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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: What is the value of (\cos^2 67^o - \sin^2 23^o) ?

Answer:

(\cos^2 67^o - \sin^2 23^o)

= (\cos^2 67^o - \sin^2 (90^o-67^o) )

= (\cos^2 67^o - \cos^2 67^o)

= 0

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Question 2: In an AP, if the common difference (d) = -4 , and the seventh term (a_7) is 4 , then find the first term.

Answer:

Given d = -4, a_7 = 4

a_n = a + (n-1) d

\Rightarrow a_7 = a + (7-1) d

\Rightarrow 4 = a + (7-1)(-4)

\Rightarrow 4 = a - 24

\Rightarrow a = 28

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Question 3: Given \triangle ABC \sim \triangle PQR , if \frac{AB}{PQ} = \frac{1}{3} , then find \frac{ar \triangle ABC}{ar \triangle PQR}

Answer:

Given \triangle ABC \sim \triangle PQR

\Rightarrow \frac{ar(\triangle ABC)}{ar(\triangle PQR)} = \Big( \frac{AB}{PQ} \Big)^2= \Big( \frac{1}{3} \Big)^2 = \frac{1}{9}

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Question 4: What is the HCF of smallest prime number and the smallest composite number ?

Answer:

Smallest composite number is = 4 = 2 \times 2

Smallest prime number is = 2 = 2 \times 1

Therefore HCF of 4 and 2 is 2

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Question 5: Find the distance of a point P(x, y) from the origin.

Answer:

Distance of a point P(x, y) from the origin O(0, 0) = \sqrt{(x-o)^2 + (y-0)^2} = \sqrt{x^2 + y^2}

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Question 6: If x = 3 is one root of the quadratic equation x^2 - 2kx - 6 = 0 , then find the value of k .

Answer:

If x = 3 is one root of the quadratic equation x^2 - 2kx - 6 = 0 than it should satisfy the equation.

\Rightarrow (3)^2 - 2k(3) - 6 = 0

9 - 6k - 6 = 0

k = \frac{1}{2}

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Section – B

Question number 7 to 12 carry 2 mark each.

Question 7: Two different dice are tossed together. Find the probability :
(i) of getting a doublet
(ii) of getting a sum 10 , of the numbers on the two dice.

Answer:

Given that two different dice are tossed.

Total number of outcomes n(S) = 6^2 = 36

(i) Let A be the event of getting a doublet.

n(A) = \{ 1,1 \},{2,2}, \{3,3 \}, \{4,4 \}, \{5,5 \}, \{6,6 \} = 6

Required probability P(A) = \frac{n(A)}{n(S)} = \frac{6}{36} = \frac{1}{6}

(ii) Let B be the event of getting a sum of 10 of the numbers of two dice.

n(B) = \{6,4 \}, \{4,6 \}, \{5,5 \} = 3

Required probability P(B) = \frac{n(B)}{n(S)} = \frac{3}{36} = \frac{1}{12}

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Question 8: Find the ratio in which P(4, m) divides the line segment joining the points A(2, 3) and B(6, -3) . Hence find m .

Answer:

Let P(4, m) divides A (2, 3) and B (6, -3) in the ratio of k:1

By section formula,

4 = \frac{k \times 6 + 1 \times 2}{k+1} 

\Rightarrow 4k + 4 = 6k + 2

\Rightarrow k = 1

Also m = \frac{1 \times (-3) + 1 \times 3}{1+1} 

\Rightarrow m = 0

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Question 9: An integer is chosen at random between 1 and 100 . Find the probability that it is :
(i) divisible by 8 .
(ii) not divisible by 8 .

Answer:

Number of integers between 1 and 100: 2, 3 , 4, 5 , \ldots , 99

Total number of outcomes n(S) = 98

(i) Numbers which are divisible by 8 are: 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96

Favorable Outcomes n(A) = 12

Probability that integer is divisible by 8, P(E) = \frac{n(A)}{n(S)} = \frac{12}{98} = \frac{6}{49}

(ii) Probability that integer is not divisible by 8

P(E') = 1 - P(E)

= 1 - \frac{6}{49}

= \frac{43}{49}

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Question 10: In Fig. 1, ABCD is a rectangle. Find the values of x and y .

