2019 CBSE (Class 10) Board Paper Solution Mathematics : 30-5-1

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 $\displaystyle ab$ .

Answer:

$\displaystyle a$ 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 $\displaystyle kx2 + 2x - 3 = 0$ .

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

Answer:

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

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

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

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

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

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 $\displaystyle n$ .

Answer:

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 $\displaystyle y$ .

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

Answer:

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

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

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

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

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

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

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

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

$\displaystyle = 0$

$\displaystyle \\$

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.

Answer:

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 $\displaystyle 2x - 3y = 5$ .

Answer:

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.

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

Answer:

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, $\displaystyle 7, 11, 15, \ldots$

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

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 $\displaystyle b$ .

Answer:

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 ?

Answer:

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.

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.

Answer:

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 $\displaystyle 1314 = 2 \times 3 \times 3 = 18$

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.

Answer:

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 ?

Answer:

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.

Answer:

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 $\displaystyle - 2$ .

Answer:

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

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 $\displaystyle k$ .

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

Answer:

$\displaystyle AP : AB = 1:3$ (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 $\displaystyle 2x - y + k = 0$

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

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

$\displaystyle \Rightarrow k = -8$

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:

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

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

Answer:

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.

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.

Answer:

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 $AC^2 = AD^2 + 3CD^2$ .

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 $\triangle ABD \sim \triangle ECF$ .

Answer:

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.

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.

Answer:

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 $DR = DS$ .

Adding the above equations,

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

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

$AB + CD = AD + BC$

$2AB = 2BC$

(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.

Answer:

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.

Answer:

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 $\pi = 3.14$ )

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.

Answer:

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

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

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

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

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.

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

Answer:

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

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

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

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

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

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

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

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

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

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

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

$\displaystyle \Rightarrow x = -a$ 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 $\displaystyle \{ - (p + q) \}$ .

Answer:

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.

Answer:

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.

Answer:

$\displaystyle \\$

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 $\sqrt{2} = 1.414$ )

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.

Answer:

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 $\triangle AED$

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

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	$x$	25	30	$y$	10	100

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

Answer:

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 $30-40$

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

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

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

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

$\Rightarrow x = 9$

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

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:

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

Answer:

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

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

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

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

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