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Universal Tutorials – X CBSE

Question Paper Universal Tutorials – X CBSE (2018–19) – Maths of 3

Full Test 01 (Set 1): Maths Date: ___________

Recommended Time: 3 hours Max. Marks: 80

General Instructions:

1) All questions are compulsory.

2) The question paper consists of 30 questions divided into four sections A, B, C and D.

3) Section A comprises 6 questions of 1 mark each.

Section B comprises 6 questions of 2 marks each.

Section C comprises 10 questions of 3 marks each.

Section D comprises 8 questions of 4 marks each.

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

5) Use of calculators is not permitted.

Section A: (Question numbers 1 to 6 carry 1 mark each)

1) On applying Euclid’s division lemma to a positive integer a and b = 4. There exist integers q and r such that a = 4q + r. Write the condition which ‘r’ must satisfy. [1]

2) If ax2 + bx + c = 0 has equal roots, what is the value of c? [1]

OR Find the values of k for which the quadratic equation 9x2 – 3kx + k = 0 has equal roots.

3) The nth term of a progression is 3n + 1. Find the 6th term. [1]

4) In fig. if DE || BC, find the value of x. [1]

5) Find the co–ordinates of a point A, where AB is the diameter of a circle whose centre is (2, –3) and B is (1, 4). [1]

6) The ratio of the length of a rod and its shadow is 1 : . Find the angle of elevation of the

sun. [1]

OR If 2 sin = 1, find the value of sec2 – cosec2 .

Section B: (Question numbers 7 to 12 carry 2 marks each)

7) Express 2717 as a product of its prime factors. [2]

OR Determine the value of p and q so that the prime factorization of 2520 is expressible as

23 3P q 7.

8) On comparing the ratios 2

1

a

a,

2

1

b

b and

2

1

c

c, find out whether the following pair of linear

equations are consistent, or inconsistent. 2

3x +

3

5y = 7; 9x – 10y = 14. [2]

9) Is –150 a term of the A.P: 11, 8, 5, 2, …… justify your answer. [2]

OR If 5

4, a, 2 are three consecutive terms of an A.P., then find the value of a.

10) If P(–1, 3), Q(1, –1) and R(5, 1) are the vertices of a triangle PQR, then find the length of the median through the vertex P. [2]

11) Two dice are thrown simultaneously. Find the probability of getting: [2]

a) A total of atleast 10 b) A doublet of even number

12) All kings, queens and aces are removed from a pack of 52 cards. The remaining cards are well shuffled and then a card is drawn from it. Find the probability that the drawn card is:

3

4 cm E

A

B C

D

3 cm

2 cm

x

E– 2


i) a black face card ii) a red card [2]

Section C: (Question numbers 13 to 22 carry 3 marks each)

13) Use Euclid’s Division Lemma to show that the cube of any positive integer is either of the form 9m, 9m + 1 or 9m + 8. [3]

14) Find the zeroes of the quadratic polynomial 6x2 – 3 – 7x and verify the relationship between the zeroes and the coefficients of the polynomial. [3]

15) Draw the graphs of the equations x = 3, x = 5 and 2x – y – 4 = 0. Also find the area of the quadrilateral formed by the lines and the x-axis. [3]

16) ABCD is a trapezium in which AB || CD.

he diagonals AC and BD intersect at O. Prove that [3]

i) AOB ~ COD

ii) If OA = 6 cm, OC = 8 cm. Find:COD)( Area

AOB)( Area

OR In Fig. ABC and DBC are on the same base BC.

If AD intersects BC at O, prove that )DBC(Ar

)ABC(Ar

=

DO

AO.

17) The point P(7, b) lies on the line joining A(–5, 2) and B(3, 6). Find the ratio and

hence find b. [3]

OR If A and B are (–2, –2) and (2, –4) respectively, find the coordinate of a point P on line

segment AB such that AP = AB.

18) From a point P, two tangents PA and PB are drawn to a circle with centre O.

If OP = diameter of the circle show that APB is equilateral. [3]

19) If sec + tan = p, show that sec – tan = p

1. Hence, find the values of cos and sin .

[3]

OR Evaluate the following:

75? tan . 60? tan 15? tan

52?cosec 38? cos3

32? sin

58? cos2 .

20) If a wire is bent into the shape of a square, then the area of the square is 81 cm2. When the same wire is bent into a semicircular shape, find the area of the semicircle. [3]

21) A wooden article was made by scooping out of a hemisphere from

each side of a solid cylinder, as shown in fig. If the height of the

cylinder is 10cm, and its base is of radius 3.5cm, find the total surface

area of the article. [3]

OR The radii of the circular ends of a frustum are 33 cm and 27 cm, its slant height is 10 cm. Find its volume and whole surface area.

