Subject

Mathematics

Class

CBSE Class 10

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Sample Papers

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 Multiple Choice QuestionsShort Answer Type

1.

The different dice are tossed together. Find the probability that the product of the two number on the top of the dice is 6.


When two dice are thrown simultaneously, the possible outcomes can be listed as:


 

1

2

3

4

5

6

1

(1, 1)

(1, 2)

(1, 3)

(1, 4)

( 1, 5)

(1, 6)

2

(2, 1)

(2, 2)

(2, 3)

(2, 4)

(2, 5)

(2, 6)

3

(3, 1)

(3, 2)

(3, 3)

(3, 4)

(3, 5)

(3, 6)

4

(4, 1)

(4, 2)

(4, 3)

(4, 4)

(4, 5)

(4, 6)

5

(5, 1)

(5, 2)

(5, 3)

(5, 4)

(5, 5)

(5, 6)

6

(6, 1)

(6, 2)

(6, 3)

(6, 4)

(6, 5)

(6, 6)


∴ Total number of possible outcomes = 36

The outcomes favourable to the event the product of the two number of the top of the dice is 6 denoted by E are (1, 6), (2, 3), (3, 2) and (6, 1)

∴ Number of favourable outcomes = 4

straight P space left parenthesis space straight E right parenthesis space equals space fraction numerator Favourable space number space of space outcomes over denominator Total space number space of space outcomes end fraction
equals space 4 over 36

space equals space 1 over 9

2773 Views

2.

The points A (4, 7), B (p, 3) and C (7, 3) are the vertices a right triangle, right-angled at B find the value of p.


Given ,
The vertices of a right triangle, such as,
A (4, 7), B (p, 3) and C (7, 3) 



In right ΔABC, using Pythagoras theorem

(AB)2 + (BC)2 = (AC)2  

⇒ [(3 - 7)2 + (p - 4)2] + [(3-3)2 +(7 - p)2] = [(3-7)2 + (7 - 4)2]

⇒ (p - 4)2 + (7 - p)2 = 9

⇒ p2 + 16 -8p + 49 + p2 - 14p = 9

⇒ 2p2 - 22p + 56 = 0

⇒ p- 11p + 28 = 0

⇒ p2 - 4p -7p + 28 =0

⇒ p(p - 4) -7(p - 4)=0

⇒ (p - 7 )(p - 4) = 0

⇒ p - 7 = 0 or p - 4 = 0

⇒ p = 7 or p = 4

Hence, the value of p is 4 or 7 

2708 Views

3.

Solve the following quadratic equation for x,

4x2 + 4bx - (a2 - b2) = 0


Given equation,
4x2 + 4bx - (a2 - b2) = 0 

⇒ 4x2  +4bx - ( a + b) (a - b) = 0

⇒ 4x2 + 2[(a + b) - (a - b)] x - (a + b)(a-b) = 0

⇒ 4x2 +2 (a + b)x - 2(a - b)x - (a + b ) (a - b) = 0

⇒ 2x [2x + (a + b)] [(a - b)[2x + (a + b)] = 0

⇒ 2x + (a + b) = 0 or 2x - (a - b) = 0

⇒ x =  - (a + b)/2 OR x = (a - b)/2

2298 Views

4.

Find the relation between x and y if the points A (x, y), B (-5, 7) and C (-4, 5) are collinear.


It is given that the points A (x, y), B (-5, 7) and C (-4, 5) are collinear.

∴ Area of triangle, ABC = 0

⇒ 1/2 [x (7 - 5) + (-5) (5 - y) + (-4) (y -7)]  = 0

⇒ 2x - 25 + 5y - 4y + 28 = 0

⇒ 2 x + y + 3 = 0

This is the required relation between x and y.

1320 Views

5.

In the given figure, PQ is a chord of a circle with centre O and PT is tangent. If ∠QPT = 60o, find ∠ PRQ.


PQ is the chord of the circle and PT is tangent.

