2019 CBSE (Class 10) Board Paper Solution Mathematics : 30-4-2

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: For what values of $\displaystyle k$ does the quadratic equation $\displaystyle 4x^2 - 12x - k = 0$ have no real roots ?

Answer:

Given equation: $\displaystyle 4x^2 - 12x - k = 0$

Comparing it with $\displaystyle ax^2 + bx + c = 0, a = 4, b = -12 \text{ and } c = -k$

When $\displaystyle b^2 - 4ac < 0$ , there are no real roots.

$\displaystyle \therefore (-12)^2 - 4 (4)(-k) <0$

$\displaystyle \Rightarrow 144 + 16k < 0$

$\displaystyle \Rightarrow k < -9$

Therefore for $\displaystyle k < -9$ , there will be no real roots.

$\displaystyle \\$

Question 2: Find the distance between the points $\displaystyle (a, b) \text{ and } (- a, - b)$ .

Answer:

Given points $\displaystyle (a, b) \text{ and } (-a, -b)$

Therefore the distance between the two points $\displaystyle = \sqrt{(-a-a)^2 + (-b-b)^2} = \sqrt{4a^2 + 4b^2} = 2 \sqrt{a^2 + b^2}$

$\displaystyle \\$

Question 3: Find a rational number between $\displaystyle \sqrt{2} \text{ and } \sqrt{7}$ .

Write the number of zeroes in the end of a number whose prime factorization is $\displaystyle 2^2 \times 5^3 \times 3^2 \times 17$ .

Answer:

$\displaystyle \sqrt{2} = 1.412 \text{ and } \sqrt{7} = 2.645$

Therefore $\displaystyle 2.5$ is between $\displaystyle \sqrt{2} \text{ and } \sqrt{7}$

$\displaystyle \therefore 2.5 = \frac{25}{10} = \frac{5}{2} \text{ is a rational number between } \sqrt{2} \text{ and } \sqrt{7}$

The given expression:

$\displaystyle 2^2 \times 5^3 \times 3^3 \times 17 = (2 \times 5) \times (2 \times 5) \times 5 \times 3^3 \times 17$

$\displaystyle = 10 \times 10 \times 5 \times 3^3 \times 17$

We know that zeros in an expression are a result of number of $\displaystyle 10's$ in it.

Hence the expression we can see that there will be $\displaystyle 2$ zeros in the given expression.

$\displaystyle \\$

Question 4: Let $\displaystyle \triangle ABC \sim \triangle DEF$ and their areas be respectively, $\displaystyle 64 \ cm^2 \text{ and } 121 \ cm^2$ . If $\displaystyle EF = 15.4$ cm, find $\displaystyle BC$ .

Answer:

Since $\displaystyle \triangle ABC \sim \triangle DEF$

$\displaystyle \frac{ar(\triangle ABC)}{ar(\triangle DEF)} = \Big( \frac{BC}{EF} \Big)^2$

$\displaystyle \Rightarrow \frac{64}{121} = \Big( \frac{BC}{EF} \Big)^2$

$\displaystyle \Rightarrow \frac{BC}{EF} = \frac{8}{11}$

$\displaystyle \Rightarrow BC = \frac{8}{11} \times 15.4 = 11.2 \text{ cm }$

$\displaystyle \\$

$\displaystyle \text{Question 5: Evaluate: } \frac{\tan 65^{\circ} }{\cot 25^{\circ} }$

Express $\displaystyle ( \sin 67^{\circ} + \cos 75^{\circ} )$ in terms of trigonometric ratios of the angle between $\displaystyle 0^{\circ} \text{ and } 45^{\circ}$ .

Answer:

$\displaystyle \frac{\tan 65^{\circ} }{\cot 25^{\circ} } = \frac{\tan 65^{\circ} }{\cot (90^{\circ} -65^{\circ} )} = \frac{\tan 65^{\circ} }{\tan 65^{\circ} } = 1$

$\displaystyle \sin 67^{\circ} + \cos 75^{\circ} = \sin (90^{\circ} -23^{\circ} ) + \cos (90^{\circ} -15^{\circ} ) = \cos 23^{\circ} + \sin 15^{\circ}$

$\displaystyle \\$

$\displaystyle \text{Question 6: Find the number of terms in the A.P. : } 18, 15 \frac{1}{2} , 13, \cdots , - 47$

Answer:

