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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: If HCF (336, 54) = 6 , find LCM (336, 54) .

Answer:

We know HCF \times LCM = product of the two numbers

\therefore 6 \times LCM = 336 \times 54

\Rightarrow LCM = \frac{336 \times 54}{6} = 3024

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Question 2: Find the nature of roots of the quadratic equation 2x^2 - 4x + 3 = 0 .

Answer:

Given equation 2x^2 - 4x + 3 = 0

Therefore Roots = \frac{4 \pm \sqrt{16 - 4 \times 2 \times 3}}{4} = \frac{4 \pm \sqrt{-8}}{4}

Therefore the nature of the roots is imaginary

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Question 3: Find the common difference of the Arithmetic Progression (A.P.) \frac{1}{a}, \frac{3-a}{3a}, \frac{3-2a}{3a} , \cdots (a \neq 0)

Answer:

Given AP: \frac{1}{a}, \frac{3-a}{3a}, \frac{3-2a}{3a} , \cdots (a \neq 0)

Common difference  d = a_2 - a_1,  a_3 - a_2, \cdots

\therefore d = \frac{3-a}{3a} - \frac{1}{a} = \frac{3 -a - 3}{3a} = - \frac{1}{3}

To confirm, we also do a_3 - a_2 = \frac{3-2a}{3a} - \frac{3-a}{3a} = \frac{3 - 2a - 3 + a}{3a} = - \frac{1}{3}

Hence d = - \frac{1}{3}

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Question 4: Evaluate : \sin^2 60^o+ 2 \tan 45^o - \cos^2 30^o

Or

If \sin A = \frac{3}{4} , calculate \sec A

Answer:

\sin^2 60^o+ 2 \tan 45^o - \cos^2 30^o

= \Big( \frac{\sqrt{3}}{2} \Big)^2 + 2 (1) - \Big( \frac{\sqrt{3}}{2} \Big)^2

= 2

Or2019-06-30_7-39-35

If \sin A = \frac{3}{4}

\therefore AB = \sqrt{4^2 - 3^2} = \sqrt{7}

\therefore \sec A = \frac{4}{\sqrt{7}} 

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Question 5: Write the coordinates of a point P on x-axis which is equidistant from the points A(- 2, 0) and B(6, 0) .

Answer:

Let P(x, 0) be equidistant from A and B

AP = BP \Rightarrow AP^2 = BP^2 2019-06-30_7-42-43.png

\Rightarrow (-2-x)^2 +(0-0)^2 = (6-x)^2 + (0-0)^2

\Rightarrow 4 + x^2 + 4x = 36 + x^2 - 12x

\Rightarrow 16x = 32 \Rightarrow x = 2

\therefore P is (2, 0)

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Question 6: In Figure 1, ABC is an isosceles triangle right angled at C with AC = 4 cm. Find the length of AB .

2019-05-22_21-21-38
Figure 1

Or

In Figure 2, DE \parallel BC . Find the length of side AD , given that AE = 1.8 cm, BD = 7.2 cm and CE = 5.4 cm.

2019-05-22_21-21-59
Figure 2

Answer:

Since \triangle ABC is isosceles

\Rightarrow AC = BC

(hypotenuse AB has to be the largest side, hence the other two sides are equal)

\therefore AB = \sqrt{4^2 + 4^2} = 4\sqrt{2} cm

Or

Since DE \parallel BC by Thales Theorem

\frac{AD}{BD} = \frac{AE}{EC} 

\Rightarrow \frac{AD}{7.2} = \frac{1.8}{5.4} 

\Rightarrow AD = \frac{1.8}{5.4} \times 7.2 = 2.4 cm

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Section – B

Question number 7 to 12 carry 2 mark each.

Question 7: Write the smallest number which is divisible by both 306 and 657 .

