CBSE > Papers > 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 2010-2011Time: 3 to 31/2 hoursM.M.: 80All 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-AQuestion 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<b2.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 cm4.5.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)
 3 443
 _5_ 1212 5147
 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 equals7.      The decimal expansion of120
 8
(A)     1                               (B)     2                               (C)     3The pair of linear equations 3x+2y=5; 2x-3y=7 have(A)    One solution                     (B)    Two solutions(C)    Many Solutions      (D)    No solution
 9
15If sec A = Cosec B = —, then A+B is equal to(A)    Zero(B)    90°(C)     <90c(D)     >90c10.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)     20.5SECTION-BQuestion numbers 11 to 18 carry 2 marks each.
 #-c
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-y12 Fig. 414.     If 7sin29+3Cos29 = 4, show that tan6=1OR„     ’15                                 (2 + 2sin9)(1-sin9)If Cot9=—, evaluate ) -f± f8          (1+Cos9)(2-2Cos6)15.     In Fig. 5, DEIIAC and DFIIAE. Prove thatFE _EC BF  BE116.     In Fig. 6, AD 1 BC and BD = -CD.3Prove 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 CQuestion 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 positive20.integer.Prove that —— is irrational.ORProve 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 — markis 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 x2-2x-15, then form a quadratic polynomial whose zeroes are (2a) and (2/3). 23.     Prove that (cosec0-sin9)(sec9-cos9) =
 24. 25.
tan9+cot9If 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)   AOarea(ADBC)   DO27.     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 ORThe 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 DQuestion numbers 29 to 34 carry 4 marks each29.     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.ORProve 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 thatsec9+tan9-1    cos 9tan9-sec9+1   1-sin9OR
 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
p2-1If 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.

Section A1.      (C)                            2.      (B)    3.      (C)    4.      (B)    5.      (C)6.       (D)                           7.       (C)     8.       (A)     9.       (B)     10.     (D)SECTION B7x5x3x2+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 numberIn 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 iopposite sides of a rectangle are equalx+y=12 …(i) andx-y=8 …(ii) Adding (i) and (ii), we get 2x=20 or x=10and y=2                                                                                                                                    ..-.     x=10, y=2                                                                                                                                   }7sin29+3cos29=4 or 3(sin29 + cos29) + 4sin26 = 41                              1=> sin29=-        sin9=-  ^9 = 30° 4                              2.-. tan9 = tan30°=-^L V3OR15cot 9=— (given) 82(1 + sin6)(1 -sine)Given expression =————————— = cot 29
 ’15x:
2(1 + cos9)(1-cos9)225v 8,64_a-      be   bd15 deiiac^ec=5a    <!)y2dfiiaebf efbdda….(ii)y16.BE  BF     CE  FE From (i) and (ii)—=— or —=—Let BD=x => CD=3x, In right triangle ADCCA2=CD2+AD2…………….. (i)and AB2 = AD2+BD2AD2 = AB2 – BD2 …….. (ii)Substituting (ii) in (i),CA2 = CD2+AB2-BD2 OR    2CA2 = 2AB2+2(9x2-x2) = 2AB2+BC2 (•.• BC=4x) =>     2CA2=2AB2+BC2V2+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 64-32i+y2SECTION C19.     Let a be a positive odd integerBy Euclid’s Division algorithm a=4q+rWhere q, r are positive integes and 0 < r<4 .-. a=4qa4q+1 a4q+2a4q+3yBut 4q and 4q+2 are both even => a is of the form 4q+1 or 4q+3Let—— = x where x is a rational number2V3 = 5xorV3=— (i)2  “w5xAs x is a rational number, so is —.-. V3 is also rational which is a contradiction as >/3 is an irrational2/3—– is irrationalOR 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 irrationalLet the speed of stream be x km/hour.-. Speed of the boat rowingupstream = (5-x) km/hour downstream = (5+x) km/hour.-. According to the question, 40    3×40— =———- => x = 2.55-x    5 + x.-. Speed of the stream = 2.5 km/hour ORLet the number of correct answers be x .-. wrong answers are (120-x) in number.-. x~(120-x) =903x= 150x=100.-. The number of correctly answered questions = 100 p(x) = x2-2x-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 x2-4x-60
 - sinG
LHS can be written assinG(1-sin2e)(1-cos26) sin6cos6sine cos6,cos6 = sine cos6 1sin26 + cos20      sin20        cos26sin6cos6   sine cos61tanO+cote
 si
ine+cos9 = ^cos9 => sine = (V2-l)cos0(V2-1)(V2+1) or sine = ^—, J} ,‘- cos6(V2+1or sine =cos6 n/2+1cos6 – sine = V2 sineZA+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                                                                                                                                           1/2x = A.M + ——– x10 = 25+0 = 25.0IfiOR
 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-cfMedian = 1+xh60+x10 = 66.43SECTION Dp(x) = 2x4+7x3-19x2-14x+30If 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+7x-15Now 2x2+7x-15 = 2x2+10x-3x-15 = (2x-3)(x+5)3.-. other two zeroes of p(x) are -and -5Correctly stated given, to prove, construction and correct figure 4x^Correct proof OR. 1Correctly stated given, to prove, construction and correct figure 4x— correct proofsec9+tan9-1    sec9+tan9-(sec29-tan29)LHS — ——————– =—————————————–tan9-sec9 + 1                         tan9-sec9 + 1(sec9+tan9)f1-sec9 + tan9l        „       „   1 + sin9J = sec9 + tan9 =(1-sec9 + tan9)                                                             cos9(1 + sin9)(1-sin9) (1-sin9)cos9ORcosec (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