Design of Sample Question Paper Mathematics, SA1 Class X (20102011)
Type of Question 
Marks per question 
Total No. of Questions 
Total Marks 
M.C.Q. 
1 
10 
10 
SAI 
2 
8 
16 
SAII 
3 
10 
30 
LA 
4 
6 
24 
TOTAL 

34 
80 
Blue Print Sample Question Paper1 SA1
Topic / Unit 
MCQ 
SA(I) 
SA(II) 
LA 
Total 
Number System 
2(2) 
1(2) 
2(6) 
 
5(10) 
Algebra 
2(2) 
2(4) 
2(6) 
2(8) 
8(20) 
Geometry 
1(1) 
2(4) 
2(6) 
1(4) 
6(15) 
Trigonometry 
4(4) 
1(2) 
2(6) 
2(8) 
9(20) 
Statistics 
1(1) 
2(4) 
2(6) 
1(4) 
6(15) 
TOTAL 
10(10) 
8(16) 
10(30) 
6(24) 
34(80) 
Sample Question Paper Mathematics First Term (SA1) Class X 20102011
Time: 3
to 3
^{1}/
_{2} hours
M.M.: 80
All questions are compulsory.
The questions 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. How ever, 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.
SectionA
Question numbers 1 to 10 are of one mark each.
1. Euclid’s Division Lemma states that for any two postive integers a and b, there exist unique integres q and r such that a=bq+r, where r must satisfy.
(A) l<r<b
(B) 0<r<b
(C) 0<r<b
(D) 0<r<b
2.
In Fig. 1, the graph of a polynomial
p(x) is shown. The number of zeroes of p(x) is
(B) 1
In Fig. 2, if DEIIBC, then x equals
(A) 6 cm (B) 8 cm
(C) 10 cm (D) 12.5 cm
4.
5.
If sin 39 = Cos (96°), where (39) and (96°) are both acute angles, then the value of 0 is
(A) 18° (B) 24° (C) 36° (D) 30°
, . 1 . . Cosec
^{2}0Sec
^{2}0 .
Given that tanG^, the value of
_{Cosec2+Sec2Q} «
(A) 1
(B) 1
(O
(D) x
will terminate after how many places of decimal?
(D) will not terminate 
In Fig. 3, AD=4 cm, BD = 3 cm and CB = 12 cm, then CotG equals
7. The decimal expansion of
120
(A) 1 (B) 2 (C) 3
The pair of linear equations 3x+2y=5; 2x3y=7 have
(A) One solution (B) Two solutions
(C) Many Solutions (D) No solution
15
If sec A = Cosec B = —, then A+B is equal to
(A) Zero
(B) 90°
(C) <90
^{c}
(D) >90
^{c}
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
SECTIONB
Question numbers 11 to 18 carry 2 marks each.
 Is 7x5x3x2+3 a composite number? Justify your answer.
 Can (x2) be the remainder on division of a polynomial p(x) by (2x+3)? Justify your answer. _{x+y}
 In Fig. 4, ABCD is a rectangle. Find the values of x and y.
^{1} xy
12 Fig. 4
14. If 7sin
^{2}9+3Cos
^{2}9 = 4, show that tan6=
1
OR
„ ’15 (2 + 2sin9)(1sin9)
If Cot9=—, evaluate
) f± f
8 (1+Cos9)(22Cos6)
15. In Fig. 5, DEIIAC and DFIIAE. Prove that
FE _EC BF BE
1
16. In Fig. 6, AD 1 BC and BD = CD.
3
Prove that 2CA
^{2}=2AB
^{2}+BC
^{2}
Write the above distribution as less than type cumulative frequency distribution. Find the mode of the following distribution of marks obtained by 80 students:
Marks obtained 
010 
1020 
2030 
3040 
4050 
Number of students 
6 
10 
12 
32 
20 
SECTION C
Question numbers 1928 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
20.
integer.
Prove that —— is irrational.
OR
Prove that f5>/2l is irrational.
21. A person rowing a boat at the rate of 5km/hour in still water, take thrice as much time in going 40 km upstream as in going 40 km downstream. Find the speed of the stream.
22.
In a competitive examination, one mark is awarded for each correct answer while — mark
is deducted for each wrong answer. Jayanti answered 120 questions and got 90 marks. How many questions did she answer correctly?
If
a, /3are zeroes of the polynomial x
^{2}2×15, then form a quadratic polynomial whose zeroes are
(2a) and (2/3).
