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SUMMATIVE ASSESSMENT – II

MATHEMATICS

Time allowed: 3 hours                                                             Maximum Marks: 80

General Instructions:

(i) All questions are compulsory

(ii) The question paper consists of 30 questions divided into four sections – A, B, C and D

(iii) Section A consists of 6 questions of 1 mark each. Section B consists of 6 questions of 2 marks each. Section C consists of 10 questions of 3 marks each. Section D consists of 8 questions of 4 marks each.

(iv) There is no overall choice. However, an internal choice has been provided in two questions of 1 mark, two questions of 2 marks, four questions of 3 marks each and three questions of 4 marks each. You have to attempt only one of the alternative in all such questions. 

(v) Use of calculator is not permitted.

SECTION – A

Question number 1 to 6 carry 1 mark each.

Question 1: The HCF of two numbers a and b is 5 and their LCM is 200 . Find the product ab .

Answer:

a and b are two numbers.

H.C.F. of a and b = 5

L.C.M. of a and b = 200

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

5 \times 200 = ab \Rightarrow  ab = 1000

So, The value of product of a and b is 1000

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

Or

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

Answer:

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

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

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

\Rightarrow 4k + 4 - 3 = 0

\Rightarrow k = - \frac{1}{4} 

Or

Given equation: 3x^2 + kx + 3 = 0

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

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

\Rightarrow k^2 = 36

\Rightarrow k = \pm 6

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

Answer:

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

Since a_n = a + (n-1)d

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

\Rightarrow 15 - 3n + 3 = 0

\Rightarrow 3n = 18

\Rightarrow n = 6

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

Or

Find the value of ( \cos 48^o - \sin 42^o) .

Answer:

\sin x + \cos y = 1

\Rightarrow \sin 30^o + \cos y = 1

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

\Rightarrow \cos y = \cos 60^o

\Rightarrow y = 60^o

Or

\cos 48^o - \sin 42^o

= \cos 48^o - \sin (90^o-48^o)

= \cos 48^o - \cos 48^o

= 0

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

Answer:

Let AB and DE are corresponding sides of the similar triangles

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

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

Hence the ratio of the corresponding sides = 5:11

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

Answer:

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

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

\Rightarrow 6-3a = 5

\Rightarrow 3a = 1

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

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Section – B

Question number 7 to 12 carry 2 mark each.

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

Or

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

Answer:

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

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

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

This means  First term + Second term = 18

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

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

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

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

Or

Let the first term = a and the common difference = d

T_n = a +(n-1) d

\therefore T_{17} = a + 16 d

T_{10} = a + 9 d

Given T_{17} = T_{10}+ 7

a + 16 d = a + 9d + 7

7d = 7

d = 1

Common difference = 1

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

Answer:

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

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

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

\Rightarrow a -1 = 1 \Rightarrow  a = 2

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

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

\Rightarrow b = 2

Hence a = 2, b = 2

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Question 9: A child has a die whose 6 faces show the letters given below :

2019-05-24_20-04-44

The die is thrown once. What is the probability of getting (i) A (ii) B ?

Answer:

Number of possible events = 6

No of A’s on dice = 3

No of B’s of dice = 2

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

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

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Question 10: Find the HCF of 612 and 1314 using prime factorisation.

Or

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

Answer:

First factorize each number

612 = 2 \times 2 \times 3 \times 3 \times 17

1314 = 2 \times 3 \times 3 \times 73

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

Or

Let a be any positive integer.

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

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

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

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

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

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

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Question 11: Cards marked with numbers 5 to 50 (one number on one card) are placed in a box and mixed thoroughly. One card is drawn at random from the box. Find the probability that the number on the card taken out is (i) a prime number less than 10 , (ii) a number which is a perfect square.

Answer:

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

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

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

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

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

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Question 12: For what value of k , does the system of linear equations
2x + 3y = 7 (k - 1) x + (k + 2) y = 3k  have an infinite number of solutions ?

Answer:

If the system of equations are ax_1 + b_1y + c_1 = 0 and ax_2 + b_2y + c_2 = 0 and they have infinitely many solutions then it satisfy the following:
\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}

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

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

From first two terms

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

From second and third term

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

From first and third term

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

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

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Section – C

Question number 13 to 22 carry 3 mark each.

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

Answer:

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

Since \sqrt{5} = \frac{a}{b}

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

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

5b^2 = a^2 = 25c^2

Dividing by 5 , this means

b^2 = 5c^2

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

Hence \sqrt{5} is irrational.