$\displaystyle \text{Given AP: } 18, 15\frac{1}{2}, 13, \ldots , -47$

$\displaystyle a = 18, d = 15.5 - 18 = -2.5$

Let $\displaystyle -47$ be the $\displaystyle n^{th}$ term

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

$\displaystyle \Rightarrow -47 = 18 + (n-1) (-2.5)$

$\displaystyle \Rightarrow -65 = -2.5 (n-1)$

$\displaystyle \Rightarrow 26 = n - 1$

$\displaystyle \Rightarrow n = 27$

Therefore $\displaystyle -47$ is the $\displaystyle 27^{th}$ term in the given AP

$\displaystyle \\$

Section – B

Question number 7 to 12 carry 2 mark each.

Question 7: A bag contains $\displaystyle 15$ balls, out of which some are white and the others are black. If the probability of drawing a black ball at random from the bag is $\displaystyle \frac{2}{3}$ , then find how many white balls are there in the bag.

Answer:

Total number of balls $\displaystyle = 15$

Let the number of black balls $\displaystyle = x$

$\displaystyle \therefore \frac{x}{15} = \frac{2}{3} \Rightarrow x = 10$

Hence the number of white balls $\displaystyle = 15 - 10 = 5$

$\displaystyle \\$

Question 8: A card is drawn at random from a pack of $\displaystyle 52$ playing cards. Find the probability of drawing a card which is neither a spade nor a king.

Answer:

Total number of cards $\displaystyle = 52$

Number of space cards $\displaystyle = 13$ (includes the king of space)

No of kings other than king of spade $\displaystyle = 3$

$\displaystyle \text{Therefore Probability } = \frac{52 - (13+3)}{52} = \frac{36}{52} = \frac{9}{13}$

$\displaystyle \\$

Question 9: Find the solution of the pair of equations :

$\displaystyle \frac{3}{x} + \frac{8}{y} = - 1; \frac{1}{x} - \frac{2}{y} = 2 , x, y \neq 0$

Find the value(s) of $\displaystyle k$ for which the pair of equations $\displaystyle kx+2y = 3, 3x+6y = 10$ has a unique solution.

Answer:

Given equation:

$\displaystyle \frac{3}{x} + \frac{8}{y} = - 1$

$\displaystyle \frac{1}{x} - \frac{2}{y} = 2$

Let $\displaystyle \frac{1}{x} = u \text{ and } \frac{1}{y} = v$

Therefore

$\displaystyle 3u + 8 v = -1$ … … … … … iii)

$\displaystyle u - 2v = 2$ … … … … … iv)

Multiplying iv) by iii) and subtracting it from iii)

$\displaystyle \hspace*{1cm} 3u + 8 v = -1 \\ (-1) \underline {\hspace*{0.2cm}3u - 6v = 6} \\ \hspace*{1.5cm} 14v = -7$

$\displaystyle v= \frac{-1}{2}$

$\displaystyle \text{Therefore } u = 2v + 2 = 2 ( \frac{-1}{2} ) + 2 = -1 +2 = 1$

$\displaystyle \text{Hence }\frac{1}{x} = 1 \Rightarrow x = 1$

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

Given equations:

$\displaystyle kx + 2y = 3$

$\displaystyle 3x + 6y = 10$

If the system of equations are $\displaystyle ax_1 + b_1y + c_1 = 0 \text{ and } ax_2 + b_2y + c_2 = 0$ and they have unique solution then it satisfy the following:

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

$\displaystyle \Rightarrow \frac{k}{3} \neq \frac{2}{6} \neq \frac{-3}{-10}$

From first two terms

$\displaystyle k \neq 1$

From First and Third term

$\displaystyle k \neq \frac{9}{10}$

$\displaystyle \therefore k \neq 1 \text{ or } k \neq \frac{9}{10} \text{for the equations to have a unique solution.}$

$\displaystyle \\$

Question 10: How many multiples of $\displaystyle 4$ lie between $\displaystyle 10 \text{ and } 205$ ?

Determine the A.P. whose third term is $\displaystyle 16 \text{ and } 7^{th}$ term exceeds the $\displaystyle 5^{th}$ term by $\displaystyle 12$ .

