Instructions:

• Please check that this question paper consists of 11 pages.
• 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.
• Please check that this question paper consists of 30 questions.
• Please write down the serial number of the question before attempting it.
• 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.

SUMMATIVE ASSESSMENT – II

MATHEMATICS

Time allowed: 3 hours                                                             Maximum Marks: 80

General Instructions:

(i) All questions are compulsory

(ii) The question paper consists of 30 questions divided into four sections – A, B, C and D

(iii) Section A consists of 6 questions of 1 mark each. Section B consists of 6 questions of 2 marks each. Section C consists of 10 questions of 3 marks each. Section D consists of 8 questions of 4 marks each.

(iv) There is no overall choice. However, an internal choice has been provided in two questions of 1 mark, two questions of 2 marks, four questions of 3 marks each and three questions of 4 marks each. You have to attempt only one of the alternative in all such questions.

(v) Use of calculator is not permitted.

SECTION – A

Question number 1 to 6 carry 1 mark each.

Question 1: The HCF of two numbers $a$ and $b$ is $5$ and their LCM is $200$. Find the product $ab$.

$a$ and $b$ are two numbers.

H.C.F. of $a$ and $b = 5$

L.C.M. of $a$ and $b = 200$

So, By Fundamental theorem of Arithmetic , we have $HCF \times LCM = a \times b$

$5 \times 200 = ab \Rightarrow ab = 1000$

So, The value of product of $a$ and $b$ is $1000$

$\\$

Question 2: Find the value of $k$ for which $x = 2$ is a solution of the equation $kx2 + 2x - 3 = 0$.

Or

Find the value(s) of $k$ for which the quadratic equation $3x^2 + kx + 3 = 0$ has real and equal roots.

Given equation: $kx^2 + 2x - 3 = 0$

If $x=2$ is a solution, then it should satisfy the given equation.

$\therefore k(2)^2 + 2(2) - 3 = 0$

$\Rightarrow 4k + 4 - 3 = 0$

$\Rightarrow k = -$ $\frac{1}{4}$

Or

Given equation: $3x^2 + kx + 3 = 0$

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

$\Rightarrow k^2 - 4(3) (3) = 0$

$\Rightarrow k^2 = 36$

$\Rightarrow k = \pm 6$

$\\$

Question 3: If in an A.P., $a = 15, d = - 3$ and $a_n = 0$, then find the value of $n$.

Given: $a = 15, d = -3$ and $a_n = 0$

Since $a_n = a + (n-1)d$

$\Rightarrow 0 = 15 + (n-1) (-3)$

$\Rightarrow 15 - 3n + 3 = 0$

$\Rightarrow 3n = 18$

$\Rightarrow n = 6$

$\\$

Question 4: If $\sin x + \cos y = 1; x = 30^o$ and $y$ is an acute angle, find the value of $y$.

Or

Find the value of $( \cos 48^o - \sin 42^o)$.

$\sin x + \cos y = 1$

$\Rightarrow \sin 30^o + \cos y = 1$

$\Rightarrow \cos y = 1-$ $\frac{1}{2}$ $=$ $\frac{1}{2}$

$\Rightarrow \cos y = \cos 60^o$

$\Rightarrow y = 60^o$

Or

$\cos 48^o - \sin 42^o$

$= \cos 48^o - \sin (90^o-48^o)$

$= \cos 48^o - \cos 48^o$

$= 0$

$\\$

Question 5: The area of two similar triangles are $25$ sq. cm and $121$ sq. cm. Find the ratio of their corresponding sides.

Let $AB$ and $DE$ are corresponding sides of the similar triangles

$\therefore$ $\frac{25}{121}$ $= \Big($ $\frac{AB}{DE}$ $\Big)^2$

$\Rightarrow$ $\frac{AB}{DE}$ $=$ $\frac{5}{11}$

Hence the ratio of the corresponding sides $= 5:11$

$\\$

Question 6: Find the value of $'a'$ so that the point $(3, a)$ lies on the line represented by $2x - 3y = 5$.

If $(3, a)$ lies on the equation $2x - 3y = 5$, it must satisfy the equation.

