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 QuestionsShort Answer Type

1.

In Fig, from an external point P, two tangents PT and PS are drawn to a circle with centre O and radius r. If OP=2r, show that  ∠ OTS = ∠ OST = 30°.


In the given figure,


Given
OP= 2r
∠OTP = 90°  (radius drawn at the point of contact is perpendicular to the tangent)
Now ,
In ΔOTP,
Sin∠ OPT = OT/OP = 1/2 = Sin 30o
⇒ ∠ OPT = 30o
therefore,
∠ TOP=60o
∴ ΔOTP is a 30o-60o-90o, right triangle.

In ΔOTS,
OT = OS   … (Radii of the same circle)
therefore,
ΔOTS is an isosceles triangle.

∴∠OTS = ∠OST … (Angles opposite to equal sides of an isosceles triangle are equal)

In ΔOTQ and ΔOSQ

OS = OT … (Radii of the same circle)
OQ = OQ ...(side common to both triangles)
∠OTQ = ∠OSQ … (angles opposite to equal sides of an isosceles triangle are
equal)

∴ ΔOTQ ≅ ΔOSQ … (By S.A.S)
∴ ∠TOQ = ∠SOQ = 60° … (C.A.C.T)
∴ ∠TOS = 120° … (∠TOS = ∠TOQ + ∠SOQ = 60° + 60° = 120°)
∴ ∠OTS + ∠OST = 180° – 120° = 60°
∴ ∠OTS = ∠OST = 60° ÷ 2 = 30°

7763 Views

2.

The 4th term of an A.P. is zero. Prove that the 25th term of the A.P. is three times its 11th term.


We have given that
4th term of an A.P.= a4 = 0
∴ a + (4 – 1)d = 0
∴ a + 3d = 0
∴ a = –3d        ….(1)

25th term of an A.P. = a25
= a + (25 – 1)d
= –3d + 24d      ….[From the equation (1)]
= 21d

3 times 11th term of an A.P. = 3a11
= 3[a + (11 – 1)d]
= 3[a + 10d]
= 3[–3d + 10d]
= 3 × 7d
= 21d

∴ a25 = 3a11

i.e., the 25th term of the A.P. is three times its 11th term.

4144 Views

3.

In Fig. , PQ is tangent at point C to a circle with centre O. If AB is a diameter and ∠CAB = 30°, find ∠PCA.


In the given figure,

In Δ ACO,
OA=OC (Radii of the same circle)
Therefore,
ΔACO is an isosceles triangle.
∠CAB = 30° (Given)
∠CAO = ∠ACO = 30° (angles opposite to equal sides of an isosceles triangle are equal)
∠PCO = 90° …(radius is drawn at the point of
contact is perpendicular to the tangent)
Now ∠PCA = ∠PCO – ∠CAO
Therefore,
∠PCA = 90° – 30° = 60°

30599 Views

4.

Let P and Q be the points of trisection of the line segment joining the points A(2, -2) and B(-7, 4) such that P is nearer to A. Find the coordinates of P and Q.


P and Q are the points of trisection of AB,
therefore, AP=PQ=QB
Thus, P divides AB internally in the ratio 1:2 and Q divide AB internally in the ratio 2:1.

straight P identical to space open parentheses fraction numerator 1 left parenthesis negative 7 right parenthesis plus 2 left parenthesis 2 right parenthesis over denominator 1 plus 2 end fraction comma fraction numerator 1 left parenthesis 4 right parenthesis plus 2 left parenthesis negative 2 right parenthesis over denominator 1 plus 2 end fraction close parentheses identical to space open parentheses fraction numerator negative 7 plus 4 over denominator 3 end fraction comma fraction numerator 4 minus 4 over denominator 3 end fraction close parentheses
identical to open parentheses fraction numerator negative 3 over denominator 3 end fraction comma 0 close parentheses space identical to open parentheses negative 1 comma 0 close parentheses
straight Q identical to space open parentheses fraction numerator 2 left parenthesis negative 7 right parenthesis plus 1 left parenthesis 2 right parenthesis over denominator 2 plus 1 end fraction comma fraction numerator 2 left parenthesis 4 right parenthesis plus 1 left parenthesis negative 2 right parenthesis over denominator 2 plus 1 end fraction close parentheses identical to open parentheses fraction numerator negative 14 plus 2 over denominator 3 end fraction comma fraction numerator 8 minus 2 over denominator 3 end fraction close parentheses
identical to open parentheses fraction numerator negative 12 over denominator 3 end fraction comma 6 over 3 close parentheses space identical to left parenthesis negative 4 comma 2 right parenthesis

8136 Views

5.

