NOTE:

(I)  Please check that this question paper consists of 15 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)

—————————————————————————————————————————————–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 sum of exponents of prime factors  in the prime factorisation of  $196$ is

(a)  $3$          (b)  $4$          (c)  $5$          (d)  $2$

$\fbox{b}$

$196 = 2 \times 2 \times 7 \times 7 = 2^2 \times 7^2$

Therefore the sum of exponents of prime factors  in the prime factorisation of  $196$ is $= 2 + 2 = 4$

$\\$

Question 2: Euclid’s division Lemma states that for two positive integers $a$ and $b$ there exists unique integer $q$ and $r$ satisfying $a = bq + r$  and

(a)  $0 < r < b$         (b)  $0 < r \leq b$          (c)  $0 \leq r < b$         (d)  $0 \leq r \leq b$

$\fbox{c}$

Euclid’s division Lemma states that for two positive integers $a$ and $b$ there exists unique integer $q$ and $r$ satisfying $a = bq + r$  and  $0 \leq r < b$

$\\$

Question 3:  The zeros of the polynomial $x^2 - 3x - m (m+3)$ are

(a)  $m, m+3$         (b)  $-m, m+3$         (c)  $m , - ( m+3)$         (d) $-m, -(m+3)$

$\fbox{b}$

$x^2 - 3x - m (m+3) = 0$

$\Rightarrow x^2 - m^2 - 3x - 3m = 0$

$\Rightarrow (x-m)(x+m) - 3 ( x+m) = 0$

$\Rightarrow (x+m) [ x- m - 3]= 0$

Therefore zeros are $-m, (m+3)$

$\\$

Question 4:  The value of k for which the system of linear equations $x + 2y = 3$ and $5x+ky+7 = 0$ is inconsistent is

(a)  $-$ $\frac{14}{3}$          (b)  $\frac{2}{5}$         (c)  $5$          (d)  $10$

$\fbox{d}$

For $x + 2y = 3$  $\Rightarrow a_1 = 1, b_1 = 2$ and $c_1 = -3$

For $5x+ky+7 = 0$  $\Rightarrow a_2 = 5, b_2 = k$ and $c_2 = 7$

For a unique solution, $\frac{a_1}{a_2}$ $\neq$ $\frac{b_1}{b_2}$

$\Rightarrow \frac{1}{5}$ $\neq$ $\frac{2}{k}$

$\Rightarrow k \neq 10$

Therefore for all values other than $10$ the given equations will have a unique solution. Hence for $k = 10$, the given equations will be inconsistent.

$\\$

Question 5:  The roots of the quadratic equation $x^2 - 0.04 = 0$ are

(a)  $\pm 0.2$         (b)  $\pm 0.02$          (c)  $0.4$         (d) $2$

$\fbox{a}$

$x^2 - 0.04 = 0 \Rightarrow (x-0.2)(x+0.2) = 0$

$\therefore x = \pm 0.2$

$\\$

Question 6:  The common difference of the AP $\frac{1}{p}$ $,$ $\frac{1-p}{p}$ $,$ $\frac{1-2p}{p}$ $,$ $\ldots$ is

(a)  $1$         (b)  $\frac{1}{p}$          (c)  $-1$         (d)  $-$ $\frac{1}{p}$

$\fbox{c}$

First term $(a) =$ $\frac{1}{p}$

Common difference $(d) =$ $\frac{1-p}{p}$ $-$ $\frac{1}{p}$ $=$ $\frac{1-p-1}{p}$ $= -1$

$\\$

Question 7:  The $n^{th}$ term of the A.P. $a, 3a, 5a, \ldots$ , is

(a)  $na$        (b)  $(2n-1)a$        (c)    $(2n+1)a$      (d) $2na$

$\fbox{b}$

First term $(a) = a$

Common difference $(d) = 3a - a = 2a$

Therefore $n^{th}$ term $= a + ( n-1) d = a + ( n-1) (2a) = (2n-1)a$

$\\$

Question 8:  The point $P$ on x-axis equidistant from the points $A(-1, 0)$  and $B(5, 0)$ is

(a)  $(2, 0)$         (b)    $(0, 2)$        (c)    $(3, 0)$        (d)  $(2, 2)$

$\fbox{a}$

Since $P$ is on x-axis, the coordinate of $P$ would be $( x, 0)$

Since $P$ is equidistant from $A$ and $B$ we get

$x =$ $\frac{-1+5}{2}$ $= 2$

Therefore coordinate of $P$ is $( 2, 0)$

$\\$

Question 9:  The coordinate of the point which is the reflection of the point$(-3, 5)$  in x -axis is

(a)    $(3, 5)$        (b)  $(3, -5)$         (c)  $(-3, -5)$         (d)  $(-3, 5)$

$\fbox{a}$

Because the reflection is on x-axis, the y coordinate would not change.

