CBSE Sample Question Paper (Solved)
[Released by CBSE, New Delhi in July 2010 for First Term (SA-I) to be held during
the academic year 2010-11 and onward]
Time : 3 to 3
1
hours
2
M.M. : 80
General Instructions
(i) All questions are compulsory.
(ii) The question paper consists of 34 questions divided into four sections A, B, C and D.
Section A comprises of 10 questions of 1 mark each, Section B comprises of 8
questions of 2 marks each, Section C comprises of 10 questions of 3 marks each and
Section D comprises of 6 questions of 4 marks each.
(iii) Question numbers 1 to 10 in Section A are multiple choice questions, where you are
to select one correct option out of the given four.
(iv) There is no overall choice. However, internal choice has been provided in 1 question
of two marks, 3 questions of three marks each and 2 questions of four marks each. You
have to attempt only one of the alternatives in all such questions.
(v) Use of calculators is not permitted.
SECTION – A
(Question numbers 1 to 10 are of one mark each.)
1. Euclid’s Division Lemma states that for any two positive integers a and b, there exist
unique integres q and r such that a = bq + r, where r must satisfy :
(a) 1 < r < b
(b) 0 < r < b
(c) 0 < r < b
(d) 0 < r < b
Sol. (c) 0 < r < b
2. In the figure, the graph of a polynomial p(x) is shown.
The number of zeroes of p(x) is :
(a) 4
(b) 1
(c) 2
(d) 3
Sol. (b) Since the graph cuts the x-axis at only one point,
hence p(x) has only one zero.
3. In the figure, if DE || BC, then x equals :
(a) 6 cm
(b) 8 cm
(c) 10 cm
(d) 12.5 cm
AD DE
=
Sol. (c) We have
[By BPT]
AB BC
2 4
⇒ = ⇒ x = 10 cm
5 x
CBSE Sample Question Paper (Solved) (Term - I)
1
4. If sin 3θ = cos (θ – 6°), where (3θ) and (θ – 6°) are both acute angles, then the value
of θ is :
(a) 18°
(b) 24°
(c) 36°
(d) 30°
Sol. (b) sin 3θ = cos{90° – (96° – θ)} = sin (96° – θ)
96°
= 24°
⇒ 3θ = 96° – θ ⇒ θ =
4
1
cosec 2 θ − sec 2 θ
5. Given that tan θ =
, the value of
is :
3
cosec 2 θ + sec 2 θ
1
1
(d) –
(a) –1
(b) 1
(c)
2
2
Sol.
(c)
cosec 2 θ − sec 2 θ
cosec 2 θ + sec 2 θ
=
1 + cot 2 θ − 1 − tan 2 θ
1 + cot 2 θ + 1 + tann 2 θ
cot 2 θ − tan 2 θ
3−
1
3
8 3 1
= × =
1
3 16 2
cot θ + tan θ + 2 3 + + 2
3
6. In the figure, AD = 4 cm, BD = 3 cm and CB = 12 cm, then cot θ equals :
=
2
3
4
4
(c)
3
(a)
Sol.
=
2
5
12
12
(d)
5
(b)
(d) AB = AD2 + BD2 = 16 + 9 cm = 5 cm
∴ cot θ =
BC 12
=
AB 5
147
will terminate after how many places of decimal?
120
(b) 2
(c) 3
(d) will not terminate
7. The decimal expansion of
(a) l
Sol.
8.
Sol.
9.
Sol.
2
147
= 1.225
120
The pair of linear equations 3x + 2y = 5; 2x – 3y = 7 have :
(a) One solution (b) Two solutions (c) Many Solutions (d) No solution
(a) Here a1 = 3, b1 = 2, c1 = 5, a2 = 2, b2 = –3, c2 = 7.
a1 b1
∵ a ≠ b hence, equations have unique solution.