23. Prove that (cosec0sin9)(sec9cos9) =
tan9+cot9
If cos9+sin9 = V2cos9, show that cos9sin9=V2 sin9 In Fig. 7, AB ± BC, FG1 BC and DEI AC. Prove that AADEAGCF
AABC and ADBC are on the same base BC and on opposite sides of BC and 0 is the point of intersections of AD and BC.
area (AABC) AO
area(ADBC) DO
27. Find mean of the following frequency distribution, using stepdeviation method:
Classes 
010 
1020 
2030 
3040 
4050 
Frequency 
7 
12 
13 
10 
8 
OR
The mean of the following frequency distribution is 25. Find the value of p. 
Classes 
010 
1020 
2030 
3040 
4050 
Frequency 
2 
3 
5 
3 
P 
28. Find the median of the following data
Classes 
010 
1020 
2030 
3040 
4050 
5060 
6070 
7080 
8090 
90100 
Frequency 
5 
3 
4 
3 
3 
4 
7 
9 
7 
8 
SECTION D
Question numbers 29 to 34 carry 4 marks each
29. Find other zeroes of the polynomial p(x) = 2x
^{4}+7x
^{3}19x
^{2}14x+30 if two of its zeroes are
4l and V2.
Prove that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.
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.
Prove that
sec9+tan91 cos 9
tan9sec9+1 1sin9
OR
sec9 cose(90°9)tan9 cot(90°9) +sin
^{2} 55° + sin
^{2} 35
^{c} tan 10° tan 20° tan 60° tan 70° tan 80°
p
^{2}1
If sec9+tan9 = p, prove that sin9 =
Draw the graphs of following equations: 2xy = 1, x+2y = 13
(i) Find the solution of the equation from the graph.
(ii) Shade the triangular region formed by the lines and the yaxis
The following table gives the production yield per hectare of wheat of 100 farms of a village:
Production yield in kg/hectare 
5055 
5560 
6065 
6570 
7075 
7580 
Number of farms 
2 
8 
12 
24 
38 
16 
Change the above distribution to more than type distribution and draw its ogive.
Answers:
Section A
1. (C) 2. (B) 3. (C) 4. (B) 5. (C)
6. (D) 7. (C) 8. (A) 9. (B) 10. (D)
SECTION B
7x5x3x2+3 =3(7x5x2+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
In case of division of a polynomial by another polynomial the degree of remainder (polynomial) is always less than that of divisor
.. (x2) can not be the remainder when p(x) is divided by (2x+3) as degree i
opposite sides of a rectangle are equal
x+y=12 …(i) andxy=8 …(ii) Adding (i) and (ii), we get 2x=20 or x=10
and y=2 .
.. x=10, y=2 }
7sin
^{2}9+3cos
^{2}9=4 or 3(sin
^{2}9 + cos
^{2}9) + 4sin
^{2}6 = 4
1 1
=> sin
^{2}9= sin9= ^9 = 30°
4 2
.. tan9 = tan30
°=^L V3
OR
15
cot 9=— (given) 8
2(1 + sin6)(1 sine)
Given expression =————————— = cot
^{2}9
2(1 + cos9)(1cos9)
225
v 8,
64
_
_{a} be bd
^{15} ^{deiiac}^ec
=5a <
^{!})
y_{2}
dfiiae
bf ef
bd
da
….(ii)
y
16.