$\Rightarrow 2(3) - 3(a) = 5$

$\Rightarrow 6-3a = 5$

$\Rightarrow 3a = 1$

$\Rightarrow a =$ $\frac{1}{3}$

$\\$

Section – B

Question number 7 to 12 carry 2 mark each.

Question 7: If $S_n$, the sum of the first $n$ terms of an A.P. is given by $S_n = 2n^2 + n$, then find its $n^{th}$ term.

Or

If the $17^{th}$ term of an A.P. exceeds its $10^{th}$ term by $7$, find the common difference.

Sum of first $n$ terms of AP, $S_n = 2n^2 + 5n$

Now choose $n =1$ and put in the above formula, First term $= 2+5 = 7$

Now put $n=2$ to get the sum of first two terms $= 2 \times 4 + 5 \times 2= 8 + 10 = 18$

This means  First term + Second term $= 18$

But first term $=7$ as calculated above Hence, second term $= 18-7 = 11$

So common difference (d) becomes, $11-7 = 4$

So the AP becomes, $7, 11, 15, \ldots$

$n^{th} term = a + (n - 1)d = 7 + (n-1)4 = 7 + 4n - 4 = 3 + 4n$

Or

Let the first term $= a$ and the common difference $= d$

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

$\therefore T_{17} = a + 16 d$

$T_{10} = a + 9 d$

Given $T_{17} = T_{10}+ 7$

$a + 16 d = a + 9d + 7$

$7d = 7$

$d = 1$

Common difference $= 1$

$\\$

Question 8: The mid-point of the line segment joining $A(2a, 4)$ and $B(-2, 3b)$ is $(1, 2a + 1)$. Find the values of $a$ and $b$.

Given points $A(2a, 4)$ and $B(-2, 3b)$

Mid Point of $AB = \Big($ $\frac{2a-3}{2}$ $,$ $\frac{4+3b}{2}$ $\Big) = (1, 2a+1)$

$\Rightarrow (a-1, 2 +$ $\frac{3}{2}$ $b) = ( 1, 2a+1)$

$\Rightarrow a -1 = 1 \Rightarrow a = 2$

Also $2 +$ $\frac{3}{2}$ $b = 2(2) +1$

$\Rightarrow$ $\frac{3}{2}$ $b = 3$

$\Rightarrow b = 2$

Hence $a = 2, b = 2$

$\\$

Question 9: A child has a die whose 6 faces show the letters given below :

The die is thrown once. What is the probability of getting (i) A (ii) B ?

Number of possible events $= 6$

No of A’s on dice $= 3$

No of B’s of dice $= 2$

i) Probability $(A) =$ $\frac{3}{6}$ $=$ $\frac{1}{2}$

ii) Probability $(B) =$ $\frac{2}{6}$ $=$ $\frac{1}{3}$

$\\$

Question 10: Find the HCF of $612$ and $1314$ using prime factorisation.

Or

Show that any positive odd integer is of the form $6m + 1$ or $6m + 3$  or  $6m + 5$, where $m$ is some integer.

First factorize each number

$612 = 2 \times 2 \times 3 \times 3 \times 17$

$1314 = 2 \times 3 \times 3 \times 73$

Therefore HCF of $612$ and $1314 = 2 \times 3 \times 3 = 18$

Or

Let $a$ be any positive integer.

Eculid’s division theorem, any positive number can be expressed as $a = bm+r$ where $m$ is the quotient, $b$ is the divisor and $r$ is the remainder and $0 \leq r < b$

Take $b = 6 \Rightarrow a = 6m + r$

Since $0 \leq r < 6$, the possible remainders are $0, 1, 2, 3, 4, 5$

That is $a$ can be $6m, 6m+1, 6m+2, 6m+3, 6m+4$ or $6m+5$

Since $a$ is odd, $a$ cannot be $6m$ or $6m+2$ or $6m+4$

Therefore any odd integer is of the form $6m+1$ or $6m+3$ or $6m+5$.

$\\$

Question 11: Cards marked with numbers $5$ to $50$ (one number on one card) are placed in a box and mixed thoroughly. One card is drawn at random from the box. Find the probability that the number on the card taken out is (i) a prime number less than $10$, (ii) a number which is a perfect square.