2
2
15
, then A + B is equal to :
If sec A = cosec B =
7
(a) Zero
(b) 90°
(c) <90°
(d) >90°
(b) sec A = cosec (90° – A) = cosec B
⇒ 90° – A = B ⇒ A + B = 90°
(c)
Sample Papers in Mathematics-X (Term - I)
10. For a given data with 70 observations the ‘less then ogive’ and the ‘more than ogive’
intersect at (20.5, 35). The median of the data is :
(a) 20
(b) 35
(c) 70
(d) 20.5
Sol. (d) Median is the x-coordinate of (20.5, 35) i.e. 20.5.
SECTION-B
(Question numbers 11 to 18 carry 2 marks each.)
11. Is 7 × 5 × 3 × 2 + 3 a composite number? Justify your answer.
Sol. 7 × 5 × 3 × 2 + 3 = 3(7 × 5 × 2 + 1) = 3 × 71 ....(i)
By Fundamental Theorem of Arithmetic, every composite number can be expressed as
product of primes in a unique way, apart from the order of factors.
∴ (i) is a composite number
12. Can (x – 2) be the remainder on division of a polynomial p(x) by (2x + 3)? Justify your
answer.
Sol. In case of division of a polynomial by another polynomial the degree of remainder
(polynomial) is always less than that of divisor
∴ (x – 2) cannot be the remainder when p(x) is divided by (2x + 3), as degree is same
13. In the figure, ABCD is a rectangle.
Find the values of x and y.
Sol. Opposite sides of a rectangle are equal
∴ x + y = 12 ... (i) and x – y ...(ii)
Adding (i) and (ii), we get 2x = 20 or x = 10
From (i), y = 2
∴ x = 10, y = 2
14. If 7 sin2θ + 3 cos2θ = 4, show that tan θ =
1
3
Sol. 7 sin2θ + 3cos2θ = 4 ⇒ 3 (sin2θ + cos2θ) + 4sin2θ = 4
1
1
⇒ 3 + 4 sin2 θ = 4 ⇒ sin2θ =
⇒ sin θ =
⇒ θ = 30°
4
2
1
∴ tanθ = tan30° =
3
OR
If cot θ =
Sol.
15
( 2 + 2 sin θ ) (1 − sin θ )
, evaluate
8
(1 + cos θ ) ( 2 − 2 cos θ )
15
[Given]
8
2(1 + sinθ) (1 − sinθ) 1 − sin 2 θ cos2 θ
=
= cot 2 θ
=
Given expression =
2
2
2(1 + cosθ) (1 − cosθ) 1 − cos θ sin θ
cotθ =
2
225
⎛ 15 ⎞
= ⎜ ⎟ =
⎝ 8⎠
64
CBSE Sample Question Paper (Solved) (Term - I)
3
FE EC
=
15. In the figure, DE || AC and DF || AE. Prove that
BF BE
BE BD
=
Sol. In ∆ABC, DE || AC ⇒
...(i)
[BPT]
EC DA
BF BD
=
...(ii)
[BPT]
In ∆BAE, DF||AE ⇒
EF DA
BE BF
CE FE
=
⇒
=
Proved.
From (i) and (ii)
EC EF
BE BF
1
16. In the figure, AD ⊥ BC and BD = CD.
3
Prove that 2CA2 = 2AB2 + BC2
Sol. Let BD = x ⇒ CD = 3x.
…(i)
In right triangle ADC, CA2= CD2 + AD2
2
2
2
And in right ∆ABD, AB = AD + BD
AD2 = AB2 – BD2
…(ii)
⇒
2
From (i) and (ii), we have, CA = CD2 + AB2 – BD2
2CA2 = 2AB2 + 2(9x2 – x2) = 2AB2 + BC2
[∵ BC = 4x]
⇒
2
2
2
2CA = 2AB + BC Proved.
⇒
17. The following distribution gives the daily income of 50 workers of a factory. Write the
distribution as less than type cumulative frequency distribution.
Daily income
(in rupees)
100 – 120
120 – 140
140 – 160
160 – 180
180 – 200
12
14
8
6
10
Number of
Workers
Sol.
