2020 CBSE (Class 10) Board Paper Solution Mathematics : 30-5-1

NOTE:

(I) Please check that this question paper consists of 23 pages.

(II) Code number given on the right hand side of the question paper should be written on the title page of the answer book by the candidate.

(III) Please check that this question paper consists of 40 questions.

(IV) Please write down the serial number of the question before attempting it.

(V) 15 minutes times has been allotted to read this question paper. The question paper will be distributed at 10:15 am. From 10:15 am to 10:30 am, the students will read the question paper only and will not write any answer on the answer book during this period.

MATHEMATICS (STANDARD) – Theory

—————————————————————————————————————————————–Time allowed: 3 hours Maximum Marks: 80 —————————————————————————————————————————————–

General Instructions:

Read the following instructions very carefully and strictly follow them:

(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: On dividing a polynomial $p(x)$ by $x^2 - 4$ , quotient and remainder are found to be $x$ and $3$ respectively. The polynomial $p(x)$ is:

(A) $3x^2 + x - 12$ (B) $x^3 - 4x + 3$ (C) $x^2 + 3x - 4$ (D) $x^3 - 4x - 3$

Answer:

$\fbox B$

We know that $p(x) = q(x). g(x) + r(x)$

Where $q(x) = x$ , $g(x) = x^2 - 4 \text{ and } r(x) = 3$

Substituting all the above values we get

$p(x) = x ( x^2 - 4) + 3$

$\Rightarrow p(x) = x^2 - 4x + 3$

$\\$

Question 2: In Figure – 1, $ABC$ is an isosceles triangle, right angled at $C$ . Therefore:

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

Answer:

$\fbox A$

Given $BC = AC$

Using pythagoras theorem

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

$\Rightarrow AB^2 = 2 AC^2$ or $AB^2 = 2 BC^2$

$\\$

Question 3: The point on the $x-$ axis which is equidistant from $(-4, 0)$ and $(10, 0)$ is

(A) $(7, 0)$ (B) $(5, 0)$ (C) $(0, 0)$ (D) $(3, 0)$

(OR)

The centre of a circle whose end points of a diameter are $(-6, 3)$ and $(6, 4)$ is

(A) $(8, -1)$ (B) $(4, 7)$

Answer:

$\fbox D$

Given points $A(-4, 0)$ and $B(10, 0)$

Let the equidistant point be $C(x, 0)$ as it is on x-axis.

Therefore $AC = BC$

$\Rightarrow \sqrt{(x+4)^2 + (0-0)^2} = \sqrt{ (10-x)^2 + (0-0)^2}$

$\Rightarrow (x+4)^2 = (10-x)^2$

$\Rightarrow x^2 + 16 + 8x = 100 + x^2 - 20 x$

$\Rightarrow 28 x = 84$

$\Rightarrow x = 3$

Therefore the required point is $(3, 0)$

$\fbox C$

Given points $A(-6, 3) \text{ and } B(6, 4)$

Let the center is $O(x, y)$

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

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

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

$\\$

Question 4: The value(s) of $k$ for which the quadratic equation $2x^2 + kx + 2 = 0$ has equal roots, is

(A) $4$ (B) $\pm 4$ (C) $- 4$ (D) $0$

Answer:

$\fbox B$

Given $2x^2 + kx + 2 = 0$

$\therefore a = 2, b = k$ and $c = 2$

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

$\therefore k^2 - 4 ( 2) ( 2) = 0$

$\Rightarrow k^2 = 16$

$\Rightarrow k = \pm 4$

$\\$

Question 5: Which of the following is not A.P.?

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

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

Answer:

$\fbox C$

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

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

Therefore this is an A.P.

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

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

Therefore this is an A.P.

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

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

Therefore this is NOT an A.P.

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

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

Therefore this is an A.P.

