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:

(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) $0$

$\fbox B$

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

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

$\displaystyle \fbox C$

For $\displaystyle - 1.2, 0.8, 2.8, \ldots$

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

Therefore this is an A.P.

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

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

Therefore this is an A.P.

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

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

Therefore this is NOT an A.P.

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

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

Therefore this is an A.P.

$\\$

Question 3:  In Figure – 1, 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:

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

$\fbox B$

$PQ = PR$

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

$\\$

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) $\sqrt{2m^2+2n^2}$

$\fbox C$

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: The degree of polynomial having zeroes -3 and 4 only is

(A) 2      (B) 1       (C) more than 3       (D) 3

$\fbox A$

Given zeros are $-3$ and $4$. This implies that the polynomial $f(x) = ( x - (-3))(x-4) = x^2 - x - 12$

Therefore the degree of the polynomial $= 2$

$\\$

Question 6: In Figure – 2, $ABC$ is an isosceles triangle, right angled at $C$. Therefore:

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

$\fbox A$

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 7:  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) $\Big(4, \frac{7}{2} \Big)$

$\fbox D$

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 $O(x, y)$

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

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

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

$\\$

Question 8: The pair of linear equations $\displaystyle \frac{3x}{2} + \frac{5y}{3} = 7$ and $9x+10y=14$ is

(A) Consistent  (B) Inconsistent

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

$\fbox B$

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

$\Rightarrow 9x + 10y = 42$

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

For $9x+10y=14$

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

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

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

Hence the equations are inconsistent.

$\\$

Question 9: 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:

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

$\fbox A$

$\triangle AOB$ 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 $\angle OBP = 90^{\circ}$

$\\$

Question 10: The radius of a sphere (in cm) whose volume is $12 \pi \ cm^3$, is

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

$\fbox C$

Volume of sphere $= 12 \pi \ cm^3$

Let the radius $= R$ cm

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

$\Rightarrow R^3 = 9$

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

$\\$

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

Question 11:  $AOBC$ is a rectangle whose three vertices are $A(0, -3), O(0, 0)$ and $B(4, 0)$. The length of its diagonals is $\underline{ \hspace{2.0cm}}$

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

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

$\therefore AB = 5$ units

$\\$

$\displaystyle \text{Question 12: In the formula } \overline{x} = \frac{\Sigma f_i u_i}{\Sigma f_i} \times h , u_i = \underline{ \hspace{2.0cm}}$

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

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

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

$\\$

Question 13:  All concentric circles are $\underline{ \hspace{2.0cm}}$ to each other.

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

All concentric circles are similar to each other.

$\\$

Question 14:  The probability of an event that is sure to happen, is $\underline{ \hspace{2.0cm}}$

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

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:  Simplest form of  $(1 - \cos^2 A) (1 + \cot^2 A) \text{ is } \underline{ \hspace{2.0cm}}$

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

$(1 - \cos^2 A) (1 + \cot^2 A)$

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

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

$\displaystyle = \frac{\sin^2 A}{\sin^2 A} = 1$

$\\$

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

Question 16:  The LCM of two numbers is $182$ and their HCF is $13$. If one of the numbers is $26$, find the other.

We know

$LCM \times HCF =$ Product of two numbers

$182 \times 13 = 26 \times x$

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

Therefore the other number $= 91$

$\\$

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 $(x^2 + 3)$ ?

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.

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

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

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

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

$\\$

Question 19:  Evaluate:  $2 \sec 30^{\circ} \times \tan 60^{\circ}$

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

$\\$

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.

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

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

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

$\\$

Section – B

Question Nos. 21 to 26 carry 2 marks each.

Question 21:  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

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

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

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

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

$= 30 + 4.167 = 34.17$

$\\$

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

OR

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

Given $ABCD$ 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

$AP = AB + BP$    … … … … … 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:  How may cubes of side 2 cm can be made from a solid cube of side 10 cm?

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 $= 2 ( lh + hb + bh)$

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

$\\$

Question 24:  In the Figure – 7, $DE \parallel AC$ and $DF \parallel AE$ Prove that: $\displaystyle \frac{BF}{BE} = \frac{FE}{EC}$

$\text{Given } DE \parallel AC \text{ and } DF \parallel AE$

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

$\text{Proof: In } \triangle ABC, DE \parallel AC$

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.

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

Similarly, in $\triangle AEB, DF \parallel AE$

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

From i) and ii)

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

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 $n$. $12^n = (2 \times 2 \times 3)^n$

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

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

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

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

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

$\displaystyle \Rightarrow \sqrt{7}$ 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 $\displaystyle A, B$ and $\displaystyle C$ are interior angles of a $\displaystyle \triangle ABC$, then show that $\displaystyle \cot \Big( \frac{B+C}{2} \Big) = \tan \Big( \frac{A}{2} \Big)$

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

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

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

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

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

Hence proved.

$\\$

Section – C

Question Nos. 27 to 34 carry 3 marks each.

Question 27:  In Figure-8, 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.

Area of the quadrant of the circle $=$

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

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 $= 56.52 - 36 = 20.52 \ cm^2$

$\\$

Question 28:  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.

Step 1: Construct $\triangle ABC$

• 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 $\displaystyle = \frac{3}{4} < 1$

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)

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

$\displaystyle \text{By construction } \frac{BC'}{BC} = \frac{3}{4}$

Therefore our construction is justified.

OR

Draw 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 $P$.

