2020 CBSE (Class 10) Board Paper Solution Mathematics : 30-2-3

NOTE:

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

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

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

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

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

MATHEMATICS (STANDARD)

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

General Instructions:

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

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

Answer:

$\displaystyle \fbox{a}$

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

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

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

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

$\displaystyle \\$

Question 2: The coordinate of the point which is the reflection of the point $\displaystyle (-3, 5)$ in x -axis is

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

Answer:

$\displaystyle \fbox{a}$

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

$\displaystyle \\$

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

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

Answer:

$\displaystyle \fbox{d}$

Using section formula, we get

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

$\displaystyle \Rightarrow 3y = 3$

$\displaystyle \Rightarrow y = 1$

$\displaystyle \\$

Question 4: The sum of exponents of prime factors in the prime factorisation of $\displaystyle 196$ is

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

Answer:

$\displaystyle \fbox{b}$

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

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

$\displaystyle \\$

Question 5: Euclid’s division Lemma states that for two positive integers $\displaystyle a$ and $\displaystyle b$ there exists unique integer $\displaystyle q$ and $\displaystyle r$ satisfying $\displaystyle a = bq + r$ and

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

Answer:

$\displaystyle \fbox{c}$

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

$\displaystyle \\$

Question 6: The zeros of the polynomial $\displaystyle x^2 - 3x - m (m+3)$ are

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

Answer:

$\displaystyle \fbox{b}$

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

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

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

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

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

$\displaystyle \\$

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

$\displaystyle \text{(a) } - \frac{14}{3} \hspace{0.5cm} \text{(b)} \frac{2}{5} \hspace{0.5cm} \text{(c)} 5 \hspace{0.5cm} \text{(d)} 10$

Answer:

$\displaystyle \fbox{d}$

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

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

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

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

$\displaystyle \Rightarrow k \neq 10$

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

$\displaystyle \\$

Question 8: The roots of the quadratic equation $\displaystyle x^2 - 0.04 = 0$ are

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

Answer:

$\displaystyle \fbox{a}$

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

$\displaystyle \therefore x = \pm 0.2$

$\displaystyle \\$

$\displaystyle \text{Question 9: The common difference of the AP } \frac{1}{p} , \frac{1-p}{p} , \frac{1-2p}{p} , \ldots$ is

$\displaystyle \text{(a) } 1 \hspace{0.5cm} \text{(b) } \frac{1}{p} \hspace{0.5cm} \text{(c) } -1 \hspace{0.5cm} \text{(d) } - \frac{1}{p}$

Answer:

$\displaystyle \fbox{c}$

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

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

$\displaystyle \\$

Question 10: The $\displaystyle n^{th}$ term of the A.P. $\displaystyle a, 3a, 5a, \ldots$ , is

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

Answer:

$\displaystyle \fbox{b}$

$\displaystyle \text{First term } (a) = a$

$\displaystyle \text{Common difference } (d) = 3a - a = 2a$

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

$\displaystyle \\$

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

Question 11: In Fig. 3, the angles of depressions from the observing positions $O_1$ and $O_2$ respectively of the object $A$ are $\underline{\hspace{2.0cm}}$

Answer:

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

Please refer to the adjoining diagram

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

$\\$

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

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

Answer:

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

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

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

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

$\Rightarrow 144 = 108 + 36$

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

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

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

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

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

$\\$

Question 13: In given Fig. 2, the length $PB = \underline{\hspace{2.0cm}}$ cm.

Answer:

$\underline{PB = 4 \ cm}$

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

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

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

Since $OP \perp AB, AP = PB$

Hence $PB = 4$ cm

$\\$

Question 14: In Fig. 1, $MN \parallel BC$ and $AM : MB = 1:2$ , then

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

Answer:

$\underline{1 : 9}$

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

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

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

Since $MN \parallel CB$

Using “Basic Proportionality Theorem” we get

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

$\\$

Question 15: The value of $\sin 32^{\circ} \cos 58^{\circ} + \cos 32^{\circ} \sin 58^{\circ} \ is \ \underline{\hspace{2.0cm}}$

$\displaystyle \text{The value of } \frac{\tan 35^{\circ}}{\cot 55^{\circ}} + \frac{\cot 78^{\circ}}{\tan 12^{\circ}} \ is \ \underline{\hspace{2.0cm}}$

Answer:

$\sin 32^{\circ} \cos 58^{\circ} + \cos 32^{\circ} \sin 58^{\circ}$

$= \sin 32^{\circ} \cos (90^{\circ} - 32^{\circ}) + \cos 32^{\circ} \sin (90^{\circ} - 32^{\circ})$

$= \sin^2 32^{\circ} + \cos^2 32^{\circ} = 1$

$\displaystyle \frac{\tan 35^{\circ}}{\cot 55^{\circ}} + \frac{\cot 78^{\circ}}{\tan 12^{\circ}}$

$\displaystyle = \frac{\tan 35^{\circ}}{\cot (90 - 35^{\circ})} + \frac{\cot 78^{\circ}}{\tan (90 - 78^{\circ})}$

$\displaystyle = \frac{\tan 35^{\circ}}{\tan 35^{\circ}} + \frac{\cot 78^{\circ}}{\cot 78^{\circ}}$

$= 2$

$\\$

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

Question 16: A die is thrown once. What is the probability of getting a prime number.