)(

)(

PBl

APl

7

3

O

A B

C D

E– 3


22) The following table shows the marks obtained by 100 students of Class X in a school during a particular academic session. Find the mode of this distribution. [3]

Marks No. of Students

Less than 10 7

Less than 20 21

Less than 30 34

Less than 40 46

Less than 50 66

Less than 60 77

Less than 70 92

Less than 80 100

Section D: (Question numbers 23 to 30 carry 4 marks each)

23) Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an

angle of 60. Justify. [4]

24) If the roots of the quadratic equation (a – b)x2 + (b – c)x + (c – a) = 0, a b are equal, then prove that b + c = 2a. [4]

OR Solve for x :

1x

3x7

3x

1x2 = 5; given that x –3, x 1.

25) Two cars start together in the same direction at the same place. The first goes at uniform speed of 10 km/h. The second goes at a speed of 8 km/h in the first hour and increases

the speed by km each succeeding hour. After how many hours will the second car

overtake the first car if both cars go nonstop? [4]

26) A man in a boat rowing away from a light house 100m high takes 3 minutes to change the

angle of elevation of the top of light house from 60 to 45. Find the speed of the boat in

m/min. ( = 1.73) [4]

27) The ratio of the areas of two similar triangles is equal to the square ratio of their corresponding sides. Prove. [4]

OR) In figure, XY || AC and XY divides triangular region

ABC into two parts of equal areas. Determine AB

AX.

28) Prove that = 1 + tan + cot . [4]

29) A juice seller serves his customers using a glass as shown in Figure.

The inner diameter of the cylindrical glass is 5 cm, but the bottom of the

glass has a hemispherical portion raised which reduces the capacity of

the glass. If the height of the glass is 10 cm, find the apparent capacity

of the glass and its actual capacity. What values does the juice seller

depict by using such a glass? (Use = ) [4]

30) Draw the more than cumulative frequency curve of the following data: [4]

Rent (Rs.) 15-25 25-35 35-45 45-55 55-65 65-75 75-85 85-95

No. of houses 5 10 20 35 30 20 13 7

OR) If the median of the distribution given below is 28.5, find the values of x and y.

Class interval 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 Total

Frequency 5 x 20 15 y 5 60

2

1

3

tan1

cot

cot1

tan

7

22

A

C Y B

X

E–

Universal Tutorials – X CBSE

Model Answer (1st page) Universal Tutorials – X CBSE (2018–19) – Maths of 17

(1st few pages form part of question paper)

Full Test 01 (Set 1): Maths (Model Answer) Date: ___________

Recommended Time: 3 hours Max. Marks: 80

Section A: (Question numbers 1 to 6 carry 1 mark each)

1) On applying Euclid’s division lemma to a positive integer a and b = 4. There exist integers q and r such that a = 4q + r. Write the condition which ‘r’ must satisfy. [1]

Ans: 0 r < 4 [1]

2) If ax2 + bx + c = 0 has equal roots, what is the value of c? [1]

Ans: Since the quadratic equation has equal roots, then

b2 – 4ac = 0 4ac = b2 c = a4

b2

. [1]

OR Find the values of k for which the quadratic equation 9x2 – 3kx + k = 0 has equal roots.

Sol: a = 9, b = – 3k, c = k

Since roots of the equation are equal

b2 – 4ac = 0

(–3k)2 – (4 9 k) = 0

9k2 – 36k = 0

k2 – 4k = 0

k(k – 4) = 0

k = 0 or k = 4

Since k = 0 is not possible for the equation

k = 4 [1]

3) The nth term of a progression is 3n + 1. Find the 6th term. [1]

Sol: an = 3n + 1 To find: 6th term, put n = 6

a6 = 3 6 + 1 = 19 a6 = 19. [1]

4) In fig. if DE || BC, find the value of x. [1]

Sol: ADE ABC [AA similarity]

BC

DE

AB

AD

x

4

5

2

2x = 20 x = 10 cm [1]

5) Find the co–ordinates of a point A, where AB is the diameter of a circle whose centre is (2, –3) and B is (1, 4). [1]

Sol: Let A be (x1, y1) Centre is the midpoint of AB

2 = and –3 =

i.e. x1 + 1 = 4 x1 = 3 y1 + 4 = –6 y1 = –10

A is (3, –10) [1]

6) The ratio of the length of a rod and its shadow is 1 : . Find the angle of elevation of the

sun. [1]

Sol: tan = = 30 Angle of elevation of sun is 30. [1]

OR If 2 sin = 1, find the value of sec2 – cosec2 .

