Maths X- CBSE Sample Questions with Answers

Design of Question Paper
Mathematics (041)
Type of Question
Marks per Question
Total no. of Questions
Total Marks
M.C.Q
1
8
8
SA-I
2
6
12
SA-II
3
10
30
LA
4
10
40
34
90
Total
The Question Paper will include value based question(s) to the extent of 3–5 marks
Weightage
S.No.
Unit No.
Topic
Weightage
1
II
Algebra
23
2
III
Geometry
17
3
IV
Trigonometry
08
4
V
Probability
08
5
VI
Coordinate Geometry
11
6
VII
Mensuration
23
Total
90
CBSE (CCE) Sample Questions 1
CBSE (CCE) Sample Questions
Multiple Choice type [1 mark] questions
1. Value(s) of k for which the quadratic equation 2x2 – kx + k = 0 has equal roots is/are
(a) 0 only
(b) 4 (c) 8 only
(d) 0, 8
2. A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q, such that
OQ = 12 cm. Length of PQ is
(a) 12 cm
(b) 13 cm
(c) 8.5 cm
(d) 119 cm
3. A card is drawn from a deck of cards numbered 1 to 52. The probability that the number on the card is a perfect square is
(a) 1
13
(b) 2
13
4. The distance of the point P(2, 3) from the x-axis is
(a) 2 units
(b) 3 units (c) 7
52
(c) 1 unit
5. The area of the circle that can be inscribed in a square of side 6 cm is
(a) 36 p cm2
(b) 18 p cm2 (c) 12 p cm2
(d) 10
13
(d) 5 units
(d) 9 p cm2
Short Answer-I type [2 marks] questions
1.
2.
3.
4.
5.
Find the nature of the roots of the quadratic equation: 3x2 – 4 3 x + 4 = 0.
For what value of k are 2k, k + 10 and 3k + 2 in AP ?
Prove that tangents drawn at the ends of a diameter of a circle are parallel.
Prove that the parallelogram circumscribing a circle is a rhombus.
12 defective pens are accidently mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it
is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.
Short Answer-II type [3 marks] questions
1. Find the roots of the following quadratic equation by factorisation: 2 x2 – 7x + 5 2 = 0.
2. In an AP, first term is 5, common difference is 3 and nth term is 50. Find the value of n and the sum of its first
n terms.
3. A contractor plans to install two slides for the children to play in a park. For the children below the age of 5 years, she prefers
to have a slide whose top is at a height of 1.5 m, and is inclined at an angle of 30° to the ground, whereas for elder children,
she wants to have a steep slide at a height of 3 m and inclined at an angle of 60° to the ground. What should be the length of
the slide in each case?
4. A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the segment of the circle.
(Use p = 3.14 and 3 = 1.73)
5. A circus tent is cylindrical up to a height of 3 m and conical above it. If the diameter of the base is 105 m and the slant height
of the conical part is 53 m, find the area of canvas used in making the tent.
Long Answer type [4 marks] questions
1. The numerator of a fraction is 2 less than the denominator. If 1 is added to both numerator and denominator, the sum of the
19
new and original fraction is . Find the original fraction.
15
2. The sum of the first n terms of an AP is given by Sn = 3n2 – 4n. Determine the AP and the 12th term.
3. Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
2 Together with® Mathematics—X
4. A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle
of elevation of the balloon from the eyes of the girl at any instant is 60°. After sometime, the angle of elevation reduces to
30° (see figure). Find the distance travelled by the balloon during the interval.
60°
88.2 m
30°
5. A sphere of diameter 6 cm is dropped into a cylindrical vessel partly filled with water. The diameter of the vessel is 12 cm.
If the sphere is completely submerged, how much will the surface of water be raised?
CBSE (CCE) Sample Questions 3
Solutions
[1 Mark]
As sum of interior angles on the same side of transversal is
180°.
⇒
PQ || RS
S
R
C
4.