$\\$

Question 6: The pair of linear equations $\displaystyle \frac{3x}{2} + \frac{5y}{3} = 7$ and $\displaystyle 9x+10y=14$ is

(A) Consistent (B) Inconsistent

Answer:

$\displaystyle \fbox B$

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

$\displaystyle \Rightarrow 9x + 10y = 42$

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

For $\displaystyle 9x+10y=14$

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

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

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

Hence the equations are inconsistent.

$\\$

Question 7: In Figure – 2, $PQ$ is tangent to the circle with centre at $O$ , at the point $B$ . If $\angle AOB =100^{\circ}$ , then $\angle ABP$ is equal to:

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

Answer:

$\fbox A$

$\triangle AOB$ is an isosceles triangle because $OA = OB$ (radius of the same circle)

$\therefore \angle OAB = \angle OBA$

Now $\angle OAB + \angle OBA + 100^{\circ} = 180^{\circ}$

$\Rightarrow \angle ABO = 40^{\circ}$

$\therefore \angle ABP = 90^{\circ} - 40^{\circ} = 50^{\circ}$ since $\angle OBP = 90^{\circ}$

$\\$

Question 8: The radius of a sphere (in cm) whose volume is $12 \pi \ cm^3$ , is

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

Answer:

$\fbox C$

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

Let the radius $= R$ cm

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

$\Rightarrow R^3 = 9$

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

$\\$

Question 9: The distance between the points $(m, -n)$ and $(-m, n)$ is

(A) $\sqrt{m^2+n^2}$ (B) $m+n$ (C) $2 \sqrt{m^2+n^2}$ (D) $\sqrt{2m^2+2n^2}$

Answer:

$\fbox C$

Given points $(m, -n)$ and $(-m, n)$

Distance between points $= \sqrt{ (-m-m)^2 + ( n - (-n))^2}$

$= \sqrt{(-2m)^2 + (2n)^2} = \sqrt{4(m^2 + n^2)} = 2 \sqrt{m^2 + n^2}$

$\\$

Question 10: In Figure – 3, from an external point $P$ , two tangents $PQ$ and $PR$ are drawn to a circle of radius $4$ cm with centre $O$ . If $\angle QPR = 90^{\circ}$ , then length of $PQ$ is:

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

Answer:

$\fbox B$

$PQ = PR$

Given $\angle QPR = 90^{\circ}$

Also $\angle OQP = \angle ORP = 90^{\circ}$

$\therefore \angle QOR = 360^{\circ} - ( 90^{\circ}+90^{\circ}+90^{\circ}) = 90^{\circ}$

Therefore $PQOR$ is a square

Hence all sides are equal.

Therefore $PQ = 4$ cm

$\\$

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

Question 11: The probability of an event that is sure to happen, is $\underline{ \hspace{2.0cm}}$

Answer:

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

The probability of an event is a number describing the chance that the event will happen. An event that is certain to happen has a probability of 1.

$\\$

Question 12: Simplest form of $\displaystyle \frac{1 + \tan^2 A}{1 + \cot^2 A}$ is $\underline{ \hspace{2.0cm}}$

Answer:

$\underline{ \hspace{0.5cm} \tan^2 A \hspace{0.5cm} }$

$\displaystyle \frac{1 + \tan^2 A}{1 + \cot^2 A} = \Bigg[ \frac{(\cos^2 A + \sin^2 A)}{\cos^2 A} \Bigg] \Bigg[ \frac{\sin^2A}{(\sin^2 A + \cos^2 A)} \Bigg] = \frac{\sin^2 A}{ \cos^2 A} = \tan^2 A$

$\\$

Question 13: $AOBC$ is a rectangle whose three vertices are $A(0, -3), O(0, 0)$ and $B(4, 0)$ . The length of its diagonals is $\underline{ \hspace{2.0cm}}$

Answer:

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

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

$\therefore AB = 5$ units

$\\$

$\displaystyle \text{Question 14: In the formula } \overline{x} = \frac{\Sigma f_i u_i}{\Sigma f_i} \times h , u_i = \underline{ \hspace{2.0cm}}$

Answer:

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

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

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

$\\$

Question 15: All concentric circles are $\underline{ \hspace{2.0cm}}$ to each other.