Sol: Given, 2 sin = 1

sin = 2

1 = sin 45º

= 45º [½]

Now sec2 – cosec2 = sec2 45º – cosec2 45º [½]

2

1x1

2

41 y

3

3

1

4 cm E

A

B C

D

3 cm

2 cm

x

E– 5

Model Answer Universal Tutorials – X CBSE (2018–19) – Maths of 17

= ( 2 )2 – ( 2 )2

= 2 – 2

= 0

Section B: (Question numbers 7 to 12 carry 2 marks each)

7) Express 2717 as a product of its prime factors. [2]

Ans:

[1]

2717 = 11 13 19 [1]

OR Determine the value of p and q so that the prime factorization of 2520 is expressible as

23 3P q 7.

Sol: Prime factorization of 2520 is given by.

2520 = 23 32 5 7 [1]

Given that 2520 = 23 3P q 7,

On comparing both the factorization.

We get, p = 2 and q = 5 [1]

8) On comparing the ratios 2

1

a

a,

2

1

b

b and

2

1

c

c, find out whether the following pair of linear

equations are consistent, or inconsistent. 2

3x +

3

5y = 7; 9x – 10y = 14. [2]

Sol: x + y = 7; 9x – 10y = 14 6

y10x9 = 7,

9x + 10y = 42, and 9x – 10y = 14

9x + 10y – 42 = 0 and 9x – 10y – 14 = 0.

a1 = 9, b1 = 10, c1 = –42 a2 = 9, b2 = –10, c2 = –14. [1]

2

1

a

a =

9

9 = 1,

2

1

b

b =

10

10

= –1

2

1

a

a

2

1

b

b unique solution (consistent) [1]

9) Is –150 a term of the A.P: 11, 8, 5, 2, …… justify your answer. [2]

Sol.: Here, the given A.P. is 11, 8, 5, 2, ……

Then, T2 – T1 = 8 – 11 = –3 d = – 3 Also, a = 11

Let, – 150 be a term of a given A.P. say, nth

Using the General Term Formula, we have Tn = a + (n – 1)d [½]

–150 = 11 + (n – 1) (–3) –150 = 11 – 3n + 3

–150 = –3n + 14 3n = 150 + 14 = 164 n = 3

164 [1]

Which is a rational number. So, our assumption is wrong since n should be a positive integer.

Hence, –150 is not a term of the given A.P. [½]

OR If 5

4, a, 2 are three consecutive terms of an A.P., then find the value of a.

Sol: Since 5

4, a, 2 are the three consecutive terms of an A.P., then

d = a – 5

4 = 2 – a [1]

2a = 2 + 5

4

10a = 10 + 4

10a = 14

2

3

3

5

11 2717 13 247 19

E– 6


a = 5

7 [1]

10) If P(–1, 3), Q(1, –1) and R(5, 1) are the vertices of a triangle PQR, then find the length of the median through the vertex P. [2]

Sol: P, Q, R are the vertices of a triangle PQR. PM is the median.

To find the length of median PM.

As M is the mid point of QR

x coordinate of point M = 2

xx 21 = 2

51 =

2

6 = 3

y coordinate of point M = 2

yy 21 = 2

11 =

2

0 = 0

M(x, y) = (3, 0) [1]

Hence, applying the distance formula

PM2 = (–1 –3)2 + (3 – 0)2 = (–4)2 + (3)2 = 16 + 9 = 25.

PM = 25 = 5units. [1]

11) Two dice are thrown simultaneously. Find the probability of getting: [2]

a) A total of atleast 10 b) A doublet of even number

Sol: a) A total of at least 10 = (4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6) = 6

P (a total of at least 10) = 6

1

36

6 [1]

b) A doublet of even numbers = (2, 2) (4, 4), (6, 6) = 3

P (a doublet of even numbers) = 12

1

36

3 [1]

12) All kings, queens and aces are removed from a pack of 52 cards. The remaining cards are well shuffled and then a card is drawn from it. Find the probability that the drawn card is:

i) a black face card ii) a red card [2]

Sol: i) No. of cards removed = 4 kings + 4 queens + 4 aces = 12

No. of cards left = 52 – 12 = 40 = sample space

No. of Face cards: 4 kings + 4 queens + 4 jacks = 12

But 4 kings and 4 queens are removed.

No. of face cards left = 4 jacks.

Out of 4 jacks, only 2 jacks (jacks of club and spade) are black face cards.