We know that the tangent to a circle is perpendicular to the radius through the point of contact.

∴ ∠ OPT = 90o

Now, Given

∠ QPT = 60o 

∴ ∠ OPQ = ∠OPT - ∠QPT

 ⇒ ∠ OPQ = 90o - 60o = 30o

In Δ OPQ,

OP= OR (Radii of the same circle)

∠ OQP = ∠ OPQ = 30o (in a triangle, equal sides have equal angles opposite to them.)

Now,

∠ OQP + ∠OPQ + ∠POQ = 180o [Angle sum property]

⇒ 30o + 30o + ∠POQ = 180o 

⇒ ∠POQ = 180o - 60o = 120o

⇒ Reflex ∠POQ = 360o -120o = 240o

We know that the angle subtended by an arc of a circle at the centre is twice the angle subtended by it any point on the remaining part of the circle.

Therefore,

Reflex ∠POQ = 2 ∠PRQ

⇒ 240o = 2 ∠PRQ

⇒ ∠PRQ = 240 / 2 = 120o 

Hence, the measure of angle PRQ is 120o.

3511 Views

6.

In an AP, if S5 + S7 = 167 and S10 = 235, then find the AP,  Where Sn denotes the sum of its first n terms.


Let a and d be the first term and the common difference of the AP, respectively

therefore space Sum space of space straight n space terms comma space straight S subscript straight n space equals space straight n over 2 space left square bracket 2 straight a space plus space left parenthesis straight n minus 1 right parenthesis straight d right square bracket
We space have
straight S subscript 5 space plus space straight S subscript 7 space equals space 167

rightwards double arrow space 5 over 2 space left parenthesis 2 straight a space plus space 4 straight d space right parenthesis space plus 7 over 2 space left parenthesis 2 straight a space plus space 6 straight d right parenthesis space equals space 167
rightwards double arrow space 5 space left parenthesis straight a space plus space 2 straight d right parenthesis space space plus 7 space left parenthesis straight a space plus 3 straight d right parenthesis space equals 167

rightwards double arrow 12 straight a space space plus space 31 straight d space equals space 167 space space.. left parenthesis straight i right parenthesis

Also comma

straight S subscript 10 space equals space 235
rightwards double arrow space 10 over 2 space left parenthesis 2 straight a space plus 9 straight d right parenthesis space equals space 23

rightwards double arrow 2 straight a space plus 9 straight d space equals space 47 space space... space left parenthesis ii right parenthesis

solving space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis comma space we space ge
straight a equals space 1 space and space straight d equals 5
Hence comma space required space AP space is space 1 comma space 6 comma space 11 space space

4564 Views

7.

In Given figure, a tower AB is 20 m high and BC, its shadow on the ground, is m20 square root of 3 long. Find the sun's altitude.


Let the sun's altitude be θ.

In ΔABC,

tan space straight theta space equals space AB over BC

rightwards double arrow space space tan space straight theta space equals space fraction numerator 20 over denominator 20 square root of 3 end fraction

rightwards double arrow space space tan space equals space fraction numerator 1 over denominator square root of 3 end fraction
rightwards double arrow space tan space straight theta space equals space tan space 30 to the power of straight o

rightwards double arrow space straight theta space equals space 30 to the power of straight o
Hence comma space the space altitude space of space Sun space is space 30 to the power of straight o
2000 Views

8.

In the given figure, two tangents RQ and RP are drawn from an external point R to the circle with centre O. If ∠PRQ = 120o, then prove that OR = PR + RQ.


Given: RQ and RP are tangents drawn from an external point R  to the circle with centre O such that ∠PRQ = 120o

To prove, PR + PQ = OR

Construction, Join OP and OQ.