2019-05-25_10-41-18
Fig. 1

Answer:

Given:

x+y = 30    … … … … … i)

x-y = 14    … … … … … ii)

Adding i) and ii) we get

2x = 44 \Rightarrow x = 22

\therefore y = 30 - 22 = 8

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Question 11: Find the sum of first 8 multiples of 3 .

Answer:

The first 8 multiples of 3 are : 3, 6, 9, 12, 15, 18, 21, 24

Number of terms (n) = 8

The first term (a) = 3

Common difference (d) = 3

S_n = \frac{n}{2} [2a + (n-1)d ]

S_8 = \frac{8}{2} [2 \times 3 + (8-1)\times 3]

S_8 = 108

Hence the sum of the first 8 multiples of 3 is 108 .

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Question 12: Given that \sqrt{2} is irrational, prove that (5 + 3 \sqrt{2} ) is an irrational number.

Answer:

Given that, \sqrt{2} is irrational

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

As 5 + 3\sqrt{2} is rational. (Assumed) They must be in the form of \frac{p}{q} where q \neq 0 , and  p and  q are co prime.

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

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

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

We know \sqrt{2} is irrational but \frac{p-5q}{3q} is a rational number

Therefore we contradict the statement that, 5+3\sqrt{2} is rational.

Hence proved that 5 + 3\sqrt{2} is irrational.

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Section – C

Question number 13 to 22 carry 3 mark each.

Question 13: If A(-2, 1), B(a, 0), C(4, b) and D(1, 2) are the vertices of a parallelogram ABCD , find the values of a and b. Hence find the lengths of its sides.

Or

If A(-5, 7), B(-4, -5), C(-1, -6) and D(4, 5) are the vertices of a quadrilateral, find the area of the quadrilateral ABCD .

Answer:

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

Mid point of BD = \Big( \frac{a+1}{2} , \frac{0+2}{2} \Big) = \Big(  \frac{a+1}{2} , 1 \Big) 2019-07-13_11-32-00

Since the mid point of AB and BD are the same

\frac{a+1}{2} =1 \Rightarrow a + 1 = 2 \Rightarrow a = 1

Similarly, \frac{1+b}{2} =1 \Rightarrow 1 + b = 2 \Rightarrow b = 1

\therefore a =1  , b = 1

\therefore AB = \sqrt{(1-(-2))^2+(0-1)^2} = \sqrt{9+1} = \sqrt{10}

BC = \sqrt{(4-1)^2+(1-0)^2} = \sqrt{9+1} = \sqrt{10}

Therefore DC = \sqrt{10} and DA = \sqrt{10}

Hence AB = BC = CD = DA \Rightarrow ABCD is a Rhombus

Or

Join AC 2019-07-13_11-29-15

\therefore Ar (Quadrilateral ABCD) = Ar (\triangle ABC) + Ar (\triangle ACD)

Note: For a \triangle with vertices (x_1, y_1), (x_2, y_2) \ \ \& \ \ (x_3, y_3) the area of the triangle is \Big| \frac{1}{2} [ x_1(y_2-y_3) + x_2(y_3-y_1)  + x_3(y_1-y_2) ] \Big|

Ar(\triangle ABC) = \Big| \frac{1}{2} [ -5(-5-(-6)) + (-4)(-6-7) + (-1)(7-9-5)) ] \Big|

= \Big| \frac{1}{2} [ -5(1) - 4(-13) + (-1)(12) ] \Big| 

= \Big|  \frac{1}{2} ( -5+ 52-12 ) \Big| 

= \Big|  \frac{1}{2} \times 35 \Big| 

= 17.5 sq. units

Ar(\triangle ACD) = \Big|  \frac{1}{2} [ -5(5-(-6)) + 4(-6-7) + (-1)(7-5)) ] \Big| 