Answer:

306 = 2 \times 3^2 \times 17

657 = 3^2 \times 73

LCM of 306 and 657 = 2 \times 3^2 \times 17 \times 73 = 22338

Hence the smallest number which is divisible by 306 and 657 is 22338

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Question 8: Find a relation between x and y if the points A(x, y), B( - 4, 6) and C( - 2, 3) are collinear.

Or

Find the area of a triangle whose vertices are given as (1, - 1) (- 4, 6) and (- 3, - 5) .

Answer:

If A , B and C are collinear, then the area of the \triangle ABC should be 0 .

Area of \triangle ABC = \frac{1}{2} \times |x_1(y_2-y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) |

\Rightarrow 0 =  \frac{1}{2} \times |x(6-3) + (-4)(3-y) + (-2)(y-6) |

\Rightarrow 0 = | 3x - 12 + 4y - 2y + 12 |

\Rightarrow 0 = |3x + 2y |

\Rightarrow 3x + 2y = 0

Or

Let the points be A(1, -1), B(-4, 6) and C(-3, -5)

Area of triangle ABC = \frac{1}{2} [x_1(y_2-y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) ]

Here x_1 = 1, y_1 = -1, x_2 = -4, y_2 = 6, x_3 = -3, y_3 = -5

Substituting in the formula above

Area of \triangle ABC = \frac{1}{2} [1(6-(-5)) + (-4) (-5 - (-1)) + (-3)(-1-6) ]

= \frac{1}{2} [ 11+ 16 + 21 ] = \frac{1}{2} \times 48 = 24 sq. units.

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Question 9: The probability of selecting a blue marble at random from a jar that contains only blue, black and green marbles is \frac{1}{5} . The probability of selecting a black marble at random from the same jar is \frac{1}{4} . If the jar contains 11 green marbles, find the total number of marbles in the jar.

Answer:

P(selecting Blue marble) = \frac{1}{5}

P(selecting Black marble) = \frac{1}{4}

P(selecting Green marble) = 1 - \frac{1}{5} - \frac{1}{4} = \frac{20-9}{20} = \frac{11}{20}

Therefore there are a total of 20 marbles in the Jar.

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Question 10: Find the value(s) of k so that the pair of equations x + 2y = 5 and 3x + ky + 15 = 0 has a unique solution.

Answer:

Pair of equations x + 2y = 5 and 3x + ky + 15 = 0

If the intersecting lines have a unique solution, then

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

\Rightarrow \frac{1}{3} \neq \frac{2}{k} \neq \frac{-5}{15}

From the first two terms, k \neq 6

From the second and the third terms, k \neq -6

Hence all values of k except 6 and -6

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Question 11: The larger of two supplementary angles exceeds the smaller by 18^o . Find the angles.

Or

Sumit is 3 times as old as his son. Five years later, he shall be two and a half times as old as his son. How old is Sumit at present ?

Answer:

Let x and y be the two supplementary angles

\therefore x + y = 180^o

Also given x = y + 18^o (given one angle is greater than the other by 18^o )

Solving the two equations

2y + 18 = 180^o \Rightarrow 2y = 162^o \Rightarrow y = 81^o

\therefore x = 81^o + 18^o = 99^o

Hence the two angles are 81^o ad 99^o

Or

Let the age of son be x years

Therefore are of Sumit is 3x

Age of son 5 years later = x + 5

Age of Sumit 5 years later = 3x + 5

Give: 3x+5 = 2.5 (x + 5)

\Rightarrow 3x + 5 = 2.5 x + 12.5

\Rightarrow 0.5 x = 7.5

\Rightarrow x = 15 years

Therefore age of Sumit = 3 \times 15 = 45 years

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Question 12: Find the mode of the following frequency distribution :

Class Interval: 25-30 30-35 35-40 40-45 45-50 50-55
Frequency: 25 34 50 42 38 14

Answer:

2019-06-25_6-57-32

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Section – C

Question number 13 to 22 carry 3 mark each.