Less than
Daily income
120
140
160
180
200
Number of workers
12
26
34
40
50
18. Find the mode of the following distribution of marks obtained by 80 students :
Marks obtained
Number of students
0 – 10
10 – 20
20 – 30
30 – 40
40 – 50
6
10
12
32
20
Sol. Here, the maximum class frequency is 32 and the class corresponding to this frequency
is 30 – 40.
Hence, modal class is 30 – 40
l = 30, h = 10, f1 = 32, f0 = 12, f2 = 20
32 − 12
⎛
⎞
f1 − f 0
∴ Mode = l + ⎜
⎟ × h = 30 + 64 − 32 × 10 = 30 + 6.25 = 36.25
⎝ 2 f1 − f 0 − f 2 ⎠
4
Sample Papers in Mathematics-X (Term - I)
SECTION C
(Question numbers 19 to 28 carry 3 marks each)
19. Show that any positive odd integer is of the form 4q + 1 or 4q + 3, where q is a
positive integer.
Sol. Let a be a positive odd integer.
By Euclid’s Division algorithm a = 4q + r, where q, r are positive integers
and 0 < r < 4 ⇒ a = 4q or 4q + 1 or 4q + 2 or 4q + 3
But, 4q and 4q + 2 are both even ⇒ a is of the form 4q + 1 or 4q + 3 Proved.
20. Prove that
Sol. Let
2 3
is irrational.
5
5x
2 3
= x, where x is a rational number ⇒ 2 3 = 5 x ⇒ 3 =
... (i)
2
5
5x
As x is a rational number, so is
2
∴ 3 is also rational, which is a contradiction as 3 is an irrational
∴
2 3
is irrational Proved.
5
Prove that (5 –
Sol. Let 5
OR
2 ) is irrational.
2 = y , where y is a rational number.
∴ 5 – y = 2 .... (i)
As y is a rational number, so is 5 – y
∴ From (i), 2 is also rational which is a contradiction as
2 is irrational
∴ 5
Proved
2 is irrational.
21. A person rowing a boat at the rate of 5 km/hour in still water, takes thrice as much
time in going 40 km upstream as in going 40 km downstream. Find the speed of the
stream.
Sol. Let the speed of stream be x km/hour
∴ Speed of the boat rowing upstream = (5 – x) km/hour
And, speed of the boat rowing downstream = (5 + x) km/hour
40
3 × 40
∴ According to the question,
=
5− x 5+ x
⇒ 200 + 40x = 600 – 120x ⇒ 160x = 400 ⇒ x = 2.5
∴ Speed of the stream = 2.5 km/hour
OR
1
In a competitive examination, one mark is awarded for each correct answer, while
2
mark is deducted for each wrong answer. Jayanti answered 120 questions and got 90
marks. How many questions did she answer correctly?
Sol. Let the number of correct answers be x
∴ No. of wrong answers are (120 – x)
CBSE Sample Question Paper (Solved) (Term - I)
5
∴2x4 + 7x3 – 19x2 – 14x + 30 = (x2 – 2)(2x2 + 7x – 15)
Now, 2x2 + 7x – 15 = 2x2 + 10x – 3x –15 = (2x – 3) (x + 5)
3
∴Other two zeroes of p(x) are
and –5
2
30. Prove that the ratio of the areas of two similar triangles is equal to the ratio of the
squares of their corresponding sides.
Sol. Given : ∆ABC ~ ∆DEF. AB, DE; AC, DF and BC, EF are corresponding sides.
area ΔABC AB2 AC 2 BC 2
=
=
=
To prove :
area ΔDEF DE 2 DF 2 EF 2
Construction : Draw AG and DH perpendiculars to BC and EF respectively.
Let AG = p and DH = q.
