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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: Find the coordinates of a point A , where AB is the diameter of a circle whose center is (2, -3) and B is the point ( 1, 4) .

Answer:2019-06-24_6-55-47

Let the coordinates of A   be (x, y)

Since (2, -3) is the center, it is also the midpoint of AB

\therefore \frac{x+1}{2} = 2 \Rightarrow x+1 = 4 \Rightarrow x = 3

Similarly, \frac{y+4}{2} = -3 \Rightarrow y+4 = -6 \Rightarrow y = -10

\therefore A is (3, -10)

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Question 2: For what value of k , the roots of the equation x^2 + 4x + k = 0 are real?

Or

Find the value of k , for which the roots of the equation 3x^2 - 10x + k = 0 are reciprocal of each other.

Answer:

Given x^2 + 4x + k = 0

For roots to be real, b^2 - 4ac = 0 \Rightarrow 4^2 - 4(1)(k) = 0 \Rightarrow k = 1

Or

Given 3x^2 - 10x +k = 0 \Rightarrow a = 3, b = -10, c = k

Roots are : \frac{10 \pm \sqrt{10^2-12k}}{6}

Therefore Roots are \frac{10 + \sqrt{100-12k}}{6} and \frac{10 - \sqrt{100-12k}}{6}

\Rightarrow \frac{10 + \sqrt{100-12k}}{6} = \frac{6}{10 - \sqrt{100-12k}}

\Rightarrow 100 - (100-12k) = 36

\Rightarrow 100 - 100 + 12 k = 36

\Rightarrow k = 3

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Question 3: Find A if \tan 2A = \cot (A-24^o)

Or

Find the value of ( \sin^2 33^o + \sin^2 57^o)

Answer:

\tan 2A = \cot (A- 24^o)

\Rightarrow \tan 2A = \tan (90^o - ( A - 24^o))

\Rightarrow 2A = 90^o -A + 24^o

\Rightarrow 3A = 114 \Rightarrow A = 38^o

Or

\sin^2 33^o + \sin^2 57^o

= \sin^2 33^o + \sin^2(90^o-33^o)

= \sin^2 33^o + \cos^2 33^o

= 1

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Question 4: How many two digits number are divisible by 3 ?

Answer:

Numbers divisible by 3 are 3, 6, 9, 12, \cdots

Lowest 2 digit number divisible by 3 is 12

The highest 2 digit number divisible by 3 is 99

So the series starts with 12 and ends with 99

Difference between the numbers is 3

So the AP will be 12, 15, 18, \cdots , 99

We need to find n

\therefore a = 12, a_n = 99, d = 3

Now a_n = a + (n-1) d

\Rightarrow 99 = 12 + (n-1) (3)  \Rightarrow 99 = 9 + 3n  \Rightarrow 90 = 3n  \Rightarrow n = 30

Therefore there are 30 two digit numbers divisible by 3

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Question 5: In Fig. 1, DE \parallel BC, AD = 1 cm and BD = 2 cm. What is the ratio of the ar( \triangle ABC) to the ar ( \triangle ADE) ?

2019-05-22_18-55-42
Fig. 1

 

Answer:

Given DC \parallel BC 2019-06-24_7-04-43

Consider \triangle ABC and \triangle ADE

\angle ADE = \angle ABC (alternate angles)

\angle A is common

\angle AED = \angle ACD ( alternate angles)

\therefore \triangle ADE \sim \triangle ABC ( AAA criterion)

Since DE \parallel BC , by Thales Theorem

\frac{AD}{DB} = \frac{AE}{EC} = \frac{1}{2}

Since triangles are similar,

\frac{ar(\triangle ABC)}{ar(\triangle ADE)} = \frac{3^2}{1} = \frac{9}{1}

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Question 6: Find a rational number between \sqrt{2} and \sqrt{3} .

Answer:

We know \sqrt{2} = 1.412 and \sqrt{3} = 1.732

Therfore 1.5 is between \sqrt{2} and \sqrt{3}

\Rightarrow 1.5 = \frac{15}{10} = \frac{3}{2}

Hence \frac{3}{2} is between \sqrt{2} and \sqrt{3}

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Section – B

Question number 7 to 12 carry 2 mark each.