Question 13: Prove that 2 + 5 \sqrt{3} is an irrational number, given that \sqrt{3} is an irrational number.

Or

Using Euclid’s Algorithm, find the HCF of 2048 and 960 .

Answer:

Let us assume that 5 +2 \sqrt{3} is a rational number.

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

\therefore 5 + 2\sqrt{3} = \frac{p}{q}

2\sqrt{3} = \frac{p}{q} - 5

2\sqrt{3} = \frac{p-5q}{q}

\sqrt{3} = \frac{p-5q}{2q}

\therefore \sqrt{3} is a rational no as \frac{p-5q}{2q} is a rational number

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

Hence 5 + 2 \sqrt{3} is an irrational number.

Or

960 ) \overline{2048} ( 2 \\ \hspace*{1cm} \underline{1920} \\ \hspace*{1cm} 128 ) \overline{960} ( 7 \\ \hspace*{2cm} \underline{896} \\ \hspace*{2cm}  64 ) \overline{128} ( 2 \\ \hspace*{3cm}  \underline{128} \\ \hspace*{3cm} \times 

Hence HCF = 64

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Question 14: Two right triangles ABC and DBC are drawn on the same hypotenuse BC  and on the same side of BC . If AC and BD intersect at P , prove that AP \times PC = BP \times DP .

Or

Diagonals of a trapezium PQRS intersect each other at the point O, PQ \parallel RS and PQ = 3RS . Find the ratio of the areas of triangles POQ and ROS .

Answer:

To prove: AP \times PC = BP \times PD 2019-06-30_7-52-16

Consider \triangle ABP and \triangle CDP

\angle BAP = \angle CDP = 90^o (given)

\angle BPA = \angle CPD ( vertically opposite angles)

\therefore \triangle ABP \sim \triangle CDP (By AA criterion)

\therefore \frac{BP}{PC} = \frac{AP}{PD} 

\Rightarrow AP \times PC = BP \times PD

Hence proved.

Or

Given PQRS is a trapezium2019-06-30_7-51-48

PQ  \parallel RS and PQ = 3 RS

Consider \triangle POQ and \triangle SRO 

\angle SOR = \angle POQ (vertically opposite angles)

\angle RSO = \angle OQP (alternate angles)

\angle SRO = \angle OPQ (alternate angles)

\therefore \triangle POQ \sim \triangle SRO  (By AAA criterion)

By property of similar triangles

\frac{ar (\triangle POQ)}{ar(\triangle SOR)} = \frac{PQ^2}{RS^2} = \Big( \frac{PQ}{RS} \Big)^2 = \Big( \frac{3}{1} \Big)^2 = \frac{9}{1}

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Question 15: In Figure 3, PQ and RS are two parallel tangents to a circle with center O  and another tangent AB with point of contact C intersecting PQ at A and RS at B . Prove that \angle AOB = 90^o .

2019-05-22_21-21-04
Figure 3

Answer:

Given: PQ \parallel RS , AB is a tangent

To prove: \angle AOB = 90^o

Join DE through O . DOE is diameter of the circle

\therefore \angle ADO = \angle BEO = 90^o 2019-06-25_7-49-32.png

\Rightarrow \angle ADO + \angle BEO = 180^o

\Rightarrow DA \parallel EB

We know that tangents to a circle from an external point are equally inclined to the line segment joining this point to the center

\therefore \angle 1 = \angle 2 and \angle 3 = \angle 4

Now DA \parallel RB and AB is transversal

\therefore (\angle 1 + \angle 2) + (\angle 3 + \angle 4) = 180^o

\Rightarrow 2 \angle 1 + 2 \angle 3 = 180^o

\Rightarrow \angle 1 + \angle 3 = 90^o

From \triangle AOB

\angle AOB + \angle 1 + \angle 3 = 180^o

\Rightarrow \angle AOB = 180^o - 90^0 = 90^o

Hence \angle AOB  = 90^o

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Question 16: Find the ratio in which the line x - 3y = 0 divides the line segment joining the points (- 2, - 5) and (6, 3) . Find the coordinates of the point of intersection.