Total number of ways to select a card $= 46$ (Cards are marked $5$ to $50$ )

i) Prime numbers less than $10$ are $5$ and $7$ only

Therefore Probability (prime number less than $10$) $=$ $\frac{2}{46}$ $=$ $\frac{1}{23}$

ii) Numbers which are perfect squares are $9, 16, 25, 36, 49$

Therefore Probability (a number which is a perfect square) $=$ $\frac{5}{46}$

$\\$

Question 12: For what value of $k$, does the system of linear equations
$2x + 3y = 7$$(k - 1) x + (k + 2) y = 3k$ have an infinite number of solutions ?

If the system of equations are $ax_1 + b_1y + c_1 = 0$ and $ax_2 + b_2y + c_2 = 0$ and they have infinitely many solutions then it satisfy the following:
$\frac{a_1}{a_2}$ $=$ $\frac{b_1}{b_2}$ $=$ $\frac{c_1}{c_2}$

For $2x + 3y = 7$  and $(k - 1) x + (k + 2) y = 3k$ have an infinite number of solutions:

$\frac{2}{k-1}$ $=$ $\frac{3}{k+2}$ $=$ $\frac{7}{3k}$

From first two terms

$2(k+2) = 3 (k-1) \Rightarrow 2k + 4 = 3k - 3 \Rightarrow k = 7$

From second and third term

$3(3k) = 7(k+2) \Rightarrow 9k = 7k + 14 \Rightarrow k = 7$

From first and third term

$2(3k) = 7 (k-1) \Rightarrow 6k = 7k - 7 \Rightarrow k = 7$

Therefore $k = 7$, the  equation with have infinite solutions.

$\\$

Section – C

Question number 13 to 22 carry 3 mark each.

Question 13: Prove that $\sqrt{5}$ is an irrational number.

Assume $\sqrt{5}$ is a rational number i.e. it can be expressed as a rational fraction of the form $\frac{a}{b}$ where $a, b$ are relatively prime numbers.

Since $\sqrt{5} =$ $\frac{a}{b}$

we have $5 =$ $\frac{a^2}{b^2}$ or $a^2 = 5 b^2$

This would imply that $a^2$ is a multiple of $5$. Since $5$ is prime, this implies $a$ is a multiple of $5$. Thus $a = 5c$ for some integer $c$, and

$5b^2 = a^2 = 25c^2$

Dividing by $5$, this means

$b^2 = 5c^2$

So $b^2$ is a multiple of $5$, and, just as it did for $a$, this means $b$ is a multiple of $5$. But $a$ and $b$ were presumed to lack a common factor other than $1$, so this is a contradiction, and the fraction $\frac{a}{b}$ for $\sqrt{5}$ must fail to exist.

Hence $\sqrt{5}$ is irrational.

$\\$

Question 14: Find all the zeroes of the polynomial $x^4 + x^3 - 14x^2 - 2x + 24$, if two of its zeroes are $2$ and $- 2$.

$f(x) = x^4 + x^3 - 14x^2 - 2x + 24$

Two zeros are $\sqrt{2}$ and $-\sqrt{2}$

Therefore $(x - \sqrt{2})$ and $(x+\sqrt{2})$ are factors of $f(x)$

$\Rightarrow (x^2 - 2)$ is a factor of $f(x)$

$x^2-2 ) \overline{x^4 + x^3 - 14x^2 - 2x + 24} ( x^2+x-12 \\ \hspace*{1cm} (-) \underline{x^4-2x^2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \hspace*{2cm}x^3-12x^2-2x+24 \\ \hspace*{1.5cm} (-) \underline{x^3-2x \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \hspace*{3.5cm}-12x^2+24 \\ \hspace*{3cm} (-) \underline{-12x^2+24 \ \ \ \ \ } \\ \hspace*{5cm}\times$

$\therefore f(x) = (x - \sqrt{2})(x + \sqrt{2})(x^2+x-12)$

$\Rightarrow f(x) = (x - \sqrt{2})(x + \sqrt{2})(x+4)(x-3)$

Therefore the other two zeros are $-4, 3$

$\\$

Question 15: Point P divides the line segment joining the points $A(2, 1)$ and $B(5, -8)$ such that $\frac{AP}{AB}$ $=$ $\frac{1}{2}$. If $P$ lies on the line $2x - y + k = 0$, find the value of $k$.