P (a black face card) = 20

1

40

2 [1]

ii) Out of 12 cards removed, there are 6 red and 6 black cards (2 each of kings, queens and aces)

Hence no. of red cards left = 26 – 6 = 20 [ Total no. of red cards = 26]

P (red cards) = 2

1

40

20 [1]

Section C: (Question numbers 13 to 22 carry 3 marks each)

13) Use Euclid’s Division Lemma to show that the cube of any positive integer is either of the form 9m, 9m + 1 or 9m + 8. [3]

Sol: Let a is any positive integer and b = 3

By Euclid’s division Algorithm,

a = bq + r where 0 r < b.

a = 3q + r where 0 r < 3. i.e. r = 0, 1, 2

a = 3q or 3q + 1 or 3q + 2 [1]

when a = 3q

a3 = 27q3 = 9(3q3) = 9m where m = 3q3

when a = 3q + 1

a3 = (3q + 1)3 = 27q3 + 27q2 + 9q + 1

= 9 (3q3 + 3q2 + q) + 1 = 9m + 1 where m = 3q3 + 3q2+q [1]

when a = 3q + 2

P(–1,3)

Q(1, –1) M R(5,1)

E– 7


a3 = (3q + 2)3 = 27q3+ 3.3q.2 (3q + 2) + 8 = 27q3 + 54q2 + 36q + 8

= 9(3q3 + 6q2 + 4q) + 8 = 9m + 8 where m = 3q3 + 6q2 + 4q [1]

cube of any positive integer is of the from 9m, 9m + 1 or 9m + 8

14) Find the zeroes of the quadratic polynomial 6x2 – 3 – 7x and verify the relationship between the zeroes and the coefficients of the polynomial. [3]

Sol: Let one zero be and the other zero be .

p(x) = 6x2 – 7x – 3 = 6x2 – 9x + 2x – 3 = 3x(2x – 3) + 1(2x – 3) = (3x + 1) (2x – 3)

For p(x) = 0, 3x + 1 = 0 or 2x – 3 = 0

x = 3

1 or x =

2

3 =

3

1, =

2

3 [1]

Sum of zeroes = + = 6

7

2

3

3

1

… (1)

Here a = 6, b = –7, c = –3

Also, sum of zeroes = 6

)7(

a

b=

6

7 … (2) (1) = (2) [½]

Now, product of zeroes = = 2

1

2

3

3

1 … (3)

Also, Product of zeroes = 2

1

6

3

a

c

… (4) [1]

(3) = (4) Hence verified. [½]

15) Draw the graphs of the equations x = 3, x = 5 and 2x – y – 4 = 0. Also find the area of the quadrilateral formed by the lines and the x-axis. [3]

Sol: 2x – y = 4

x 2 3 5

y 0 2 6

Quadrilateral is like trapezium whose parallel sides

are 2 units and 6 units. Distance between parallel

sides is 2 units. So, area of trapezium

= 2

1 (Sum of parallel sides)

(Distance between parallel sides) [Graph 2]

= 2

1 (2 + 6) 2 = 8 sq. units [1]

16) ABCD is a trapezium in which AB || CD.

he diagonals AC and BD intersect at O. Prove that [3]

i) AOB ~ COD

ii) If OA = 6 cm, OC = 8 cm. Find:COD)( Area

AOB)( Area

Sol: ABCD is a trapezium AB || CD

Diagonals AC and BD intersect at O.

OA = 6 cm, OC = 8 cm.

i) In ΔAOB and ΔCOD.

AOB = COD [V.O.A] [fig ½]

OAB = OCD [Alt. angles]

Δ AOB ∼ ΔCOD [By AA similarity] [1]

ii) ΔAOB ∼ ΔCOD

COD)( Area

AOB)( Area

=

16

9

64

36

8

6

OC

OA

OC

OA22

2

2

[1½]

x

y

–1

1 2 3 4

(5, 6)

(3, 2)

1

2

3

4

5

6

(2, 0)

x=5 x=3

2x-y=4

0 5 6

O

A B

C D

O

A B

C D

M

N

E– 8


OR In Fig. ABC and DBC are on the same base BC.

If AD intersects BC at O, prove that )DBC(Ar

)ABC(Ar

=

DO

AO.

Sol: Given: Two triangles ABC and DBC in which

i) The base BC is common

ii) AD and BC intersect at O

To prove: )DBC(ar

)ABC(ar

=

DO

AO

Construction:

i) Through A, draw AM BC ii) Through D, draw DN BC [fig ½]

Proof: In triangles AMO and DNO, we have

AMO= DNO [By construction, each 90]

and AOM = DON [Vert. Opp. s]

By ‘AA’ Similarity Criteria

AMO DNO [1]

Their corresponding sides are proportional.

i.e. DO

AO =

DN

AM [CPST] … (1) [½]

Now, )DBC(ar

)ABC(ar

=

DNBC2

1

AMBC2

1

= DN

AM [ (Area of ) =

2

1 Base Altitude]

Hence proved = DO

AO [From (1)] [1]

17) The point P(7, b) lies on the line joining A(–5, 2) and B(3, 6). Find the ratio and

hence find b. [3]

Sol: Let P divide the line AB in the ratio k : 1 [where k = ]

x coordinate of P 7 = 7 =

7(k + 1) = 3k – 5 7k + 7 = 3k – 5 7k – 3k = – 5 – 7 = –12 [½]

4k = –12 k = – k = – 3 …. (i) [1]

As k is negative, point P divides AP, externally as shown where =

Also, y coordinate of P

b = b = = [½]

But k = – 3 [from (i) above]

b = = = = 8 k = – 3 and b = 8 [1]

OR If A and B are (–2, –2) and (2, –4) respectively, find the coordinate of a point P on line

segment AB such that AP = AB.

