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 $\displaystyle a$ and $\displaystyle b$ is $\displaystyle 5$ and their LCM is $\displaystyle 200$. Find the product $\displaystyle ab$.

$\displaystyle a$ and $\displaystyle b$ are two numbers.

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

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

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

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

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

$\displaystyle \\$

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

Or

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

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

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

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

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

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

Or

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

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

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

$\displaystyle \Rightarrow k^2 = 36$

$\displaystyle \Rightarrow k = \pm 6$

$\displaystyle \\$

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

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

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

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

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

$\displaystyle \Rightarrow 3n = 18$

$\displaystyle \Rightarrow n = 6$

$\displaystyle \\$

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

Or

Find the value of $\displaystyle ( \cos 48^{\circ} - \sin 42^{\circ})$.

$\displaystyle \sin x + \cos y = 1$

$\displaystyle \Rightarrow \sin 30^{\circ} + \cos y = 1$

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

$\displaystyle \Rightarrow \cos y = \cos 60^{\circ}$

$\displaystyle \Rightarrow y = 60^{\circ}$

Or

$\displaystyle \cos 48^{\circ} - \sin 42^{\circ}$

$\displaystyle = \cos 48^{\circ} - \sin (90^{\circ}-48^{\circ})$

$\displaystyle = \cos 48^{\circ} - \cos 48^{\circ}$

$\displaystyle = 0$

$\displaystyle \\$

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

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

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

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

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

$\displaystyle \\$

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

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

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

$\displaystyle \Rightarrow 6-3a = 5$

$\displaystyle \Rightarrow 3a = 1$

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

$\displaystyle \\$

Section – B

Question number 7 to 12 carry 2 mark each.

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

Or

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

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

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

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

This means First term + Second term $\displaystyle = 18$

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

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

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

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

Or

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

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

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

$\displaystyle T_{10} = a + 9 d$

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

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

$\displaystyle 7d = 7$

$\displaystyle d = 1$

Common difference $\displaystyle = 1$

$\displaystyle \\$

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

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

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

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

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

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

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

$\displaystyle \Rightarrow b = 2$

Hence $\displaystyle a = 2, b = 2$

$\displaystyle \\$

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 $\displaystyle = 6$

No of A’s on dice $\displaystyle = 3$

No of B’s of dice $\displaystyle = 2$

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

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

$\displaystyle \\$

Question 10: Find the HCF of $\displaystyle 612$ and $\displaystyle 1314$ using prime factorization.

Or

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

First factorize each number

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

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

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

Or

Let $\displaystyle a$ be any positive integer.

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

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

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

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

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

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

$\displaystyle \\$

Question 11: Cards marked with numbers $\displaystyle 5$ to $\displaystyle 50$ (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 $\displaystyle 10$, (ii) a number which is a perfect square.

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

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

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

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

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

$\displaystyle \\$

Question 12: For what value of $\displaystyle k$, does the system of linear equations

$\displaystyle 2x + 3y = 7$ , $\displaystyle (k - 1) x + (k + 2) y = 3k$ have an infinite number of solutions ?

If the system of equations are $\displaystyle ax_1 + b_1y + c_1 = 0$ and $\displaystyle ax_2 + b_2y + c_2 = 0$ and they have infinitely many solutions then it satisfy the following:

$\displaystyle \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$

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

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

From first two terms

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

From second and third term

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

From first and third term

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

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

$\displaystyle \\$

Section – C

Question number 13 to 22 carry 3 mark each.

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

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

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

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

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

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

Dividing by $\displaystyle 5$, this means

$\displaystyle b^2 = 5c^2$

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

Hence $\displaystyle \sqrt{5}$ is irrational.

$\displaystyle \\$

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

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

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

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

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

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

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

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

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

$\displaystyle \\$

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

Or

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

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

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

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

$\displaystyle \Rightarrow 3AP = AP + PB$

$\displaystyle \Rightarrow 2AP = PB$

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

$\displaystyle \Rightarrow AP : PB = 1:2$

Applying section formula

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

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

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

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

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

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

$\displaystyle \Rightarrow k = -8$

Or

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

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

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

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

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

$\displaystyle \Rightarrow 2p - 10 = 0$

$\displaystyle \Rightarrow p = 5$

$\displaystyle \\$

Question 16: Prove that:

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

Or

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

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

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

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

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

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

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

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

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

$\displaystyle =$ RHS. Hence proved.

Or

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

Squaring both sides

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

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

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

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

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

Hence proven.

$\displaystyle \\$

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 $\displaystyle A$ takes food for $\displaystyle 25$ days, he has to pay Rs. $\displaystyle 4,500$, whereas a student $\displaystyle B$ who takes food for $\displaystyle 30$ days, has to pay Rs. $\displaystyle 5,200$. Find the fixed charges per month and the cost of food per day.

Let the fixed charges $\displaystyle = x$

Let the charges for food per day $\displaystyle = y$

Therefore for Student A:

$\displaystyle x + 25 y = 4500$ … … … … … i)

Therefore for Student B:

$\displaystyle x + 30 y = 5200$ … … … … … ii)

Solving i) and ii)

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

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

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

$\displaystyle \\$

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)

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

$\\$

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.

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

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

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

$= 77 \ cm^2$

$\\$

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

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

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

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

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

$= 50 + 8.20 = 58.20$

$\\$

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

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

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

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

Or

Volume of sand remains the same

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

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

$\Rightarrow r = 36$ 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 $\displaystyle 360$ km at a uniform speed. If the speed had been $\displaystyle 5$ km/hr more, it would have taken $\displaystyle 1$ hr less for the same journey. Find the speed of the train.

Or

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

Distance traveled $\displaystyle = 360$ km

Let the original speed $\displaystyle = x$ km/hr

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

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

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

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

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

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

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

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

Hence the original speed $\displaystyle = 40$ km/hr

Or

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

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

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

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

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

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

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

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

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

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

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

$\displaystyle \\$

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

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

Given $\displaystyle S_p = q$

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

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

Also $\displaystyle S_q = p$

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

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

Subtracting ii) from i)

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

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

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

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

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

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

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

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

$\displaystyle \\$

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: $\displaystyle AC^2 = AB^2 + BC^2$

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

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

Proof: From $\displaystyle \triangle PQR$

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

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

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

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

$\displaystyle \therefore AC = PR$ … … … … … iii)

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

$\displaystyle AB = PQ$ (by construction)

$\displaystyle BC = QR$ (by construction)

$\displaystyle AC = PR$ (from iii)

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

$\displaystyle \therefore \angle B = \angle Q$

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

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

$\displaystyle \\$

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

$\displaystyle \\$

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 airplane 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 $3600 \sqrt{3}$ meters, find the speed of the airplane.

In $\triangle ABC$

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

$\therefore AE = AB - 20 = 30 \text{ m }$

In $\triangle AED$

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

Or

In $\triangle ABC$

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

$\therefore AC = 3600$ m

In $\triangle AED$

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

$\therefore AE = 3600 \times 3 = 10800 \text{ m }$

Hence $CE = 10800 - 3600 = 7200 \text{ m }$

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

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

$\\$

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

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

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

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

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

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

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

$\Rightarrow x = 9$

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

$\\$

Question 29: Prove that:

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

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

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

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

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

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

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

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

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

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

$=$ RHS. Hence proved.

$\\$

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 $\pi = \frac{22}{7}$ )

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

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

$h = 24 \text{ cm }$

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

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

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

$= 23728.57 cm^3 = 23.723$ liters

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

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

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

$= 3351.07 \ cm^2$

$= 3.351 \ m^2$