Or

For what value of $p$, are the points $(2, 1), (p, -1)$ and $(-1, 3)$ collinear ?

$AP : AB = 1:3$ (trisects)

$\frac{AP}{AB}$ $=$ $\frac{1}{3}$

$\Rightarrow \frac{AP}{AP+PB}$ $=$ $\frac{1}{3}$

$\Rightarrow 3AP = AP + PB$

$\Rightarrow 2AP = PB$

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

$\Rightarrow AP : PB = 1:2$

Applying section formula

Coordinates of $P = \Big($ $\frac{1 \times 5 + 2 \times 2}{1 + 2}$ $,$ $\frac{1 \times (-8) + 2 \times (1)}{1 + 2}$ $\Big)$

$\Rightarrow P = \Big($ $\frac{5+4}{3}$ $,$ $\frac{-8+2}{3}$ $\Big)$

$\Rightarrow P = (3, -2)$

Since $P(3, -2)$ lies on $2x - y + k = 0$

$\therefore 2(3) - (-2) + k = 0$

$\Rightarrow 6 + 2 + k = 0$

$\Rightarrow k = -8$

Or

Given point $(2, 1), (p, -1)$ and $(-1, 3)$

If the points are collinear are of the $\triangle ABC = 0$

$\therefore 0 =$ $\frac{1}{2}$ $[ x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) ]$

$\Rightarrow 2(-1-3) + p(3-1) -1(1+1) = 0$

$\Rightarrow -8+2p-2=0$

$\Rightarrow 2p - 10 = 0$

$\Rightarrow p = 5$

$\\$

Question 16: Prove that:

$\frac{\tan \theta }{1 - \tan \theta }$ $-$ $\frac{\cot \theta }{1 - \cot \theta }$ $=$ $\frac{\cos \theta - \sin \theta }{\cos \theta + \sin \theta }$

Or

If $\cos \theta + \sin \theta = \sqrt{2} \cos \theta$  show that $\cos \theta -\sin \theta = \sqrt{2} \sin \theta$

LHS $=$ $\frac{\tan \theta }{1 - \tan \theta }$ $-$ $\frac{\cot \theta }{1 - \cot \theta }$

$=$ $\frac{\sin \theta \cos \theta }{\cos \theta (\cos \theta - \sin \theta )}$ $-$ $\frac{\cos \theta \sin \theta }{\sin \theta (\sin \theta - \cos \theta )}$

$=$ $\frac{\sin \theta }{ (\cos \theta - \sin \theta )}$ $-$ $\frac{\cos \theta }{(\sin \theta - \cos \theta )}$

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

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

$= -$ $\frac{(\sin \theta - \cos \theta)(\sin \theta + \cos \theta)}{(\sin \theta - \cos \theta)(\sin \theta - \cos \theta)}$

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

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

$=$ RHS. Hence proved.

Or

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

Squaring both sides

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

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

Add $\sin^2 \theta$ on both sides

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

$\Rightarrow \sqrt{2} \sin\theta = \cos \theta - \sin \theta$

Hence proven.

$\\$

Question 17: A part of monthly hostel charges in a college hostel are fixed and the remaining depends on the number of days one has taken food in the mess. When a student $A$ takes food for $25$ days, he has to pay Rs. $4,500$, whereas a student $B$ who takes food for $30$ days, has to pay Rs.  $5,200$. Find the fixed charges per month and the cost of food per day.

Let the fixed charges $= x$

Let the charges for food per day $= y$

Therefore for Student A:

$x + 25 y = 4500$   … … … … … i)

Therefore for Student B:

$x + 30 y = 5200$   … … … … … ii)

Solving i) and ii)

$5y = 700 \Rightarrow y = 140$ Rs.

$\therefore x = 4500 - 25(140) = 1000$ Rs.

Therefore the fixed charges are $1000$ Rs. and food cost per day is $140$ Rs.