Sol: AP = AB 7AP = 3AB = 3(AP + PB)

7AP = 3AP + 3PB 7AP – 3AP = 3PB

4AP = 3PB m = 3, n = 4 [1]

)(

)(

PBl

APl

n

m

112

k

xkx

1

)5()3(

k

k

4

12

PB

AP

1

3

112

k

yky

1

2)6(

k

k

1

26

k

k

13

2)3(6

2

218

2

16

7

3

7

3

4

3

PB

AP

A P B

E– 9


p(x) = [1]

p(y) = [1]

p(x, y) = .

18) From a point P, two tangents PA and PB are drawn to a circle with centre O.

If OP = diameter of the circle show that APB is equilateral. [3]

Sol: PA and PB are tangents

PA = PB, Also, OA = OB = r

And OAP = OBP = 90º [radius tangent relation]

In OAP and OBP

OA = OB = r

PA = PB [Equal tangents]

PO = PO [Common]

OAP OBP OPA = OPB = [CPCT] [1]

Now in right angle OAP, sin =

sin = [ OP = diameter = 2r given]

sin = = sin 30º = 30º.

APB = + = 2 = 2 30º = 60º. [1]

Also, PA = PB PAB = PBA = x [Angles opposite to equal sides are equal]

In PAB: x + x + 2 = 180º [Angle sum property]

2x + 2 30 = 180 2x = 180 – 60 = 120

x = = 60º PAB = PBA = APB = 60º

APB is an equilateral triangle. [1]

19) If sec + tan = p, show that sec – tan = p

1. Hence, find the values of cos and sin .

[3]

Sol: sec + tan = p (i)

p

1 =

tan sec

) tan (sec

tan sec

1 [½]

p

1 =

tan sec

tan sec

tan sec22

i.e p

1= sec – tan (ii) [½]

Adding (i) and (ii)

We get sec = p2

1p

p

1p

2

1 2

Subtracting eq (ii) from eq (i) we get

tan = p2

1p

p

1p

2

1 2

cos = sec

1 [1]

cos = 1p

2p2

and tan = p2

1p2

p2

1p

cos

sin 2

7

2

7

86

43

)2(4)2(3

nm

nxmx 12

7

20

7

812

43

)2(4)4(3

nm

nymy 12

7

20,

7

2

OP

OA

hypotenuse

opposite

r2

r

2

1

2

120

P

A

B

O

r

r

x

x

E– 10


sin = cos . p2

1p2 =

1p

1p

p2

1p

1p

p22

22

2

[1]

OR Evaluate the following:

75? tan . 60? tan 15? tan

52?cosec 38? cos3

32? sin

58? cos2 .

Sol:

75? tan . 60? tan 15? tan

52?cosec 38? cos3

32? sin

58? cos2

=

)5?1 (90? tan . 15? tan 60? tan

)38? (90?cosec 38? cos3

58? (90? sin

58? cos2 [1]

= 13

132

)5?1 cot . 15? tan 3

38?sec 38? cos3

58? cos

58? cos2

= 2 – 1 = 1 [2]

[ tan 60º = , cos sec = 1, tan cot = 1]

20) If a wire is bent into the shape of a square, then the area of the square is 81 cm2. When the same wire is bent into a semicircular shape, find the area of the semicircle. [3]

Sol: Side of square = = 9 cm

Perimeter of square = 4 9 = 36 = Perimeter of semicircle. [1]

r + 2r = 36 or r = 36 r = = 7 cm [1]

Area of semicircle = r2 = 7 7 = 77 cm2 [1]

21) A wooden article was made by scooping out of a hemisphere from

each side of a solid cylinder, as shown in fig. If the height of the

cylinder is 10cm, and its base is of radius 3.5cm, find the total surface

area of the article. [3]

Sol: Total SA (TSA) of the article = curved SA (CSA) of the cylinder

+ CSA of the upper HS + CSA of the lower HS

TSA of article = 2rh + 2r2 + 2r2 [1]

= 2rh + 4r2 = 2r (h + 2r)

= 2 3.5 (10 + 2 3.5)

[1]

= 2 3.5 17 = 374 cm2

TSA of article = 374 cm2 [1]

OR The radii of the circular ends of a frustum are 33 cm and 27 cm, its slant height is 10 cm. Find its volume and whole surface area.