Answer:

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

All concentric circles are similar to each other.

$\\$

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

Question 16: Find the sum of the first $100$ natural numbers.

Answer:

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

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

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

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

$\\$

Question 17: In Figure – 4, the angle of elevation of the top of a tower from a point $C$ on the ground, which is $30$ m away from the foot of the tower, is $30^{\circ}$ . Find the height of the tower.

Answer:

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

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

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

$\\$

Question 18: The LCM of two numbers is $182$ and their HCF is $13$ . If one of the numbers is $26$ , find the other.

Answer:

We know

$LCM \times HCF =$ Product of two numbers

$182 \times 13 = 26 \times x$

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

Therefore the other number $= 91$

$\\$

Question 19: Form a quadratic polynomial, the sum and product of whose zeroes are $(-3)$ and $2$ respectively.

Can $(x^2 - 1)$ be a remainder while dividing $x^4 - 3x^2 + 5x - 6$ by $(x^2 + 3)$ ?

Answer:

Let $\alpha$ and $\beta$ are two roots of equation

Therefore $\alpha + \beta = - 3$

and $\alpha \beta = 2$

A quadratic equation can be written in the form

$p(x) = x^2 - (\ sum \ of \ roots\ ) x + product \ of \ roots$

$\Rightarrow p(x) =x^2 - ( \alpha + \beta) x + \alpha \beta$

$\Rightarrow p(x) = x^2 + 3x + 2$

$(x^2 -1)$ cannot be the remainder because the degree of the remainder should be less than the degree of the divisor

Long division

$\begin{array}{r l l} x^2+3 ) & \overline{ x^4 - 3x^2 + 5x - 6 } & ( x^2 - 6 \\ (-) & x^4 + 3x^2 & \\ \hline& \hspace{0.5cm} -6x^2 + 5x - 6 & \\ (-) & \hspace{0.5cm} -6x^2 - 18 & \\ \hline & \hspace{2.0cm} 5x + 12 & \end{array}$

The remainder is $5x+12$

$\\$

$\displaystyle \text{Question 20: Evaluate: } \frac{2 \tan 45^{\circ} \times \cos 60^{\circ}}{\sin 30^{\circ}}$

Answer:

$\displaystyle \frac{2 \tan 45^{\circ} \times \cos 60^{\circ}}{\sin 30^{\circ}} = \frac{2 \times 1 \times ( \frac{1}{2} )}{( \frac{1}{2} )} = 2$

$\\$

Section – B

Question Nos. 21 to 26 carry 2 marks each.

Question 21: In the Figure – 5, $DE \parallel AC \text{ and } DF \parallel AE$ Prove that: $\displaystyle \frac{BF}{BE} = \frac{FE}{EC}$

Answer:

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

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

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

Note: line drawn parallel to one side of the triangle, intersects the other two sides in distinct points, then it divides the other two sides in the same ratio.

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

$\text{Similarly, in } \triangle AEB, DF \parallel AE$

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

From i) and ii)

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

Hence proved.

$\\$

Question 22: Show that $5+ 2 \sqrt{7}$ is an irrational number, where $\sqrt{7}$ is given to be an irrational number

Check whether $12^n$ can end with the digit $0$ for any natural number $n$ .

Answer:

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

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

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

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

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

$\displaystyle \Rightarrow \sqrt{7}$ is a rational number. But it is given that $\sqrt{7}$ is an irrational number.

Hence our assumption that $5 + 2\sqrt{7}$ is a rational number is wrong.

Hence $5 + 2 \sqrt{7}$ is an irrational number.

For any number to end with a $0$ or $5$ , must be divisible by $5$ .

$12 = 2 \times 2 \times 3 = 2^2 \times 3$

$\Rightarrow 12^n = (2^2 \times 3)^n = 2^{2n} \times 3^n$

This factorization does not contain any term of $5$ .