Prove that the points (3, 0), (6, 4) and (-1, 3) are the vertices of a right-angled isosceles triangle.


Given, the points (3, 0), (6, 4) and (-1, 3) are the vertices of a right-angled isosceles triangle.
Therefore, 
Let A (3,0), B (6,4) and C (-1,3) be the given points.
Now,

AB space equals space square root of left parenthesis 6 minus 3 right parenthesis squared plus left parenthesis 4 plus 0 right parenthesis squared end root space equals space square root of 3 squared plus 4 squared end root space equals space square root of 9 plus 16 end root space equals square root of 25
BC equals square root of negative left parenthesis 1 minus 6 right parenthesis squared plus left parenthesis 3 minus 4 right parenthesis squared end root space equals square root of left parenthesis negative 7 right parenthesis squared plus left parenthesis negative 1 right parenthesis squared end root space equals space square root of 49 plus 1 end root space equals square root of 50
AC equals square root of left parenthesis negative 1 minus 3 right parenthesis squared plus left parenthesis 3 minus 0 right parenthesis squared end root space equals space square root of left parenthesis negative 4 right parenthesis squared plus 3 squared space end root space equals space square root of 16 plus 9 end root space equals square root of 25
Since space AB equals BC
therefore
AB squared equals left parenthesis square root of 25 right parenthesis squared space equals 25
BC squared space equals left parenthesis square root of 50 right parenthesis end root squared space equals 50
AC squared space equals space left parenthesis square root of 25 right parenthesis squared space equals 25
therefore space AB squared plus AC squared space equals space BC squared
Thus comma space increment ABC space is space right space minus angled space isosceles space triangle

6944 Views

6.

If -5 is a root of the quadratic equation 2x2  + px – 15 = 0 and the quadratic equation p(x2 + x)k = 0 has equal roots, find the value of k.


We Given -5 is a root of the quadratic equation 2x2+px-15 =0
, -5 satisfies the given equation.
∴ 2 (-5)2 +p(-5)-15=0
50-5p-15 =0
35-5p=0
5p=35 ⇒ p=7

Substituting p=7 in (x2+x)+k =0, we get
7(x2+x)+k =0
7x2+7x+k=0
The roots of the equation are equal
∴ Discriminant =b2-4ac =0
Here, a= 7, b=7,c=k
b2-4ac=0
∴(7)2-4(7)(k) =0
49-28k =0
28k=49
k= 49/28 
=7/4

3820 Views

7.

A ladder leaning against a wall makes an angle of 60° with the horizontal. If the foot of the ladder is 2.5 m away from the wall, find the length of the ladder.


Let AB be the ladder and CA be the wall.

The ladder makes an angle of 60o with the horizontal.
∴ ΔABC is a 30o-60o-90o, right triangle.


Given: BC = 2.5 m, ∠ABC = 60°
AB = 5 cm and ∠BAC = 30°


From Pythagoras Theorem, we have
AB2 = BC2 + CA2
52 = (2.5)2 + (CA)2
(CA)2 = 25 – 6.25 = 18.75 m
Hence, length of the ladder is square root of 18.75 end root space almost equal to 4.33 straight m

4627 Views

8.

In Fig.2, a quadrilateral ABCD is drawn to circumscribe a circle, with centre O, in such a way that the sides AB, BC, CD and DA touch the circle at the points P, Q, R and S respectively. Prove that AB + CD = BC + DA.




In the Given figure, since tangents are drawn from an exterior point to a circle are equal in length,
AP = AS ….(1)
BP = BQ ….(2)
CR = CQ ….(3)
DR = DS ….(4)
Adding equations (1), (2), (3) and (4), we get

AP + BP + CR + DS = AS + BQ + CQ + DS
(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
AB + CD = AD + BC
AB + CD = BC + DA

Hence proved

2810 Views

9.

For what value of k will k+9, 2k-1 and 2k+7 are the consecutive terms of an A.P?


If k+9, 2k-1 and 2k+7 are the consecutive terms of A.P, then the common difference will be the same.
∴ (2k – 1) – (k + 9) = (2k + 7) – (2k – 1)

 k – 10 = 8

∴ k = 18

6286 Views

10.

A card is drawn at random from a well -shuffled fled pack of 52 playing cards. Find the probability of getting neither a red card nor a queen.


There space are space 26 space red space cards space including space 2 space red space queens.
Two space more space queens space with space 26 space red space cards space will space be space space 26 plus 2 space equals space 28
therefore space Probability space of space getting space straight a space red space card space or space straight a space queen space equals space 28 over 52
therefore Probability space of space getting space neither space straight a space red space card space nor space space queen space equals
1 minus 28 over 52 space equals space 24 over 52 equals 6 over 13

1.

3046 Views