Sol: Given: r1 = 33 cm, r2 = 27 cm, l = 10 cm.

l = l2 = h2 + (r1 – r2)2 (10)2 = h2 + (33 – 27)2

100 = h2 + 62 = h2 + 36. h2 = 100 – 36 = 64 h = = 8 cm.

V = h (r + r1r2 + r ) = h [½]

V = 8 = 8 (62 + 3 33 27)

= 8 (36 + 2673) = 8 2709 = 22,704 cm3 [1]

Volume = 22,704 cm3.

Whole surface area = curved surface area + upper circular area + bottom circular area.

= (r1 + r2) l + r + r = [½]

3

81

2

7

22

36

736

2

1

2

1

7

22

7

22

7

22

2212 rrh

64

3

1 21

22

3

1 212

21 rr3rr

3

1

7

22 2733327332

3

1

7

22

3

1

7

22

3

1

7

22

21

22 2

22121 rrl rr

10

r = 3.5

E– 11


= = [600 + 1089 + 729]

= 2418 = 7599.42 cm2. [1]

22) The following table shows the marks obtained by 100 students of Class X in a school during a particular academic session. Find the mode of this distribution. [3]

Marks No. of Students

Less than 10 7

Less than 20 21

Less than 30 34

Less than 40 46

Less than 50 66

Less than 60 77

Less than 70 92

Less than 80 100

Sol: The given data can be written as:

C.I. (Marks) c.f fi. (No. of Students)

0–10 7 7

10–20 21 14 (21 – 7)

20–30 34 13 (34 – 21)

30–40 46 f0 12 (46 – 34)

M.C 40–50 66 f1 20 (66 – 46)

50–60 77 f2 11 (77 – 66)

60–70 92 15 (92 – 77)

70–80 100 8 (100 – 92)

[Table 1½]

Maximum frequency = 20

Modal class (M. C) is 40 – 50

l (lower limit of M.C) = 40, f1 (frequency of M.C) = 20, f0 (frequency of preceding M.C) = 12 f2 (frequency of succeeding M.C)= 11, h (width of M.C) = 10

Mode = l +

201

01

fff2

ff h [½]

Mode = 40 +

10

1112202

1220 = 40 +

17

80 = 40 + 4.7

= 44.7 marks [1]

Section D: (Question numbers 23 to 30 carry 4 marks each)

23) Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an

angle of 60. Justify. [4]

Sol: [Steps of Construction:

Step I: Draw a circle with centre O and radius 5 cm.

Step II: Draw any diameter AOB.

Step III: Draw a radius OC such that BOC = 60

Step IV: At C, we draw CM OC and at A, we draw AN OA.

Step V: Let the two perpendiculars intersect each other at P.

Then, PA and PC are required tangents.]

Justification:

Since OA is the radius, so PA has to be a tangent to the circle.

Similarly, PC is also tangent to the circle. [Const. 3]

APC = 360 – (OAP + OCP + AOC)

= 360 – (90 + 90 + 120) = 360 – 300 = 60

7

22 22)27(33102733

7

22

7

22

A B

C

P

M N

O

600

E– 12


Hence, tangents PA and PC are inclined to each other at an angle of 60. [Justification 1]

24) If the roots of the quadratic equation (a – b)x2 + (b – c)x + (c – a) = 0, a b are equal, then prove that b + c = 2a. [4]

Sol: (a – b)x2 + (b – c)x + (c – a) = 0 A = a – b, B = b – c, C = c – a

As the roots are equal; D = 0 i.e. B2 – 4AC = 0 [1]

(b – c)2 – 4(a – b) (c – a) = 0 b2 + c2 – 2bc – 4ac + 4a2 + 4bc – 4ab = 0 [1]

4a2 + b2 + c2 – 4ab + 2bc – 4ac = 0 (2a – b – c)2 = 0 [1]

b + c = 2a [1]

OR Solve for x :

1x

3x7

3x

1x2 = 5; given that x –3, x 1.

