CLASS X (2019-20) MATHEMATICS BASIC(241) SAMPLE PAPER …

Mathematics Basic X

Sample Paper 11 Solved

CLASS X (2019-20) MATHEMATICS BASIC(241)

SAMPLE PAPER-11

.in

Time : 3 Hours

Maximum Marks : 80

General Instructions :

(i) All questions are compulsory.

(ii) The questions paper consists of 40 questions divided into four sections A, B, C and D.

(iii) Section A comprises of 20 questions of 1 mark each. Section B comprises of 6 questions of 2 marks each. Section

C comprises of 8 questions of 3 marks each. Section D comprises of 6 questions of 4 marks each.

(iv) There is no overall choice. However, an internal choices have been provided in two questions of 1 mark each, two

questions of 2 marks each, three 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.

(v) Use of calculators is not permitted.

Section A

Q.1-Q.10 are multiple choice questions. Select the most appropriate answer from the given options.

1. In the adjoining figure, +BAC = 90c and AD = BC ,

then

[1]

(a) BC $ CD = BC 2 (c) BD $ CD = AD2 Ans : (c) BD $ CD = AD2

(b) AB $ AC = BC 2 (d) AB $ AC = AD2

a

+BAC = 90c

and

AD = BC

`

TDBA + TDAC

(AA similarity)

AD DC

=

BD AD

AD2 = BD # DC

2. In TABC + TEDF and TABC is not similar to

TDEF , then which of the following is not true? [1]

(a) BC $ EF = AC $ DE (c) BC $ DE = AB $ EF

(b) AB $ AC = BC 2 (d) BC $ DE = AB $ FD

Ans : (c) BC $ DE = AB $ EF

a

TABC + TEDF

AB ED

=

BC DF

=

AC EF

BC $ ED ! AB $ EF

3. The value of 3 tan226c - 3 cosec264c is

[1]

(a) 0

(b) 3

(c) - 3

(d) - 1

Ans : (c) - 3

3 tan2 26c - 3 sec2 26c = 3 tan2 26c - 3 sec2 26c

(a cosec 64c = sec 26c) = - 3^sec2 26c - 3 sec2 26ch =- 3

4.

If

cos A

=

3 5

,

then

the

value

of

tan A

is

[1]

(a)

3 4

(b)

4 5

(c)

4 3

Ans

:

(c)

4 3

(d)

5 4

cos A

=

3 5

Now,

sec A

=

5 3

tan A = sec2A - 1 =

=

4 3

b

5 3

2

l

-

1

5. If the ratio of the circumferences of two circles is 4 : 9,

then the ratio of their areas is

[1]

(a) 9 : 4

(b) 4 : 9

(c) 2 : 3

(d) 16 : 81

Ans : (d) 16 : 81

2r1 2r2

=

4 9

r1 r2

=

4 9

Now,

r12 r22

=

r12 r2

=

b

4 9

2

l

=

16 81

6. If the sector of a circle of diameter 10 cm subtends an

angle of 144c at the centre, then the length of the arc

of the sector is

[1]

(a) 2 cm

(b) 4 cm

(c) 5 cm

(d) 6 cm

Ans : (b) 4 cm

2r = 10 cm

r = 5 cm

Arc length

=

360

#

2r

=

144 360

#

2

#

5

= 4 cm

To Get 20 Solved Paper Free PDF by whatsapp add +91 89056 29969 in your class Group

Page 1

Mathematics Basic X

Sample Paper 11 Solved

cbse.online

7. If the classes of a frequency distribution are 1 - 10

, 11 - 20, 21 - 30, .......... , 61 - 70, then the upper

limit of the class 11 - 20 is

[1]

(a) 20

(b) 21

(c) 19.5

(d) 20.5

Ans : (d) 20.5

Given classes are discontinuous.

Adjustment factor = 0.5 By converting the given data into continuous classes, we get 0.5 - 10.5, 10.5 - 20.5, 20.5 - 30.5, ..........