Hence there is no value of $n \in N$ for which $12^n$ ends with digit $0$ or $5$

$\\$

Question 23: If $\displaystyle A, B$ and $\displaystyle C$ are interior angles of a $\displaystyle \triangle ABC$ , then show that $\displaystyle \cos \Big( \frac{B+C}{2} \Big) = \sin \Big( \frac{A}{2} \Big)$

Answer:

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

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

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

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

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

Hence proved.

$\\$

Question 24: In Figure – 6, a quadrilateral $ABCD$ is drawn to circumscribe a circle. Prove that $AB + CD = BC + AD$ .

In Figure – 7, find the perimeter of $\triangle ABC$ , if $AP = 12$ cm.

Answer:

Given $ABCD$ is a quadrilateral

The quadrilateral touches the circle at $P, Q, R$ and $S$

To prove: $AB + CD = AD + BC$

Proof:

We know that the lengths of the tangents drawn from an external point to a circle are equal. Therefore

$AP = AS$ … … … … … i) $BP = BQ$ … … … … … ii)

$CR = CQ$ … … … … … iii) $DR = DS$ … … … … … iv)

Adding i), ii), iii) and iv) we get

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

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

$\Rightarrow AB + CD = AD + BC$

Hence proved.

$AP = AB + BP$ … … … … … i)

$AQ = AC + CQ$ … … … … … ii)

We know that the lengths of the tangents drawn from an external point to a circle are equal. Therefore

$BP = BD$ and $CD = CQ$ . Also $AP = AQ$

Therefore we can write i) as

$AP = AB + BD$ … … … … … iii)

and similarly ii) as $AQ = AC + DC$ … … … … … iv)

Adding iii) and iv) we get

$AP + AQ = AB + ( BD + DC) + AC = AB + BC + AC =$ Perimeter of the $\triangle ABC$

Therefore the perimeter of $\triangle ABC = 12 + 12 = 24$ cm

$\\$

Question 25: Find the mode of the following distribution.

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

Answer:

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

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

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

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

$= 30 + 4.167 = 34.17$

$\\$

Question 26: $2$ cubes, each of volume $125 \ cm^3$ , are joined end to end. Find the surface area of the resulting cuboid.

Answer:

Volume of cube $= 125 \ cm^3$

Therefore the side of the cube $= 5 \ cm^3$

Dimension of the new cuboid

$l = 10$ cm $h = 5$ cm $b = 5$ cm

Therefore surface area of cuboid $= 2 ( lh + hb + bh)$

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

$\\$

Section – C

Question Nos. 27 to 34 carry 3 marks each.

Question 27: A fraction becomes $\frac{1}{3}$ when $1$ is subtracted from the numerator and it becomes $\frac{1}{4}$ when $8$ is added to its denominator. Find the fraction.

The present age of a father is three years more than three times the age of his son. Three years hence the father’s age will be $10$ years more than twice the age of the son. Determine their present ages.

Answer:

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

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

$\Rightarrow 3x - 3 = y$

$\Rightarrow 3x - y = 3$ … … … … … i)

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

$\Rightarrow 4x = y + 8$

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

Subtracting i) from ii) we get

$\begin{array}{r r} & 4x-y = 8 \\ (-) & 3x - y = 3 \\ \hline & x = 5 \end{array}$

Substituting in i)

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

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

Let the present age of the son $= x$ years

Therefore the present age of father $= 3x + 3$

After $3$ years

$(3x+3) + 3 = 2 ( x+3) + 10$

$\Rightarrow 3x + 6 = 2x + 16$

$\Rightarrow x = 10$ years

Therefore present age of father $= 3(10) + 3 = 33$ years and the present age of son is $10$ years.

$\\$

Question 28: Use Euclid Division Lemma to show that the square of any positive integer is either of the form $3q \ or \ 3q + 1$ for some integer $q$ .