Question 7: Find the HCF of 1260 and 7344   using Euclid’s algorithm.

Or

Show that every positive odd integer is of the form (4q+1) or (4q+3) , where q is some integer.

Answer:

1260 ) \overline{7344} ( 5 \\ \hspace*{1cm} \underline{6300} \\ \hspace*{1cm} 1044 ) \overline{1260} ( 1 \\ \hspace*{2cm} \underline{1044} \\ \hspace*{2cm}  216 ) \overline{1044} ( 4 \\ \hspace*{3cm}  \underline{864} \\ \hspace*{3cm} 180 ) \overline{216} ( 1 \\ \hspace*{4cm}  \underline{36} \\ \hspace*{4cm}  36 ) \overline{180} ( 5 \\ \hspace*{4.5cm} \underline{180} \\ \hspace*{4.75cm} \times

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

\therefore 180 = 36 \times 5 + 0

So HCF of 1260 and 7344 is 36

Or

Let a be any positive integer.

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

Take b = 4 \Rightarrow a = 4q + r

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

That is a can be 4q, 4q+1, 4q+2 or 4q+3

Since a is odd, a cannot be 4q or 4q+2

Therefore any odd integer is of the form 4q+1 or 4q+3 .

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Question 8: Which term of AP   3, 15, 27, 39, \cdots will be 120 more than its 21^{st} term?

Or

If S_n , the sum of the first n terms of an AP is given by S_n = 3n^2 - 4n , find the n^{th} term.

Answer:

Given AP is 3, 15, 27, 39, \cdots

Here a = 3, d = 15-3 = 12

Therefore 21^{st} term is given by

T_{21} = a + (21-1)d = 3 + 30 \times 12 = 243

Required term = 243 + 120 = 363

Let it be the n^{th} term, then

T_n = 363

\Rightarrow a + (n-1)d = 363

\Rightarrow 3 + (n-1) \times 12 = 363

\Rightarrow 12n = 372

\Rightarrow n = 31

Hence 31^{st} term is the required term.

Or

Given S_n = 3n^2 - 4n

Put n = 1 , S_1 = 3(1)^2 - 4(1) = -1

n = 2 , S_1 = 3(2)^2 - 4(2) = 12-8 = 4

n = 3 , S_1 = 3(3)^2 - 4(3) = 27 - 12 = 15

S_1 = -1 \Rightarrow a = -1

S_2 - S_1 = 4 - ( -1) = 5

S_3 - S_2 = 15 - 4 = 11

Hence the common difference is d = 11 - 5 = 6

Therefore n^{th} term t_n = a + (n-1) d = -1 + ( n-1) 6

\Rightarrow t_n = 6n - 7

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Question 9: Find the ratio in which the segment joining the points (1, -3) and ( 4, 5)   is divided by x-axis . Also find the coordinates of this point on x-axis .

Answer:

Given Points A(1, -3) and B(4, 5)

Let the point P(x, 0) be the point which divides the segment joining A and B in the ratio m:1

The using the section formula, we get

(x, 0) = \Big( \frac{m \times 4 + 1 \times 1}{m+1} , \frac{m \times 5 + 1 \times (-3)}{m+1} \Big)

\Rightarrow 0 = \frac{m \times 5 + 1 \times (-3)}{m+1}

\Rightarrow 5m = 5

\Rightarrow m = \frac{3}{5}

Therefore P divides (1, -3) and (4, 5) in the ratio 3:5

\therefore x = \frac{m \times 4 + 1 \times 1}{m+1}

= \frac{\frac{3}{5} \times 4 + 1 \times 1}{\frac{3}{5}+1}

= \frac{12+5}{3+5} = \frac{17}{8}

Therefore the point is \Big( \frac{17}{8} , 0 \Big)

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Question 10: A game consists of tossing a coin three times and noting the outcome each time. If getting the same result in all the tosses is success, find the probability of losing the game.

