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CBSE Sample Paper of Mathematics for Class X

Design of Sample Question Paper Mathematics, SA-1 Class X (2010-2011)
Type of QuestionMarks per questionTotal No. of QuestionsTotal Marks
M.C.Q.11010
SA-I2816
SA-II31030
LA4624
TOTAL3480
Blue Print Sample Question Paper-1 SA-1
Topic / UnitMCQSA(I)SA(II)LATotal
Number System2(2)1(2)2(6)-5(10)
Algebra2(2)2(4)2(6)2(8)8(20)
Geometry1(1)2(4)2(6)1(4)6(15)
Trigonometry4(4)1(2)2(6)2(8)9(20)
Statistics1(1)2(4)2(6)1(4)6(15)
TOTAL10(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
  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-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 obtained0-1010-2020-3030-4040-50
Number of students610123220
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 x2-2×-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:
Classes0-1010-2020-3030-4040-50
Frequency71213108
ORThe mean of the following frequency distribution is 25. Find the value of p.
Classes0-1010-2020-3030-4040-50
Frequency2353P
G
“1
28.     Find the median of the following data
Classes0-1010-2020-3030-4040-5050-6060-7070-8080-9090-100
Frequency5343347978
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/hectare50-5555-6060-6565-7070-7575-80
Number of farms2812243816
Change the above distribution to more than type distribution and draw its ogive.

Answers:

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 incomeLess than
120140160180200
Number of works1226344050
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) = x2-2×-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 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
Classes0-1010-2020-3030-4040-50
Class marks (x.)515253545
Frequency (fi)71213108
a- xi”25 di = -i—10-2-1012
fidi-14-1201016
Ifi = 50, I fidi = 0                                                                                                                                           1/2x = A.M + ——– x10 = 25+0 = 25.0IfiOR
Classes0-1010-2020-3030-4040-50
Frequency (fi)2353P
Class mark (x.)515253545
fixi104512510545p
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+7×-15Now 2x2+7×-15 = 2x2+10x-3×-15 = (2×-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
ClassesFrequencyCumulative Frequency(More than type)
50-55250 or more than 50100
55-60855 or more than 5598
60-651260 or more than 6090
65-702465 or more than 6578
70-753870 or more than 7054
75-801675 or more than 7516