Sol:

1x

3x7

3x

1x2 = 5

Let y3x

1x

y

1

1x

3x

Equation: 2y – 5

y

7

5y

7y2 2

2y2 – 7 = 5y [1]

2y2 – 5y – 7 = 0 2y2 + 2y – 7y – 7 = 0

2y (y + 1) – 7 (y + 1) = 0 (y + 1) (2y – 7)

y = – 1 or 2y – 7 = 0 y = –1 or 2y = 7

y = 2

7 [1]

Case – I: When y = – 1,

13x

1x

x – 1 = – 1 (x + 3) x – 1 = – x – 3 x + x = – 3 + 1

2x = – 2 x = 12

2 [1]

Case – II: When y = 2

7

2

7

3x

1x

7 (x + 3) = 2 (x – 1) 7x + 21 = 2x – 2 7x – 2x = – 2 – 21

5x = –23 x = 5

23 x = –1 or

5

23 [1]

25) Two cars start together in the same direction at the same place. The first goes at uniform speed of 10 km/h. The second goes at a speed of 8 km/h in the first hour and increases

the speed by km each succeeding hour. After how many hours will the second car

overtake the first car if both cars go nonstop? [4]

Sol: Let the second car overtake the first car after n hours.

Distance covered by the 1st car in each hour = 10 km.

Distances covered by 1st car in n hours = 10n km. [½]

Increase in speed of the 2nd car per hour = km

Distance covered in 1st hour, 2nd hour, 3rd hour are 8, 8 , 9,…

This is an A.P. with first term, a = 8 and common difference, d = [½]

Distance covered by the 2nd car in n hours is given by Sn = 8 + 8 + 9 + …

Using Sum Formula of an A.P., we have Sn = [2a + (n 1)d]

2

1

2

1

2

1

2

1

2

1

2

n

E– 13


Sn = = = km [1]

At the time of overtaking, both cars have travelled the same distance

= 10n n2 + 31n = 40n n2 9n = 0 n(n 9) = 0 [½]

Either n = 0 or n = 9 [1]

Rejecting n = 0, which corresponds to the time, of their we get n = 9

Hence, the second car overtakes the first car after 9 hours [½]

26) A man in a boat rowing away from a light house 100m high takes 3 minutes to change the

angle of elevation of the top of light house from 60 to 45. Find the speed of the boat in

m/min. ( = 1.73) [4]

Sol: Let AB be the light house and C, D be two position of the boat.

In ABD, = cot 45 = 1 BD = 100 [1]

In ABC, = cot 60 = BC = = [1]

CD = BD – BC = 100 – = = m [1]

Time = 3 min

Speed = = 3 = = m/min = 14 m/min. [1]

27) The ratio of the areas of two similar triangles is equal to the square ratio of their corresponding sides. Prove. [4]

Sol: Given: ABC ~ PQR

To Prove:

Construction: Draw AD BC, PS QR.

Proof: ar(ABC) = BC AD

ar(PQR) = QR PS [Area of triangle = base height]

––– (i) [1]

In ADB and PSQ,

B = Q [Given ABC ~ PQR]

ADB = PSQ [By construction, each 90] [1]

ADB PSQ [AA corollary]

[Corresponding sides of similar triangles are proportional]

But [Corresponding sides of ABC PQR]

–– (ii)

[i and ii] [1]

[As ABC PQR]

Hence, = [1]

2

1)1()8(2

2n

n

2

31

2

nn

4

312 nn

4

312 nn

3

AB

BD

100

BD

AB

BC

100

BC

3

1

3

100

3

3100

3

3100

3

173300

3

127

T

D

3

127

3

127

3

1

9

127

9

1

2

2

2

2

2

2)

PR

AC

QR

BC

PQ

AB

PQR)( Area

ABC( Area

2

1

2

1

2

1

PSQR

ADBC

PSQR

ADBC

PQRar

ABCar

2

12

1

)(

)(

PQ

AB

PS

AD

QR

BC

PQ

AB

QR

BC

PS

AD

2

2

)(

)(

QR

BC

QRQR

BCBC

PQRar

ABCar

PR

AC

QR

BC

PQ

AB

PQRar

ABCar

2

2

2

2

2

2

PR

AC

QR

BC

PQ

AB

A

B D C

45 60

100m

A

B CD

P

Q R

S

E– 14


E– 15


OR) In figure, XY || AC and XY divides triangular region

ABC into two parts of equal areas. Determine AB

AX.

Sol: given XY || AC.

Area (trapezium AXYC) = area ∆BXY, to determine AB

AX

Now, area (Trapezium AXYC) = area (∆BXY)

BXY)(

AXYC) (trapezium

area

area =

1

1

BXY

AXYC) (trapezium

area

area + 1 = 1 + 1 (adding ‘1’ both sides)

BXY

BXY of area AXYC) (trapezium

area

area = 2

)(

)(

BXYarea

BACarea

= 2 … (1) [1]

In ∆BAC and ∆BXY, XY || AC (given)

BXY = BAC (corresponding angles) and BYX = BCA (corresponding angles)

∆BXY ~ ∆BAC (By AA similarity)

)(

)(

BXYarea

BACarea

=

2

2

BX

AB (Areas of 2 similar triangles are proportional to square of corresponding

sides)

From (1) and (2), 2

2

BX

AB = 2 [1]