= \Big|  \frac{1}{2} [ -5(11) + 4(-13) + (-1)(2) ] \Big| 

= \Big|  \frac{1}{2} ( -55- 52-2 ) \Big| 

= \Big|  \frac{1}{2} \times (-109) \Big| 

= 54.5 sq. units

\therefore Ar (Quadrilateral ABCD) = 17.5 + 54.5 = 77 sq. units

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Question 14: Find all zeroes of the polynomial (2x^4 - 9x^3 + 5x^2 + 3x - 1) if two of its zeroes are (2 + \sqrt{3}) and (2 - \sqrt{3}) .

Answer:

Given (2+\sqrt{3})   and (2-\sqrt{3}) are zeros of the polynomial p(x) = 2x^4 - 9x^3+5x^2+3x-1

Therefore (x - 2-\sqrt{3})(x - 2+\sqrt{3}) = x^2 -4x+1 is a factor of the polynomial p(x) Hence

x^2 -4x+1 ) \overline{2x^4 - 9x^3+5x^2+3x-1} ( 2x^2-x-1 \\ \hspace*{1.6cm} (-) \underline{2x^4-8x^3+2x^2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \hspace*{2cm}-x^3+3x^2+3x-1 \\ \hspace*{1.5cm} (-) \underline{-x^3+4x^2-x \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \hspace*{3.5cm}-x^2+4x-1 \\ \hspace*{3cm} (-) \underline{-x^2+4x-1 \ \ \ \ \ } \\ \hspace*{5cm}\times

Therefore 2x^2-x-1 is a factor of p(x)

2x^2-x-1 = 2x^2 - 2x + x - 1

= 2x(x-1) + (x-1)

= (x-1)(2x+1)

\Rightarrow (x-1) and (2x+1) are factors of p(x)

Hence x = 1 and x = - \frac{1}{2}  are the other two zeros of the given polynomial 2x^4 - 9x^3+5x^2+3x-1

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Question 15: Find HCF and LCM of 404 and 96 and verify that HCF \times LCM = Product of the two given numbers.

Answer:

96 = 2^5 \times 3

404 = 2^2 \times 101

Therefore HCF of 96 and 404 = 2^2 = 4

Also LCM of 96 and 404 = 2^5 \times 3 \times 101 = 9696

Product of two numbers = 96 \times 404 = 38784

HCF \times LCM = 4 \times 9696 = 38784

Hence Product of two numbers = HCF \times LCM

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Question 16: Prove that the lengths of tangents drawn from an external point to a circle are equal.

Answer:

2019-07-13_12-19-35.pngGiven: Let the circle with center O

Let P be an external point from which tangents PQ and PR are drawn as shown in the diagram

To prove: PQ = PR

Construction: Join OQ, OR and OP

Proof: As PQ is tangent OQ \perp PQ . Therefore \angle OQP = 90^o

Similarly, As PR is tangent OR \perp PR . Therefore \angle ORP = 90^o

(Note: Tangents at any point on a circle is perpendicular to the radius through the point of contact)

In \triangle OQP and \triangle ORP

\angle OQP = \angle ORP = 90^o

OP is common

OQ = OR   (radius of the same circle)

\therefore \triangle OQP \cong \triangle ORP

\Rightarrow PQ = PR

Therefore both tangents are equal in length.

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Question 17:  Prove that the area of an equilateral triangle described on one side of the square is equal to half the area of the equilateral triangle described on one of its diagonal.

Or

If the area of two similar triangles are equal, prove that they are congruent.

Answer:

Let the side of the square = a 2019-07-13_11-50-11

Therefore the sides of the equilateral \triangle ABE = a as well

Now BD = \sqrt{a^2 + a^2} = \sqrt{2} a

Therefore the sides of the equilateral \triangle BDF = \sqrt{2} a

Since the triangles are equilateral, each of the angles are 60^o . Hence by AAA criterion, the two triangles are similar.

