Subject

Mathematics

Class

CBSE Class 10

Pre Boards

Practice to excel and get familiar with the paper pattern and the type of questions. Check you answers with answer keys provided.

Sample Papers

Download the PDF Sample Papers Free for off line practice and view the Solutions online.

 Multiple Choice QuestionsMultiple Choice Questions

1.

If the difference between the circumference and the radius of a circle is 37cm, then using π = 227 , the circumference (in cm) of the circle is:

  • 154

  • 44

  • 14

  • 7


B.

44

Let r be the radius of the circle,

From the given information, we have

2πr - r =37 cm r ( 2π - 1 ) = 37 cm r  2 x 227 - 1 = 37 cm   r x 377 = 37 cm r = 7 cm Circumference of the circle = 2πr = 2 x 227 x 7 cm = 44 cm


 Multiple Choice QuestionsShort Answer Type

2.

Solve the following quadratic equation for x;43x2 + 5x - 23 = 0 


43 x2 + 5x -23 = 0 43 x2  + 8x - 3x - 23 = 04x ( 3 x + 2 ) - 3 (3 x + 2) = 0(4x - 3) (3x + 2 ) = 0 x = 34   or  x = -23


 Multiple Choice QuestionsMultiple Choice Questions

3.

In fig., the area of triangle ABC (in sq. units) is

  • 15

  • 10

  • 7.5

  • 2.5


C.

7.5

From the figure, the coordinates of A, B, and C are (1,3), (-1, 0) and (4, 0)

respectively.

Area ofABC       = 12 1(0-0) + (-1)(0-3) +(3-0)       =12 0+3+12       =12 15       = 7.5 sq. units


4.

The probability of getting an even number, when a die is thrown once, is

  • 12

  • 13

  • 16

  • 56


A.

12

S = { 1, 2, 3, 4, 5, 6 }

let event E be defined as 'getting an even number'.

n(E) = { 2, 4, 6 }

  P(E) = Number of favourable outcomesNumber of possible outcomes =36                  = 12


5.

In fig., PA and PB are two tangents drawn from an external point P to a circlewith centre C and radius 4 cm. If PA PB, then the length of each tangent is

  • 3

  • 4

  • 5

  • 6


B.

4

AP PB      (Given)CAAP,  CB  BP  (Since radius is perpendicular to tangent)AC= CB= radius of the circle
Therefore, APBC is a square having side equal to 4 cm.
Therefore, length of each tangent is 4 cm.


6.

In fig., a circle with centre O is inscribed in a quadrilateral ABCD such that, it touches the sides BC, AB, AD and CD at points P, Q, R and S respectively, If AB = 29 cm, AD = 23 cm, B = 90o and DS = 5 cm, then the radius of thecircle (in cm) is 

  • 11

  • 18

  • 6

  • 15


A.

11

Given: Ab, BC,CD,and AD tangents to the circle with centre O at

Q,P,S and R respectively.

AB=29 cm, AD=23 cm, DS=5 cm and B=900

Construction: Join PQ.

We know that, the lengths of the tangents drawn from an external

point to a circle are equal.

DS=DR=5 cm

AR= AD - DR= 23 cm- 5 cm= 18 cm 

AQ=AR= 18 cm

 QB = AB - AQ = 29 cm - 18 cm= 11 cm

QB = BP = 11 cm

In InPQBPQ2 = QB2 +BP2 =(11 cm)2 +(11 cm)2 = 2x(11 cm)2PQ=112 cm    ..........(1)INOPQ,PQ2 = OQ2 +OP2 = r2 +r2 = 2r2  (112)2 = 2r2 121 = r2r = 11 cmThus, the radius of the circleis 11 cm.


7.

The angle of depression of a car, standing on the ground, from the top of a 75 m high tower, is 30o. The distance of the car from the base of the tower (in m.) is

  • 253

  • 503

  • 753

  • 150


C.

753

Let AB be the tower of height 75 m and C be position of the car.

In 

ABC,cot 30°= ACABAC=AB cot 30°AC=75m x 3AC=753mThus, the distance of the car from the base of the tower is 753 m. 


8.

A box contains 90 discs, numbered from 1 to 90. If one disc is drawn at random from the box, the probability that it bears a prime-number less than 23,is

  • 790

  • 1090

  • 445

  • 989


C.

445

S = { 1, 2, 3,........90 }

n(s) = 90

The prime number less than 23 are 2, 3, 5, 7, 11, 13, 17 and 19

Let event E be defined as 'getting a prime number less than 23'.

n(E) = 8

 p(E) =Number of favourable outcomesNumber of possible outcomes = 890               =445


9.

The common difference of AP 13q. 1-6q3q, 1-12q3q, ........ is;1 1 6q 1 12q, , 3q 3q 3q
... is:

  • q

  • -q

  • -2

  • 2


C.

-2

 Comman difference = 1 - 6q3q - 13q = 1-6q-13q = -6q3q = -2


 Multiple Choice QuestionsShort Answer Type

10.

How many three-digit natural numbers are divisible by 7?


Three digit numbers divisible by 7 are 

105, 112, 119, .......994

This is an AP with first term (a) =105 and comman difference (d)= 7

an be the last term.an  = a+ (n - 1 )994 = 105 + (n - 1 ) ( 7 )7 ( n - 1 ) =889n - 1 = 127n = 128Thus, there are 128 three-digit natural terms that are divisible by 7.