$\\$

Question 10:  If the point $P(6, 2)$ divides the line segment joining $A(6,5)$ and $B(4, y)$  in the ratio $3:1$, then the value of $y$ is

(a)  $4$         (b)    $3$        (c)    $2$       (d) $1$

$\fbox{d}$

Using section formula, we get

$2 =$ $\frac{3y+ 5}{3+1}$

$\Rightarrow 3y = 3$

$\Rightarrow y = 1$

$\\$

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

Question 11:  In Fig. 1, $MN \parallel BC$ and $AM : MB = 1:2$, then

$\frac{ar (\triangle AMN)}{ar (\triangle ABC)}$ $= \underline{\hspace{2.0cm}}$

$\underline{1 : 9}$

Given $\frac{AM}{MB}$ $=$ $\frac{1}{2}$

Let $AM = x \Rightarrow MB = 2x$

$\therefore AB = AM + MB = 3x$

Since $MN \parallel CB$

Using “Basic Proportionality Theorem” we get

$\frac{ar( \triangle AMN) }{ar( \triangle ABC) }$ $=$ $\frac{AM^2}{AB^2}$ $=$ $\frac{x^2}{9x^2}$ $=$ $\frac{1}{9}$

$\\$

Question 12: In given Fig. 2, the length $PB = \underline{\hspace{2.0cm}}$  cm.

$\underline{PB = 4 \ cm}$

$\angle APO = 90^{\circ}$

$\therefore \triangle AOP$ is a right angled triangle

$\therefore AP = \sqrt{5^2 - 3^2} = \sqrt{16} = 4$ cm

Since $OP \perp AB, AP = PB$

Hence $PB = 4$ cm

$\\$

Question 13:  In $\triangle ABC, AB = 6\sqrt{3}$ cm, $AC = 12$ cm  and $BC = 6$ cm, then $\angle B = \underline{\hspace{2.0cm}}$

OR

Two triangles are similar  if their corresponding sides are $\underline{\hspace{2.0cm}}$

$\underline{\angle B = 90^{\circ}}$

We have to first check if the $\triangle ABC$ is a right angled triangle.

If it is a right angled triangle, then Pythagoras theorem should hold

$\Rightarrow 12^2 = ( 6\sqrt{3})^2 + (6)^2$

$\Rightarrow 144 = 108 + 36$

$\Rightarrow 144 = 144$. Therefore the $\triangle ABC$ is a right angled triangle.

$\therefore \angle B = 90^{\circ}$

OR

Two triangles are similar if there corresponding sides are $\underline{proportional}$.

If  $\triangle ABC$ and $\triangle PQO$ are similar, then

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

$\\$

Question 14:  The value of $( \tan 1^{\circ} \tan 2^{\circ} \ldots \tan 89^{\circ})$  is equal to $\underline{\hspace{2.0cm}}$

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

$\tan 1^{\circ} \tan 2^{\circ} \ldots \tan 89^{\circ}$

$= \tan 1^{\circ} \tan 2^{\circ} \ldots \tan 44^{\circ} \tan 45^{\circ} \tan 46^{\circ} \ldots\tan 89^{\circ}$

$= \tan 1^{\circ} \tan 2^{\circ} \ldots \tan 44^{\circ} \tan 45^{\circ} \tan (90^{\circ} - 44^{\circ}) \ldots \tan (90^{\circ} - 1^{\circ})$

$= \tan 1^{\circ} \tan 2^{\circ} \ldots \tan 44^{\circ} \tan 45^{\circ} \cot 44^{\circ} \ldots \cot 1^{\circ}$

$= \tan 45^{\circ} = 1$

$\\$

Question 15: In Fig. 3, the angles of depressions  from the observing positions $O_1$ and $O_2$  respectively  of the object $A$  are $\underline{\hspace{2.0cm}}$

$\underline{30^{\circ} \& 45^{\circ}}$

Therefore the angle of depression are $30^{\circ} \& 45^{\circ}$.

$\\$

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

Question 16:  If  $\sin A + \sin^2 A = 1$,  then find the value of the expression $(\cos^2 A + \cos^4 A)$.

Given $\sin A + \sin^2 A = 1$

$\Rightarrow \sin A = 1 - \sin^2 A$

$\Rightarrow \sin A = \cos^2 A$

Now $\cos^2 A + \cos^4 A$ $= \cos^2 A ( 1 + \cos^2 A)$

$= \sin A ( 1 + \sin A )$  $= \sin A + \sin^2 A$ $= 1$

Hence $(\cos^2 A + \cos^4 A) = 1$.