Therefore

\frac{Ar (\triangle ABE}{Ar (\triangle BDF)} = \Big( \frac{AB}{BD} \Big)^2 = \Big( \frac{a}{\sqrt{2}a} \Big)^2 = \frac{1}{2}

Hence proven.

Or

Given: \triangle ABC and \triangle DEF 2019-07-13_12-09-20.png

Also \triangle ABC \sim \triangle DEF and Ar(\triangle ABC) = Ar(\triangle DEF)

To Prove: \triangle ABC \cong \triangle DEF

Since \triangle ABC \sim \triangle DEF we know

\frac{Ar (\triangle ABC )}{Ar (\triangle DEF)} = \Big( \frac{BC}{EF} \Big)^2 = \Big( \frac{AB}{DE} \Big)^2 =\Big( \frac{AC}{DF} \Big)^2

1 = \Big( \frac{BC}{EF} \Big)^2 = \Big( \frac{AB}{DE} \Big)^2 =\Big( \frac{AC}{DF} \Big)^2

Therefore we get

1 = \Big( \frac{BC}{EF} \Big)^2 \Rightarrow \frac{BC}{EF} = 1 \Rightarrow BC = EF

1 = \Big( \frac{AB}{DE} \Big)^2 \Rightarrow \frac{AB}{DE} = 1 \Rightarrow AB = DE

1 = \Big( \frac{AC}{DF} \Big)^2 \Rightarrow \frac{AC}{DF} = 1 \Rightarrow AC = DF

Therefore \triangle ABC \cong \triangle DEF by SSS criterion.

Hence proved.

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Question 18: A plane left 30  minutes late than its scheduled time and in order to reach the destination 1500 km away in time, it had to increase its speed by 100 km/h from the usual speed. Find its usual speed.

Answer:

Distance to travel = 1500 km

Let the usual speed = x km/hr

Increased speed = (x+100) km/hr

\therefore \frac{1500}{x+100} + \frac{1}{2} = \frac{1500}{x}

\frac{3000 + x + 100}{2(x+100)} = \frac{1500}{x}

3000x + x^2 + 100x = 3000x + 300000

x^2 + 100x - 300000 = 0

x^2 - 500x + 600x - 300000 = 0

x( x- 500) + 600(x-500) = 0

(x-500)(x+600) = 0

x = 500 km/hr or x = -600 km/hr (this is not possible as speed cannot be negative)

Therefore the usual speed = 500 km/hr

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Question 19: The table below shows the salaries of 280 persons:

Salary (in thousand Rs.) No.  of persons
5-10 49
10-15 133
15-20 63
20-25 15
25-30 6
30-35 7
35-40 4
40-45 2
45-50 1

Answer:

Class Interval Frequency (f) Cumulative Frequency (cf)
5 – 10 49 49
10 – 15 133 182
15 – 20 63 245
20 – 25 15 260
25 – 30 6 266
30 – 35 7 273
35 – 40 4 277
40 – 45 2 279
45 – 50 1 280

cf = 133, f = 133, L = 10, N = 280 \Rightarrow  \frac{N}{2} = 140, C = 5

Median (M) = L + \frac{\frac{N}{2} - cf}{f} \times C = 10 + \Big( \frac{140 - 49}{133} \Big) \times 5 = 10 + 3.42 = 13.42

Therefore median salary = 13.42 \times 1000 =13420 Rs.

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Question 20: A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in Fig. 2. If the height of the cylinder is 10 cm and its base is of radius 3.5 cm. Find the total surface area of the article.

2019-05-25_10-19-00
Fig. 2

Or

A heap of rice is in the form of a cone of base diameter 24 m and height 3.5 m. Find the volume of the rice. How much canvas cloth is required to just cover the heap ?