Proof:
∠OPR = ∠OQR = 90o
(Radius of the circle is perpendicular to the tangent to the circle through the point contact)

We Know that the centre lies on the bisector of the angle between the two tangents.
So, ∠PRO = ∠QRO = 1/2 ∠PRQ = 60o

Now, In ΔPRO,

Cos space 60 to the power of straight o space equals space PR over OR

rightwards double arrow space 1 half space equals space PR over OR

rightwards double arrow space equals space PR space equals space 1 half OR space.... left parenthesis straight i right parenthesis
Similarly comma space in space increment QRO comma
RQ space equals space 1 half space OR space space... space left parenthesis 2 right parenthesis
Adding space left parenthesis 1 right parenthesis space and space left parenthesis 2 right parenthesis comma space we space get

PR space plus space RQ space equals space 1 half space OR space plus 1 half OR space equals space OR

Hence space Proved.

2137 Views

9.

If the quadratic equation haspx squared minus 2 square root of 5px plus 15 equals 0 two roots, then find the value of p.


The given quadratic equation haspx squared minus 2 square root of 5px plus 15 equals 0 two equal roots.
therefore,

by using the discriminant method

straight b squared space minus space 4 ac space equals space 0 space.. left parenthesis straight i right parenthesis

straight b space equals space minus 2 square root of 5 straight p

straight a space space equals space straight p

straight c space equals space 15

thus comma space putting space the space value space in space equation space left parenthesis straight i right parenthesis
left parenthesis negative 2 square root of 5 straight p end root right parenthesis squared minus 4 space straight x space straight p space straight x 15 space equals 0

rightwards double arrow space left parenthesis negative 2 square root of 5 straight p right parenthesis squared minus 4 space straight x space straight p space straight x 15 space equals 0


⇒  20p2 - 60p = 0

⇒  20p (p - 3) = 0

⇒  p = 0 or p - 3 = 0

⇒  p = 0 or p = 3

p cannot be zero.

Hence, the value of p  is 3. 

12206 Views

10.

In the given figure, a triangle ABC is drawn to circumscribe of radius 3 cm, such that the segments RD and DC are respectively of length 6 cm and 9 cm. If the area of ΔABC is 54 cm2, then find lengths of sides AB and AC.


Suppose the given circle touch the sides AB and AC of the triangle at point F and E, respectively. Let the length of the line segment AF is x cm.


We know that the length of tangents drawn from an external point to a circle are equal.

In ΔABC,

CE = CD = 9 cm (Tangents of the circle from point C)

BF = BD = 6 cm (Tangents on the circle from point B)

AE = AF + FB = x + 6

BC = BD + DC = 6 + 9 = 15

CA = CE + EA  = 9 + x

Area space of space increment OBC space equals space 1 half space straight x space BC space straight x space OD space
space equals space 1 half space straight x space 15 space straight x space 3 space equals space 45 over 2

Area space of space increment OCA space equals space 1 half space straight x space AC space straight x space OE space
equals 1 half space straight x space left parenthesis straight x plus 9 right parenthesis space straight x space 3 space equals space 3 over 2 left parenthesis straight x space plus space 9 right parenthesis

Area space of space increment OAB space equals space 1 half space straight x space AB space straight x space OF
equals 1 half space straight x space left parenthesis space straight x space plus space 6 right parenthesis space straight x space 3 space space equals space 3 over 2 space left parenthesis space straight x space plus space 6 right parenthesis

Area space of space increment ABC space equals space Area space of space increment OBC space plus increment OCA space plus Area space of space increment OAB
space 54 space equals space 45 over 2 space plus 3 over 2 space left parenthesis straight x space plus space 9 right parenthesis space plus 3 over 2 space left parenthesis space straight x space plus space 6 right parenthesis

rightwards double arrow space 54 space equals space 45 space plus space 3 straight x space

rightwards double arrow space 3 space straight x space equals space 54 space minus space 45

rightwards double arrow space straight x space equals space 3

Therefore comma

AB space equals space straight x plus space 6 space equals space 3 space plus space 6 space equals space 9 space cm

and space AC space equals space 9 space plus space straight x space equals space 9 space plus space 3 space equals space 12 space cm

2947 Views