$\\$

Question 17: In Fig. 4 is a sector of a circle of radius $10.5$ cm. Find the perimeter of the sector. $\Big($ Take $\pi =$ $\frac{22}{7}$ $\Big)$

Given: radius $= OA = OB = 10.5$ cm

Central angle $( \theta) = 60^{\circ}$

We know, if the radius of the circle is $r$ and the length of the arc is $l$, then

$l =$ $\frac{\theta}{180}$ $\pi r$

$\Rightarrow l =$ $\frac{60}{180}$ $\times$ $\frac{22}{7}$ $\times 10.5 = 11$ cm

Therefore perimeter of the sector $= 10.5 + 11 + 10.5 = 32$ cm

$\\$

Question 18:  If a number $x$ is chosen at random  from the numbers $-3, -2, -1, 0, 1, 2, 3$ then find the probability of $x^2 < 4$.

OR

What is the probability  that a randomly taken leap year has $52$ Sundays?

Total number of cases $n(s) = 7$

If $x^2 < 4$, then $x$ can be either $-1, 0, 1$ only.

Therefore number of favorable events $n(f) = 3$

Therefore the probability $n(E) =$ $\frac{n(f)}{n(s)}$ $=$ $\frac{3}{7}$

OR

There are $52$ weeks in a way which means there will be $52$ Sundays in a year.

There are $365$ days in a year but we have $366$ days in a leap year.

There are $7$ days in a week.

If we multiply the weeks by the days we have $52 \times 7$ which equals to $364$. This means that there are $2$ extra days in a leap year which will make it $366$.

The probability of having $52$ Sundays in a leap year is thus: the remaining two days can be any of this formation:

Sunday-Monday, Monday-Tuesday, Tuesday-Wednesday, Wednesday-Thursday, Thursday-Friday, Friday-Saturday, Saturday-Sunday.

However, to get $52$ Sundays in a leap year, none of the remaining two days must be a Sunday. Therefore, out of the $7$ combinations above, that can be only realized $5$ out of $7$ times. The connection “Sunday-Monday and Saturday-Sunday” most be scraped off.

The probability of having $52$ Sundays in a leap year is therefore $\frac{5}{7}$

$\\$

Question 19: Find the class – marks  of the classes $10-25$ and $35-55$.

Given: classes $10-25$ and $35-55$

We know,  Class Marks $(x) =$ $\frac{lower \ limit + upper \ limit}{2}$

For Class interval $(10-25)$   lower limit $= 10$ ,   upper limit $= 25$

Therefore Class Marks of $10-15 =$ $\frac{10+25}{2}$ $= 17.5$

Similarly, For Class interval $(35-55)$   lower limit $= 35$ ,   upper limit $= 55$

Therefore Class Marks of $35-55 =$ $\frac{35+55}{2}$ $= 45$

$\\$

Question 20: A die is thrown once. What is the probability of getting a prime number.

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

Therefore the number of possible outcomes $n(s) = 6$

No of prime numbers on the dice are $2, 3, 5$

Therefore the number of favorable outcomes $n(f) = 3$

Hence the probability of getting a prime number $n (E) =$ $\frac{n(f)}{n(s)}$ $=$ $\frac{3}{6}$ $=$ $\frac{1}{2}$

$\\$

Section – B

Question Nos. 21 to 26 carry 2 marks each.

Question 21:  A teacher asked $10$ of his students to write a polynomial in one variable on a paper and then to hand over the paper. The following were the answers given by the students:

$2x+3$,    $3x^2 + 7x + 2$,    $4x^3 + 3x^2 + 2$,    $x^3 + \sqrt{3x} + 7$,    $7x + \sqrt{7}$,    $5x^3 - 7x + 2$,    $2x^2 + 3 -$ $\frac{5}{x^2}$,    $5x -$ $\frac{1}{2}$,    $ax^3 + bx^2 +cx + d$,    $x +$ $\frac{1}{x}$

(i)  How many of the above $10$ are not polynomials?

(ii) How many of the above are quadratic polynomials?

Definitions:

• A polynomial is an expression consisting of variables and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponents of variables
• A quadratic polynomial, a polynomial of degree 2, or simply a quadratic, is a polynomial function with one or more variables in which the highest-degree term is of the second degree.

$2x+3$ : This is a polynomial.

$3x^2 + 7x + 2$ : This is a polynomial. This is also a quadratic polynomial as the highest-degree term is of the second degree.

$4x^3 + 3x^2 + 2$ : This is a polynomial.

$x^3 + \sqrt{3x} + 7 \Rightarrow x^3 + (3x)^{\frac{1}{2}} + 7$ : This is NOT a polynomial as one of the exponent is not an integer.

$7x + \sqrt{7}$ : This is a polynomial.

$5x^3 - 7x + 2$ :  This is a polynomial.

$2x^2 + 3 -$ $\frac{5}{x^2}$ : This is NOT a polynomial as one of the exponent is a negative integer.

$5x -$ $\frac{1}{2}$ : This is a polynomial.