Answer:

Since the multiples of $\displaystyle 4$ lie between $\displaystyle 10 \text{ and } 205$ are $\displaystyle 12, 16, 20, \ldots , 204$

$\displaystyle \therefore a = 12, d = 16-12 = 4$

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

$\displaystyle \Rightarrow 204 = 12 + (n-1) (4)$

$\displaystyle \Rightarrow 192 = (n-1)(4)$

$\displaystyle \Rightarrow 48 = n - 1$

$\displaystyle \Rightarrow n = 49$

Therefore there are $\displaystyle 49$ multiples of $\displaystyle 4$ between $\displaystyle 10 \text{ and } 205$

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

Given $\displaystyle T_3 = 16$

$\displaystyle \therefore a + (3 - 1) d = 16$

$\displaystyle \Rightarrow a + 2d = 16$ … … … … … i)

$\displaystyle T_7 = a + 6d$

$\displaystyle T_5 = a + 4d$

Given $\displaystyle T_7 = T_5 + 12$

$\displaystyle \Rightarrow a + 6d = a + 4d + 12$

$\displaystyle \Rightarrow 2d = 12$

$\displaystyle \Rightarrow d = 6$

Therefore form i) we get $\displaystyle a = 16 - 2( 6) = 4$

Therefore the AP is $\displaystyle 4, 10, 16, 22, \ldots$

$\displaystyle \\$

Question 11: Use Euclid’s division algorithm to find the HCF of $\displaystyle 255 \text{ and } 867$ .

Answer:

$\displaystyle 255 ) \overline{867} ( 3 \\ \hspace*{0.7cm} \underline{765} \\ \hspace*{0.8cm} 102 ) \overline{255} ( 2 \\ \hspace*{1.5cm} \underline{204} \\ \hspace*{1.7cm} 51 ) \overline{102} ( 2 \\ \hspace*{2.25cm} \underline{102} \\ \hspace*{2.5cm} \times$

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

$\displaystyle \therefore 867 = 51 \times 17 + 0$

So HCF of $\displaystyle 867 \text{ and } 255$ is $\displaystyle 51$

$\displaystyle \\$

Question 12: The point $\displaystyle R$ divides the line segment $\displaystyle AB$ , where $\displaystyle A( 4, 0) \text{ and } B(0, 6)$ such that $\displaystyle AR = \frac{3}{4} AB$ . Find the coordinates of $\displaystyle R$ .

Answer:

Given points $\displaystyle A(-4, 0) \text{ and } B(0, 6)$

$\displaystyle AR = \frac{3}{4} AB$

$\displaystyle \Rightarrow \frac{AR}{AB} = \frac{3}{4}$

$\displaystyle \Rightarrow \frac{AR}{AR + RB} = \frac{3}{4}$

$\displaystyle \Rightarrow 4AR = 3AR + 3RB$

$\displaystyle \Rightarrow \frac{AR}{RB} = \frac{3}{1}$

Using sections formula, $\displaystyle R(x, y) = \Big( \frac{3 \times 0 + 1 \times (-4)}{3+1} , \frac{3 \times 6 + 1 \times 0}{3+1} \Big) = \Big( -1, \frac{9}{2} \Big)$

$\displaystyle \\$

Section – C

Question number 13 to 22 carry 3 mark each.

Question 13: Prove that : $\displaystyle ( \sin \theta + 1 + \cos \theta) ( \sin \theta - 1 + \cos \theta) . \sec \theta \ \mathrm{cosec} \theta = 2$

$\displaystyle \sqrt{\frac{\sec \theta -1}{\sec \theta +1}} + \sqrt{\frac{\sec \theta +1}{\sec \theta -1}} = 2 \ \mathrm{cosec} \theta$

Answer:

LHS $\displaystyle = ( \sin \theta + 1 + \cos \theta) ( \sin \theta - 1 + \cos \theta) . \sec \theta \ \mathrm{cosec} \theta$

$\displaystyle = [ (\sin \theta + \cos \theta )^2 -1 ] \sec \theta \ \mathrm{cosec} \theta$

$\displaystyle = [ \sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta -1] \sec \theta \ \mathrm{cosec} \theta$

$\displaystyle = [ 1 + 2 \sin \theta \cos \theta -1 ] \sec \theta \ \mathrm{cosec} \theta$

$\displaystyle = 2 \sin \theta \cos \theta \sec \theta \ \mathrm{cosec} \theta$

$\displaystyle = 2\ \frac{\sin \theta \cos \theta }{\sin \theta \cos \theta }$

$\displaystyle = 2$ . Hence proved.