Answer:

Possible outcomes of tossing a coin three times is 2 \times 2 \times 2 = 8

i.e. Total number of events = 8

The favorable events are HHH and TTT

Therefore Probability of success is \frac{2}{8} = \frac{1}{4}

Hence the probability of losing is 1 - \frac{1}{4} = \frac{3}{4}

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Question 11: A die is thrown once. Find the probability of getting a number which i) is a prime number ii) lies between 2 and 6 .

Answer:

The possible outcomes when a dice is thrown one are 1, 2, 3, 4, 5, 6

Therefore Total No. of events = 6

No of prime events = 3 ( 2, 3, 5 are prime numbers  )

No of events when the number lies between 2 and 6 = 3

Therefore

i) Probability ( is a prime number ) = \frac{3}{6} = \frac{1}{2} 

ii) Probability ( number lies between 2 and 6) = \frac{3}{6} = \frac{1}{2} 

Note: 1 only has one positive divisor (1 itself  ) . So it is not a prime number. A prime number is a positive integer whose positive divisor are exactly 1 and the number itself.

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Question 12: Find c if the system of equations cx+3y + ( 3-c) = 0 ; 12x + cy - c = 0 has infinitely many solutions.

Answer:

Given: cx + 3y + (3-c) = 0 and 12x + cy - c = 0

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}

From the given system

a_1 = c, b_1 = 3, c_1 = 3-c

a_2 = 12, b_2 = c, c_2 = -c

Therefore

\frac{c}{12} = \frac{3}{c} = \frac{3-c}{-c}

Taking the first two

c^2 = 36 \Rightarrow c = \pm 6    … … … … … (i)

Taking the last two

-3c = 3c - c^2 \Rightarrow c^2 = 6c \Rightarrow c = 0, 6    … … … … … (ii)

From (i) and (ii) we get c = 6

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Section – C

Question number 13 to 22 carry 3 mark each.

Question 13: Prove that \sqrt{2} is an irrational number.

Answer:

Assume \sqrt{2} is a rational number

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

Since \sqrt{2} = \frac{b}{a} 

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

b^2 must be even and since b^2 is even, b should be even

Let b = 2c

We have 4c^2 = 2a^2 \Rightarrow a^2 = 2c^2

Since 2c^2 is even, a^2 is even and since a^2 is even, a must be even

However two even numbers cannot be relatively prime so \sqrt{2} cannot be expresses as a rational fraction.

Hence \sqrt{2} is irrational.

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Question 14: Find the value of k such that the polynomial x^2 - ( k+6)x + 2 ( 2k-1) has sum of its zeros  equal to half of their products.

Answer:

Polynomial is x^2 - (k+6)x + 2(2k-1) = 0

Let \alpha and \beta be the zeros of the equation

On comparing with ax^2 + bx+ c = 0 we get

a = 1, b = -(k+6) c = 2(2k-1)

Therefore \alpha + \beta = - \frac{b}{a} = \frac{k+6}{1}

\Rightarrow \alpha + \beta  = k+6    … … … … … i)

\alpha . \beta = \frac{c}{a} = \frac{2(2k-1)}{1}

\Rightarrow \alpha . \beta = 4k-2    … … … … … ii)

Given \alpha + \beta  = \frac{1}{2} \alpha . \beta

\Rightarrow k+6 = \frac{1}{2} (4k-2)

\Rightarrow k+6 = 2k - 1 \Rightarrow k = 7

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Question 15: A father’s age is three times the sum of the ages of his two children. After 5 years his age will be two times the sum of their ages. Find the present age of the father.

Or

A fraction becomes \frac{1}{3} when 2 is subtracted from the numerator  and it becomes \frac{1}{2} when 1 is subtracted from the denominator. Find the fraction.

Answer:

Let the ages of sons be x and y

Given that father’s age is 3 times the ages of his two sons

Therefore present age of father = 3 (x+y)    … … … … … i)

Given 5 years hence, father’s are will be twice the sum of the ages of his sons

Father’s age after 5 years = 3(x+5) + 5

Ages of his sons 5 year hence = (x+5) and (y+5)

\therefore 3(x+y) + 5 = 2 [ (x+5) + (y+5) ]

\Rightarrow 3x + 3y + 5 = 2x + 10 + 2y + 10

\Rightarrow x + y = 15    … … … … … ii)

Using i) and ii) we get that the age of father is 3 ( 15) = 45 years.