8. If the probability of an event is p , then the probability

of its complementary event will be

[1]

(a) p - 1

(b) p

(c) 1 - p Ans : (c) 1 - p

(d)

1

-

1 p

a

P^E h = 1 - P^Eh

P^E h = 1 - p

9. For some integer q , every odd integer is of the form [1]

(a) q

(b) q + 1

(c) 2q

(d) 2q + 1

Ans : (d) 2q + 1

Every odd integer = 2q + 1

for

q dI

10. If the base area of a cone is 51 cm2 and its volume is

85 cm3, then its vertical height is

[1]

(a) 3.5 cm

(b) 4 cm

(c) 4.5 cm

(d) 5 cm

Ans : (d) 5 cm

Vertical Height

=

3

# 85 cm3 51 cm2

= 5 cm

(Q.11-Q.15) Fill in the blanks.

11. If the ratio of the corresponding sides of two similar

triangles is 7 : 11, then the ratio of their corresponding

altitudes is ..........

[1]

Ans : For two similar triangles,

The ratio of areas = Square of the ratio of their corresponding sides Also, The ratio of areas = Square of the ratio of their corresponding altitudes

` Ratio of altitudes = Ratio of sides

= 7 : 11

12. The perimeter of a semicircular protactor of diameter

14 cm is ..........

[1]

Ans :

Given,

Diameter,

2r = 14 cm

r =7

Perimeter = r + 2r

=

22 7

#

7

+

14

= 36 cm

or

If the area of a circle is 616 cm2, then its circumference is ..........

Ans :

As we know that,

Area of circle = r2

Now,

r2 = 616 cm2

r2

=

616 22

#

7

r2 = 196

r = 14 cm

` Circumference = 2r

=

2

#

22 7

#

14

= 88 cm

13. Ogive is a curve which represents continuous ..........

frequency distribution graphically.

[1]

Ans : Cumulative

14. The value of sec 23c is ..........

[1]

cosec 67c

Ans :

sec 23c cosec 67c

=

sec 23c cosec ^90c - 23ch

=

sec 23c sec 23c

=

1

15. A line intersecting a circle in two points is called a

..........

[1]

Ans : Secant

(Q.16-Q.20) Answer the following 16. Given that LCM (91, 26) = 182, find HCF (91, 26). [1]

Ans :

a HCF # LCM = 91 # 26

HCF # 182 = 91 # 26

HCF

=

91 # 26 182

= 13

or

Does

the

rational

number

441 22 $ 57 $

72

has

a

terminating

or a non-terminating decimal representation?

Ans :

4419 22 $ 57 $ 72

=

22

9 $

57

1

17. If 1 is a zero of the polynomial p^x h = ax2 - 3^a - 1hx - 1

, then find the value of a .

[1]

Ans :

Given 1 is a zero of, p^x h = ax2 - 3^a - 1hx - 1 p^1h = 0

a # 12 - 3^a - 1h # 1 - 1 = 0 a - 3a + 3 - 1 = 0 - 2a = - 2 a =1

Download 20 Solved Sample Papers pdfs from cbse.online or .in

Page 2

Mathematics Basic X

Sample Paper 11 Solved

.in

18. The nth term of an AP is 7 - 4n . Find its common

difference.

[1]

Ans :

Given,

an = 7 - 4n

a2 = 7 - 4 # 2 =- 1

and

a1 = 7 - 4 # 1 = 3

` Common difference,

d = a2 - a1 =- 1 - 3 =- 4

19. In the given figure, DE z BC and AD = 1 cm, BD

= 2 cm. What is the ratio of the area of TABC to the

area of TADE ?

[1]

and

Hence, and

a1 a2

=

b1 b2

=

c1 c2

2m - 1 3

=

n

3 -

1

=

- -

5 2

2m - 1 3

=

- -

5 2

- 4m + 2 = - 15

m

=

17 4

3 n-1

=

- -

5 2

- 5n + 5 = - 6

n

=

11 5

m

=

17 4

n

=

11 5

22. Find the roots of the quadratic equations:

[2]

4 3 x2 + 5x - 2 3 = 0

Ans :

Ans :

Area of TABC Area of TADE

=

b

AB AD

2

l

=

b

AD + BD AD

2

l

=

b

1

+ 1

2

2

l

=

9 1

(a TABC + TADE )

` Area of TABC : Area of TADE = 9 : 1

20. In the given figure, PA and PB are tangents to the

circle drawn from an external point P . CD is a third

tangent touching the circle at Q . If PB = 10 cm and

CQ = 2 cm, what is the length of PC ?

