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.
Advertisement

 Multiple Choice QuestionsLong Answer Type

21.

Construct a triangle ABC with side BC = 7 cm, ∠B = 45°, ∠A = 105°. Then construct another triangle whose sides are 3/4 times the corresponding sides of the Δ ABC.


22.

Two different dice are thrown together. Find the probability that the numbers obtained have


(i) even sum, and


(ii) even product


23.

In the given figure, XY and X’Y’ are two parallel tangents to a circle with centre O and another tangents AB with point of contact C, is intersecting XY at A and X’Y’ at B. Prove that ∠AOB = 90°.

                  


24.

In a rain–water harvesting system, the rain-water from a roof of 22 m × 20 m drains into a cylindrical tank having diameter of base 2 m and height 3.5m. If the tank is full, find the rainfall in cm. Write your views on water conservation


Advertisement
25.

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


26.

If the ratio of the sum of the first n terms of two A.Ps is  (7n + 1) : (4n + 27), then find the ratio of their 9th terms.


27.

Solve for x:

x - 12x + 1 + 2x + 1x - 1 = 2,    Where x -12, 1


28.

A takes 6 days less than B to do a work. If both A and B working together can do it in 4 days, how many days will B take to finish it?


Advertisement
29.

From the top of a tower, 100 m high, a man observe two cars on the opposite sides of the tower and in same straight line with its base, with its base, with angles of depression 30° and 45°. Find the distance between the cars.
[Take 3 = 1.732]


Advertisement

30.

In the given figure, O is centre of the circle with AC = 24 cm, AB = 7 cm and ∠BOD = 90°. Find the area of the shaded region.

                         


AC = 24 cm,  AB = 7 cm

                                   

Since BC is the diameter of the circle,

So, BAC = 90°In right BAC,BC2 = AC2 = AB2BC2 = 242 + 72BC2 =625 BC = 25So, the radius of the circle = OC = 12.5 cmArea of the shaded region = Area of the circle - Area of BAC - Area of sector CD= πr2 - 12 x AB X AC - θ360 X   πr2 = 227 x 12.5 x 12.5 - 12 x 7 x 24 - 90360 x 227 x 12.5 x 12.5                  ..........(BOD = 90°   COD = 90° )= 491.07 - 84 - 122.77= 284.3 cm2  ( approximately)Hence, the area of the shaded region is 284.3 cm2 approximately.


Advertisement
Advertisement