Answer:

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

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

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

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

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

$\\$

Question 17: If a number $x$ is chosen at random from the numbers $-3, -2, -1, 0, 1, 2, 3$ then find the probability of $x^2 < 4$ .

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

Answer:

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

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

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

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

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

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

There are $7$ days in a week.

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

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

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

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

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

$\\$

Question 18: If $\sin A + \sin^2 A = 1$ , then find the value of the expression $(\cos^2 A + \cos^4 A)$ .

Answer:

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

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

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

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

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

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

$\\$

Question 19: Find the area of the sector of a circle of radius $6$ cm whose central angle is $30^{\circ}$ $($ Take $\pi = 3.14)$

Answer:

Given: radius $= OA = OB = 6 \ cm$

Central angle $( \theta) = 30^{\circ}$

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

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

$\displaystyle \Rightarrow l = \frac{30}{180} \times 3.14 \times 6 = 3.14 \ cm$

Therefore perimeter of the sector $= 6 + 3.14 + 6 = 15.14 \ cm$

$\\$

Question 20: Find the class – marks of the classes $20-50$ and $35-60$ .

Answer:

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

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

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

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

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

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

$\\$

Section – B

Question Nos. 21 to 26 carry 2 marks each.

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

$\displaystyle 2x+3 , \hspace{0.5cm} 3x^2 + 7x + 2 , \hspace{0.5cm} 4x^3 + 3x^2 + 2 , \hspace{0.5cm} x^3 + \sqrt{3x} + 7 , \hspace{0.5cm} 7x + \sqrt{7} , \\ \\ 5x^3 - 7x + 2 , \hspace{0.5cm} 2x^2 + 3 - \frac{5}{x^2} , \hspace{0.5cm} 5x - \frac{1}{2} , \hspace{0.5cm} ax^3 + bx^2 +cx + d , \hspace{0.5cm} x + \frac{1}{x}$

Answer the following questions:

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

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

Answer:

Definitions:

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

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

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

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

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

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

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

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

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

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

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

Therefore:

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

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

$\\$

Question 22: A child has a die whose six faces show the letters shown below:

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

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

Answer:

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

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

i) The probability of getting $A$

$n(f) = 2$

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

ii) The probability of getting $D$

$n(f) = 1$

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

$\\$

Question 23: In Fig. 4, $ABC$ and $DBC$ are two triangles on the same base $BC$ , If $AD$ intersects $BC$ at $O$ , show that

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

In Fig. 5, if $AD \perp BC$ , then prove that $AB^2 + CD^2 = BD^2 + AC^2$

Answer:

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

To prove:

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

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

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

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

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

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

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

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

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

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

Therefore from iii) and iv) we get

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

Given: $AD \perp BC$

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

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

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

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

Subtracting ii) from i) we get

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

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

Hence proved.

$\\$

$\displaystyle \text{Question 24: Prove that } 1 + \frac{\cot^2 \alpha}{1 + \mathrm{cosec} \alpha} = \mathrm{cosec} \alpha$

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

Answer:

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

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

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

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

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

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

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

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

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

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

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

$\\$

Question 25: Find the mode for the following frequency distribution:

Class	15-20	20-25	25-30	30-35	35-40	40-45
Frequency	3	8	9	10	3	2

Answer:

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

$\therefore 12-16$ is the modal class

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

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

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

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

$\\$

Question 26: From a solid right circular cylinder of height 14 cm and base radius 6 cm, a right circular code of the same height and same base radius is removed. Find the volume of the remaining solid.

Answer:

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

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

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

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

$\Rightarrow r^3 = 8$

$\Rightarrow r = 2$ cm

Therefore the height of the cylinder is $2$ cm

$\\$

Section – C

Question Nos. 27 to 34 carry 3 marks each.

Question 27: If a circle touches the side $BC$ of a triangle $ABC$ at $P$ and the extended sides $AB$ and $AC$ at $Q$ and $R$ respectively, prove that

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

Answer:

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

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

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

Also $AQ = AR$

$\Rightarrow AQ = AC + CR$

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

Also $AQ = AB + BQ$

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

Adding i) and ii) we get

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

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

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

$\\$

Question 28: The area of a circular playground is $22176 \ cm^2$ . Find the cost of fencing this ground at the rate of Rs. $50$ per meter.