[1]

Ans :

PC = PA - CA = PB - CQ = ^10 - 2h cm = 8 cm

Section B

21. Determine the values of m and n , so that the following

system of linear equations has infinite number of

solutions:

[2]

^2m - 1hx + 3y - 5; 3x + ^n - 1hy - 2 = 0

Ans :

For infinite number of solutions,

4 3 x2 + 5x - 2 3 = 0

4 3 x2 + 8x - 3x - 2 3 = 0

4x^ 3 x + 2h - 3 ^ 3 + 2h = 0

^ 3 x + 2h^4x - 2 h = 0

3x+2 =0

or

4x - 3 = 0

x = - 2 or 3

3 4

or

Solve for x : 6x + 7 - ^2x - 7h = 0

Ans :

6x + 7 - ^2x - 7h = 0

...(i)

6x + 7 = ^2x - 7h2

6x + 7 = 4x2 - 28x + 49

4x2 - 34x + 42 = 0

2x2 - 17x + 21 = 0

^2x - 3h^x - 7h = 0

x = 32, 7

Putting,

x

=

3 2

in

equation (i)

We get,

6

#

3 2

+7

- b2

#

3 2

- 7l

=

0

4 - ^-4h = 0 8 = 0 i.e. False

Putting, x = 7 in equation (i) We get,

6 # 7 + 7 - ^2 # 7 - 7h = 0

7 - 7 = 0 i.e. True

`

x =7

To Get 20 Solved Paper Free PDF by whatsapp add +91 89056 29969 in your class Group

Page 3

Mathematics Basic X

Sample Paper 11 Solved

cbse.online

23. A bag contains 5 red, 8 green and 7 white balls.

One ball is drawn at random from the bag. Find the

probability of getting:

[2]

(i) a white ball or a green ball.

(ii) neither a green ball nor a red ball.

Ans :

Total number of possible outcomes = 5 + 8 + 7 = 20

(i) Number of favourable outcomes

= Number of white balls or number of green balls = 7 + 8 = 15

` P (a white or a green ball)

=

15 20

=

3 4

(ii) Number of favourable outcomes

= Number of white balls

= 7

` P (neither green nor red ball)

=

7 20

24. Without using trigonometric tables, find the value

of:

[2]

cos 70c sin 20c

+

cos

57c

$

cosec

33c

-

2

cos

60c

Ans :

+C = +B (angles opp. equal sides of a T are equal)

In TABD and TECF ,

+ABD = +EFC

(a +B = +C , proved above)

and +ADB = +EFC

` TABD + TECF

`

AB EC

=

AD EF

(each = 90c) (by AA similarity criterion)

AB # EF = AD # EC , as required.

26. Two circles touch externally. The sum of their areas is

58 cm2 and the distance between their centres is 10

cm. Find the radii of the two circles.

[2]

Ans :

Let r1 cm and r2 cm be the radii of two circles,

cos sin

70c 20c

+

cos

57c

$

cosec

33c

-

2

cos

60c

=

cos 70c sin^90c - 70ch

+

cos

57c

$

cosec ^90c

-

57ch

-

2

#

1 2

=

cos cos

70c 70c

+

cos

57c

$

sec

57c

-

1

ca

sec

=

1 cos

m

=

1

+

cos cos

57c 57c

-

1

=1+1-1 =1

25. In the adjoining figure, E is a point on the side CB

produced of an isosceles triangle ABC with AB = AC

. If AD = BC and EF = AC , prove that:

[2]

Ans :

AB # EF = AD # EC

Given, TABC is an isosceles triangle with AB = AC

Then, and

r1 + r2 = 10 r12 + r22 = 58

r12 + r22 = 58

...(i) ...(ii)

Substituting the value of r2 from (i) in (ii), we get,

r12 + ^10 - r1h2 = 58

2r12 - 20r1 + 42 = 0

r12 - 10r1 + 21 = 0

^r1 - 7h^r1 - 3h = 0

r1 = 7, 3

If,

r1 = 7 ,

Then,

r2 = 3

and if

r1 = 3,

Then,

r2 = 7

Hence, the radii of two circles are 7 cm and 3 cm.

or

A student takes a rectangular piece of paper 30 cm long and 21 cm wide. Find the area of the biggest circle that can be cut from the paper. Also find the area of the paper left after cutting out the circle.