Answer:

Let P(x, y) divides point A(-2, -5) and B (6, 3) in the ratio k : 1

Using section formula: P(x, y) = \Big( \frac{mx_2+nx_1}{m+n} , \frac{my_2+ny_1}{m+n} \Big)

A(-2, -5) and B(6, 3) in the ratio of k:1

\Rightarrow x = \frac{6k-2}{k+1}  and \Rightarrow y = \frac{3k-5}{k+1}

Since P(x, y) lies on the line x-3y = 0

\frac{6k-2}{k+1} - 3 \Big(  \frac{3k-5}{k+1}  \Big) = 0

\Rightarrow 6k - 2 - 9k + 15 = 0

\Rightarrow -3k +13 = 0

\Rightarrow k = \frac{13}{3}

Therefore P divided A and B in the ratio of 13:3

\therefore x = \frac{6(\frac{13}{3})-2}{\frac{13}{3}+1} = \frac{(26-2)\times 3}{16} = \frac{9}{2}

y = \frac{3(\frac{13}{3})-5}{(\frac{13}{3})+1} = \frac{(13-5) \times 3}{16} = \frac{3}{2}

Hence P( \frac{9}{2} , \frac{3}{2} ) which is the point of intersection.

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Question 17: Evaluate: \Big( \frac{3 \sin 43^o}{\cos 47^o} \Big)^2  - \frac{\cos 37^o  \mathrm{cosec} 53^o}{\tan 5^o \tan 25^o \tan 45^o \tan 65^o \tan 85^o}

Answer:

\Big( \frac{3 \sin 43^o}{\cos 47^o} \Big)^2  - \frac{\cos 37^o  \mathrm{cosec} 53^o}{\tan 5^o \tan 25^o \tan 45^o \tan 65^o \tan 85^o}

= \Big( \frac{3 \sin 43^o}{\cos (90-43)^o} \Big)^2  - \frac{\cos 37^o  \frac{1}{\sin (90-37)^o } }{\tan 5^o \tan 25^o .1. \tan (90-25)^o \tan (90-5)^o}

= \Big( \frac{3 \sin 43^o}{\sin 43^o} \Big)^2  - \frac{\cos 37^o  \frac{1}{\cos 37^o } }{\tan 5^o \tan 25^o .1. \cot 25^o \cot 5^o}

= 3^2 -1 = 9-1 = 8

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Question 18: In Figure 4, a square OABC is inscribed in a quadrant OPBQ . If OA = 15 cm, find the area of the shaded region. (Use \pi = 3.14 )

2019-05-22_21-07-18
Figure 4

Or

In Figure 5, ABCD is a square with side 2 \sqrt{2} cm and inscribed in a circle. Find the area of the shaded region. (Use \pi = 3.14 )

2019-05-22_21-02-55
Figure 5

Answer:

OA = OB = 15 cm (since OABC is a square)

\therefore OB = \sqrt{15^2 + 15^2} = 15 \sqrt{2} = Radius

Area of square OABC = 15 \times 15 = 225 \  cm^2

Area of quadrant = \frac{1}{4} (\pi (15 \sqrt{2})^2 ) = 353.25 \ cm^2

Shaded area = 353.25 - 225 = 128.25 \  cm^2

Or

Area of square = 2\sqrt{2} \times 2\sqrt{2} = 8 \ cm^2

Diameter AC = \sqrt{(2\sqrt{2})^2 + (2\sqrt{2})^2} = \sqrt{8+8} = 4 \ cm

Therefore Radius of circle = 2 \ cm

Area of circle = \pi (2)^2 = 3.14 \times 4 = 12.56 \ cm^2

Therefore shaded area = 12.56 - 8 = 4.56 \ cm^2

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Question 19: A solid is in the form of a cylinder with hemispherical ends. The total height of the solid is 20 cm and the diameter of the cylinder is 7 cm. Find the total volume of the solid. (Use \pi = \frac{22}{7} )