$\displaystyle \text{LHS } = \sqrt{\frac{\sec \theta -1}{\sec \theta +1}} + \sqrt{\frac{\sec \theta +1}{\sec \theta -1}}$

$\displaystyle = \frac{(\sqrt{\sec \theta - 1})^2 + (\sqrt{\sec \theta + 1})^2}{\sqrt{(\sec \theta + 1)(\sec \theta - 1)}}$

$\displaystyle = \frac{\sec \theta - 1 + \sec \theta + 1}{\sqrt{\sec^2 \theta -1}}$

$\displaystyle = \frac{2}{\cos \theta } \sqrt{\frac{1}{\frac{1}{\cos^2 \theta }-1}}$

$\displaystyle = \frac{2}{\cos \theta } \sqrt{\frac{\cos^2 \theta}{1- \cos^2 \theta}}$

$\displaystyle = 2 \ \frac{\cos \theta}{\cos \theta } \sqrt{\frac{1}{\sin^2 \theta}}$

$\displaystyle = \frac{2}{\sin \theta}$

$\displaystyle = 2 \ \mathrm{cosec} \theta$

Hence proved.

$\displaystyle \\$

Question 14: In what ratio does the point $\displaystyle P(- 4, y)$ divide the line segment joining the points $\displaystyle A(- 6, 10) \text{ and } B(3, - 8)$ ? Hence find the value of $\displaystyle y$ .

Find the value of $\displaystyle p$ for which the points $\displaystyle (- 5, 1), (1, p) \text{ and } (4, - 2)$ are collinear.

Answer:

Let $\displaystyle P$ divides $\displaystyle AB$ in the ratio of $\displaystyle k:1$

$\displaystyle \therefore P(-4, y) = \Big( \frac{3k-6}{k+1} , \frac{-8k + 10}{k + 1} \Big)$

$\displaystyle \Rightarrow -4 = \frac{3k-6}{k+1}$

$\displaystyle \Rightarrow -4k - 4 = 3k - 6$

$\displaystyle \Rightarrow 7k = 2$

$\displaystyle \Rightarrow k = \frac{2}{7}$

Therefore ratio is $\displaystyle 2:7$

$\displaystyle \therefore y = \frac{-8k + 10}{k + 1}$

$\displaystyle = \frac{-8(\frac{2}{7}) + 10}{(\frac{2}{7}) + 1}$

$\displaystyle = \frac{-16+70}{2 + 7}$

$\displaystyle = \frac{54}{9} = 6$

Given point $\displaystyle A(-5, 1), B(1, p) \text{ and } C(4, -2)$

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

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

$\displaystyle \Rightarrow -5(p+2) + 1(-2-1) + 4(1-p) = 0$

$\displaystyle \Rightarrow -5p - 10 - 3 + 4 - 4p = 0$

$\displaystyle \Rightarrow -9p - 9 = 0$

$\displaystyle \Rightarrow p = -1$

$\displaystyle \\$

Question 15: $ABC$ is a right triangle in which $\angle B = 90^o$ . If $AB = 8$ cm and $BC = 6$ cm, find the diameter of the circle inscribed in the triangle.

Answer:

$AC = \sqrt{AB^2 + BC^2} = \sqrt{8^2 + 6^2} = 10$ cm

Area of $\triangle ABC =$ Area of $\triangle AOB +$ Area of $\triangle BOC +$ Area of $\triangle AOC$

$\displaystyle \frac{1}{2} \times BC \times AB = \frac{1}{2} \times AB \times r +$ $\frac{1}{2} \times BC \times r = \frac{1}{2} \times AC \times r$

$\Rightarrow 6 \times 8 = 8r + 6r + 10r$

$\Rightarrow 48 = 24r$

$\Rightarrow r = 2 cm$

$\\$

Question 16: In Figure 1, $BL$ and $CM$ are medians of a $\triangle ABC$ right-angled at $A$ . Prove that $4 (BL^2 + CM^2) = 5 BC^2$ .

Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.