Answer:

As per Euclid’s Division Lemma, if $a$ and $b$ are two positive integers, than

Let $a$ be the positive integer and $b = 3$

Therefore $a = 3m + r$ where $0 \leq r < b$ and $m$ is any positive integer.

Therefore $r$ can be either $0, 1$ or $2$ .

If $r = 0$ , then $a = 3m$

$\therefore a^2 = (3m)^2 = 3 \times 3m^2 = 3q$ where $q = 3m^2$

If $r = 1$ , then $a = 3m + 1$

$\therefore a^2 = ( 3m+1)^2 = 9m^2 + 6m + 1$

$\Rightarrow a^2 = 3(3m^2 + 2m) + 1 = 3q+1$ where $q = 3m^2 + 2m$

If $r = 2$ , then $a = 3m + 2$

$\therefore a^2 = ( 3m+2)^2 = 9m^2 + 12m + 4$

$\Rightarrow a^2 = 3(3m^2 + 4m+1) + 1 = 3q+1$ where $q = 3m^2 + 4m + 1$

Therefore the square of any positive number is of the form $3q$ or $3q+1$ for some integer $q$

$\\$

Question 29: Find the ratio in which the $y-$ axis divides the line segment joining the points $(6, -4)$ and $(-2, -7)$ . Also find the point of intersection.

Show that the points $(7, 10), (-2, 5)$ and $(3, -4)$ are vertices of an isosceles right triangle.

Answer:

Given y-axis divides the line segment.

Therefore the point $= ( 0, y)$

Applying section formula

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

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

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

Comparing

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

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

$\displaystyle \Rightarrow k = 3$

Therefore the ratio is $\displaystyle 3:1$

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

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

Given three vertices of a triangle

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

Similarly, $BC = \sqrt{ (3-(-2))^2 + ( -4-5)^2} = \sqrt{25+ 81} = \sqrt{106}$

And $CA = \sqrt{ (7-3)^2 + ( 10-(-4))^2} = \sqrt{16+ 196} = \sqrt{212}$

Since $AB = BC, \triangle ABC$ is an isosceles triangle. … … … … … i)

Applying Pythagoras theorem

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

$\Rightarrow (\sqrt{106})^2 + (\sqrt{106})^2 = (\sqrt{212})^2$

$\Rightarrow 106 + 106 = 212$

Therefore $\triangle ABC$ is a right angled triangle. … … … … … ii)

Hence from i) and ii), $\triangle ABC$ is an isosceles right angled triangle.

$\\$

$\displaystyle \text{Question 30: Prove that } \sqrt{ \frac{1+\sin A}{1 - \sin A} } = \sec A + \tan A$

Answer:

LHS $\displaystyle = \sqrt{ \frac{1+\sin A}{1 - \sin A} }$

$\displaystyle = \sqrt{ \frac{1+\sin A}{1 - \sin A} \times \frac{1+\sin A}{1 + \sin A} }$

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

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

$\displaystyle = \frac{1+\sin A}{\cos A}$

$\displaystyle = \sec A + \tan A = \text{RHS. Hence proved.}$

$\\$

Question 31: For an A.P., it is given that the first term $a = 5$ , common difference $d = 3$ and the $n^{th}$ term $a_n = 50$ . Find $n$ and sum of first $n$ terms $S_n$ of the A.P.

Answer:

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

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

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

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

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

$\Rightarrow 15 = n - 1$

$\Rightarrow n = 16$

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

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

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

$\\$

Question 32: Construct a $\triangle ABC$ with sides $BC = 6$ cm, $AB = 5$ cm and $\angle ABC = 60^{\circ}$ . Then construct a triangle whose sides are $\frac{3}{4}$ of the corresponding sides of $\triangle ABC$ .

Draw a circle of radius 3.5 cm. Take a point $P$ outside the circle at a distance of 7 cm from the center of the circle and construct a pair of tangents to the circle from that point.