Answer:2019-06-30_7-57-54

Radius of hemisphere = 3.5 \ cm

Therefore height of cylinder = 20-3.5 -3.5 = 13 \ cm

Therefore total volume = 2 \times ( \frac{1}{2} \times \frac{4}{3} \times \pi (3.5)^2 ) + \pi (3.5)^2 \times 13

= 179.67 + 500.5 = 680.17 \ cm^2

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Question 20: The marks obtained by 100 students in an examination are given below :

Marks: 30-35 35-40 40-45 45-50 50-55 55-60 60-65
Number of Students: 14 16 28 23 18 8 3

Answer:

Marks ( Class interval) No. of Students (Frequency) Mean Value $latex (x) $ fx
30-35 14 32.5 455
35-40 16 37.5 600
40-45 28 42.5 1190
45-50 23 47.5 1092.5
50-55 18 52.5 945
55-60 8 57.5 460
60-65 3 62.5 187.5
\Sigma f = 110 \Sigma fx = 4930

Mean = \frac{\Sigma fx}{\Sigma f} = \frac{4930}{110} = 44.818

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Question 21: For what value of k, is the polynomial f(x) = 3x^4 - 9x^3 + x^2 + 15x + k  completely divisible by 3x^2 - 5 ?

Or

Find the zeroes of the quadratic polynomial 7y^2 - \frac{11}{3} y - \frac{2}{3} and verify the relationship between the zeroes and the coefficients.

Answer:

f(x) divisible by P(x) \Rightarrow   the remainder of \frac{f(x)}{P(x)} = 0

3x^2-5 ) \overline{3x^4 - 9x^3 + x^2 + 15x + k} ( x^2-3x+2 \\ \hspace*{1cm}(-) { 3x^4-5x^2} \\ \hspace*{2cm}  \overline{-9x^3 + 6x^2 + 15x+ k} \\ \hspace*{2cm}(-) \underline{ -9x^3+15x} \\ \hspace*{3.5cm}  {6x^2+k}  \\ \hspace*{3cm} (-) \underline{6x^2-10} \\ \hspace*{4cm} k+10 

Remainder should be 0 \Rightarrow k+10 = 0 \Rightarrow k = -10

(3x^2-5) and (x^2 - 3x+2) are factors of f(x)

\Rightarrow (\sqrt{3} x - \sqrt{5}) and (\sqrt{3} x + \sqrt{5}) are factors

\Rightarrow x = \frac{\sqrt{5}}{\sqrt{3}}  or - \frac{\sqrt{5}}{\sqrt{3}}

Also x^2 - 3x + 2 = (x-2)(x-1) are factors \Rightarrow x = 2, 1

Hence \frac{\sqrt{5}}{\sqrt{3}} , - \frac{\sqrt{5}}{\sqrt{3}} , 2, 1 are zeros of f(x)

Or

Given polynomial P(y) = y^2 - \frac{11}{3} y - \frac{2}{3} = 0

\Rightarrow 21y^2 - 11y - 2 = 0

\Rightarrow 7y(3y-2) + (3y-2) = 0

\Rightarrow (3y-2)(7y+1) = 0

\Rightarrow y = \frac{2}{3} or y = - \frac{1}{7}

Comparing P(y) with ax^2 + bx + c =0 we get a = 7, b = - \frac{11}{3} and c = - \frac{2}{3}

Sum of zeros = - \frac{b}{a} = - (- \frac{11}{3} ) ( \frac{1}{7} ) = \frac{11}{21}    … … … … … i)

Sum of roots = \frac{2}{3} + \Big(  \frac{-1}{7} \Big) = \frac{11}{21}    … … … … … ii)

Therefore i) = ii)

Products of zeros = \frac{c}{a} = \frac{-\frac{2}{3}}{7} = \frac{-2}{21}    … … … … … iii)

Product of roots = \frac{2}{3} \times \Big( \frac{-1}{7} \Big) = \frac{-2}{21}    … … … … … iv)

Therefore iii) = iv)

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Question 22: Write all the values of p for which the quadratic equation x^2 + px + 16 = 0 has equal roots. Find the roots of the equation so obtained.