Answer:

Given $BL$ and $CM$ are medians

$\displaystyle \Rightarrow BM = MA \Rightarrow AM = \frac{1}{2} AB$

$\displaystyle \Rightarrow CL = AL \Rightarrow AL = \frac{1}{2} AC$

From

$\triangle ABC, BC^2 = AC^2 + AB^2$ … … … … … i)

$\triangle BAL, BL^2 = AL^2 + AB^2$ … … … … … ii)

$\triangle ACM, CM^2 = AC^2 + AM^2$ … … … … … iii)

Adding ii) and iii)

$BL^2 + CM^2 = AL^2 + AB^2 + AC^2 + AM^2$

$\displaystyle BL^2 + CM^2 = \Big( \frac{1}{2} AC \Big)^2 + AB^2 + AC^2 + \Big( \frac{1}{2} AB \Big)^2$

$\displaystyle BL^2 + CM^2 = \frac{1}{4} AC^2 + AB^2 + AC^2 + \frac{1}{4} AB^2$

$\displaystyle BL^2 + CM^2 = \frac{1}{4} (AC^2 + AB^2) + AB^2 + AC^2$

$4 (BL^2 + CM^2) = 5 BC^2$

Hence Proved.

Given: $ABCD$ is a rhombus

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

We know that the diagonals of a rhombus bisect at right angles

Therefore from $\triangle AOB, AB^2 = AO^2 + OB^2$

$\displaystyle \Rightarrow AB^2 = \Big( \frac{AC}{2} \Big)^2 + \Big( \frac{BD}{2} \Big)^2$

$\Rightarrow 4 AB^2 = AC^2 + BD^2$

Since $AB = BC = CD = DA$

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

Hence proved.

$\\$

Question 17: In Figure 2, two concentric circles with centre $O$ , have radii $21$ cm and $42$ cm. If $\angle AOB = 60^o$ , find the area of the shaded region.

Answer:

$\displaystyle \text{Area of shaded area } = \frac{300}{360} \pi \times (42)^2 - \frac{300}{360} \pi \times (21)^2$

$\displaystyle = \frac{5}{6} \pi (42^2 - 21^2)$

$\displaystyle = \frac{5}{6} \times \frac{22}{7} \times 1323$

$= 3465 \ cm^2$

$\\$

Question 18: Calculate the mode of the following distribution :

Class:	10-15	15-20	20-25	25-30	30-35
Frequency:	4	7	20	8	1

Answer:

$\\$

Question 19: A cone of height $\displaystyle 24$ cm and radius of base $\displaystyle 6$ cm is made up of modelling clay. A child reshapes it in the form of a sphere. Find the radius of the sphere and hence find the surface area of this sphere.

A farmer connects a pipe of internal diameter $\displaystyle 20$ cm from a canal into a cylindrical tank in his field which is $\displaystyle 10$ m in diameter and $\displaystyle 2$ m deep. If water flows through the pipe at the rate of $\displaystyle 3$ km/hr, in how much time will the tank be filled ?

Answer:

$\displaystyle \text{Volume of cone } = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (6)^2 (24) = 288 \pi \ cm^3$

Let the radius of the sphere be $\displaystyle R$

$\displaystyle \therefore \frac{4}{3} \pi R^3 = 288 \pi$

$\displaystyle \Rightarrow R^3 = 3 \times 72 = 216$

$\displaystyle \Rightarrow R = 6$ cm

$\displaystyle \text{Surface are of the sphere } = 4 \pi R^2 = 4 \times \frac{22}{7} \times 36 = 452.57 \ cm^2$

$\displaystyle \text{Rate of flow of water } = \frac{3 \times 1000}{60} \frac{m}{min} = 50 \frac{m}{min}$

$\displaystyle \text{Volume of tank } = \pi r^2 h = \pi (5)^2 \times 2 = 50 \pi m^3$

$\displaystyle \text{Volume of water flowing per minute } = \pi (0.1)^2 \times 50 \frac{m^3}{min} = 0.5 \pi \frac{m^3}{min}$

$\displaystyle \text{Therefore the time taken to fill the tank } = \frac{50 \pi m^3}{0.5 \pi \frac{m^3}{min} } = 100 \text{ minutes}$

$\displaystyle \\$

Question 20: Prove that $\displaystyle 2 + 3 \sqrt{3}$ is an irrational number when it is given that $\displaystyle \sqrt{3}$ is an irrational number.

Answer:

Let us assume that $\displaystyle 2 +3 \sqrt{3}$ is a rational number.