$ax^3 + bx^2 +cx + d$ : This is a polynomial.

$x +$ $\frac{1}{x}$ : This is NOT a polynomial as one of the exponent is a negative integer.

Therefore:

(i)  How many of the above $10$ are not polynomials? –  Three

(ii) How many of the above are quadratic polynomials? – One

$\\$

Question 22:  In Fig. 5, $ABC$ and $DBC$  are two triangles on the same base $BC$, If $AD$ intersects $BC$ at $O$, show that

$\frac{ar(\triangle ABC)}{ar(\triangle DBC)}$ $=$ $\frac{AO}{DO}$

OR

In Fig. 6, if $AD \perp BC$, then prove that $AB^2 + CD^2 = BD^2 + AC^2$

Given: $\triangle ABC$ and $\triangle DBC$ have common bases.

To prove:

$\frac{ar(\triangle ABC)}{ar(\triangle DBC)}$ $=$ $\frac{AO}{DO}$

Construction: Draw $AE \perp BC$ and $DF \perp BC$

Proof:  $ar(\triangle ABC) =$ $\frac{1}{2}$ $\times BC \times AE$     … … … … … i)

Similarly, $ar(\triangle DBC) =$ $\frac{1}{2}$ $\times BC \times DF$     … … … … … ii)

Therefore $\frac{ar(\triangle ABC)}{ar(\triangle DBC)}$ $=$ $\frac{AE}{DF}$

Consider $\triangle AOE$ and $\triangle DOF$

$\angle AEO = \angle DFO = 90^{\circ}$

$\angle AOE = \angle DOF$  (vertically opposite angle)

$\therefore \triangle AOE \sim \triangle DOF$  (by AA similarity criterion)

$\therefore$ $\frac{AE}{DF}$ $=$ $\frac{AO}{DO}$ $=$ $\frac{OE}{OF}$     … … … … … iv)

Therefore from iii) and iv) we get

$\frac{ar(\triangle ABC)}{ar(\triangle DBC)}$ $=$ $\frac{AO}{DO}$

OR

Given: $AD \perp BC$

To prove: $AB^2 + CD^2 = BD^2 + AC^2$

Proof: Since $AD \perp BC, \angle ADC = \angle ADB =90^{\circ}$

In $\triangle ADB, AB^2 = AD^2 + BD^2$     … … … … … i)

Similarly, in $\triangle ADC, AC^2 = AD^2 + CD^2$    … … … … … ii)

Subtracting ii) from i) we get

$AB^2 - AC^2 = BD^2 - CD^2$

$\Rightarrow AB^2 + CD^2 = AC^2 + BD^2$

Hence proved.

$\\$

Question 23:  Prove that $1 +$ $\frac{\cot^2 \alpha}{1 + \mathrm{cosec} \alpha}$ $= \mathrm{cosec} \alpha$

OR

Show that $\tan^4 \theta + \tan^2 \theta = \sec^4 \theta - \sec^2 \theta$

LHS $= 1 +$ $\frac{\cot^2 \alpha}{1 + \mathrm{cosec} \alpha}$

$= 1 +$ $\frac{\cos^2 \alpha}{\sin^2 \alpha} \Big( \frac{ \sin \alpha}{\sin \alpha + 1}$ $\Big)$

$=1 +$ $\frac{\cos^2 \alpha}{\sin \alpha (\sin \alpha+1)}$

$=$ $\frac{\sin^2 \alpha + \sin \alpha + \cos^2 \alpha }{\sin \alpha ( \sin \alpha + 1)}$

$=$ $\frac{1+ \sin \alpha}{\sin \alpha ( \sin \alpha + 1)}$

$=$ $\frac{1}{\sin \alpha}$ $= \mathrm{cosec} \alpha =$ RHS. Hence proved.

OR

LHS $= \tan^4 \theta + \tan^2 \theta$

$= \tan^2 \theta ( \tan^2 \theta + 1)$

Since $1 + \tan^2 \theta = \sec^2 \theta$

$= (\sec^2 \theta - 1) \sec^2 \theta$

$= \sec^2 \theta - \sec^2 \theta =$ RHS. Hence proved.

$\\$

Question 24:  The volume of the right circular cylinder with its height equal to the radius  is $25$ $\frac{1}{7}$ $cm^3$. Find the height of the cylinder. $($ Use $\pi =$ $\frac{22}{7}$ $)$

Volume of the cylinder $= 25$ $\frac{1}{7}$ $\ cm^3$

Let the radius $= r$. Therefore height $= r$

Therefore $\frac{22}{7}$ $\times r^2 \times r = 25$ $\frac{1}{7}$

$\Rightarrow$ $\frac{22}{7}$ $r^3 =$ $\frac{176}{7}$

$\Rightarrow r^3 = 8$

$\Rightarrow r = 2$ cm

Therefore the height of the cylinder is $2$ cm

$\\$

Question 25:  A child has a die whose six faces show the letters shown below:

$\fbox{A}$      $\fbox{B}$      $\fbox{C}$      $\fbox{D}$      $\fbox{E}$      $\fbox{A}$

The Die is thrown once. What is the probability of getting (i) $A$   (ii) $D$  ?