Answer:

Radius of the cylinder (r) = 3.5 cm Height (h) = 10 cm

Curved surface area of cylinder = 2 \pi r h = 2 \pi \times 3.5 \times 10 = 70 \pi \ cm^2

Surface area of two hemispheres = 2 [ \frac{1}{2} \times 4 \pi r^2 ]  = 4\pi \times 3.5 \times 3.5 = 49 \pi \ cm^2

Therefore total surface area = 70 \pi + 49 \pi = 119 \times \frac{22}{7} = 374 \ cm^2

Or

Diameter = 24 m \Rightarrow Radius (r) = 12 m Height (h) = 3.5 m

Volume of Rice = \frac{1}{3} \pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times 12 \times 12 \times 3.5 = 22 \times 2 \times 12 = 528 \ m^3

Slant height (l) = \sqrt{12^2 + 3.5^2} = \sqrt{156.25} = 12.5 m

Curved surface area of the heap = \pi r l = \frac{22}{7} \times 12 \times 12.5 = 471.43 \ m^2

Hence Volume of Rice = 528 \ m^3 and Canvas required to cover the heap = 471.43 \ m^2

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Question 21: Find the area of the shaded region in Fig. 3, where arcs drawn with centres A, B, C and D intersect in pairs at mid-points P, Q, R and S of the sides AB, BC, CD  and DA  respectively of a square ABCD of side 12 cm. [Use \pi = 3.14 ]

2019-05-25_10-33-48
Fig. 3

Answer:

Given square ABCD of side 12 cm

Area of square = 12 \times 12 = 144 \ cm^2

Area of APS = \frac{1}{4} (\pi \times 6 \times 6) = 9\pi \ cm^2

Therefore total area enclosed in arc’s = 4 \times 9\pi = 36 \pi = 36 \times 3.14 = 113.04 \ cm^2

Therefore shaded area = 144 - 113.04 = 30.96 \ cm^2

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Question 22: If 4 \tan \theta = 3 , evaluate \frac{4 \sin \theta - \cos \theta + 1}{4 \sin \theta + \cos \theta - 1}

Or

If \tan 2A = \cot (A - 18^o) , where 2A is an acute angle, find the value of A .

Answer:

Given 4 \tan \theta = 3 \Rightarrow \tan \theta = \frac{3}{4}  \Rightarrow \sin \theta = \frac{3}{5}  \Rightarrow \cos \theta = \frac{4}{5}

Therefore

\frac{4 \sin \theta - \cos \theta + 1}{4 \sin \theta + \cos \theta - 1}  = \frac{4 \times \frac{3}{5} - \frac{4}{5} + 1}{4 \times \frac{3}{5} + \frac{4}{5} - 1}  = \frac{12 - 4 + 5}{12 + 4 -5}  = \frac{13}{11}

Or

\tan 2A = \cot (A - 18^o)

\Rightarrow \tan 2A = \tan [ 90- (A - 18^o) ]

\Rightarrow \tan 2A = \tan ( 108^o - A)

\Rightarrow 2A = 108^o - A

\Rightarrow A = 36^o

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Section – D

Question number 23 to 30 carry 4 mark each.

Question 23: As observed from the top of a 100 m high light house from the sea-level, the angles of depression of two ships are 30^o and 45^o . If one ship is exactly behind the other on the same side of the light house, find the distance between the two ships. [Use \sqrt{3} = 1.732 ]

Answer:

2019-07-14_12-13-11In \triangle ABC:

\frac{100}{BC} =\tan 45^o = 1 \Rightarrow BC = 100 m

In \triangle ABD:

\frac{100}{BD} =\tan 30^o = 1 \Rightarrow BC = 100 \sqrt{3} m

\therefore DC = 100 \sqrt{3} -100 = 173.2 - 100 = 73.2 m

Therefore the distance between the ships is 73.2 m.