Answer:

Step 1: Construct $\triangle ABC$

Draw base $BC$ of side $6$ cm
Draw $\angle B = 60^{\circ}$
Taking $B$ as a center, $5$ cm as radius, draw an arc. Let the point where arc intersect the ray be point $A$
Join $AC$
This completes the construction of $\triangle ABC$

$\displaystyle \text{Now we need to make a triangle which is } \frac{3}{4} \text{times its size.}$

$\displaystyle \text{Therefore the scale factor } = \frac{3}{4} < 1$

Step 2:

Draw any ray $BX$ making an acute angle with $BC$ on the side opposite to vertex $A$
Mark 4 points – $B_1, B_2, B_3$ and $B_4$ so that $BB_1 = B_1B_2 = B_2B_3 = B_3B_4$
Join $B_4 C$ and draw a line through $B_3$ parallel to $B_4C$ to intersect $BC$ at $C'$
Draw a line through $C'$ parallel to the line $AC$ to intersect $BA$ at $A'$
Therefore $\triangle A'BC'$ is the required triangle

Now consider $\triangle ABC$ and $\triangle A'BC'$

$\angle B$ is common

Since $A'C' \parallel AC$

$\angle A'C'B = \angle ACB$

$\therefore \triangle ABC \sim \triangle A'BC'$ ( AA similarity)

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

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

Therefore our construction is justified.

Draw a circle of radius $3.5$ cm

Mark any point $P$ at a distance of $7$ cm from the center

Join $OP$ and locate the midpoint $M$

Taking $M$ as a center draw a circle of radius $3.5$ cm

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

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

$\\$

Question 33: Read the following passage and answer the questions given at the end :

Diwali Fair

A game in a booth at a Diwali Fair involves using a spinner first. Then, if the spinner stops on an even number, the player is allowed to pick a marble from a bag. The spinner and the marbles in the bag are represented in Figure – 8.

Prizes are given, when a black marbles is picked. Shweta plays the same once.

(i) What is the probability that she will be allowed to pick a marble from the bag?

(ii) Suppose she is allowed to pick a marble from the bag, what is the probability of getting a prize, when it is given that the bag contains $20$ balls out of which 6 are black?

Answer:

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

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

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

ii) Number of black marbles $= 6$

Number of white marbles $= 14$

Therefore total number of marbles $= 20$

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

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

$\\$

Question 34: In Figure-9, a square $OPQR$ is inscribed in a quadrant $OAQB$ of a circle. If the radius of circle is $6 \sqrt{2}$ cm, find the area of the shaded region.

Answer:

Area of the quadrant of the circle $=$

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

Now $a^2 + a^2 = (6\sqrt{2})^2$

$\Rightarrow 2a^2 = 36 \times 2 \Rightarrow a = 6$ cm

Therefore area of square $OPQR = 36 \ cm^2$

Hence the shaded area $= 56.52 - 36 = 20.52 \ cm^2$

$\\$

Section – D

Question Nos. 35 to 40 carry 3 marks each.

Question 35: Obtain other zeroes of the polynomial $p(x) = 2x^4 - x^3 - 11x^2 + 5x + 5$ if two of its zeores are $\sqrt{5}$ and $- \sqrt{5}$ .

What minimum must be added to $2x^3 - 3x^2 + 6x + 7$ so that the resulting polynomial will be divisible by $x^2 - 4x + 8$ ?

Answer:

Given $p(x) = 2x^4 - x^3 - 11x^2 + 5x + 5$

Given that $\sqrt{5}$ and $-\sqrt{5}$ are zeros of $p(x)$

This means that $( x - \sqrt{5})$ and $(x+\sqrt{5})$ are factors of $p(x)$ .