$\\$

Question 18: In $\triangle ABC, \angle B = 90^o$ and $D$ is the mid-point of $BC$. Prove that $AC^2 = AD^2 + 3CD^2$.

Or

In Figure 1, $E$ is a point on $CB$ produced of an isosceles $\triangle ABC$, with side $AB = AC$. If $AD \perp BC$ and $EF \perp AC$, prove that $\triangle ABD \sim \triangle ECF$.

Given: $BC = DC$

From $\triangle ABC$

$AB^2 + BC^2 = AC^2 \Rightarrow AB^2 = AC^2 - BC^2$   … … … … … i)

From $\triangle ABD$

$AB^2 + BD^2 = AD^2 \Rightarrow AB^2 = AD^2 - BD^2$   … … … … … ii)

From i) and ii)

$AC^2 - BC^2 = AD^2 - BD^2$

$\Rightarrow AC^2 = AD^2 + BC^2 - BD^2$

$\Rightarrow AC^2 = AD^2 + (BC-BD)(BC+BD)$

$\Rightarrow AC^2 = AD^2 + CD (2CD + CD)$

$\Rightarrow AC^2 = AD^2 + + 2 CD^2$

Hence proved.

Or

Given:  $AB = AC \Rightarrow \angle ABC = \angle ACB$

$AD \perp BC$ and $EF \perp AC$

In $\triangle ABD$ and $\triangle ECF$

$\angle ADB = \angle EFC = 90^o$ (given)

$\angle ABD = \angle ECF$

$\therefore \triangle ABD \sim \triangle ECF$ ( By AA criterion)

$\therefore \frac{AB}{EC}$ $=$ $\frac{AD}{EF}$ (corresponding sides)

$\\$

Question 19: Prove that the parallelogram circumscribing a circle is a rhombus.

Given: $ABCD$ be a parallelogram circumscribing a circle with center $O$.

To prove: $ABCD$ is a rhombus.

We know that the tangents drawn to a circle from an exterior point are equal in length.

Therefore, $AP = AS, BP = BQ, CR = CQ$ and $DR = DS$.

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

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

$AB + CD = AD + BC$

$2AB = 2BC$

(Since, $ABCD$ is a parallelogram so $AB = DC$ and $AD = BC$)

$AB = BC$

Therefore, $AB = BC = DC = AD$.

Hence, $ABCD$ is a rhombus.

$\\$

Question 20: In Figure 2, three sectors of a circle of radius $7$ cm, making angles of $60^o$$80^o$ and $40^o$ at the centre are shaded. Find the the area of the shaded region.

$=$ $\frac{80}{360}$ $\pi (7)^2 +$ $\frac{60}{360}$ $\pi (7)^2 +$ $\frac{40}{360}$ $\pi (7)^2$

$=$ $\frac{180}{360}$ $\pi (7)^2$

$=$ $\frac{1}{2}$ $\times$ $\frac{22}{7}$ $\times 7 \times 7$

$= 77 \ cm^2$

$\\$

Question 21: The following table gives the number of participants in a yoga camp :

 Age (in years): 20-30 30-40 40-50 50-60 60-70 No. of participants: 8 40 58 90 83

Find the modal age of the participants.

First we find the modal class of the given data which is the highest participant class i.e.

Modal class $= 50 -60$

Formula to find Mode is

$M = l +$ $\frac{f_1 - f_0}{2f_1 - f_0-f_2}$ $\times h$

$l = 50, \ f_1 = 90, \ f_0 = 58, \ f_2 = 83, \ h = 10$

$\therefore M = 50 +$ $\frac{90 - 58}{2(90) - 58 - 83}$ $\times 10$

$= 50 +$ $\frac{320}{39}$

$= 50 + 8.20 = 58.20$

$\\$

Question 22: A juice seller was serving his customers using glasses as shown in Figure 3. The inner diameter of the cylindrical glass was $5$ cm but bottom of the glass had a hemispherical raised portion which reduced the capacity of the glass. If the height of a glass was $10$ cm, find the apparent and actual capacity of the glass. (Use $\pi = 3.14$)

Or

A girl empties a cylindrical bucket full of sand, of base radius $18$ cm and height $32$ cm on the floor to form a conical heap of sand. If the height of this conical heap is $24$ cm, then find its slant height correct to one place of decimal.