Proof : Since ∆ABC ~ ∆DEF
⇒
⇒
AB
AC BC
=
=
DE
DF EF
AB2
DE 2
=
Now,
area ΔABC
=
area ΔDEF
Since
and
⇒
∠B =
∠AGB =
∆ABG ~
AC 2
DF 2
=
BC 2
EF 2
…(i)
…(ii)
1
BC × p
BC p
2
=
×
…(iii)
1
EF
q
EF × q
2
∠E
[∆ABC ~ ∆DEF]
∠DHE = 90°
∆DEH
p
p
AB
BC
=
=
[From (i)]
⇒
q
q
DE
EF
p
Substituting for
in (iii), we get
q
BC BC BC 2
area ΔABC AB2 AC 2 BC 2
area ΔABC
=
×
=
⇒
=
=
=
Proved.
EF EF EF 2
area ΔDEF DE 2 DF 2 EF 2
area ΔDEF
OR
⇒
Prove that in a triangle, if the square of one side is equal to the sum of the squares
of the other two sides, then the angle opposite to the first side is a right angle.
Sol. Given : ∆ABC in which AC2
= AB2 + BC2.
CBSE Sample Question Paper (Solved) (Term - I)
9
To prove : ∠ABC = 90°.
Construction :At A, draw ∠BAP = 90°.
Mark AD = BC. Join DB.
Proof : In ∆BAD, DB2 = AB2 + AD2
…(i)
2
2
2
In ∆ABC, AC = AB + BC
…(ii)
But,
BC = AD
∴ (i) becomes, DB2 = AB2 + BC2 …(iii)
From (ii) and (iii), we get
AC2 = DB2
⇒
AC = DB
…(iv)
s
Consider ∆ ABC and ABD
AB = AB
BC = AD
and
AC = DB
⇒ ∆ABC ≅ ∆BAD
⇒ ∠ABC = ∠BAD
⇒ ∠ABC = 90°
sec θ + tan θ − 1
cos θ
=
31. Prove that
tan θ − sec θ + 1 1 − sin θ
Sol. LHS =
=
[Pythagoras Theorem]
[Given]
[By construction]
[Common]
[Consutruction]
[From (iv)]
[SSS congruency]
[CPCT]
[∵∠BAD = 90°]
Proved.
secθ + tanθ 1 sec θ + tan θ (sec 2 θ tan 2 θ)
=
tanθ secθ + 1
tan θ sec θ + 1
(secθ + tanθ) [1 secθ + tanθ]
1
sinθ 1 + sin θ
= sec θ + tan θ =
+
=
(1 secθ + tanθ)
cos θ cos θ
cos θ
(1 + sinθ ) (1 sinθ)
1 − sin 2 θ
cos θ
=
=
=
= RHS
(1 sinθ) cos θ
(1 − sin θ) cos θ 1 sin θ
Proved.
OR
Evaluate :
sec ‚ cos ec (90° − ‚ ) − tan θ cot (90° − ‚ ) + sin 2 55° + sin 2 35°
tan 10° tan 20° tan 60° tan 70° tan 80°
Sol. cosec (90° – θ) = secθ, cot (90° – θ) = tanθ, sin 55° = sin (90 – 35)° = cos 35°
tan 80° = tan(90 – 10)° = cot 10°, tan 70° = tan (90 – 20)° = cot 20°, tan 60° =
Given expression becomes
(sec 2 θ tan 2 θ) + (sin 2 35° + cos2 35°)
tan 10° cot 10° tan 20° cot 20° 3
32. If secθ + tanθ = p, prove that sinθ =
10
=
1+1
3
=
3
2
3
p2 − 1
p2 + 1
Sample Papers in Mathematics-X (Term - I)
Change the above distribution to more than type distribution and draw its ogive.
Sol.
Classes
Frequency
Cumulative Frequency
(More than type)
50–55
2
50 or more than 50
100
55–60
8
55 or more than 55
98
60–65
12
60 or more than 60
90
65–70
24
65 or more than 65
78
70–75
38
70 or more than 70
54
75–80
16
75 or more than 75
16
Now plot the points (50, 100), (55, 98), (60, 90), (65, 78), (70, 54) and (75, 16) on
a graph paper to get the required ogive.
12
Sample Papers in Mathematics-X (Term - I)