Answer:

Area $= 22176 \ cm^2$

Let the radius $= r \ cm$

$\therefore \pi r^2 = 22176$

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

$\Rightarrow r = 84 \ cm$

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

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

$\\$

Question 29: If the mid point of the line segment joining the points $A(3, 4)$ and $B ( k, 6)$ , is $P(x, y)$ and $x+y-10= 0$ , find the value of $k$ .

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

Answer:

If $P$ is the mid point then

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

Also $x + y - 10 = 0$

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

$\Rightarrow k+3 = 10$

$\Rightarrow k = 7$

Since $M$ and $N$ are mid points

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

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

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

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

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

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

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

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

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

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

$\displaystyle = 12$ sq. units

$\\$

Question 30: In Fig. 6, if $\triangle ABC \sim \triangle DEF$ and their sides of the lengths (in cm) are marked along them, then find the length of the sides of each of the triangles.

Answer:

Since $\triangle ABC \sim \triangle DEF$

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

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

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

$\Rightarrow 2x - 1 = 9$

$\Rightarrow x = 5$

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

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

$\\$

Question 31: If $2x+y = 3$ and $4x-y = 19$ find the value of $(5y-2x)$

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

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

Answer:

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

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

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

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

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

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

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

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

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

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

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

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

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

$x = 1$ or $x = 2$

$\\$

$\displaystyle \text{Question 32: Which term of the A.P. } 20, 19 \frac{1}{4} , 18 \frac{1}{2} , 17 \frac{3}{4} , \ldots \\ \\ \text{is the first negative term.}$

Find the middle term of the A.P. $7, 13, 19, \ldots , 247$ .

Answer:

$\displaystyle \text{Given AP: }20, 19 \frac{1}{4} , 18 \frac{1}{2} , 17 \frac{3}{4} , \ldots$

First term $( a) = 20$

$\displaystyle \text{Common difference }( d) = 19 \frac{1}{4} - 20 = - \frac{3}{4}$

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

$\displaystyle \Rightarrow a_n = 20 + ( n-1) (- \frac{3}{4} )$

$\displaystyle \Rightarrow a_n = 20 + \frac{3}{4} - \frac{3n}{4} = \frac{83 - 3n}{4}$

Let’s see which term be closest to $0$

$\therefore 83 - 3n = 0$

$\Rightarrow n = 27.66$

Therefore the $28^{th}$ terms would be the first negative term

Given AP: $7, 13, 19, \ldots , 247$

First term $( a) = 7$

Common difference $(d) = 13 - 7 = 6$

Let $247$ be the $n^{th}$ term

Therefore $247 = 7 + ( n - 1 ) (6)$

$\Rightarrow 240 = 6 ( n - 1)$

$\Rightarrow n -1 = 40$

$\Rightarrow n = 41$

Therefore the middle term is $21^{st}$ term

$\therefore a_{21} = 7 + ( 21 - 1 ) ( 6)$

$\Rightarrow a_{21} = 7 + 120$

$\Rightarrow a_{21} = 127$

$\\$

Question 33: Water in a canal $6$ m wide and $1.5$ m deep, is flowing with the speed of $10$ km/hr. How much area will it irrigate in $30$ minutes, if $8$ cm standing water is required.

Answer:

Canal is in the shape of cuboid: breadth $= 6$ m and height $= 1.5$ m

$\displaystyle \text{Speed of water }= 10 km/hr = \frac{1000}{6} \ m/min$

$\displaystyle \text{Volume of water flow }= 6 \times 1.5 \times \frac{1000}{6} = 1500 \frac{m^3}{min}$

Volume of water flow in $30$ minutes $= 30 \times 1500 = 45000 \ m^3$

$\displaystyle \text{Therefore the Area irrigated }= \frac{45000}{0.08} = 562500 \ m^2$

$\\$

Question 34: Show that:

$\displaystyle \frac{ \cos^2 ( 45^{\circ} + \theta) + \cos^2 ( 45^{\circ} - \theta)}{\tan (60^{\circ}+\theta) \tan (30^{\circ} - \theta)} = 1$

Answer:

$\displaystyle \text{LHS }= \frac{ \cos^2 ( 45^{\circ} + \theta) + \cos^2 ( 45^{\circ} - \theta)}{\tan (60^{\circ}+\theta) \tan (30^{\circ} - \theta)}$

$\displaystyle = \frac{ \cos^2 ( 45^{\circ} + \theta) + \sin^2 ( 90^{\circ} - 45^{\circ} + \theta)}{\tan (60^{\circ}+\theta) \cot (90 - 60^{\circ} + \theta)}$

$\displaystyle = \frac{ \cos^2 ( 45^{\circ} + \theta) + \sin^2 ( 45^{\circ} + \theta)}{\tan (60^{\circ}+\theta) \cot (60^{\circ} + \theta)}$

$\displaystyle = \frac{1}{1} = 1 = \ RHS.$

Hence proved.