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Question 24: The diameters of the lower and upper ends of a bucket in the form of a frustum of a cone are 10 cm and 30 cm respectively. If its height is 24 cm, find :
(i) The area of the metal sheet used to make the bucket.
(ii) Why we should avoid the bucket made by ordinary plastic ? [Use \pi = 3.14 ]

Answer:

2019-07-14_12-50-26.png

Diameter of upper end of bucket =30 cm

Radius of the upper end of the frustum of cone ( r_2) = 15 cm

Diameter of lower end of bucket = 10 cm

radius of the lower end of the frustum of cone ( r_1) = 5 cm

Height of the frustum of Cone (h) = 24 cm

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

= \frac{1}{3} \times 3.14 \times 24 \times ( 5^2 + 15^2 + 5 \times 15)

= 8164 \ cm^3 = 8.164 liters

l = \sqrt{h^2 + (r_2 - r_1)^2} = \sqrt{24^2 + (15-5)^2} = \sqrt{24^2 + 10^2} = 26 cm

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

= 3.14   ( 5^2 + (5+15)\times 26 ) 

= 1711.3 \ cm^2

Hence, the Area of metal sheet used to make the bucket is = 1711.3 \ cm^2

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Question 25: Prove that: \frac{\sin A - 2 \sin^3 A }{2 \cos^3 A - \cos A} =  \tan A

Answer:

LHS = \frac{\sin A - 2 \sin^3 A}{2 \cos^3 A - \cos A}

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

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

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

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

= \tan A

= RHS. Hence Proved.

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Question 26: The mean of the following distribution is 18 . Find the frequency f of the class 19 - 21 .

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

Or

The following distribution gives the daily income of 50 workers of a factory:

Daily Income (in Rs) 100-120 120-140 140-160 160-180 180-200
Number of workers 12 14 8 6 10

Convert the distribution above to a less than type cumulative frequency distribution and draw its ogive.

Answer:

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

Mean = \frac{\Sigma f_ix_i }{40 + f } = \frac{704 + 20f}{40 + f}

\Rightarrow 720 + 18f = 704 + 20f

\Rightarrow 2f = 16

\Rightarrow f = 8

Or

Daily wages in Rs (less than) Frequency (f_i) Cumulative Frequency
Less than 120 12 12
Less than 140 14 26
Less than 160 8 34
Less than 180 6 40
Less than 200 10 50

2019-07-14_12-36-43.png

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Question 27: A motor boat whose speed is 18 km/hr in still water takes 1 hr more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream.

Or

A train travels at a certain average speed for a distance of 63 km and then travels at a distance of 72 km at an average speed of 6 km/hr more than its original speed. If it takes 3 hours to complete total journey, what is the original average speed ?

Answer:

Distance traveled by the boat = 24 km

Speed of the boat = 18 km/hr

Let the speed of the stream = x km/hr

Therefore speed of boat upstream = (18 - x) km/hr

Also speed of boat downstream = (18 + x) km/hr

Hence: \frac{24}{18-x} - \frac{24}{18+x} = 1

\Rightarrow 432 + 24 x  - (432 - 24x) = 324 - x^2

\Rightarrow 48 x = 324 - x^2

\Rightarrow x^2 + 48x - 324 = 0

\Rightarrow x^2 + 54x - 6x - 324 = 0

\Rightarrow x(x+54)-6(x+54) = 0

\Rightarrow (x+54)(x-6) = 0

\Rightarrow x = 6 km/hr or x = -54 km/hr (not possible as speed cannot be negative)

Therefore the speed of the stream is 6 km/hr

Or

Let the original speed is x km/hr

Time taken to travel 63 km = \frac{63}{x}  hrs

New speed = (x+6) km/hr

Time taken to travel 72 km  = \frac{72}{x+6}

\therefore \frac{63}{x} + \frac{72}{x+6} = 3

\Rightarrow 63x + 378 + 72x = 3x^2 + 18x

\Rightarrow 3x^2 - 117 x - 378 = 0

\Rightarrow x^2 - 39x - 126 = 0

\Rightarrow x^2 - 42 x + 3x - 126 =0

\Rightarrow x(x-42)+3(x-42) = 0

\Rightarrow (x-42)(x+3) = 0

\Rightarrow x = 42 km/hr or x = -3 km/hr (not possible as speed cannot be negative)

Hence original speed = 42 km/hr

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Question 28: The sum of four consecutive numbers in an AP is 32 and the ratio of the product of the first and the last term to the product of two middle terms is 7 : 15 . Find the numbers.