Answer:

Given quadratic equation is: x^2 + px + 16 = 0

If the roots are equal then the discriminant is 0

i.e. b^2 - 4ac = 0

\Rightarrow p^2 - 4 (1) (16) = 0

\Rightarrow p^2 = 64

\Rightarrow p = \pm 8

When p = 8, x^2 + 8x + 16 = 0

\Rightarrow (x+4)(x+4) = 0

\Rightarrow x = -4

When p = -8, x^2 - 8x + 16 = 0

\Rightarrow (x-4)(x-4) = 0

\Rightarrow x = 4

Therefore the roots of equation are 4, -4

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Section – D

Question number 23 to 30 carry 4 mark each.

Question 23: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then prove that the other two sides are divided in the same ratio.

Answer:

Given: In \triangle ABC, DE intersects AB and AC at D and E respectively.2019-06-30_8-07-41.png

To Prove: \frac{AD}{DB} = \frac{AE}{EC}

Construct: DM \perp AC and EN \perp AB

Proof:

Area of \triangle ADE = \frac{1}{2} \times AD \times EN

Area of \triangle BDE = \frac{1}{2}  \times BD \times EN

\therefore \frac{ar(\triangle ADE)}{ar(\triangle BDE)}   = \frac{AD}{BD}    … … … … … i)

Now ar(\triangle BDE) = ar(\triangle CDE)    … … … … … ii)

(Because they have the same base DE and are between the same parallel)

Similarly, \frac{ar(\triangle ADE)}{ar(\triangle CDE)} = \frac{AE}{EC}    … … … … … iii)

From i) , ii) and iii) we get \frac{AD}{DB} = \frac{AE}{EC} . Hence proved.

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Question 24: Amit, standing on a horizontal plane, finds a bird flying at a distance of 200 m from him at an elevation of 30^o . Deepak standing on the roof of a 50 m high building, finds the angle of elevation of the same bird to be 45^o . Amit and Deepak are on opposite sides of the bird. Find the distance of the bird from Deepak.

Answer:2019-06-30_8-15-47.png

\frac{h}{200} = \sin 30^o = \frac{1}{2}

\Rightarrow h = 100 m

\therefore \frac{100-50}{BD} = \sin 45^o = \frac{1}{\sqrt{2}}

\therefore BD = 50\sqrt{2}

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Question 25: A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 \ cm^3 of iron has approximately 8 gm mass. (Use \pi = 3.14 )

Answer:2019-06-30_8-21-23.png

Volume of pole = \pi (12)^2 \times 220 + \pi (8)^2 \times 60

= 31680 \pi + 3840 \pi

= 35520 \times 3.14

= 111532.8 \ cm^3

Weight of the pole = 111532.8 \ cm^3 \times 8 \frac{gm}{cm^3} = 892262.4 \ gm = 892.26 \ kg

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Question 26: Construct an equilateral \triangle ABC with each side 5 cm. Then construct another triangle whose sides are \frac{2}{3} times the corresponding sides of \triangle ABC .

Or

Draw two concentric circles of radii 2 cm and 5 cm. Take a point P on the outer circle and construct a pair of tangents PA and PB to the smaller circle. Measure PA .