Probability $n(E) =$ $\frac{n(f)}{n(s)}$

Number of faces in the dice $n(s) = 6$

i) The probability of getting $A$

$n(f) = 2$

$\therefore$ Probability $=$ $\frac{2}{6}$ $=$ $\frac{1}{3}$

ii) The proobability of getting $D$

$n(f) = 1$

$\therefore$ Probability $=$ $\frac{1}{6}$

$\\$

Question 26:  Compute the mode for the following frequency distribution:

 Size of the Items (in cm) 0-4 4-8 8-12 12-16 16-20 20-24 24-28 Frequency 5 7 9 17 12 10 6

Here the maximum frequency is $17$ and the corresponding class is $12-16$

$\therefore 12-16$ is the modal class

$\therefore l = 12, \hspace{0.5cm} h = 4 \hspace{0.5cm} f= 17, \hspace{0.5cm} f_1 = 9 , \hspace{0.5cm} f_2 = 12$

Mode $= l +$ $\frac{f-f_1}{2f - f_1 - f_2}$ $\times h$

$= 12 +$ $\frac{17-9}{2 \times 17 - 9 - 12}$ $\times 4$

$= 12 +$ $\frac{8}{13}$ $\times 4 = 12 + 2.4 = 14.4$

$\\$

Section – C

Question Nos. 27 to 34 carry 3 marks each.

Question 27:  If $2x+y = 23$ and $4x-y = 19$ find the value of $(5y-2x)$

and $\Big($ $\frac{y}{x}$ $- 2 \Big)$

OR

Solve for $x :$ $\frac{1}{x+4}$ $-$ $\frac{1}{x-7}$ $=$ $\frac{11}{30}$ $, x \neq -4, 7$

Given equations:  $2x+y = 23$ and $4x-y = 19$

Adding the two equations we get $6x = 42 \Rightarrow x = 7$

Substituting it back we get $y = 23 - 2(7) = 23 - 14 = 9$

Hence $5y-2x = 5(9) - 2(7) = 45 - 14 = 31$

$\frac{y}{x}$ $- 2 =$ $\frac{9}{7}$ $-2 = -$ $\frac{5}{7}$

OR

$\frac{1}{x+4}$ $-$ $\frac{1}{x-7}$ $=$ $\frac{11}{30}$ $, x \neq -4, 7$

$\Rightarrow 30[x-7 - x -4] = 11[ (x+4)(x-7)]$

$\Rightarrow 30[-11] = 11[ x^2 - 3x- 28]$

$\Rightarrow x^2 - 3x - 28 + 30 = 0$

$\Rightarrow x^2 - 3x + 2 = 0$

$\Rightarrow x^2 - x - 2x + 2 = 0$

$\Rightarrow x(x-1) - 2 ( x - 1) = 0$

$\Rightarrow (x-1)(x-2) = 0$

$x = 1$ or $x = 2$

$\\$

Question 28:  Show that the sum of all terms of an A.P. whose first term is $a$  and the second term is $b$ and the last term is $c$  is equal to $\frac{(a+c)(b+c-2a)}{2(b-a)}$

OR

Solve the equation: $1+ 4+ 7 + 10 + \ldots + x = 287$

First term $= a$

Second term $= b$

Therefore Common difference $(d) = b - a$

We know $n^{th}$ term $= c$

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

$\therefore c = a + ( n - 1) ( b-a)$

$\Rightarrow$ $\frac{c-a}{b-a}$ $= n - 1$

$\Rightarrow n =$ $\frac{c-a}{b-a}$ $+ 1$

$\Rightarrow n =$ $\frac{c-a+b-a}{b-a}$

$\Rightarrow n =$ $\frac{c+b-2a}{b-a}$

We know that sum of $n$ terms $=$ $\frac{n}{2}$ $[ first \ term + last \ term]$

$=$ $\frac{c+b-2a}{2(b-a)}$ $(a+c)$

Hence proved.