Answer:

Let the four consecutive terms of the AP be a, a+d, a+2d and a+3d

Given: a + ( a+d) + ( a+2d) + ( a+3d) = 32

\Rightarrow 4a + 6d = 32

\Rightarrow 2a + 3d = 16

\Rightarrow d = \frac{16-2a}{3}    … … … … … i)

Also \frac{a(a+3d)}{(a+d)(a+2d)} = \frac{7}{15}

\Rightarrow 15a^2 + 45ad = 7a^2 + 21 ad + 14d^2

\Rightarrow 8a^2 + 24ad - 14 d^2 = 0

\Rightarrow 4a^2 + 12 ad - 7d^2 = 0

\Rightarrow 4a^2 + 14ad - 2ad - 7d^2 = 0

\Rightarrow 2a(2a-d) + 7d(2a-d) = 0

(2a-d)(2a+7d) = 0

Substituting from i)

\Big(2a- \frac{16-2a}{3} \Big)\Big(2a+7\times \frac{16-2a}{3} \Big) = 0

\Rightarrow (6a - 16 + 2a) (6a + 112 - 14a) = 0

\Rightarrow (8a-16)(-8a+112) = 0

\Rightarrow a = 2 or a = 14

When a = 2, d = \frac{16-2 \times 2}{3} = \frac{12}{3} = 4

The the first four terms are 2, 6, 10, 14

When a = 14, d = \frac{16-2 \times 14}{3} = \frac{-12}{3} = -4

The the first four terms are 14, 10, 6, 2

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Question 29: Draw a triangle ABC with BC = 6 cm, AB = 5 cm and \angle ABC = 60^o . Then construct a triangle whose sides are \frac{3}{4} of the corresponding sides of the \triangle ABC .

Answer:

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Question 30: In an equilateral \triangle ABC , D is a point on side BC such that BD = \frac{1}{3} BC . Prove that: 9(AD)^2 = 7(AB)^2

Or

Prove that, in a right triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides.

Answer:

Given: \triangle ABC is an equilateral triangle.2019-07-14_10-35-07.png

\Rightarrow AB = BC = CA

Also BD = \frac{1}{3} BC

To prove: 9 AD^2 = 7 AB^2

Construction: Draw AE \perp BC

Consider \triangle AEB and \triangle AEC

AE is common

\angle AEB = \angle AEC = 90^o

AB = AC (equilateral triangle)

\triangle AEB \cong \triangle AEC (By RHS criterion)

\therefore BE = EC

\Rightarrow BE = \frac{BC}{2}

\Rightarrow BD + DE = \frac{BC}{2}

\Rightarrow \frac{BC}{3} + DE = \frac{BC}{2}

\Rightarrow DE = \frac{BD}{6}

Using Pythagoras theorem,

In \triangle AEB

AB^2 = AE^2 + BE^2 \Rightarrow AE^2 = AB^2 - BE^2    … … … … … i)

In \triangle AED

AD^2 = AE^2 + DE^2 \Rightarrow AE^2 = AD^2 - DE^2    … … … … … ii)

From i) and ii)

AB^2 - BE^2 = AD^2 - DE^2

\Rightarrow AB^2 - \Big( \frac{BC}{2} \Big)^2 = AD^2 - \Big( \frac{BC}{6} \Big)^2

\Rightarrow AB^2 - \frac{BC^2}{4} = AD^2 - \frac{BC^2}{36}

\Rightarrow AB^2 = AD^2 + \frac{8}{36} BC^2

AB^2 = AD^2 + \frac{2}{9} BC^2

But BC = AB

\therefore AB^2 - \frac{2}{9} AB^2 = AD^2

\Rightarrow 7AB^2 = 9AD^2

Hence proved.

Or

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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