Therefore $( x - \sqrt{5}) \times (x+\sqrt{5}) = (x^2 - 5)$ is a factor of $p(x)$

Long division

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

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

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

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

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

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

Long division

$\begin{array}{r l l} x^2 - 4x + 8) & \overline{2x^3 - 3x^2 + 6x + 7} & (2x+5 \\ (-) & 2x^3 - 8x^2 + 16x & \\ \hline & \hspace{1.0cm} 5x^2-10x+7 & \\ (-) & \hspace{1.0cm}5x^2 - 20x + 40 & \\ \hline & \hspace{2.0cm} 10x - 33 & \end{array}$

So add $( -10x+33)$ to the polynomial $2x^3 - 3x^2 + 6x + 7$ for it to be completely divisible by $x^2 - 4x + 8$

$\\$

Question 36: Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Answer:

Given $\triangle ABC \sim \triangle PQR$

To prove:

$\displaystyle \frac{ar(\triangle ABC)}{ar(\triangle PQR)} = \frac{AB^2}{PQ^2} = \frac{BC^2}{QR^2} = \frac{AC^2}{PR^2}$

Construction: Draw $\displaystyle AM \perp BC$ and $\displaystyle PN \perp QR$

$\displaystyle \text{Area of } \triangle ABC = \frac{1}{2} \times BC \times AM$ … … … … … i)

$\displaystyle \text{Area of } \triangle PQR = \frac{1}{2} \times QR \times PN$ … … … … … ii)

Dividing i) by ii)

$\displaystyle \frac{ar(\triangle ABC)}{ar(\triangle PQR)} = \frac{BC \times AM}{QR \times PN}$ … … … … … iii)

Consider $\displaystyle \triangle ABM$ and $\displaystyle \triangle PQN$

$\displaystyle \angle B = \angle Q$ as $\displaystyle \triangle ABC \sim \triangle PQR$

$\displaystyle \angle M = \angle N = 90^{\circ}$

$\displaystyle \therefore \triangle ABM \sim \triangle PQN$ (By AA similarity)

$\displaystyle \therefore \displaystyle \frac{AB}{PQ} = \frac{AM}{PN}$

From i)

$\displaystyle \frac{ar(\triangle ABC)}{ar(\triangle PQR)} = \frac{BC \times AM}{QR \times PN}$

$\displaystyle \Rightarrow \frac{ar(\triangle ABC)}{ar(\triangle PQR)} = \frac{BC }{QR } \times \frac{AB}{PQ}$ … … … … … iv)

Since $\displaystyle \triangle ABC \sim \triangle PQR$

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

Substituting in iv)

$\displaystyle \Rightarrow \frac{ar(\triangle ABC)}{ar(\triangle PQR)} = \frac{AB }{PQ } \times \frac{AB}{PQ}$ $\displaystyle = \frac{AB^2}{PQ^2}$

Similarly, we can prove

$\displaystyle \frac{ar(\triangle ABC)}{ar(\triangle PQR)} = \frac{AB^2}{PQ^2} = \frac{BC^2}{QR^2} = \frac{AC^2}{PR^2}$

Hence proved.

$\\$

Question 37: Sum of the areas of two squares is $544 \ m^2$ . If the difference of their perimeter is $32$ m, find the sides of the two squares.

A motor boat whose speed is $18$ km/h in still water takes $1$ hour more to go $24$ km upstream than to return downstream to the same spot. Find the speed of the stream.

Answer:

Let the sides of the two square be $x$ and $y$

$\therefore x^2 + y^2 = 544$ … … … … … i)

Perimeter of the two square will be $4x$ and $4y$

$\therefore 4x - 4y = 32$

$\Rightarrow x - y = 8$ … … … … … ii)

Substituting ii) in i) we get

$x^2 + ( x-8)^2 = 544$

$\Rightarrow x^2 + x^2 + 64 - 16x = 544$

$\Rightarrow 2x^2 - 16x - 480 = 0$

$\Rightarrow x^2 - 8x - 240 = 0$

$\Rightarrow x^2 - 20 x + 12 x - 240 = 0$

$\Rightarrow x( x - 20) + 12 ( x- 20) = 0$

$\Rightarrow (x-20)(x+ 12) =0$

$\Rightarrow x = 20$ or $x = - 12$ (this is not possible)