OR

Given $1+ 4+ 7 + 10 + \ldots + x = 287$

First term $(a) = 1$

Common difference $(d) = 4-1 = 3$

$S_n =$ $\frac{n}{2}$ $[ 2a + ( n-1) d ]$

$\Rightarrow$ $\frac{n}{2}$ $[ 2(1) + ( n-1) (3) ] = 287$

$\Rightarrow$ $\frac{n}{2}$ $[ 3n - 1 ] = 287$

$\Rightarrow 3n^2 - n - 574 = 0$

$\Rightarrow 3n^2 - 42 n + 41 n - 574 = 0$

$\Rightarrow 3n ( n - 14) + 41( n - 14) = 0$

$\Rightarrow (n-14)(3n+41) = 0$

$\Rightarrow n = 14 \ or \ n = -$ $\frac{41}{3}$  [this is not possible]

Hence $n = 14$

We know $x$ is the $14^{th}$ term

$\therefore x = T_{14} = 1 + ( 14-1) (3) = 40$

$\\$

Question 29:  In a flight of $600$ km, an aircraft was slowed down due to bad weather. The average speed of the trip was reduced by $200$ km/hr and the time of  the flight increased by $30$ minutes. Find the duration of the flight.

Distance traveled $= 600$ km

Let the average speed $= x$ km/hr

Therefore $\frac{600}{x-200}$ $-$ $\frac{600}{x}$ $=$ $\frac{1}{2}$

$\Rightarrow 600 \Big[$ $\frac{x-x+20}{(x-200)x}$ $\Big] =$ $\frac{1}{2}$

$\Rightarrow x^2 - 200 x - 24 000 = 0$

$\Rightarrow (x-600)(x+400) = 0$

$\Rightarrow x = 600$ km/hr  or $x = - 400$ km/hr [this is not possible as speed cannot be negative]

Therefore normal duration of flight $=$ $\frac{600}{600}$ $= 1$ hr

$\\$

Question 30:  If the mid point of the line segment joining the points $A(3, 4)$ and $B ( k, 6)$, is $P(x, y)$ and $x+y-10= 0$, find the value of $k$.

OR

Find the area of triangle $ABC$  with $A ( 1, -4)$  and the mid points of the sides  through $A$ being $( 2, -1)$  and $(0, -1)$.

If $P$ is the mid point then

$x =$ $\frac{k+3}{2}$ and $y =$ $\frac{4+6}{2}$ $= 5$

Also $x + y - 10 = 0$

$\Rightarrow$ $\frac{k+3}{2}$ $+5-10 = 0$

$\Rightarrow k+3 = 10$

$\Rightarrow k = 7$

OR

Since $M$ and $N$ are mid points

$\frac{x_2+ 1}{2}$ $= 2 \Rightarrow x_2 = 3$

$\frac{y_2+ (-4)}{2}$ $= -1 \Rightarrow y_2 = 2$

Hence $B$ is $( 3, 2)$

$\frac{x_3+ 1}{2}$ $= 0 \Rightarrow x_3 = -1$

$\frac{y_3-4}{2}$ $= -1 \Rightarrow y_3 = 2$

Hence $C$ is $( -1, 2)$

Now the area of a triangle where we know all the three vertices

$=$ $\frac{1}{2}$ $[ x_1 ( y_2 - y_3) + x_2 ( y_3 - y_1) +x_3 ( y_1 - y_2) ]$

$=$ $\frac{1}{2}$ $[ 1( 2-2) + 3 ( 2+4) - 1 ( -4 - 2) ]$

$=$ $\frac{1}{2}$ $[18+6]$

$= 12$ sq. units

$\\$

Question 31:  In Fig. 7, if $\triangle ABC \sim \triangle DEF$  and their sides of the lengths (in cm) are marked along them, then find the length of the sides of each of the triangles.

Since $\triangle ABC \sim \triangle DEF$

Therefore $\frac{AB}{DE}$ $=$ $\frac{BC}{EF}$ $=$ $\frac{AC}{DF}$

$\Rightarrow$ $\frac{2x-1}{18}$ $=$ $\frac{2x+2}{3x+9}$ $=$ $\frac{3x}{6x}$

$\Rightarrow$ $\frac{2x-1}{18}$ $=$ $\frac{1}{2}$

$\Rightarrow 2x - 1 = 9$

$\Rightarrow x = 5$

Therefore $AB = 2( 5) -1 = 9, \hspace{0.5cm} BC = 2(5) + 2 = 12 , \hspace{0.5cm} AC = 3(5) = 15$

Similarly, $EF = 3(5) + 9 = 24, \hspace{0.5cm} DF = 6(5) = 30$

$\\$

Question 32:  If a circle touches the side $BC$ of a triangle $ABC$ at $P$and the extended sides $AB$ and $AC$ at $Q$ and $R$ respectively, prove that

$AQ =$ $\frac{1}{2}$ $(BC + CA + AB)$

Given:  A circle touching the side $BC$ of $\triangle ABC$ at $P$ and $AB, AC$ produced at $Q$ and $R$ respectively.

We know that tangents drawn from an external point to the same circle are equal.