Answer:

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Question 27: Change the following data into ‘less than type’ distribution and draw its ogive :

Class Interval: 30-40 40-50 50-60 60-70 70-80 80-90 90-100
Frequency: 7 5 8 10 6 6 8

Answer:

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Question 28: Prove that:  \frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \sec \theta \ \mathrm{cosec} \theta

Or

Prove that: \frac{\sin \theta}{\cot \theta + \mathrm{cosec} \theta } = 2 + \frac{\sin \theta}{\cot \theta - \mathrm{cosec} \theta}

Answer:

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

= \frac{\tan \theta}{1- \frac{1}{\tan \theta}} + \frac{\frac{1}{\tan \theta}}{1 - \tan \theta}

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

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

= \frac{1 - \tan^3 \theta}{\tan \theta (1 - \tan \theta)}

= \frac{1 + \tan^2 \theta + \tan \theta}{\tan \theta }

= \frac{\sec^2 \theta + \tan \theta}{\tan \theta }

= \sec^2 \theta \cot \theta + 1

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

= \frac{1}{\cos \theta \sin \theta} + 1

= 1 + \sec \theta \ \mathrm{cosec} \theta

Or

LHS = \frac{\sin \theta}{\cot \theta+ \mathrm{cosec} \theta}

= \frac{\sin \theta}{\frac{\cos \theta}{\sin \theta} + \frac{1}{\sin \theta}}

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

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

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

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

= 1- \cos \theta

RHS = 2 + \frac{\sin \theta}{\cot \theta - \mathrm{cosec} \theta}

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

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

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

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

= 2 - ( 1 + \cos \theta)

= 1 - \cos \theta

= LHS

Hence Proved.

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Question 29: Which term of the Arithmetic Progression -7, -12, -17, -22, \cdots will be - 82 ? Is -100 any term of the A.P. ? Give reason for your answer.

Or

How many terms of the Arithmetic Progression 45, 39, 33, \cdots must be taken so that their sum is 180 ? Explain the double answer.

Answer:

Given AP: -7, -12, -17, -22, \cdots

a_1 = -7

Common difference = -12 - (-7) = -5

The n^{th} term of AP

T_n = a + (n-1)d

\Rightarrow -82 = -7 + (n-1)(-5)

\Rightarrow -82 = -7 - 5n + 5

\Rightarrow -82 = -2 - 5n

\Rightarrow -80 = -5n

\Rightarrow n = 16

Hence -82 is the 16^{th} term

Since the numbers in the given AP are not factors of 5, - 100 will not be part of the AP.

Or

Given AP is 45, 39, 33, \cdots

a_1 = 45

Common difference d = 39 - 45 = - 6

S_n = 180

\frac{n}{2} (2a+(n-1)d) = 180

\frac{n}{2} ( 2 \times 45 + (n-1)(-6)) = 180

n(90-6n + 6) = 360

n(96-6n) = 360

n(16-n) = 60

n^2 - 16n + 60 = 0

(n-10)(n-6) = 0

n = 10 or n = 6

We get double answer because the sum of the 7^{th} to 10^{th} term is 0 .

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Question 30: In a class test, the sum of Arun’s marks in Hindi and English is 30 . Had he got 2 marks more in Hindi and 3 marks less in English, the product of the marks would have been 210 . Find his marks in the two subjects.

Answer:

Let Arun’s marks in Hindi = x

Let Arun’s marks in English = y

\therefore x + y = 30    … … … … … i)

Also (x+2)(y-3) = 210    … … … … … ii)

From i) y = 30 - x    … … … … … iii)

Substituting in ii)

(x+2)(30-x-3) = 210

\Rightarrow (x+2)(27-x) = 210

\Rightarrow 27x + 54 - x^2 - 2x = 210

\Rightarrow 25x-x^2= 156

\Rightarrow x^2 - 25x + 156 = 0

\Rightarrow (x-12)(x-13) = 0

x = 12 or 13

Substituting in iii), when x = 12, y = 30-12 = 18

When x = 13 , y = 30-13 = 17

Hence his marks in Hindi are 12 or 13 . When he gets 12 in Hindi, he gets 18 in English. When he gets 13 in Hindi, he gets 17 in English.

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