The rational number is in the form of $\displaystyle \frac{p}{q}$

$\displaystyle \therefore 2 + 3\sqrt{3} = \frac{p}{q}$

$\displaystyle 3\sqrt{3} = \frac{p}{q} - 2$

$\displaystyle 3\sqrt{3} = \frac{p-2q}{q}$

$\displaystyle \sqrt{3} = \frac{p-2q}{3q}$

$\displaystyle \therefore \sqrt{3} \text{ is a rational no as } \frac{p-2q}{3q} \text{is a rational number}$

But it is given that $\displaystyle \sqrt{3}$ is an irrational number which contradicts our initial assumption.

Hence $\displaystyle 2 + 3 \sqrt{3}$ is an irrational number.

$\displaystyle \\$

Question 21: Sum of the areas of two squares is $\displaystyle 157 \ m^2$ . If the sum of their perimeters is $\displaystyle 68$ m, find the sides of the two squares.

Answer:

Let the sides of the two squares be $\displaystyle a$ and $\displaystyle b$ respectively

$\displaystyle \therefore a^2 + b^2 = 157$ … … … … … i)

Also $\displaystyle 4a + 4 b = 68$

$\displaystyle \Rightarrow a + b = 17$ … … … … … ii)

From i) $\displaystyle (a+b)^2 - 2ab = 157$

$\displaystyle 17^2 - 2ab = 157$

$\displaystyle 289 - 157 = 2ab$

$\displaystyle ab = 66$ … … … … … iii)

From ii) $\displaystyle b = 17 - a$

$\displaystyle \therefore a (17-a) = 66$

$\displaystyle 17a - a^2 -= 66$

$\displaystyle a^2 - 17a + 66 = 0$

$\displaystyle (a-11) (a-6) =0$

$\displaystyle a =11$ or $\displaystyle a = 6$

When $\displaystyle a = 11, b = 17 - 11 = 6$

When $\displaystyle a = 6, b = 17 - 6 = 11$

Hence the sides of the two squares are $\displaystyle 6$ units and $\displaystyle 11$ units

$\displaystyle \\$

Question 22: Find the quadratic polynomial, sum and product of whose zeroes are $\displaystyle -1$ and $\displaystyle - 20$ respectively. Also find the zeroes of the polynomial so obtained.

Answer:

Sum of zeros: $\displaystyle \alpha + \beta = 1$

Product of zeros: $\displaystyle \alpha \beta = -12$

The quadratic polynomial is of the form:

$\displaystyle x^2 -$ (sum of zeros) $\displaystyle x +$ (product of zeros) $\displaystyle = 0$

$\displaystyle x^2 - x - 12 = 0$

$\displaystyle \Rightarrow \alpha + \beta = 1 \Rightarrow \beta = (1-\alpha)$

$\displaystyle \alpha \beta = -12$

$\displaystyle \Rightarrow \alpha (1 - \alpha ) = -12$

$\displaystyle \Rightarrow \alpha - \alpha^2 = -12$

$\displaystyle \Rightarrow \alpha ^2 - \alpha - 12 = 0$

$\displaystyle \Rightarrow (\alpha -4)(\alpha +3) = 0$

$\displaystyle \Rightarrow \alpha = 4$ or $\displaystyle -3$

When $\displaystyle \alpha = 4, \beta = 1-4 = -3$

When $\displaystyle \alpha = -3, \beta = 1 - (-3) = 4$

Therefore zeros are $\displaystyle (4, -3)$

$\displaystyle \\$

Section – D

Question number 23 to 30 carry 4 mark each.

Question 23: A plane left $\displaystyle 30$ minutes later than the scheduled time and in order to reach its destination $\displaystyle 1500$ km away on time, it has to increase its speed by $\displaystyle 250$ km/hr from its usual speed. Find the usual speed of the plane.

Find the dimensions of a rectangular park whose perimeter is $\displaystyle 60$ m and area $\displaystyle 200 \ m^2$ .