1.(d)For equal roots, D = 0 ⇒ (– k) – 4 × 2 × k = 0
2
⇒ k2 – 8k = 0 ⇒ k(k – 8) = 0 ⇒ k = 0, 8
2.(d) In right angled DOPQ,
OQ2 =OP2 + PQ2
⇒ (12)2 =(5)2 + PQ2
D
B
⇒
144 =25 + PQ2 ⇒ PQ2 = 144 – 25
⇒
PQ = 144 – 25 cm = 119 cm
P
5 cm
O
Q
12 cm
3.(c) Favourable cases for a perfect
square number are 1, 4, 9, 16, 25, 36 and 49.
⇒ Number of favourable cases = 7
7
52
4.(b)Distance of P(2, 3) from the x-axis is equal to the
magnitude of y-coordinate, i.e. 3 units.
Probability =
6 cm
5.(d) diameter of a circle = 6 cm
\
\
radius =3 cm
Area =p(3)2cm2
PA = PD, QA = QB, RB = RC and SC = SD
[Length of the tangents drawn from a point outside the circle
is equal]
Consider, PQ + RS =PA + AQ + RC + SC
=PD + QB + RB + SD
=(PD + SD) + (QB + RB)
=PS + QR
⇒
2PQ =2PS[Q PQRS is a parallelogram
\ PQ =RS and PS = QR]
⇒
PQ =PS
As adjacent sides of the parallelogram are equal, so
parallelogram PQRS is a rhombus.
5. Total pens = 132 (good) + 12 (defective) = 144
Probability of a good pen =
6 cm
\
= 9p cm2
Q
A
P
132 11
=
144 12
[3 Marks]
1. Consider equation
2 x2 – 7x + 5 2 = 0
⇒ 2 x2 – 5x – 2x + 5 2 = 0
[2 Marks]
⇒ x( 2 x – 5) – 2 ( 2 x – 5) = 0
1. Given equation is
⇒(x –
3x2 – 4 3 x + 4 = 0
\ D = (–4 3 ) 2 – 4 × 3 × 4 = 48 – 48 = 0
As D = 0, roots are real and equal.
2. If 2k, k + 10 and 3k + 2 are in AP, then
(k + 10) – 2k =(3k + 2) – (k + 10)
⇒
– k + 10 =2k – 8 ⇒ 3k = 18 ⇒ k = 6
3. PR is diameter, O is the centre, PQ and RS are tangents at
P and R respectively.
R
S
2 )( 2 x – 5) = 0
⇒ Either x –
⇒
2 = 0 or
2x–5=0
x = 2 or
2. Given, = 5, d = 3 and an = 50.
5
2
Now, an =a + (n – 1)d
⇒
50 =5 + (n – 1)3 ⇒ 45 = 3n – 3
⇒
3n =48 ⇒ n = 16
16
S16 = [2 × 5 + (16 – 1)3]
2
=8[10 + 45] = 8 × 55 = 440
Hence,
O
3. For children below five years:
Q
\
∠ORS =∠OPQ = 90°
[Q radius is
perpendicular to the tangent at the point of contact]
Now ∠PRS + ∠RPQ =90° + 90° = 180°
4 Together with® Mathematics—X
Let length of slide be x m.
Now, in right angled DAOB,
x
= cosec 30°
1.5
⇒ x = 2 × 1.5 m = 3 m
B
1.5 m
P
O
x
30°
A
For elder children:
Area of canvas used= curved surface area of cylinder
+ curved surface area of cone
= 2prh + prl
R
Let length of slide be y m.
y
\
= cosec 60°
3
2
⇒
y = 3 ×
=2 3
3
= 2 × 1.732 m = 3.464 m
3m
In right angled DPQR,
Q
y
= 2p d
60°
= p d
=
P
4. Draw OL ^ AB
In DAOL and DBOL
AO =BO [radii]
OL =OL
[common]
∠ALO =∠BLO
[each 90°]
\
DAOL @ DBOL
[RHS]
∠AOL =∠BOL
[c.p.c.t.]
Now, ∠AOB = 120°, ∠AOL =
1
∠AOB = 60°
2
O
B
P
Also, AL = BL ⇒ [c.p.c.t.]