Or

Let the fraction be \frac{x}{y} 

\therefore \frac{x-2}{y} = \frac{1}{3}

\Rightarrow 3x - 6 = y

\Rightarrow 3x - y = 6  … … … … … i)

Also \frac{x}{y-1} = \frac{1}{2}

\Rightarrow 2x = y - 1

\Rightarrow 2x - y = -1    … … … … … ii)

Subtracting ii) from i) we get x = 5

From ii) y = 2x+1 = 2(5) + 1 = 11

Hence the fraction is \frac{5}{11}

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Question 16: Find the point on y-axis which is equidistant from the point (5, -2) and (-3, 2) .

Or

The line segment joining the points A(2, 1) and B(5, -8) is trisected at points P and Q such that P is nearer to A . If P also lies on the line given by 2x-y+k = 0 , find the value of k .

Answer:

Let P(0, y) be equidistant from A(5, -2) and B(-3, 2)

\therefore AP = \sqrt{(5-0)^2 + (-2-y)^2} = \sqrt{25 + (-2-y)^2}

BP = = \sqrt{(-3-0)^2 + (2-y)^2} = \sqrt{9 + (2-y)^2}

Since AP = BP \Rightarrow AP^2 = BP^2

\therefore 25 + (-2-y)^2 = 9 + (2-y)^2

\Rightarrow 25 + 4 + y^2 + 4y = 9 + 4 + y^2 - 4y

\Rightarrow 29 + 8y = 13

\Rightarrow 8y = - 16

\Rightarrow y = -2

Hence the point is (0, -2)

Or

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

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Question 17: Prove that (\sin \theta + \mathrm{cosec} \theta)^2 + ( \cos \theta + \sin \theta)^2 = 7 + \tan^2 \theta + \cot^2 \theta .

Or

(1 + \cot A - \mathrm{cosec} A)( 1 + \tan A + \sec A) = 2

Answer:

To Prove: ( \sin \theta +  \mathrm{cosec} \theta )^2 + (\cos \theta + \sec \theta)^2 = 7 + \tan^2 \theta + \cot^2 \theta

LHS = ( \sin \theta +  \mathrm{cosec} \theta )^2 + (\cos \theta + \sec \theta)^2

= \sin^2 \theta + 2 \sin \theta  \ \mathrm{cosec} \theta + \mathrm{cosec}^2 \theta + \cos^2 \theta + \sec^2 \theta + 2 \cos \theta \ \sec \theta

= 1 + 2 + 2 +  \mathrm{cosec}^2 \theta + \sec^2 \theta

= 5 + ( 1+ \cot^2 \theta) + ( 1 + \tan^2 \theta)

= 7 + \tan^2 \theta + \cot^2 \theta

= RHS. Hence proved.

Or

To Prove: (1 + \cot A -  \mathrm{cosec} A)(1 + \tan A + \sec A) = 2

LHS = (1 + \cot A -  \mathrm{cosec} A)(1 + \tan A + \sec A)

= \Big( 1 + \frac{\cos A}{\sin A} - \frac{1}{\sin A} \Big ) \Big(1 + \frac{\sin A}{\cos A} + \frac{1}{\cos A} \Big)

= \Big( \frac{\sin A + \cos A - 1}{\sin A} \Big)  \Big( \frac{\cos A + \sin A + 1}{\cos A} \Big)

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

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

= \frac{ 2\sin A \ \cos A  }{\sin A \ \cos A} 

= 2 = RHS. Hence proved.

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Question 18: In Fig. 2, PQ is a chord of length 8 cm of a circle of radius 5 cm and center O . The tangent at P and Q intersect at point T . Find the length of TP .