Hence $BP = BQ$ and $CP = CR$

Also $AQ = AR$

$\Rightarrow AQ = AC + CR$

$\Rightarrow AQ = AC + CP$     … … … … … i)

Also $AQ = AB + BQ$

$\Rightarrow AQ = AB + BP$      … … … … … ii)

Adding i) and ii) we get

$2 AQ = AB + AC + ( BP + CP)$

$\Rightarrow 2 AQ = AB + AC + BC$

$\Rightarrow AQ =$ $\frac{1}{2}$ $(AB + AC + BC)$

$\\$

Question 33:  If  $\sin \theta + \cos \theta = \sqrt{2}$, prove that $\tan \theta + \cot \theta = 2$

Given $\sin \theta + \cos \theta = \sqrt{2}$

Squaring both sides

$\sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta = 2$

$\Rightarrow 2 \sin \theta \cos \theta = 1$

$\Rightarrow \sin \theta \cos \theta =$ $\frac{1}{2}$   … … … … … i)

Now, LHS $= \tan \theta + \cot \theta$

$=$ $\frac{\sin \theta}{\cos \theta}$ $+$ $\frac{\cos \theta}{\sin \theta}$

$=$ $\frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta}$

$=$ $\frac{1}{\sin \theta \cos \theta}$

$=$ $\frac{1}{\frac{1}{2}}$

$= 2 =$ RHS. Hence proved.

$\\$

Question 34:  The area of a circular playground is $22176 \ cm^2$. Find the cost of fencing this ground  at the rate of Rs. $50$ per meter.

Area $= 22176 \ cm^2$

Let the radius $= r$ cm

$\therefore \pi r^2 = 22176$

$\Rightarrow r^2 =$ $\frac{7}{22}$ $\times 22176 = 7056$

$\Rightarrow r = 84$ cm

Circumference $= 2 \pi r = 2 \times$ $\frac{22}{7}$ $\times 84 = 528$ cm

Therefore cost of fencing $= 528 \times$ $\frac{50}{100}$ $= 264$ Rs.

$\\$

Section – D

Question Nos. 35 to 40 carry 3 marks each.

Question 35: Prove that $\sqrt{5}$ is an irrational number.

Let’s prove this by the method of contradiction.

Say, $\sqrt{5}$ is a rational number.

Therefore It can be expressed in the form $\frac{p}{q}$ where $p, q$ are co-prime integers.

$\Rightarrow \sqrt{5} =$ $\frac{p}{q}$

$\Rightarrow 5=$ $\frac{p^2}{q^2}$ {Squaring both the sides}

$\Rightarrow 5q^2=p^2$     … … … … … i)

$\Rightarrow p^2$ is a multiple of $5$. {Euclid’s Division Lemma}

$\Rightarrow p$ is also a multiple of $5$. {Fundamental Theorm of arithmetic}

$\Rightarrow p = 5m$

$\Rightarrow p^2 = 25 m^2$     … … … … … ii)

From equations i) and ii), we get,

$5q^2=25m^2$

$\Rightarrow q^2=5m^2$

$\Rightarrow q^2$ is a multiple of $5$. {Euclid’s Division Lemma}

$\Rightarrow q$ is a multiple of $5$.{Fundamental Theorm of Arithmetic}

Hence, $p,q$ have a common factor $5$. This contradicts that they are co-primes.

Therefore, $\frac{p}{q}$ is not a rational number. This proves that $\sqrt{5}$ is an irrational number.

$\\$

Question 36:  It takes $12$ hours to fill a swimming pool using two pipes. If the pipe of the larger diameter is used for $4$ hours and the pipe of the smaller diameter for $9$ hours, only half of the pool can be filled. How long would it take for each of the pipes to fill the pool separately.

Let the pipe with the larger diameter and the smaller diameter be pipe $A$ and $B$ respectively.

Let Pipe $A$ works at $x$ liters/hours and Pipe $B$ works at $y$ liters/hours.

Therefore the capacity of the pool $= 12x + 12y$ liters

Given, $4x + 9y =$ $\frac{1}{2}$ $(12x + 12y)$

$\Rightarrow 4x+9y = 6x + 6 y$

$\Rightarrow 3y = 2x$

For larger pipe to fill the swimming pool

$=$ $\frac{12x+12y}{x}$ $= 12 + 12$ $\frac{y}{x}$ $= 12+ 12 \times$ $\frac{2}{3}$ $= 12 + 8 = 20$ hours

For smaller pipe to fill the swimming pool

$=$ $\frac{12x+12y}{y}$ $= 12$ $\frac{x}{y}$ $+ 12 = 12 \times$ $\frac{3}{2}$ $+ 12 = 18 + 12 = 30$ hours

$\\$

Question 37:  Draw a circle of $2$ cm radius with center $O$ and take a point $P$ outside the circle such that $OP$ is $6.5$ cm. From $P$, draw two tangents to the circle.