Apparent volume $= \pi r^2 h = \pi (2.5)^2 \times 10 = 62.5 \pi = 196.25 \ cm^3$

Actual capacity $= 196.25 -$ $\frac{1}{2}$ $[$ $\frac{4}{3}$ $\pi R^3 ]$

$= 196.25 -$ $\frac{2}{3}$ $\times 3.14 \times (2.5)^3$

$= 196.25 - 32.71 = 163.542 \ cm^3$

Or

Volume of sand remains the same

$\therefore \pi (18)^2 (32) =$ $\frac{1}{3}$ $\pi r^2 (24)$

$\Rightarrow r^2 =$ $\frac{18 \times 18 \times 32 \times 3}{24}$ $= 696$

$\Rightarrow r = 36$ cm

Slant height $(l) = \sqrt{36^2 + 24^2} = \sqrt{1872} = 43.27$ cm

$\\$

Section – D

Question number 23 to 30 carry 4 mark each.

Question 23: A train travels $360$ km at a uniform speed. If the speed had been $5$ km/hr more, it would have taken $1$ hr less for the same journey. Find the speed of the train.

Or

Solve for $x :$ $\frac{1}{a+b+x}$ $=$ $\frac{1}{a}$ $+$ $\frac{1}{b}$ $+$ $\frac{1}{x}$ $; a \neq b, x \neq 0, x \neq -(a+b)$

Distance traveled $= 360$ km

Let the original speed $= x$ km/hr

$\therefore$ $\frac{360}{x}$ $-$ $\frac{360}{x+5}$ $= 1$

$\Rightarrow 360(x+5) - 360x = x(x+5)$

$\Rightarrow 360x + 1800 -360x = x^2 + 5x$

$\Rightarrow x^2 + 5x - 1800 = 0$

$\Rightarrow x^2 + 45x - 40 x - 1800 = 0$

$\Rightarrow x(x+45) - 40(x+45) = 0$

$\Rightarrow (x+45)(x-40) = 0$

$\Rightarrow x = 40$ km/hr or $-45$ km/hr (this is not possible as speed cannot be negative)

Hence the original speed $= 40$ km/hr

Or

$\frac{1}{a+b+x}$ $=$ $\frac{1}{a}$ $+$ $\frac{1}{b}$ $+$ $\frac{1}{x}$

$\Rightarrow \frac{1}{a+b+x}$ $=$ $\frac{bx + ax + ab}{abx}$

$\Rightarrow abx = (a+b+x)(ax + bx + ab)$

$\Rightarrow abx = a^2x + abx + a^2b + b^2x + abx + ab^2 + ax^2 + bx^2 + abx$

$\Rightarrow a^2x + b^2x + 2abx + a^2b + ab^2 + (a+b)x^2 = 0$

$\Rightarrow (a+b)x^2 + (a+b)^2x + ab(a+b)=0$

$\Rightarrow x^2 + (a+b)x + ab = 0$

$\Rightarrow x^2 + ax + bx + ab = 0$

$\Rightarrow x(x+a) + b(x+a) = 0$

$\Rightarrow (x+a)(x+b) = 0$

$\Rightarrow x = -a$ or $x = -b$

$\\$

Question 24:  If the sum of the first $p$ terms of an A.P. is $q$ and the sum of the first $q$ terms is $p$; then show that the sum of the first $(p + q)$ terms is $\{ - (p + q) \}$.