Download 20 Solved Sample Papers pdfs from cbse.online or .in

Page 4

Mathematics Basic X

Sample Paper 11 Solved

.in

Ans :

Given, `

Length of paper = 30 cm, Breadth of paper = 21 cm

Area of paper = ^30 # 21h cm2 = 630 cm2

Now, diameter of the biggest circle which can cut from the given paper is 21 cm.

So,

2r = 21 cm

r

=

21 2

cm

Area of circle = r2

=

22 7

#

21 2

#

21 2

= 346.5 cm2

Now, area of the paper left after cutting out the circle = ^630 - 346.5h cm2 = 283.5 cm2

a2 x

= 1

x = a2

Putting x = a2 in (i), we get,

a2 a2

-

b2 y

= 0

1

-

b2 y

= 0

1

=

b2 y

Hence, and

y = b2 x = a2 y = b2

or

The sum of the numerator and denominator of a fraction is 4 more than twice the numerator. If the numerator and denominator both increased by 3, they are in the ratio 2 : 3. Determine the fraction.

Ans :

Section C

Let

the

fraction

be

x y

.

Then according to given,

27. Prove that:

[3]

x + y = 2x + 4

sec sec

A A

- +

1 1

+

Ans :

sec A + 1 sec A - 1

= 2 cosec A

x-y+4 =0

and

x+ 3 y+3

=

2 3

...(i)

LHS =

sec sec

A A

- +

1 1

+

sec A + 1 sec A - 1

= ^sec A - 1h + ^sec A + 1h sec A + 1 $ sec A - 1

= 2 sec A = 2 sec A

sec2A - 1

tan2 A

=

2 sec A tan A

=

2

#

1 cos

A

#

cos sin

A A

=

2 sin A

= 2 cosec A

=

RHS

28. Solve the following pair of linear equations:

[3]

a2 x

-

b2 y

= 0;

a2b x

+

b2a y

= a + b, x ! 0, y ! 0

Ans :

Given,

a2 x

-

b2 y

= 0

and

a2b x

+

b2a y

=a+b

Multiplying (i) by a , we get,

a3 x

-

b2a y

= 0

Adding (ii) and (iii), we get,

a2b x

+

a3 x

=a+b

a2 x

^b

+

a

h

=

a+b

a2 x

=

a a

+ +

b b

...(i) ...(ii) ...(iii)

3x + 9 = 2y + 6 3x - 2y + 3 = 0

...(ii)

Multiplying (i) by 2 and subtracting from (ii), we get, 3x - 2y + 3 - 2^x - y + 4h = 0 3x - 2y + 3 - 2x + 2y - 8 = 0 x-5 =0 x =5

Putting x = 5 in (i), we get, 5-y+4 =0 y =9

Hence,

the

fraction

is

5 9

.

29.

Show that

1 2

and

- 3 2

are the zeroes of the polynomial

4x2 + 4x - 3 and verify the relationship between zeroes

and coefficients of the polynomial.

[3]

Ans :

Let, Then,

and

f^x h = 4x2 + 4x - 3

f

b

1 2

l

=

4

#

b

1 2

2

l

+

4

#

1 2

-3

=

4

#

1 4

+

2

-

3

=1+2-3 =0

f

b

-3 2

l

=

4

#

b

-3 2

2

l

+

4

#

b

-3 2

l

-

3

=

4

#

9 4

-

6

-

3

=9-9 =0

To Get 20 Solved Paper Free PDF by whatsapp add +91 89056 29969 in your class Group

Page 5

 ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download