2019-05-22_18-56-36
Fig. 2

 

Answer:

TP = TQ (Length of the tangents from an external point to a circle are equal)

In \triangle TPQ TP = TQ 2019-06-24_7-12-03

OT is the bisector of \angle PTQ

\therefore OT \perp PQ

Since OT \perp PQ

PR = RQ ( perpendicular from center to a chord bisects the chord)

\therefore PR = QR = \frac{1}{2} PQ = 4 cm

In Right triangle ORP

OP^2 = OR^2+ PR^2

\Rightarrow 5^2 = OR^2 + 4^2

\Rightarrow OR^2 = 25 - 16 = 9

\Rightarrow OR = 3 cm

Let TP = x

In Right triangle PRT

TP^2 = PR^2 + RT^2

\Rightarrow x^2 = 4^2 + RT^2    … … … … … i)

Since TP is a tangent, OP \perp TP

\angle OPT = 90^o

In Right triangle OPT

OT^2 = OP^2 + TP^2

\Rightarrow OT^2 = 5^2 + x^2

\Rightarrow (OR+RT)^2 = 5^2 + x^2

\Rightarrow (3+RT)^2 = 5^2 + x^2

\Rightarrow 9 + RT^2 + 6 RT = 25 + x^2

From i) x^2 = 16+RT^2

9 + RT^2 + 6 RT = 25 + 16+RT^2

\Rightarrow 6RT = 32

\Rightarrow RT = \frac{32}{6} = \frac{16}{3} 

\therefore x^2 = 16 + \Big( \frac{16}{3} \Big)^2 = \frac{400}{9} 

\Rightarrow x = \frac{20}{3} 

Hence TP is \frac{20}{3}  cm

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Question 19: In Fig. 3, \angle ACB = 90^o and CD \perp AB , prove that CD^2 = BD \times AD

2019-05-22_19-02-29
Fig. 3

 

Or

If P and Q are point of sides CA and CB respectively, of \triangle ABC , right angled at C , prove that (AQ^2+BP^2) = ( AB^2 + PQ^2)

Answer:

Given: \angle ACB = 90^o, CD \perp AB

In \triangle CAD, CA^2 = CD^2 + AD^2    … … … … … i)

In \triangle CDB, CB^2 = CD^2 + DB^2    … … … … … ii)

Adding i) and ii)

CA^2 + CB^2 = 2 CD^2 + AD^2 + DB^2

\Rightarrow AB^2 = 2 CD^2 + AD^2 + DB^2

\Rightarrow AB^2 - AD^2 = DB^2 + 2CD^2

\Rightarrow ( AB - AD)(AB+AD) - DB ^2 = 2CD^2

\Rightarrow DB(AB + AD) - DB^2 = 2 CD^2

\Rightarrow DB [ AB + AD - DB ] = 2 CD^2

\Rightarrow CD^2 = DB. AD . Hence proved.

Or

AQ^2 = AC^2 + CQ^2    … … … … … i)2019-06-24_7-15-41.png

PB^2 = PC^2 + CB^2    … … … … … ii)

Adding i) and ii)

AQ^2 + PB^2 = AC^2 + CQ^2 + PC^2 + CB^2

\Rightarrow AQ^2 + PB^2 = AB^2 + PQ^2 . Hence proved.

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Question 20: Find the area of the shaded region in Fig. 4, if ABCD is a rectangle with sides 8 cm and 6 cm and O is the center of the circle. (Take \pi = 3.14 )

2019-05-22_19-10-37
Fig. 4

Answer:

\angle ADC =90^o

(angle subtended by the diameter on the circumference is 90^o )

\therefore ABCD is a rectangle

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

Therefore the radius of the circle = 5 cm

Area of the circle = \pi (5)^2 = 25\pi

Area of ABCD = 6 \times 8 = 48 \ cm^2

Hence the shaded area = 25 \pi - 48 = 30.5 \ cm^2

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Question 21: Water in a canal, 6 m wide and 1.5 m deep, is flowing at the speed of 10 km/hour. How much area will it irrigate in 30 minutes; if $latex 8 cm standing water is needed?