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

Answer:

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

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

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

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

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

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

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

Therefore the other two zeros are -4, 3

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

Or

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

Answer:

2019-06-24_7-10-31

 

AP : AB = 1:3 (trisects)

\frac{AP}{AB} = \frac{1}{3} 

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

\Rightarrow 3AP = AP + PB

\Rightarrow 2AP = PB

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

\Rightarrow AP : PB = 1:2

Applying section formula

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

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

\Rightarrow P = (3, -2)

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

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

\Rightarrow 6 + 2 + k = 0

\Rightarrow k = -8

Or

Given point (2, 1), (p, -1) and (-1, 3)

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

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

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

\Rightarrow -8+2p-2=0

\Rightarrow 2p - 10 = 0

\Rightarrow p = 5

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Question 16: Prove that:

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

Or

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

Answer:

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

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

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

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

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

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

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

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

= RHS. Hence proved.

Or

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

Squaring both sides

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

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

Add \sin^2 \theta on both sides

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

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

Hence proven.

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Question 17: A part of monthly hostel charges in a college hostel are fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 25 days, he has to pay Rs. 4,500 , whereas a student B who takes food for 30 days, has to pay Rs.  5,200 . Find the fixed charges per month and the cost of food per day.

Answer:

Let the fixed charges = x

Let the charges for food per day = y

Therefore for Student A:

x + 25 y = 4500    … … … … … i)

Therefore for Student B:

x + 30 y = 5200    … … … … … ii)

Solving i) and ii)

5y = 700 \Rightarrow y = 140 Rs.

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

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

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Question 18: In \triangle ABC, \angle B = 90^o and D is the mid-point of BC . Prove that AC^2 = AD^2 + 3CD^2 .

Or

In Figure 1, E is a point on CB produced of an isosceles \triangle ABC , with side AB = AC . If AD \perp BC and EF \perp AC , prove that \triangle ABD \sim \triangle ECF .

2019-05-25_8-52-56
Figure 1

Answer:

Given: BC = DC 2019-07-10_20-56-21

From \triangle ABC

AB^2 + BC^2 = AC^2 \Rightarrow AB^2 =  AC^2 - BC^2    … … … … … i)

From \triangle ABD

AB^2 + BD^2 = AD^2 \Rightarrow AB^2 = AD^2 - BD^2    … … … … … ii)

From i) and ii)

AC^2 - BC^2 = AD^2 - BD^2

\Rightarrow AC^2 = AD^2 + BC^2 - BD^2

\Rightarrow AC^2 = AD^2 + (BC-BD)(BC+BD)

\Rightarrow AC^2 = AD^2 + CD (2CD + CD)

\Rightarrow AC^2 = AD^2 + + 2 CD^2

Hence proved.

Or

Given:  AB = AC \Rightarrow \angle ABC = \angle ACB 2019-05-25_8-52-56

AD \perp BC and EF \perp AC

In \triangle ABD and \triangle ECF

\angle ADB = \angle EFC = 90^o (given)

\angle ABD = \angle ECF

\therefore \triangle ABD \sim \triangle ECF ( By AA criterion)

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

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Question 19: Prove that the parallelogram circumscribing a circle is a rhombus.

Answer:2019-07-10_20-52-12

Given: ABCD be a parallelogram circumscribing a circle with center O .

To prove: ABCD is a rhombus.

We know that the tangents drawn to a circle from an exterior point are equal in length.

Therefore, AP = AS, BP = BQ, CR = CQ and DR = DS .

Adding the above equations,

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

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

AB + CD = AD + BC

2AB = 2BC

(Since, ABCD is a parallelogram so AB = DC and AD = BC )

AB = BC

Therefore, AB = BC = DC = AD .

Hence, ABCD is a rhombus.

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Question 20: In Figure 2, three sectors of a circle of radius 7 cm, making angles of 60^o 80^o and 40^o at the centre are shaded. Find the the area of the shaded region.

2019-05-25_8-14-52
Figure 2

Answer:

Area of the shaded region

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

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

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

= 77 \ cm^2

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Question 21: The following table gives the number of participants in a yoga camp :

Age (in years): 20-30 30-40 40-50 50-60 60-70
No. of participants: 8 40 58 90 83

Find the modal age of the participants.

Answer:

First we find the modal class of the given data which is the highest participant class i.e.

Modal class = 50 -60

Formula to find Mode is

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

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

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

= 50 + \frac{320}{39}

= 50 + 8.20 = 58.20

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Question 22: A juice seller was serving his customers using glasses as shown in Figure 3. The inner diameter of the cylindrical glass was 5 cm but bottom of the glass had a hemispherical raised portion which reduced the capacity of the glass. If the height of a glass was 10 cm, find the apparent and actual capacity of the glass. (Use \pi = 3.14 )

2019-05-25_8-05-38
Figure 3

Or

A girl empties a cylindrical bucket full of sand, of base radius 18 cm and height 32 cm on the floor to form a conical heap of sand. If the height of this conical heap is 24 cm, then find its slant height correct to one place of decimal.