$\therefore x = 20$ m

Hence $y = 20-8 = 12 \ m$

Speed of the boat in still water $= 18$ km/hr

Let the speed of the stream $= s$

Therefore speed of the boat upstream $= ( 18-s)$ km/hr

Similarly, the speed of the boat down stream $= (18+s)$ km/hr

Time taken upstream $=$ Time taken down stream $+ 1$

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

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

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

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

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

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

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

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

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

$\\$

Question 38: A solid toy is in the form of a hemisphere surmounted by a right circular cone of same radius. The height of the cone is $10$ cm and the radius of the base is $7$ cm. Determine the volume of the toy. Also find the area of the colored sheet required to cover the toy. $\Big( \ Use \ \pi = \frac{22}{7} \ and \ \sqrt{149} = 12.2 \Big)$

Answer:

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

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

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

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

$\displaystyle = 1232 \ cm^3$

Area of colored sheet required to cover the toy

$=$ CSA of cone $+$ CSA of hemisphere

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

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

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

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

$= 576.4 \ cm^2$

$\\$

Question 39: A statue $1.6$ m tall, stands on the top of a pedestal.From a point on the ground, the angle of elevation of the top of the statue is $60^{\circ}$ and from the same point the angle of elevation of the top of the pedestal is $45^{\circ}$ . Find the height of the pedestal. $($ Use $\sqrt{3} = 1.73)$

Answer:

Please refer to the adjoining diagram

In $\triangle ABC$

$\displaystyle \frac{1.6+h}{PB} = \tan 60^{\circ}$

$\displaystyle \Rightarrow \frac{1.6+h}{PB} = \sqrt{3}$

$\displaystyle \Rightarrow PB = \frac{1.6+h}{\sqrt{3}}$ … … … … … i)

Similarly, in $\triangle ABP$

$\displaystyle \frac{h}{PB} = \tan 45^{\circ} = 1$

$\Rightarrow h = PB$ … … … … … ii)

Therefore from i) and ii) we get

$\displaystyle h = \frac{1.6+h}{\sqrt{3}}$

$\Rightarrow (\sqrt{3} -1) h = 1.6$

$\displaystyle \Rightarrow h = \frac{1.6}{\sqrt{3}-1} = \frac{1.6 (\sqrt{3}+1)}{3 - 1} = 0.8 \times (1.73+1) = 2.184 \ m$

Therefore the height of the pedestal $= 2.184 \ m$

$\\$

Question 40: For the following data, draw a ‘less than’ ogive and hence find the median of the distribution.

Age in years	0-10	10-20	20-30	30-40	40-50	50-60	60-70
Number of persons	5	15	20	25	15	11	9

The distribution given below shows the number of wickets taken by bowlers in one-day cricket matches. Find the mean and the median of the numbers of wickets taken.

Number of wickets	20-60	60-100	100-140	140-180	180-220	220-260
Number of bowlers	7	5	16	12	2	3

Answer:

Less than frequency distribution

Age	No of persons	Class	Cumulative Frequency
0-10	5	Less than 10	5
10-20	15	Less than 20	20
20-30	20	Less than 30	40
30-40	25	Less than 40	65
40-50	15	Less than 50	80
50-60	11	Less than 60	91
60-70	9	Less than 70	100

Now plotting less than ogive $(10, 5), (20, 20), (30, 40), (40, 65), (50, 80), (60, 91), (70, 100)$

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

Therefore Median $= 34$ (from graph)

Number of wickets	Number of bowlers $(f)$	$x_i$	$\displaystyle u_i = \frac{x_i-a}{h}$	$u_if_i$
20-60	7	40	-3	-21
60-100	5	80	-2	-10
100-140	16	120	-1	16
140-180	12	160	0	0
180-220	2	200	1	2
220-260	3	240	2	6
	$\Sigma f = 45$			$\Sigma u_if_i = - 39$