Answer:

Speed of water = \frac{10 \ km}{hr} = \frac{10 \times 1000 \ m}{60 \ min} = \frac{1000 \ m}{6 \ min} 

Rate of flow = 6 \times 1.5 \ m^2 \times \frac{1000 \ m}{6 \ min} = \frac{1500 \ m^3}{min} 

Therefore volume of water in 30 mins = 1500 \frac{m^3}{min} \times 30 \ min = 45000 \ m^3

Let Area of irrigation = x

\therefore x \times \frac{8}{100} \ m = 45000 m^3

x = \frac{45000 \times 100}{8} \ m^2 = 562500 \ m^2

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Question 22: Find the mode of the following frequency distribution

Class 0-10 10-20 20-30 30-40 40-50 50-60 60-70
Frequency 8 10 10 16 12 6 7

Answer:

2019-06-23_19-51-23

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Section – D

Question number 23 to 30 carry 4 mark each.

Question 23: Two water taps can together can fill in a tank in 1 \frac{7}{8} hours. The tap with longer diameter  takes 2 hours less than the tap with the smaller one to fill the tank separately. Find the time in which each tap can fill the tank separately.

Or

A boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours it can go 40 km upstream and 55 km downstream. Determine the speed of the stream and that of the boat in still water.

Answer:

Let the time taken by the smaller tap to fill up the tank completely = x hours

So the volume of tank filled by the smaller tap in 1 hours = \frac{1}{x}

Also it is given that the time taken by the larger tap is 2 hours less

Time taken by larger tap to fill the tank completely = x - 2

So the volume of tank filled by the larger tap in 1 hours = \frac{1}{x-2}

Given that both the taps fill the tank in 1 \frac{7}{8} hours = \frac{15}{8} hours

Tank filled by smaller tap in \frac{15}{8} hours = \frac{15}{8x}

Similarly, tank filled by larger tap in \frac{15}{8} hours = \frac{15}{8(x-2)}

Therefore \frac{15}{8x} + \frac{15}{8(x-2)} = 1

15[8(x-2)] + 15[8x] = 8x[8(x-2)]

120x - 240 + 120 x = 64 x^2 -16 x

240 x - 240 = 64x^2-16x

64x^2 - 256x + 240 = 0

4x^2 - 16x + 15 = 0

x = \frac{16 \pm \sqrt{16^2 - 4 \times 4 \times 15}}{8} = \frac{16 \pm 4}{8} 

\therefore x = 2.5 hours or 1.5 hours

x cannot be 1.5 hours as the larger tap takes 2 hours less.

Hence x = 2.5 hours

Therefore small tap will fill the tank in 2.5 hours and the larger tap will fill the tank in 0.5 hours.

Or

Let the speed of the boat in still water = x km/hr

Let the speed of the stream = y km/hr

Speed of the boat downstream = (x+y) km/hr

Sped of the boat upstream = ( x- y) km/hr

Boat goes 30 km upstream and 44 km downstream in 10 hours

\Rightarrow \frac{30}{x-y} + \frac{44}{x+y} = 10     … … … … … … i)

Boat goes 40 km upstream and 55 km downstream in 13 hours

\Rightarrow \frac{40}{x-y} + \frac{55}{x+y} = 13     … … … … … … ii)

Let \frac{1}{x-y} = u and \frac{1}{x+y} = v

Substituting in i) and ii)

30u + 44v = 10     … … … … … … iii)

40u + 55v = 13   … … … … … … iv)

From iii) we get u = \frac{10-44v}{30}

Substituting in iv) we get

40( \frac{10-44v}{30} ) + 55v = 13

\Rightarrow 40-176v+ 165v = 39  \Rightarrow 11v = 1  \Rightarrow v = \frac{1}{11}

Substituting in iii) we get

30 u + 44 ( \frac{1}{11} ) = 10  \Rightarrow 30u + 4 = 10  \Rightarrow u = \frac{6}{30} = \frac{1}{5}

Now solving for x and y

\frac{1}{x-y} = \frac{1}{5}  \Rightarrow x - y = 5     … … … … … … v)

\frac{1}{x+y} = \frac{1}{11}  \Rightarrow x + y = 11     … … … … … … vi)

Adding v) and vi) we get 2x = 16 \Rightarrow x = 8 km/hr

From vi) y = 11-8 = 3 km/hr

Therefore speed o boat in still water is 8 km/hr and speed of stream is 3 km/hr

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Question 24: If the sum of the first four terms of an AP is 40 and that of the first 14 terms is 280 . Find the sum of the first n terms.