Answer:

2019-07-10_20-46-47.pngApparent volume = \pi r^2 h = \pi (2.5)^2 \times 10 = 62.5 \pi = 196.25 \ cm^3

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

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

= 196.25 - 32.71 = 163.542 \ cm^3

Or

Volume of sand remains the same

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

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

\Rightarrow r = 36 cm

Slant height (l) = \sqrt{36^2 + 24^2} = \sqrt{1872} = 43.27 cm

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Section – D

Question number 23 to 30 carry 4 mark each.

Question 23: A train travels 360 km at a uniform speed. If the speed had been 5 km/hr more, it would have taken 1 hr less for the same journey. Find the speed of the train.

Or

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

Answer:

Distance traveled = 360 km

Let the original speed = x km/hr

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

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

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

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

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

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

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

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

Hence the original speed = 40 km/hr

Or

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

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

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

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

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

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

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

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

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

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

\Rightarrow x = -a or x = -b

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

Answer:

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

Given S_p = q

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

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

Also S_q = p

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

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

Subtracting ii) from i)

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

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

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

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

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

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

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

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

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Question 25: In a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then prove that the angle opposite to the first side is a right angle.

Answer:

Given: AC^2 = AB^2 + BC^2

To prove: \angle B = 90^o

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

Proof: From \triangle PQR

PR^2 = PQ^2 + QR^2 (Pythagoras theorem)

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

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

AC^2 = PR^2 from i) and ii)

\therefore AC = PR    … … … … … iii)

Now in \triangle ABC and \triangle PQR

AB = PQ (by construction)

BC = QR (by construction)

AC = PR (from iii)

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

\therefore \angle B = \angle Q

But \angle Q = 90^o by construction

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

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

Answer:

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Question 27: A boy standing on a horizontal plane finds a bird flying at a distance of 100 m from him at an elevation of 30^o . A girl standing on the roof of a 20 m high building, finds the elevation of the same bird to be 45^o . The boy and the girl are on the opposite sides of the bird. Find the distance of the bird from the girl. (Given \sqrt{2} = 1.414 )

Or

The angle of elevation of an aeroplane from a point A on the ground is 60^o . After a flight of 30 seconds, the angle of elevation changes to 30^o . If the plane is flying at a constant height of 3600 \sqrt{3} metres, find the speed of the aeroplane.

Answer:2019-07-10_20-29-45

In \triangle ABC

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

\therefore AE = AB - 20 = 30 m

In \triangle AED

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

Or

In \triangle ABC 2019-07-10_20-35-47

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

\therefore AC = 3600 m

In \triangle AED

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

\therefore AE = 3600 \times 3 = 10800 m

Hence CE = 10800 - 3600 = 7200 m

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

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

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Question 28: Find the values of frequencies x and y in the following frequency distribution table, if N = 100 and median is 32 .

Marks: 0-10 10-20 20-30 30-40 40-50 50-60 Total
No. of Students 10 x 25 30 y 10 100

Or

For the following frequency distribution, draw a cumulative frequency curve (ogive) of ‘more than type’ and hence obtain the median value.

Class: 0-10 10-20 20-30 30-40 40-50 50-60 Total
Frequency: 5 10 20 23 17 11 9

Answer:

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

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

For the above distribution, median class is 30-40

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

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

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

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

\Rightarrow x = 9

From i) y=25-9=16

Or

Class Interval (More than) Frequency
More than 0 100
More than 10 95
More than 20 80
More than 30 60
More than 40 37
More than 50 20
More than 60 9

 

Class Interval (Less than) Frequency
Less than 10 5
Less than 20 20
Less than 30 40
Less than 40 63
Less than 50 80
Less than 60 91
Less than 70 100

2019-07-10_10-13-38

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Question 29: Prove that:

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

Answer:

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

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

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

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

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

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

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

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

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

= RHS. Hence proved.

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Question 30: An open metallic bucket is in the shape of a frustum of a cone. If the diameters of the two circular ends of the bucket are 45 cm and 25 cm and the vertical height of the bucket is 24 cm, find the area of the metallic sheet used to make the bucket. Also find the volume of the water it can hold. (Use \pi =  \frac{22}{7} )

Answer:2019-07-10_20-27-54

r_1 = \frac{25}{2} = 12.5 cm

r_2 = \frac{45}{2} = 22.5 cm

h = 24 cm

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

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

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

= 23728.57 cm^3 = 23.723 liters

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

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

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

= 3351.07 \ cm^2

= 3.351 \ m^2

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