AB =2AL
In right triangle ALO,
⇒
AL
OL
=sin 60° and
= cos 60°
OA
OA
1
3
AL =12 $
cm and OL = × 12 cm
2
2
AL = 6 3 cm and OL = 6 cm
\
AB =2 × 6 3 = 12 3 cm
⇒
Area of segment APB
= <
120°
1
× p (12) 2 – × 12 3 × 6F cm2
360°
2
= [48 × 3.14 – 36 × 1.73] cm2
= 150.72 – 62.28 = 88.44 cm
5.
59 × 105 22 2
m = 9735 m2
×
2
7
1. Let numerator = x
\ denominator = x + 2
x
Fraction =
x+2
According to given question,
...(i)
x+1
x
19
=
+
(x + 2) + 1 x + 2 15
cm °
12 60
L
105
n [6 + 53] m2
2
[4 Marks]
⇒
A
105
105
n × 3 + pd
n × 53
2
2
x+1
x
19
= +
x+3 x+2
15
⇒15[(x + 1) (x + 2) + x(x + 3)] = 19(x + 3)(x + 2)
⇒ 15[x2 + 3x + 2 + x2 + 3x] = 19(x2 + 5x + 6]
⇒
30x2 + 90x + 30 = 19x2 + 95x + 114
⇒
11x2 – 5x – 84 = 0
⇒
11x2 – 33x + 28x – 84 = 0
⇒
11x(x – 3) + 28(x – 3) = 0
⇒
(11x + 28) (x – 3) = 0
28
⇒ x = –
[rejected]
or x = 3.
11
3
3
\ Fraction is
= [From (i)]
3+2 5
2. Sn = 3n2 – 4n
S1 = a1 = 3 – 4 = – 1
S2 = a1 + a2 = 3(2)2 – 4 × 2 = 4
S3 = a1 + a2 + a3 = 3(3)2 – 4 × 3 = 15
\ a2 = S2 – S1 = 4 + 1 = 5
a3 = S3 – S2 = 15 – 4 = 11
\ AP is – 1, 5, 11, ...
2
Here, first term (a) = – 1 and common difference (d) = 6
\ a12 = – 1 + (12 – 1) (6) = – 1 + 66 = 65.
53 m
D
3. To show: ∠AOB + ∠COD = 180°
3m
105 m
and ∠BOC + ∠AOD = 180°.
R
S
C
O
Q
Construction: Join OP, OQ, OR and OS
Proof: In DAPO and ASO
A
P
B
CBSE (CCE) Sample Questions 5
AP =AS
[Length of tangents from point outside the circle are equal]
OA =OA
[common]
OP =OS
[radii]
\
DAOP @ DAOS
[SSS congruence rule]
1
⇒ ∠PAO = ∠SAO ⇒ ∠OAP = ∠DAB
2
1
1
Similarly, ∠OBP = ∠ABC, ∠OCR = ∠BCD
2
2
1
and ∠ODR = ∠ADC
2
Now, ∠AOB + ∠COD = (180° – ∠OAP – ∠OBP )
After sometime, balloon is at E, ∠EAQ = 30°.
In right triangle APC
AP
=cot 60°
CP
1
87
⇒ AP = 87 ×
=
m
3
3
In right triangle AQE
...(i)
+ (180° – ∠OCR – ∠ODR)
1
(∠DAB + ∠ABC + ∠BCD + ∠CDA)
2
[Using (i)]
= 360° –
1
× 360° = 180°
2
Similarly, we can show
= 360° –
⇒
\
AQ
=cot 30°
EQ
AQ =87 3 m
PQ =CE = AQ – AP = 87 3 –
=87 e
3–1
= 87 ×
3
2
o
m = 58 3 m
3
\ Distance travelled is 58 3 m.
5. Let water is raised by h cm.
∠BOC + ∠AOD =180°
4. We have to find CE.
h
C
60°
A
12 cm
88.2 m
30°
P
Q
1.2 m
B
6 cm
E
D
Here, CP = EQ = (88.2 – 1.2) m = 87 m
When balloon is at C, ∠CAP = 60°.
6 Together with® Mathematics—X
F
\ Volume of water raised = volume of sphere
⇒
4
p(6)2 × h = p(3)3
3
36h =36 ⇒ h = 1 cm
\ Water is raised by 1 cm.
87
3

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