iCBSE.com will not be updated for some time, meanwhile visit cbse.nic.in or cbseacademic.com
bullet CBSE > > Cbse Sample Paper 2011 Mathematics Class

CBSE Sample Paper of Mathematics for Class X

Design of Sample Question Paper Mathematics, SA-1 Class X (2010-2011)
Type of Question Marks per question Total No. of Questions Total Marks
M.C.Q. 1 10 10
SA-I 2 8 16
SA-II 3 10 30
LA 4 6 24
TOTAL 34 80
Blue Print Sample Question Paper-1 SA-1
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 (SA-1) Class X 2010-2011 Time: 3 to 31/2 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. Section-A 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
(D)
(A)
(C)
p(x) is shown. The number of zeroes of p(x) is (B)     1
L p(X)
Fig. 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 (9-6°), where (39) and (9-6°) are both acute angles, then the value of 0 is (A)     18°                          (B)     24°                          (C)     36°                          (D)     30° ,   .   1                           .     . Cosec20-Sec20 . Given that tanG^, the value of Cosec2+Sec2Q « (A)    -1 (B)     1 (O     (D)     -x
6.
(A) (C)
(B) (D)
3 4 4 3
_5_ 12 12 5 147
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
8.
(A)     1                               (B)     2                               (C)     3 The pair of linear equations 3x+2y=5; 2x-3y=7 have (A)    One solution                     (B)    Two solutions (C)    Many Solutions      (D)    No solution
9.
15 If sec A = Cosec B = —, then A+B is equal to (A)    Zero (B)    90° (C)     <90c (D)     >90c 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 SECTION-B Question numbers 11 to 18 carry 2 marks each.
#-c
  1. Is 7x5x3x2+3 a composite number? Justify your answer.
  2. Can (x-2) be the remainder on division of a polynomial p(x) by (2x+3)? Justify your answer.        x+y
  3. In Fig. 4, ABCD is a rectangle. Find the values of x and y.
1 x-y 12 Fig. 4 14.     If 7sin29+3Cos29 = 4, show that tan6= 1 OR „     ’15                                 (2 + 2sin9)(1-sin9) If Cot9=—, evaluate ) -f± f 8          (1+Cos9)(2-2Cos6) 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 2CA2=2AB2+BC2
18.
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 0-10 10-20 20-30 30-40 40-50
Number of students 6 10 12 32 20
SECTION C Question numbers 19-28 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 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 x2-2×-15, then form a quadratic polynomial whose zeroes are (2a) and (2/3).
23.     Prove that (cosec0-sin9)(sec9-cos9) =
24. 25.
tan9+cot9 If cos9+sin9 = V2cos9, show that cos9-sin9=V2 sin9 In Fig. 7, AB ± BC, FG1 BC and DEI AC. Prove that AADE-AGCF
26.
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.
Prove that
area (AABC)   AO area(ADBC)   DO 27.     Find mean of the following frequency distribution, using step-deviation method:
Classes 0-10 10-20 20-30 30-40 40-50
Frequency 7 12 13 10 8
OR The mean of the following frequency distribution is 25. Find the value of p.
Classes 0-10 10-20 20-30 30-40 40-50
Frequency 2 3 5 3 P
G
“1
28.     Find the median of the following data
Classes 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100
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) = 2x4+7x3-19x2-14x+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+tan9-1    cos 9 tan9-sec9+1   1-sin9 OR
Evaluate
sec9 cose(90°-9)-tan9 cot(90°-9) +sin2 55° + sin2 35c tan 10° tan 20° tan 60° tan 70° tan 80°
p2+1
p2-1 If sec9+tan9 = p, prove that sin9 = Draw the graphs of following equations: 2x-y = 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 y-axis The following table gives the production yield per hectare of wheat of 100 farms of a village:
Production yield in kg/hectare 50-55 55-60 60-65 65-70 70-75 75-80
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 .-. (x-2) 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) andx-y=8 …(ii) Adding (i) and (ii), we get 2x=20 or x=10 and y=2                                                                                                                                    . .-.     x=10, y=2                                                                                                                                   } 7sin29+3cos29=4 or 3(sin29 + cos29) + 4sin26 = 4 1                              1 => sin29=-        sin9=-  ^9 = 30° 4                              2 .-. tan9 = tan30°=-^L V3 OR 15 cot 9=— (given) 8 2(1 + sin6)(1 -sine) Given expression =————————— = cot 29
’15x:
2(1 + cos9)(1-cos9) 225 v 8, 64 _a-      be   bd 15 deiiac^ec=5a    <!) y2 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 CA2=CD2+AD2…………….. (i) and AB2 = AD2+BD2 AD2 = AB2 – BD2 …….. (ii) Substituting (ii) in (i), CA2 = CD2+AB2-BD2 OR    2CA2 = 2AB2+2(9x2-x2) = 2AB2+BC2 (•.• BC=4x) =>     2CA2=2AB2+BC2 V2+V2
Daily income Less than
120 140 160 180 200
Number of works 12 26 34 40 50
18.     Modal Class = 30-40 .-. Mode = 30+32 12x10 = 30+6.25 = 36.25 64-32 i+y2 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 5-V2 = y, where y is a rational number ••• 5-y = V2       …… (i) As y is a rational number, so is 5-y .-. from (i), -J2 is also rational which is a contradiction as -Jl is irrational .-. 5-V2 is irrational Let the speed of stream be x km/hour .-. Speed of the boat rowing upstream = (5-x) km/hour downstream = (5+x) km/hour .-. According to the question, 40    3×40 — =———- => x = 2.5 5-x    5 + x .-. Speed of the stream = 2.5 km/hour OR Let the number of correct answers be x .-. wrong answers are (120-x) in number .-. x~(120-x) =90 3x = 150 x=100 .-. The number of correctly answered questions = 100 p(x) = x2-2×-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 x2-4×-60
- sinG
LHS can be written as sinG (1-sin2e)(1-cos26) sin6cos6 sine cos6 ,cos6 = sine cos6 1 sin26 + cos20      sin20        cos26 sin6cos6   sine cos6 1 tanO+cote
si
ine+cos9 = ^cos9 => sine = (V2-l)cos0 (V2-1)(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 0-10 10-20 20-30 30-40 40-50
Class marks (x.) 5 15 25 35 45
Frequency (fi) 7 12 13 10 8
a- xi”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 0-10 10-20 20-30 30-40 40-50
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) = 2x4+7x3-19x2-14x+30 If two zeroes of p(x) are -J2 and -V2 .-. (x+V2~)(x-^) or x2-2 is a factor of p(x) p(x) – (x2-2) = [2x4+7x3-19x2-14x+30] h- (x2-2) = 2x2+7×-15 Now 2x2+7×-15 = 2x2+10x-3×-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+tan9-1    sec9+tan9-(sec29-tan29) LHS — ——————– =—————————————– tan9-sec9 + 1                         tan9-sec9 + 1 (sec9+tan9)f1-sec9 + tan9l        „       „   1 + sin9 J = sec9 + tan9 = (1-sec9 + tan9)                                                             cos9 (1 + sin9)(1-sin9) (1-sin9)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)
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