Answer:

$\displaystyle \text{Let the speed of the plane } = x \frac{km}{hr}$

$\displaystyle \text{New speed of the plane } = (x + 250) \frac{km}{hr}$

$\displaystyle \therefore \frac{1500}{x} - \frac{1500}{x+250} = \frac{30}{60}$

$\displaystyle \frac{1500x + 1500 \times 250 - 1500 x}{x(x+250)} = \frac{1}{2}$

$\displaystyle 1500 \times 250 \times 2 = x(x+250)$

$\displaystyle x^2 + 250 x = 750000$

$\displaystyle x^2 + 250x - 750000=0$

$\displaystyle x^2 +1000x - 750x - 750000 = 0$

$\displaystyle x(x+1000) - 750(x+1000) = 0$

$\displaystyle (x+1000)(x-750)=0$

$\displaystyle x = 750$ or $\displaystyle -1000$ ( not possible as speed cannot be negative)

$\displaystyle \therefore x = 750 \frac{km}{hr}$

Let length $\displaystyle = l$ and breadth $\displaystyle = b$

$\displaystyle \therefore lb = 200$ … … … … … i)

$\displaystyle 2(l+b) = 60$

$\displaystyle l+b = 30$ … … … … … ii)

From ii) $\displaystyle b = (30-l)$

Substituting in i)

$\displaystyle l(30-l) = 200$

$\displaystyle 30l - l^2 = 200$

$\displaystyle l^2 - 30l + 200 = 0$

$\displaystyle l^2 - 20l - 10l +200 = 0$

$\displaystyle l(l-20) - 10(l-20) = 0$

$\displaystyle (l-10)(l-20) = 0$

When $\displaystyle l = 10$ m, $\displaystyle b = 30 - 10 = 20$ m

When $\displaystyle l = 20$ m, $\displaystyle b = 30 -20 = 10$ m

Hence the dimensions are $\displaystyle 10$ m by $\displaystyle 20$ m

$\displaystyle \\$

Question 24: Find the value of $\displaystyle x$ , when in the A.P. given below $\displaystyle 2 + 6 + 10 + \cdots + x = 1800$ .

Answer:

First term of AP $\displaystyle = a = 2$

Common difference $\displaystyle = d = 6-2 = 4$

$\displaystyle \text{Sum of n terms } = \frac{n}{2} [ 2a + (n-1)d]$

$\displaystyle \frac{n}{2} [ 2 \times 2 + (n-1)(4)] = 1800$

$\displaystyle n [ 2 + (n-1) 2 ] = 1800$

$\displaystyle n [ 2n + 2 - 2] = 1800$

$\displaystyle 2n^2 = 1800$

$\displaystyle n^2 = 900$

$\displaystyle \Rightarrow n = 30$ ( total number of terms)

$\displaystyle x$ is the $\displaystyle n^{th}$ term

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

$\displaystyle x = 2 + (30-1) (4)$

$\displaystyle x = 118$

$\displaystyle \\$

$\displaystyle \text{Question 25: If } \sec \theta + \tan \theta = m , \text{ show that } \frac{m^2 -1 }{m^2 + 1} = \sin \theta$

Answer:

$\displaystyle \sec \theta + \tan \theta = m$

$\displaystyle \Rightarrow \sec^2 \theta + \tan^2 \theta + 2 \sec \theta \ \tan \theta = m^2$

$\displaystyle \Rightarrow \sec^2 \theta + \tan^2 \theta + 2 \sec \theta \ \tan \theta - 1 = m^2 - 1$

$\displaystyle \because \tan^2 \theta = \sec^2 \theta - 1$

$\displaystyle \Rightarrow 2 \tan^2 \theta + 2 \sec \theta \ \tan \theta = m^2 - 1$

$\displaystyle \Rightarrow 2 \tan \theta (\tan \theta + \sec \theta ) = m^2 -1$ … … … … … i)

Similarly,

$\displaystyle \sec^2 \theta + \tan^2 \theta + 2 \sec \theta \ \tan \theta + 1 = m^2 + 1$

$\displaystyle \because \sec^2 \theta = \tan^2 \theta + 1$

$\displaystyle \Rightarrow 2 \sec^2 \theta + 2 \sec \theta \ \tan \theta = m^2 + 1$

$\displaystyle \Rightarrow 2 \sec \theta (\tan \theta + \sec \theta ) = m^2 +1$ … … … … … ii)

Dividing i) by ii) we get

$\displaystyle \frac{m^2-1}{m^2+1} = \frac{2 \tan \theta (\tan \theta + \sec \theta) }{2 \sec \theta (\tan \theta + \sec \theta)}$

$\displaystyle \Rightarrow \frac{m^2-1}{m^2+1} = \sin \theta$

$\displaystyle \\$

Question 26: In $\triangle ABC$ (Figure 3), $AD \perp BC$ . Prove that $AC^2 = AB^2 + BC^2 - 2BC \times BD$