2

2

AB

BX =

2

1

AB

BX=

2

1 =

2

1 [½]

AB

AXAB =

2

1 [ BX = AB – AX]

AB

AB –

AB

AX =

2

1 1 –

AB

AX =

2

1 [½]

AB

AX = 1 –

2

1 =

2

12 =

22

122

AB

AX =

2

22 [½+½]

28) Prove that = 1 + tan + cot . [4]

Sol: LHS = = + = + [1]

= = [1]

= = [1]

= =

= tan + cot + 1 = RHS [1]

tan1

cot

cot1

tan

tan1

cot

cot1

tan

sincos

cossin

1

cossin

sincos

1

sincossin

cossin

cossincos

sincos

)sin(cossin

cos

)cos(sincos

sin 22

)cos(sinsin

cos

)cos(sincos

sin 22

)cos(sincossin

cossin 33

)cos(sincossin

)coscossin)(sincos(sin 22

cossin

cossincossin 22

cossin

cossin

cossin

cos

cossin

sin 22

A

C Y B

X

A

C Y B

X

E– 16


29) A juice seller serves his customers using a glass as shown in Figure.

The inner diameter of the cylindrical glass is 5 cm, but the bottom of the

glass has a hemispherical portion raised which reduces the capacity of

the glass. If the height of the glass is 10 cm, find the apparent capacity

of the glass and its actual capacity. What values does the juice seller

depict by using such a glass? (Use = ) [4]

Sol: Radius of cylinder = radius of HS = r =

Height of cylinder = h = 10 cm

Apparent Vol. of the glass = r2h = 10 = 196.43

V1 = Apparent Vol. of the glass = 196.43 cm3 [1½]

V = Actual volume of the glass = Apparent volume of the glass – volume of the H.S.

V = V1 – r3 = 196.43 – = 196.43 – 32.74 = 163.69 cm3

V1 = apparent volume ≈ 196.43 cm3, V = Actual Volume ≈ 163.69 cm3. [1½] Value show: Dishonesty, as the juice seller is giving less quantity of juice to the customers. [1]

30) Draw the more than cumulative frequency curve of the following data: [4]

Rent (Rs.) 15-25 25-35 35-45 45-55 55-65 65-75 75-85 85-95

No. of houses 5 10 20 35 30 20 13 7

Sol: We prepare the cumulative frequency table by more than method [tab 1+graph 3]

Class Frequency More than lower limit

Cumulative frequency

15 – 25 5 15 140

25 – 35 10 25 135

35 – 45 20 35 125

45 – 55 35 45 105

55 – 65 30 55 70

65 – 75 20 65 40

75 – 85 13 75 20

85 – 95 7 85 7

Coordinates: (15, 140), (25, 135), (35, 125),

(45, 105), (55, 70), (65, 40),

(75, 20) and (85, 7)

OR) If the median of the distribution given below is 28.5, find the values of x and y.

Class interval 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 Total

Frequency 5 x 20 15 y 5 60

7

22

2

5

7

22

2

5

2

5

3

2

3

2

7

22

2

5

2

5

2

5

5 cm

10 cm

10

20

30

40

50

60

15 25 35

45 55 65

Cu

mu

lati

ve

Fre

qu

en

cy

Lower Limits

O 75 8

5

70

80

90

100

110

120

130

140 (15,140)

(25,135)

(35,125)

(45,105)

(55, 70)

(65, 40)

(75, 20)

(85, 7)

95

E– 17


Sol:

C.I Frequency C.f.

0 – 10 5 5

10 – 20 x 5 + x = cf

20 – 30 20 = f 25 + x

30 – 40 15 40 + x

40 – 50 y 40 + x + y

50 – 60 5 45 + x + y = n

f1 = 45 + x + y

[Table 1]

Given: 45 + x + y = 60 = n

x + y = 60 – 45 = 15 ……… (1) [½]

Median = 28.5 [given]

By using formula: Median = l + h [½]

Here, median = 28.5 which falls in class interval 20 – 30. Hence median class (M.C) = 20 – 30

l = lower limit = 20, h = width = 30 – 20 = 10

n = 60 (given) = = 30

f = frequency of median class = 20

c.f = cumulative frequency of the preceding median class = 5 + x

Median = l + h 28.5 = 20 + 10

28.5 – 20 = 8.5 = 8.5 2 = 25 – x

17 = 25 – x x = 25 – 17 = 8 [1]

From (1): x + y = 15 8 + y = 15 y = 15 – 8 = 7 x = 8, y = 7. [1]

f

cfn

2

2

n

2

60

f

cfn

220

)5(30 x

2

530 x

2

25 x

e universal tutorials x cbse · 2019-02-24 · e– 2 question paper universal tutorials – x cbse...

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