Answer:

S_n = sum of n terms of an A.P. = \frac{n}{2} [ 2a + (n-1)d]

S_4 \Rightarrow 40 = \frac{4}{2} [2a + 3d]  =  2a + 3d = 20    … … … … … i)

S_{14} \Rightarrow 280 = \frac{14}{2} [2a + 13d]  = 2a + 13d = 40    … … … … … ii)

solving i) and ii)

gives a = 7     and d= 2

so S_n = \frac{n}{2} [ 2(7) + (n-1)2]  = 7n + n^2 - n = n^2 + 6n

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Question 25: Prove that \frac{\sin A - \cos A + 1}{\sin A + \cos A + 1} = \frac{1}{\sec A - \tan A}

Answer:

LHS = \frac{\sin A - \cos A + 1}{\sin A + \cos A - 1}

= \frac{\sin A - \cos A + 1}{\sin A + \cos A - 1} \times \frac{\sin A + \cos A + 1}{\sin A + \cos A + 1} 

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

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

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

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

= \frac{\sin A + 1}{\cos A} 

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

= \tan A + \sec A

= (\sec A + \tan A) \Big( \frac{\sec A - \tan A}{\sec A - \tan A}  \Big)

= \frac{\sec^2 - \tan^2 A}{\sec A - \tan A} 

= \frac{1}{\sec A - \tan A} 

= RHS. Hence proved.

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Question 26: A man in a boat is rowing away from the light house at 100 m high takes 2 minutes to change the angle of elevation of the top of the light house from 60^o to 30^o . Find the speed of the boat in meters per minute. [ Use \sqrt{3} = 1.732 ]

Or

Two poles of equal height are standing opposite each other on either side of the road, which is 80 m wide. From a point between them on the road, the angle of elevation of the top of the poles are 60^o to 30^o respectively. Find the height of the poles and the distance of the point from the poles.

Answer:2019-06-24_7-20-22.png

\tan 60^o = \frac{AB}{BD} = \sqrt{3}

\therefore BD = \frac{AB}{\sqrt{3}} = \frac{100}{\sqrt{3}} 

\tan 30^o = \frac{AB}{CB} = \frac{1}{\sqrt{3}}

\therefore CB = \sqrt{3} AB = \sqrt{3} (100)

\therefore  CD = CB - DB = 100\sqrt{3} - \frac{100}{\sqrt{3}} = \frac{200}{\sqrt{3}}

Therefore Speed = \frac{distance}{time} = \frac{\frac{200}{\sqrt{3}}}{2} = \frac{100}{\sqrt{3}}  m/min

Or

\tan 60^o = \frac{h}{EB} = \sqrt{3} \Rightarrow EB = \frac{h}{\sqrt{3}}

\tan 30^o = \frac{h}{CE} = \frac{1}{\sqrt{3}} \Rightarrow CE = \sqrt{3} h 2019-06-24_7-25-27.png

Given CE + EB = 80

h\sqrt{3} + \frac{h}{\sqrt{3}} = 80

\Rightarrow h = \frac{80 \sqrt{3}}{3 + 1} = 20 \sqrt{3} m

\therefore CE = \sqrt{3} (20 \sqrt{3}) = 60 m

Hence EB = \frac{20 \sqrt{3}}{\sqrt{3}} = 20 m

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Question 27: Construct a \triangle ABC in which CA is 6 cm, AB = 5 cm and \angle BAC = 45^o . The construct a triangle whose sides are \frac{3}{5} of the corresponding sides of \triangle ABC .