OR

Construct a triangle with sides $5$ cm, $6$ cm and $7$ cm and then construct another triangle whose sides are $\frac{3}{4}$ times the corresponding sides of the first triangle.

Draw a circle of radius $2.0$ cm

Mark any point $P$ at a distance of $6.5$ cm from the center

Join $OP$ and locate the midpoint $M$

Taking $M$ as a center draw a circle of with $OM$ as the radius.

The two circles intersect at $A$ and $B$

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

OR

Step 1: First construct the $\triangle ABC$. Draw $AB = 5$ cm. The taking $A$ as a center, draw an arc with $6$ cm radius. Similarly draw an arc of $7$ cm radius with $B$ as a center. Join $A$ and $B$ to point of intersection $C$. This is the $\triangle ABC$.

Step 2: Draw a ray $AX$ making an acute angle with $AB$ on the opposite site of vertex $C$.

Step 3: Mark 4 points $A_1, A_2, A_3, A_4$  where $AA_1 = A_1A_2 = A_2A_3 = A_3A_4$

Step 4: Join $A_4B$ and draw a line through $A_3$ parallel to $A_4B$ to intersect $AB$ extended at $B'$

Step 5: Draw a line through $B'$ parallel to line $BC$ to intersect $AC$ at extended $C'$

$\\$

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

Let the height of the tower be $h$ m

In $\triangle CBP$

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

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

In $\triangle ABP$

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

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

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

Substituting in i) we get

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

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

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

$\\$

Question 39:  Find the area of the shaded region in Fig. 8, if $PQ = 24$ cm, $PR = 7$ cm, and $O$ is the center of the circle.

OR

Find the curved surface area of the frustum of a cone, the diameters of whose circular ends are $20$ m and $6$ m and its height is $24$ m.

$\angle RPQ = 90^{\circ}$     (Diameter or a circle subtends a right angle on any point of the circumference of the circle)

$\therefore QR = \sqrt{7^2 + 24^2} = \sqrt{625} = 25$

$\therefore$ Radius $=$ $\frac{25}{2}$

$\therefore$ Area of semi circle $=$ $\frac{1}{2}$ $(\pi r^2) =$ $\frac{1}{2}$ $\times$ $\frac{22}{7}$ $\times \Big($ $\frac{25}{2}$ $\Big)^2 = 245.54 cm^2$

Area of $\triangle PQR =$ $\frac{1}{2}$ $\times 7 \times 24 = 84 \ cm^2$

Therefore the shaded area $= 245.54 - 84 = 161.54 \ cm^2$

OR

Curved Surface Area $= \pi l ( R + r)$ where $l =$ slant height, $R$ is the larger radius, $r$ is the smaller radius

$l = \sqrt{24^2 + 7^2} =\sqrt{625} = 25$

$\therefore CSA =$ $\frac{22}{7}$ $\times 25 \times 26 = 2042.86 \ cm^2$

$\\$

Question 40: The mean of the frequency distribution is $18$ . The frequency f in the class interval $19-21$ is missing. Determine $f$.

 Class Interval 11-13 13-15 15-17 17-19 19-21 21-23 23-25 Frequency 3 6 9 13 f 5 4

OR

The following table gives production yield per hectare  of wheat  of $100$ farms of a village

 Production yield 40-45 45-50 50-55 55-60 60-65 65-70 No. of farms 4 6 16 20 30 24

Change the distribution to a ‘more than’ type distribution and draw an ogive.

 Class Interval Frequency $(f_i)$$(f_i)$ Mean $(x_i)$$(x_i)$ $f_i x_i$$f_i x_i$ 11-13 3 12 36 13-15 6 14 84 15-17 9 16 144 17-19 13 18 234 19-21 $f$$f$ 20 $20f$$20f$ 21-23 5 22 110 23-25 4 24 96 Total $\Sigma f_i = 40 + f$$\Sigma f_i = 40 + f$ $\Sigma f_ix_i = 704+20f$$\Sigma f_ix_i = 704+20f$

Mean $=$ $\frac{\Sigma f_ix_i}{\Sigma f_i}$

$\Rightarrow 18 =$ $\frac{704 + 20f}{40+f}$

$\Rightarrow 720+ 18 f = 704 + 20 f$

$\Rightarrow 16 = 2f$

$\Rightarrow f = 8$

OR

 Production Yield Number of Farms $(f_i)$$(f_i)$ Production Yield  (x-axis) Number of Farms  (y-axis) 40-45 4 More than 40 100 45-50 6 More than 45 96 50-55 16 More than 50 90 55-60 20 More than 55 74 60-65 30 More than 60 54 65-70 24 More than 65 24 $\Sigma f_i = 100$$\Sigma f_i = 100$

Now plot $(40, 100), (45, 96), ( 50, 90), (55, 74), (60, 54), (65, 24)$