Answer:

Given $AD \perp BC$

In $\triangle ABD \ \ \ \ \ AB^2 = AD^2 + BD^2 \Rightarrow AD^2 = AB^2 -BD^2$ … … … … … i)

In $\triangle ADC \ \ \ \ \ AC^2 = AD^2 + DC^2 \Rightarrow AD^2 = AC^2 -DC^2$ … … … … … ii)

Frim i) and ii)

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

$\Rightarrow AC^2 = AB^2 + DC^2 - BD^2$

$\Rightarrow AC^2 = AB^2 + (DC - BD)(DC + BD)$

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

$\Rightarrow AC^2 = AB^2 + BC [(BC-BD)-BD]$

$\Rightarrow AC^2 = AB^2 + BC[BC-2BD]$

$\Rightarrow AC^2=AB^2 + BC^2 - 2 BC \times BD$

Hence Proved.

$\\$

Question 27: A moving boat is observed from the top of a $150$ m high cliff moving away from the cliff. The angle of depression of the boat changes from $60^o$ to $45^o$ in $2$ minutes. Find the speed of the boat in m/min.

There are two poles, one each on either bank of a river just opposite to each other. One pole is $60$ m high. From the top of this pole, the angle of depression of the top and foot of the other pole are $30^o$ and $60^o$ respectively. Find the width of the river and height of the other pole.

Answer:

$\displaystyle \frac{AB}{DB} = \tan 45^o = 1 \Rightarrow DB = AB = 150 \text{ m }$

$\displaystyle \frac{AB}{BC} = \tan 60^o = \sqrt{3} \Rightarrow BC = \frac{150}{\sqrt{3}} = 50 \sqrt{3}$

$\therefore DC = DB - CB = (150 - 50\sqrt{3} )$ m

$\displaystyle \text{Therefore speed } = \frac{distance}{time} = \frac{150 - 50\sqrt{3}}{2} \frac{m}{min}$

$\displaystyle = 75 - 25(1.732) \frac{m}{min}$

$\displaystyle = 31.7 \frac{m}{min}$

$\displaystyle \frac{AB}{CB} = \tan 60^o = \sqrt{3}$

$\displaystyle \Rightarrow CB = \frac{60}{\sqrt{3}} = 20\sqrt{3} \text{ m }$

(width of the river)

$\displaystyle \frac{AE}{20\sqrt{3}} = \tan 30^o \Rightarrow AE = \frac{20\sqrt{3}}{\sqrt{3}} = 20 \text{ m }$

Hence the height of the pole is $60-20 = 40 \text{ m }$

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Question 28: Construct a triangle with sides $5$ cm, $6$ cm and $7$ cm and then another triangle whose sides are $\frac{3}{5}$ of the corresponding sides of the first triangle.

Answer:

$\\$

Question 29: Calculate the mean of the following frequency distribution :

Class:	10-30	30-50	50-70	70-90	90-110	110-130
Frequency:	5	8	12	20	3	2

The following table gives production yield in kg per hectare of wheat of 100 farms of a village :

Production Yield

(kg/hectare):

40-45

45-50

50-55

55-60

60-65

65-70

Number of Farms:

Change the distribution to a ‘more than type’ distribution, and draw its ogive.

Answer:

Class Interval	Frequency ( $f_i$ )	$x_i$	$f_ix_i$
10-30	5	20	100
30-50	8	40	320
50-70	12	60	720
70-90	20	80	1600
90-110	3	100	300
110-130	2	120	2400
	$\Sigma f_i = 50$		$\Sigma f_ix_i = 3280$

$\displaystyle \text{Mean } = \frac{\Sigma f_ix_i}{\Sigma f_i} = \frac{3280}{50} = 65.6$

Production Yield	Cumulative Frequency
More than 40	100
More than 45	96
More than 50	90
More than 55	74
More than 60	54
More than 65	24

$\\$

Question 30: A container opened at the top and made up of a metal sheet, is in the form of a frustum of a cone of height $16$ cm with radii of its lower and upper ends as $8$ cm and $20$ cm respectively. Find the cost of milk which can completely fill the container, at the rate of Rs. $50$ per liter. Also find the cost of metal sheet used to make the container, if it costs Rs. $10$ per $100 \ cm^2$ . (Take $\pi = 3.14$ )