$\\$

$\displaystyle \text{Question 29: Prove that } \frac{2 \cos^3 \theta - \cos \theta}{\sin \theta - 2 \sin^3 \theta} = \cot \theta$

$\displaystyle \text{LHS } = \frac{2 \cos^3 \theta - \cos \theta}{\sin \theta - 2 \sin^3 \theta}$

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

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

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

$\displaystyle = \frac{\cos \theta }{\sin \theta } = \cot \theta$

$\\$

Question 30:  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.

$\displaystyle \text{Let the fraction be } \frac{x}{y}$

$\displaystyle \text{Therefore } \frac{x-1}{y} = \frac{1}{3}$

$\Rightarrow 3x - 3 = y$

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

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

$\Rightarrow 4x = y + 8$

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

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

$\displaystyle \text{Therefore the fraction is } \frac{5}{12}$

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 31:  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 $q$.

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 $q$

$\\$

Question 32:  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.

Given y-axis divides the line segment.

Therefore the point $= ( 0, y)$

Applying section formula

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

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

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

Comparing

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

$\displaystyle \Rightarrow -2k + 6 = 0$

$\displaystyle \Rightarrow k = 3$

Therefore the ratio is $\displaystyle 3:1$

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

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

OR

Given three vertices of a triangle

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

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 33:  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.

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

$n^{th}$ term $= a_n = 50$

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

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

$\Rightarrow 45 = 3(n-1)$

$\Rightarrow 15 = n - 1$

$\Rightarrow n = 16$

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

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

Hence $n = 16, S_{16} = 440$

$\\$

Question 34:  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.

(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?

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

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

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

ii) Number of black marbles $= 6$

Number of white marbles $= 14$

Therefore total number of marbles $= 20$

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

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

$\\$

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.

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 $+ 1$

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

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

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

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

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

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

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

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

Therefore the speed of the stream $= 6$ km/hr

$\\$

Question 36:  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.  $\Big( \ Use \ \pi = \frac{22}{7} \ and \ \sqrt{149} = 12.2 \Big)$

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

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

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

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

$\displaystyle = 1232 \ cm^3$

Area of colored sheet required to cover the toy

$\displaystyle =$ CSA of cone $\displaystyle +$ CSA of hemisphere

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

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

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

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

$\displaystyle = 576.4 \ cm^2$

$\\$

Question 37:  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

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 $(10, 5), (20, 20), (30, 40), (40, 65), (50, 80), (60, 91), (70, 100)$

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

Therefore Median $= 34$ (from graph)

OR

 Number of wickets Number of bowlers $(f)$$(f)$ $x_i$$x_i$ $\displaystyle u_i = \frac{x_i-a}{h}$$\displaystyle u_i = \frac{x_i-a}{h}$ $u_if_i$$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$$\Sigma f = 45$ $\Sigma u_if_i = - 39$$\Sigma u_if_i = - 39$

Assumed mean $= 160$

Class size $= 40$

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

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

$\Rightarrow \overline{x} = 160 - 34.67$

$\Rightarrow \overline{x} = 125.33$

To find median

 Number of wickets $( C_i )$$( C_i )$ Number of bowlers $(f)$$(f)$ $C_i f_i$$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

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

Median class $= 100-140$

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

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

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

$\displaystyle = 100 + \frac{105}{4}$

$= 100 + 26.25$

$= 126.25$

$\\$

Question 38:  From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a $20$ m high building are $45^{\circ}$ and $60^{\circ}$ respectively. Find the height of the tower. $($ Use $\sqrt{3} = 1.73)$

Let the height of the tower be $h$ m

In $\triangle CBP$

$\displaystyle \tan 60^{\circ} = \frac{BC}{BP}$

$\displaystyle \Rightarrow \sqrt{3} = \frac{20 + h}{BP}$     … … … … … i)

In $\triangle ABP$

$\displaystyle \tan 45^{\circ} = \frac{AB}{BP}$

$\displaystyle \Rightarrow 1 =$ $\frac{20}{BP}$

$\Rightarrow BP = 20$     … … … … … ii)

Substituting in i) we get

$\displaystyle \sqrt{3} = \frac{20 + h}{20}$

$\Rightarrow 20\sqrt{3} = 20 + h$

$\Rightarrow h = 20 ( \sqrt{3} -1 ) = 20 (1.73 - 1) = 20 \times 0.73 = 14.6$ m

$\\$

Question 39: Prove that in a right angled triangle the square of hypotenuse is equal to the sum of square of other two sides.

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

To Prove: $AC^2=AB^2+BC^2$

Construction: draw perpendicular $BD$ onto the side $AC$ .

Proof:

We know that if a perpendicular is drawn from the vertex of a right angle of a right angled triangle to the hypotenuse, than triangles on both sides of the perpendicular are similar to the whole triangle and to each other.

We have

$\triangle ADB \sim \triangle ABC$. (by AA similarity)

Therefore, $\displaystyle \frac{AD}{AB} = \frac{AB}{AC}$

(In similar Triangles corresponding sides are proportional)

$AB^2=AD \times AC$     … … … … … i)

Also, $\triangle BDC \sim \triangle ABC$

$\displaystyle \text{Therefore, } \frac{CD}{BC} = \frac{BC}{AC}$

(in similar Triangles corresponding sides are proportional)

Or, $BC^2 = CD \times AC$      … … … … … ii)

Adding the equations i) and ii) we get,

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

$AB^2+BC^2=AC(AD+CD)$

From the figure $AD + CD = AC$

$AB^2+BC^2=AC \times AC$

Therefore, $AC^2=AB^2+BC^2$

This theorem is known as Pythagoras theorem.

$\\$

Question 40:  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 $x^2 - 4x + 8$ ?

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

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

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

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

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

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

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

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 $x^2 - 4x + 8$