Let $a$ be the first term and $d$ be the common difference of the AP

Given $S_p = q$

$\Rightarrow$ $\frac{p}{2}$ $[ 2a + (p-1)d ] = q$

$\Rightarrow 2ap + p(p-1)d = 2q$   … … … … … i)

Also $S_q = p$

$\Rightarrow$ $\frac{q}{2}$ $[ 2a + (q-1)d ]= p$

$\Rightarrow 2aq + q(q-1)d = 2p$   … … … … … ii)

Subtracting ii) from i)

$2a(p-q) + [ p(p-1) - q(q-1) ] d = 2(q-p)$

$\Rightarrow 2a(p-q) + [ p^2 - p - q^2 + q ] d = -2(p-q)$

$\Rightarrow 2a(p-q) + [ (p-q) (p+q) - (p-q) ]d= -2(p-q)$

$\Rightarrow 2a(p-q) + (p-q)[ p+q - 1 ]d= -2(p-q)$

$\Rightarrow 2a + [ p+q-1] d= -2$

Now $S_{p+q} =$ $\frac{p+q}{2}$ $[2a + (p+q-1) d ]$

$\Rightarrow S_{p+q} =$ $\frac{p+q}{2}$ $(-2)$

$\Rightarrow S_{p+q} = -(p+q)$

$\\$

Question 25: In a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then prove that the angle opposite to the first side is a right angle.

Given: $AC^2 = AB^2 + BC^2$

To prove: $\angle B = 90^o$

Construction: $\triangle PQR$ is a right angled at $Q$ such that $PQ = AB$ and $QR = BC$

Proof: From $\triangle PQR$

$PR^2 = PQ^2 + QR^2$ (Pythagoras theorem)

$PR^2 = AB^2 + BC^2$ (by construction)   … … … … … i)

But $AC^2 = AB^2 + BC^2$ (given)   … … … … … ii)

$AC^2 = PR^2$ from i) and ii)

$\therefore AC = PR$   … … … … … iii)

Now in $\triangle ABC$ and $\triangle PQR$

$AB = PQ$ (by construction)

$BC = QR$ (by construction)

$AC = PR$ (from iii)

$\therefore \triangle ABC \cong \triangle PQR$ (by SSS criterion)

$\therefore \angle B = \angle Q$

But $\angle Q = 90^o$ by construction

$\therefore \angle B = 90^o$. Hence proved.

$\\$

Question 26: Construct an isosceles triangle whose base is $8$ cm and altitude $4$ cm and then another triangle whose sides are $\frac{3}{4}$ times the corresponding sides of the isosceles triangle.

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Question 27: A boy standing on a horizontal plane finds a bird flying at a distance of $100$ m from him at an elevation of $30^o$. A girl standing on the roof of a $20$ m high building, finds the elevation of the same bird to be $45^o$. The boy and the girl are on the opposite sides of the bird. Find the distance of the bird from the girl. (Given $\sqrt{2} = 1.414$)

Or

The angle of elevation of an aeroplane from a point $A$ on the ground is $60^o$. After a flight of $30$ seconds, the angle of elevation changes to $30^o$. If the plane is flying at a constant height of $3600 \sqrt{3}$ metres, find the speed of the aeroplane.

In $\triangle ABC$

$\frac{AB}{100}$ $= \sin 30^o =$ $\frac{1}{2}$ $\Rightarrow AB = 50$ m

$\therefore AE = AB - 20 = 30$ m

In $\triangle AED$

$\therefore$ $\frac{30}{AD}$ $= \sin 45^o =$ $\frac{1}{\sqrt{2}}$ $\Rightarrow AD = 30\sqrt{2} = 42.3$ m

Or

In $\triangle ABC$

$\frac{3600\sqrt{3}}{AC}$ $= \tan 60^o = \sqrt{3}$

$\therefore AC = 3600$ m

In $\triangle AED$

$\frac{3600\sqrt{3}}{AE}$ $= \tan 30^o =$ $\frac{1}{\sqrt{3}}$

$\therefore AE = 3600 \times 3 = 10800$ m

Hence $CE = 10800 - 3600 = 7200$ m

$\therefore Speed =$ $\frac{Distance}{Time}$ $=$ $\frac{7200}{30} \frac{m}{sec}$

$\Rightarrow Speed = 240$ $\frac{m}{sec}$

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Question 28: Find the values of frequencies $x$ and $y$ in the following frequency distribution table, if $N = 100$ and median is $32$.

 Marks: 0-10 10-20 20-30 30-40 40-50 50-60 Total No. of Students 10 $x$ 25 30 $y$ 10 100

Or

For the following frequency distribution, draw a cumulative frequency curve (ogive) of ‘more than type’ and hence obtain the median value.

