  In Figure – 8, ABCD  is a parallelogram, A semicircle with centre O  and the diameter AB  has been drawn and it passes through D . If AB = 12  cm and OD \perp AB , then find, the area of the shaded region. (\ use \ \pi = 3.14)

2020-07-19_16-31-28
Figure – 8

Answer:2020-07-19_16-36-03

ABCD is a parallelogram.

AB = 12 cm = diameter

Radius = 6 cm

Area of shaded = ar(parallelogram) - ar(quadrant)

= AB \times OD - \frac{1}{4} \pi (6)^2

= 12 \times 6 - \frac{1}{4} \times 3.14 \times 36

= 72 - 28.26 = 43.74 \ cm^2

\\

Question 31:  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 – 9.

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

Figure - 9
Figure – 9

(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\%

\\

Question 32:  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.

\\

Question 33:  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.

\\

Question 34:  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

\\

Section – D

Question Nos. 35 to 40 carry 3 marks each.

Question 35:  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

\\

Question 36:  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

\\

Question 37:  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

\\

Question 38:  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

\\

Question 39:  In a cylindrical vessel of radius 10  cm, containing some water, 9000  small spherical balls are dropped which are completely immersed in water which raises the water level. If each spherical ball is of radius 0.5  cm, then find the rise in the level of water in the vessel.

Answer:

Volume of cylinder = \pi R^2 h

Volume of the spherical balls = \frac{4}{3} \pi r^3

Radius of cylinder = 10 cm

Radius of the spherical balls = 0.5 cm

Let the level rises by h when 6000 spherical balls are dropped in the cylinder

Therefore

\pi (10)^2 h = 9000 \times \frac{4}{3} \pi (0.5)^3

\Rightarrow h = \frac{9000 \times 4 \times (0.5)^3}{100 \times 3}

\Rightarrow  h = 15 cm

Rise in the level of water in vessel = 15 cm.

\\

Question 40:  If a line is drawn parallel to one side of a triangle to intersect other two sides at distinct points, prove that other two sides are divided in the same ratio.

Answer:

2020-07-19_17-46-52Theorem: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points , then the other two sides are divided in the same ratio. [ Basic Proportionality Theorem Or Thales Theorem ]

In \triangle ABC , DE intersects AB at D and  AC at M

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

Construction: Join BE and DC

Draw 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 \times EN}{BD \times EN} = \frac{AD}{BD}      … … … … …  i)

Again,

Area of \triangle ADE = \frac{1}{2} \times AE \times DM

Area of \triangle CDE = \frac{1}{2} \times EC \times DM

\therefore \frac{ar(\triangle ADE)}{ar(\triangle CDE)} = \frac{AE \times DM}{EC \times DM} = \frac{AE}{EC}      … … … … …  ii)

We see that \triangle BDE and \triangle CDE are on the same base DE and between the two parallels BC and DE .

\therefore ar( \triangle BDE) = ar(\triangle CDE)    … … … … …  iii)

From i), ii) and iii)

\frac{AD}{BD} = \frac{AE}{EC}

Hence proved.