BE BF CE FE From (i) and (ii)—=— or —=—
Let BD=x => CD=3x, In right triangle ADC
CA
^{2}=CD
^{2}+AD
^{2}…………….. (i)
and AB
^{2} = AD
^{2}+BD
^{2}
AD
^{2} = AB
^{2} – BD
^{2} …….. (ii)
Substituting (ii) in (i),
CA
^{2} = CD
^{2}+AB
^{2}BD
^{2} OR 2CA
^{2} = 2AB
^{2}+2(9x
^{2}x
^{2}) = 2AB
^{2}+BC
^{2} (•.• BC=4x) => 2CA
^{2}=2AB
^{2}+BC
^{2}
V2+V2
Daily income 
Less than 
120 
140 
160 
180 
200 
Number of works 
12 
26 
34 
40 
50 
18. Modal Class = 3040
.. Mode = 30+
^{32} ^{12}x10 = 30+6.25 = 36.25 6432
i+y_{2}
SECTION C
19. Let a be a positive odd integer
By Euclid’s Division algorithm a=4q+r
Where q, r are positive integes and
0 < r<4 .. a=4qa4q+1 a4q+2a4q+3
y
But 4q and 4q+2 are both even => a is of the form 4q+1 or 4q+3
Let—— = x where x is a rational number
2V3 = 5xorV3=— (i)
2 “
^{w}
5x
As x is a rational number, so is —
.. V3 is also rational which is a contradiction as >/3 is an irrational
2/3
—– is irrational
OR Let 5V2 = y, where y is a rational number
••• 5y = V2 …… (i)
As y is a rational number, so is 5y
.. from (i),
J2 is also rational which is a contradiction as
Jl is irrational
.. 5V2 is irrational
Let the speed of stream be x km/hour
.. Speed of the boat rowing
upstream = (5x) km/hour downstream = (5+x) km/hour
.. According to the question, 40 3×40
— =——— => x = 2.5
5x 5 + x
.. Speed of the stream = 2.5 km/hour OR
Let the number of correct answers be x .. wrong answers are (120x) in number
.. x~(120x) =90
3x
= 150
x=100
.. The number of correctly answered questions = 100 p(x) = x
^{2}2×15 …(i)
As
a, /3 are zeroes of (i), => a+/3 = 2 and a/3
= 15 zeroes of the required polynomial are
2a and a/3
.. sum of zeroes = 2(a+/3) = 4 Product of zeroes = 4 (15) = 60
.. The required polynomial is x
^{2}4×60
LHS can be written as
sinG
(1sin
^{2}e)(1cos
^{2}6) sin6cos6
sine cos6
,cos6 = sine cos6 1
sin
^{2}6 + cos
^{2}0 sin
^{2}0 cos
^{2}6
sin6cos6 sine cos6
1
tanO+cote
ine+cos9 = ^cos9 => sine = (V2l)cos0
(V21)(V2
_{+}1) or sine = ^—
, J} ,—
‘ cos6
(V2
_{+}1
or sine =
cos6 n/2+1
cos6 – sine = V2 sine
ZA+ZC = 90°
Also ZA+Z2=90° => ZC=Z2 Similarly zA=Z1
.. a’s ADE and GCF are equiangular .. a ADE ~ a GCF
Classes 
010 
1020 
2030 
3040 
4050 
Class marks (x.) 
5 
15 
25 
35 
45 
Frequency (fi) 
7 
12 
13 
10 
8 
a ^{x}i”^{25} di = i—
10 
2 
1 
0 
1 
2 
fidi 
14 
12 
0 
10 
16 
Ifi = 50, I fidi = 0
^{1}/
_{2}
x = A.M + ——– x10 = 25+0 = 25.0
Ifi
OR
Classes 
010 
1020 
2030 
3040 
4050 
Frequency (fi) 
2 
3 
5 
3 
P 
Class mark (x.) 
5 
15 
25 
35 
45 
fixi 
10 
45 
125 
105 
45p 
Ifi = 13+p, I fixi = 285+45p Mean = 25 (given)
.. 25x(13+p) = 285+45p => 20p = 40 => p=2
cf
Median = 1+
xh
60+
x10 = 66.43
SECTION D
p(x) = 2x
^{4}+7x
^{3}19x
^{2}14x+30
If two zeroes of p(x) are
J2 and V2
.. (x+V2~)(x^) or x
^{2}2 is a factor of p(x)
p(x) – (x
^{2}2) = [2x
^{4}+7x
^{3}19x
^{2}14x+30] h (x
^{2}2) = 2x
^{2}+7×15
Now 2x
^{2}+7×15 = 2x
^{2}+10x3×15 = (2×3)(x+5)
3
.. other two zeroes of p(x) are and 5
Correctly stated given, to prove, construction and correct figure 4x^
Correct proof OR
. 1
Correctly stated given, to prove, construction and correct figure 4x— correct proof
sec9+tan91 sec9+tan9(sec
^{2}9tan
^{2}9)
LHS — ——————– =—————————————–
tan9sec9 + 1 tan9sec9 + 1
(sec9+tan9)f1sec9 + tan9l „ „ 1 + sin9
^{J} = sec9 + tan9 =
(1sec9 + tan9) cos9
(1 + sin9)(1sin9) (1sin9)cos9
OR
cosec (90° – 9) = sec 9, cot (90° – 9) = tan 9, sin 55° = cos 35 tan 89° = cot 10°, tan 70° = cot 20°, tan 60° = &
23
Classes 
Frequency 
Cumulative Frequency 
(More than type) 
5055 
2 
50 or more than 50 
100 
5560 
8 
55 or more than 55 
98 
6065 
12 
60 or more than 60 
90 
6570 
24 
65 or more than 65 
78 
7075 
38 
70 or more than 70 
54 
7580 
16 
75 or more than 75 
16 
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