Answer:

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Question 28: A bucket open at the top is in the form of a frustum of a cone with a capacity 12308.08 \ cm^3 . The radii of the top and bottom of the circular ends of the bucket and 20 cm and 12 cm respectively. Find the height of the bucket, and also the area of the metal sheet  used in making it. ( Use \pi = 3.14 )

Answer:2019-06-24_7-29-49.png

Volume = 12308.80 \ cm^3

r_1 = 20 cm  r_2 = 12 cm  h = ?

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

\Rightarrow 12308.80 = 3.14 \times \frac{h}{3} ( 20^2 + 12^2 + 20 \times 12)

\Rightarrow h = 12308.80 \times \frac{3}{3.14 \times 784} = 15 cm

Therefore height of bucket = 15 cm

l = \sqrt{h^2 + (r_1 - r_2)^2} = \sqrt{15^2 + 8^2} = 17 cm

Therefore surface are of the metal sheet used

= curved surface area + base area

= \pi ( r_1 + r_2) l + \pi r^2

= 3.14 \times 32 \times 17 + 3.14 \times 12^2 = 2160.32 \ cm^2

Hence the height of the bucket is 15 cm and surface are of the metal sheet is 2160.32 \ cm^2

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Question 29: Prove that in a right angle triangle, the square of the hypotenuse is equal to the sum of square of the other two sides.

Answer:

Given: A right angled \triangle ABC , right angled at B

To prove: AC^2 = AB^2 + BC^2 2019-06-24_7-33-29.png

Draw: BD \perp AC

Proof: In \triangle ADB \& \angle ABC

\angle ADB = \angle ABC = 90^o

\angle BAD = \angle CAB (common angle)

\therefore \triangle ADB \sim \triangle ABC ( by AA criterion)

Therefore \frac{AD}{AB} = \frac{AB}{AC}  (corresponding sides are proportional)

AB^2 = AD \times AC    … … … … … i)

Similarly, \triangle BDC \sim \triangle ABC

and BC^2 = CD \times AC    … … … … … ii)

Adding i) and ii)

AB^2 + BC^2 = AD \times AC + CD \times AC

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

\Rightarrow AB^2 + BC^2 = AC^2 . Hence proved.

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Question 30: If the median of the following frequency distribution is 32.5 , find the values of f_1 and f_2 .

Class 0-10 10-20 20-30 30-40 40-50 50-60 60-70 Total
Frequency f_1 5 9 12 f_2 3 2 40

Or

The marks obtained by 100 students in a class in an examination are given below.

Marks No. of Students
0-5 2
5-10 5
10-15 6
15-20 8
20-25 10
25-30 25
30-35 20
35-40 18
40-45 4
45-50 2

Answer:

Class Interval Frequency Cumulative Frequency
0-10 f_1 f_1
10-20 5 f_1 + 5
20-30 9 f_1+ 14
30-40 12 f_1 + 26
40-50 f_2 f_1+f_2+ 26
50-60 3 f_1+f_2+ 29
60-70 2 f_1+f_2+ 31
N=40

Given, Median = 32.5

The median class = 30-40

L = 30, h = 40-30 = 10, f= 12, F = 14 + f_1

Median = l + \frac{\frac{N}{2} - F}{f} \times h

\Rightarrow 32.5 = 30 + \frac{20-(14+f_1)}{12} \times 10

\Rightarrow 32.5 - 30 = \frac{20-(14+f_1)}{12} \times 10

\Rightarrow 2.5 = \frac{6 - f_1}{12} \times 10

\Rightarrow 2.5 = \frac{6 - f_1}{6} \times 5

\Rightarrow 2.5 \times 6 = (6-f_1) \times 5

\Rightarrow 3 = 6 - f_1

\Rightarrow f_1 = 3

Given sum of frequencies = 40

f_1 + f_2 + 31 = 40

f_2 + 34 = 40

f_2 = 6

Hence f_1 = 3 and f_2 = 6

Or

i)

Marks Number of Students
Less than 5 2
Less than 10 7
Less than 15 13
Less than 20 21
Less than 25 31
Less than 30 56
Less than 35 76
Less than 40 94
Less than 45 98
Less than 50 100

2019-06-23_20-03-23

ii) N = 100 , \frac{N}{2} = 50

Hence Median = 29

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