Subject

Mathematics

Class

CBSE Class 10

Pre Boards

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Sample Papers

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 Multiple Choice QuestionsShort Answer Type

1.

Prove that the tangents drawn at the end points of a chord of a circle make equal angles with the chord.


                      

 

Let AB be a chord of circle with centre O.

Let AP and BP be two tangents at A and B respectively.

Suppose the tangents meet at point P. Join OP.

Suppose OP meets AB at C.

   Now in, PCA and PCB,PA = PB      .............(Tangents from an external point are equal)APC = BPC      ........(PA and PB are equally inclined to OP)PC = PC          ..........( common)Hence, PAC  PBC       ......( by SAS congruence  criterion) PAC = PBC


2.

A line intersects the y-axis and x-axis at the points P and Q respectively. If (2, -5) is the mid-point of PQ, then find the coordinates of P and Q.


Since a line is  intersecting  Y-axis at P and X- axis at Q,

Coordinates of P = (0,y) and coordinates of Q = (x,0)

Let R be the midpoint of PQ.

 

Then, co-ordinates of R = 0 + x2, y + 02 = ( 2,-5) x2, y2 = (2, -5) x2 = 2 and  y2 = -5  x = 4  and  y = -10Hence, co-ordinates of P are (0, -10)  and  co-ordinates of Q are (4,0).


 Multiple Choice QuestionsLong Answer Type

3.

For what value of n, are the nth terms of two A.Ps.  63, 65, 67,…. and 3, 10, 17,….. equal?


For A.P.  63, 65, 67,........, we have

First term = 63  and common difference = 65-63 = 2

Hence, nth term = an = 63 + (n-1) 2

an = 63 + 2n - 2 = 2n + 61

 

For A.P.  3, 10, 17,........, we have 

First term = 3 and  common difference = 10-3 = 7

Hence,  ntn term = an = 3 + (n-1) 7

an = 3 + 7n - 7 = 7n - 4The two A.Ps. will have identical nth term, if an = an 2n + 61 = 7n - 45n = 65n = 13


 Multiple Choice QuestionsShort Answer Type

4.

If the angle between two tangents drawn from an external point P to a circle of radius a and centre O, is 60°, then find the length of OP.


                  

In the figure, PA and PB are two tangents from an external point P to a circle with centre O and radius = a

APB = 60°    ( given )APQ = 30°   ( tangents are equally inclined to the line                                  joining the point and the centre. )Now, OA  APIn right angled triangle OAP,sin 30° = OAOP 12 = aOPOP = 2a


5.

What is the common difference of an A.P. in which a21- a7= 84?


Let a be the first term and d be the common difference of the given A.P.

a21 - a7 = 84a + 20d - a + 6d = 84a + 20d -a - 6d = 84 14d = 84 d = 6Hence, the common difference is 6.


6.

The probability of selecting a rotten apple randomly from a heap of 900 apples is 0.18. What is the number of rotten apples in the heap?


Let the total number of rotten apples in a heap = n

Total number of apples in a heap = 900

probability of selecting a rotten apple from a heap = 0.18

Now.

P(selecting a rotten apple) = Number of rotten applesTotal number of apples0.18 = n900n = 0.18 x 900n = 162Hence, the number of rotten apples is 162.


7.

If a tower 30 m high, casts a shadow 10 3m long on the ground, then what is the angle of elevation of the sun?


                        

Let AB be the tower and BC  be its shadow.

AB = 30 m, BC = 103 mIn ABC,tanθ = ABBCtanθ = 30103 tanθ =33tan θ = 3  But, tan 60° =3  θ = 60°Thus, the angle of elevation of sun is 60°. 


 Multiple Choice QuestionsLong Answer Type

8.

Find the value of p, for which one root of the quadratic equation px2 – 14x + 8= 0 is 6 times the other.


Given, px2 -14x +8 =0Here, a = p,  b =  -14,  c = 8Let α and β be the roots of the given quadratic equation.Then,  β = 6αNow, sum of the roors = -baα + β = -(-14)pα + β = 14pα + 6 α = 14p 7α = 14p α = 2p  ...........(i)Product of the roots = ca αβ = 8pα x 6α = 8p 6α2 = 8p  3α2 = 4p3 x 2p2= 4p  ............From (i)3 x 4p2 = 4pp = 3


 Multiple Choice QuestionsShort Answer Type

9.

If the distances of P(x, y) from A(5, 1) and B(-1, 5) are equal, then prove that  3x = 2y.


Given, P(x,y) is equidistant from  A(5,1)  and B(-1,5)

Now,  AP = BP

5-x2 + 1-y2  = -1-x2 + 5-y2  5-x2 + 1-y2 = -1-x2 + 5-y225 + x2 - 10x + 1 + y2 -2y = 1 + x2 + 2x + 25 + y2 -10yx2 + y2 -10x -2y +26 = x2 +y2 +2x -10y +26-10x -2y = -10y + 2y-12x = -8y3x = 2y     .........(Dividing throughout by -4)


10.

A circle touches all the four sides of a quadrilateral ABCD. Prove that 

AB + CD = BC + DA


                       

 

Since tangents drawn from an external point to a circle are equal in length, we have

AP = AS    ........(i)

BP = BQ   ........(ii)

CR = CQ   ........(iii)

DR = DS   ........(iv)

Adding (i), (ii), (iii), (iv), we get 

 

    AP + BP + CR + DR = AS + BQ + CQ + DS

 

  ( AP + BP ) + ( CR + DR ) = ( AS + DS ) + ( BQ + CQ )

 

⇒  AB + CD = AD + BC

 

⇒  AB + CD = BC + DA