$\\$

Section – D

Question Nos. 35 to 40 carry 3 marks each.

Question 35: The mean of the frequency distribution is $18$ . The frequency f in the class interval $19-21$ is missing. Determine $f$ .

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

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

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

Answer:

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

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

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

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

$\Rightarrow 16 = 2f$

$\Rightarrow f = 8$

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

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

$\\$

Question 36: From a point on the ground, the angles of elevation of the bottom and top of the tower fixed at the top of a $20$ m high building are $45^{\circ}$ and $60^{\circ}$ respectively. Find the height of the tower.

Answer:

Let the height of the tower be $h$ m

In $\triangle CBP$

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

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

In $\triangle ABP$

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

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

$\Rightarrow BP = 20$ … … … … … ii)

Substituting in i) we get

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

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

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

$\\$

Question 37: It takes 12 hours to fill a swimming pool using two pipes. If the pipe of the larger diameter is used for four hours and the pipe of the smaller diameter for 9 hours, only half of the pool can be filled. How long would it take for each of the pipes to fill the pool separately.

Answer:

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

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

Therefore the capacity of the pool $= 12x + 12y \text{ liters }$

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

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

$\displaystyle \Rightarrow 3y = 2x$

For larger pipe to fill the swimming pool

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

For smaller pipe to fill the swimming pool

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

$\\$

Question 38: Prove that $\sqrt{5}$ is an irrational number.

Answer:

Let’s prove this by the method of contradiction.

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

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

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

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

$\Rightarrow 5q^2=p^2$ … … … … … i)

$\Rightarrow p^2$ is a multiple of $5 \text{ (Euclid's Division Lemma)}$

$\Rightarrow p$ is also a multiple of $5 \text{ . (Fundamental Theorem of arithmetic)}$

$\Rightarrow p = 5m$

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

From equations i) and ii), we get,

$5q^2=25m^2$

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

$\Rightarrow q^2$ is a multiple of $5 \text{ (Euclid's Division Lemma)}$

$\Rightarrow q$ is a multiple of $5 \text{ . (Fundamental Theorem of arithmetic)}$

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

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

$\\$

Question 39: Draw a circle of radius $3.5$ cm. From a point $P$ , $6$ cm from its center, draw two tangents to the circle.

Construct a $\triangle ABC$ with $AB = 6$ cm, $BC = 5$ cm and $\angle B = 60^{\circ}$ . Now construct another triangle whose sides are $\frac{2}{3}$ times the corresponding sides of $\triangle ABC$ .

Answer:

Draw a circle of radius $3.5$ cm

Mark any point $P$ at a distance of $6.0$ cm from the center

Join $OP$ and locate the midpoint $M$

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

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

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

Step 1: First construct the $\triangle ABC$ . Draw $AB = 6 cm. Taking$ latex A $ as a center, draw an angle of $60^{\circ}$ and draw a line through the point to make a line. Then taking an arc of $5$ cm draw an arc to intersect the line and mark point $C$ . Join $B$ to $C$ . This is your $\triangle ABC$ .

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

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

Step 4: Join $A_3B$ and draw a line through $A_2$ parallel to $A_3B$ to intersect $AB$ extended at $B'$

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

$\triangle AB'C'$ is $\frac{2}{3}$ of $\triangle ABC$

$\\$

Question 40: A solid is in the shape of a hemisphere surmounted by a cone. If the radius of hemisphere and the base radius of cone is $7$ cm and height of the cone is $3.5$ cm, find the volume of the solid. $\displaystyle ( Take \pi = \frac{22}{7} )$

Answer: 2020-08-01_14-00-29

$\displaystyle \text{We know: Volume of a sphere } = \frac{4}{3}$ $\pi r^3$ where $r$ is the radius of the sphere

$\displaystyle \text{Volume of a cone } = \frac{1}{3} \pi r^2 h$

where $r$ radius of the base of the cone and $h$ is the height of the cone.

$\displaystyle \text{Volume of hemisphere } = \frac{1}{2} \Big( \frac{4}{3} \pi \times 7^3 \Big) = \frac{2}{3} \times \frac{22}{7} \times 7^3 = \frac{2156}{3} \ cm^2$

$\displaystyle \text{Volume of cone } = \frac{1}{3} \pi \times 7^2 \times 3.5 = \frac{1}{3} \times \frac{22}{7} \times 7^2 \times 3.5 = \frac{539}{3} \ cm^3$

$\displaystyle \text{Therefore total volume } = \frac{2156}{3} + \frac{539}{3} = \frac{2695}{3} = 898.33 \ cm^3$

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2020 CBSE (Class 10) Board Paper Solution Mathematics : 30-2-3

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