 Class: 0-10 10-20 20-30 30-40 40-50 50-60 Total Frequency: 5 10 20 23 17 11 9

 Class Interval Frequency ( $f$ ) Cumulative Frequency ($cf$ ) 0 – 10 10 10 10 – 20 $x$ $10+x$ 20 – 20 25 $35+x$ 30 – 40 30 $65+x$ 40 – 40 $y$ $65+x+y$ 50 – 60 10 $75+x+y$

$75+x+y = 100 \Rightarrow x+y = 25$   … … … … … i)

For the above distribution, median class is $30-40$

$\frac{N}{2} = 50, f = 30, C = 10, N = 100, cf= 35+x, L = 30$

Median $(M) = L + \Big($ $\frac{\frac{N}{2} -cf}{f}$ $\Big) \times C$

$\Rightarrow 32 = 30 +$ $\frac{50-35-x}{30}$ $\times 10$

$\Rightarrow 2 =$ $\frac{15-x}{3}$

$\Rightarrow x = 9$

From i) $y=25-9=16$

Or

 Class Interval (More than) Frequency More than 0 100 More than 10 95 More than 20 80 More than 30 60 More than 40 37 More than 50 20 More than 60 9

 Class Interval (Less than) Frequency Less than 10 5 Less than 20 20 Less than 30 40 Less than 40 63 Less than 50 80 Less than 60 91 Less than 70 100

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Question 29: Prove that:

$\frac{(1 + \cot \theta + \tan \theta )( \sin \theta - \cos \theta )}{\sec^3 \theta - \ cosec^3 \ \theta }$ $=$ $\sin^2 \theta \cos^2\theta$

LHS $=$ $\frac{(1 + \cot \theta + \tan \theta )( \sin \theta - \cos \theta )}{\sec^3 \theta - \ cosec^3 \ \theta }$

$=$ $\frac{(1 + \frac{\cos \theta}{\sin \theta} + \frac{\sin \theta}{\cos \theta} )( \sin \theta - \cos \theta )}{\frac{1}{\cos^3 \theta} - \frac{1}{\sin^3 \theta} }$

$=$ $\frac{(\sin \theta \cos \theta + \cos^2 \theta + \sin^2 \theta)(\sin \theta - \cos \theta)(\sin^3 \theta \cos^3 \theta)}{\sin \theta \ \cos \theta (\sin^3 \theta - \cos^3 \theta)}$

$=$ $\frac{(\sin \theta \cos \theta + 1)(\sin \theta - \cos \theta)(\sin^2 \theta \cos^2 \theta)}{(\sin^3 \theta - \cos^3 \theta)}$

Since $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$

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

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

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

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

$=$ RHS. Hence proved.

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Question 30: An open metallic bucket is in the shape of a frustum of a cone. If the diameters of the two circular ends of the bucket are $45$ cm and $25$ cm and the vertical height of the bucket is $24$ cm, find the area of the metallic sheet used to make the bucket. Also find the volume of the water it can hold. (Use $\pi =$ $\frac{22}{7}$ )

$r_1 =$ $\frac{25}{2}$ $= 12.5$ cm

$r_2 =$ $\frac{45}{2}$ $= 22.5$ cm

$h = 24$ cm

Volume of bucket $=$ $\frac{\pi}{3}$ $h ({r_1}^2 + {r_2}^2 + r_1r_2 )$

$=$ $\frac{1}{3}$ $\times$ $\frac{22}{7}$ $\times 24 \times ( 12.5^2 + 22.5^2 + 12.5 \times 22.5)$

$=$ $\frac{176}{7}$ $\times$ $\frac{3775}{4}$

$= 23728.57 cm^3 = 23.723$ liters

$l = \sqrt{h^2 + (r_2 - r_1)^2} = \sqrt{24^2 + (22.5-12.5)^2} = \sqrt{24^2 + 10^2} = 26$ cm

Therefore surface area $= \pi {r_1}^2 + \pi (r_1 + r_2)l$

$=$ $\frac{22}{7}$ $\Big( 12.5^2 + (12.5+22.5)\times 26 \Big)$

$= 3351.07 \ cm^2$

$= 3.351 \ m^2$

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