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 3
1/
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
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 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 . . Cosec
20-Sec
20 .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 equals7. The decimal expansion of120
(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
15If sec A = Cosec B = —, then A+B is equal to(A) Zero(B) 90°(C) <90
c(D) >90
c10.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-BQuestion numbers 11 to 18 carry 2 marks each.- Is 7x5x3x2+3 a composite number? Justify your answer.
- Can (x-2) 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 x-y12 Fig. 414. If 7sin
29+3Cos
29 = 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 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 | 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 x
2-2x-15, then form a quadratic polynomial whose zeroes are
(2a) and (2/3).
23. Prove that (cosec0-sin9)(sec9-cos9) =
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
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) 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 |
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) = 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.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
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
2-1If 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.
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 }7sin
29+3cos
29=4 or 3(sin
29 + cos
29) + 4sin
26 = 41 1=> sin
29=- sin9=- ^9 = 30°
4 2.-. tan9 = tan30
°=-^L V3OR15cot 9=— (given) 82(1 + sin6)(1 -sine)Given expression =————————— = cot
29
2(1 + cos9)(1-cos9)225
v 8,64_
a- be bd
15 deiiac^ec
=5a <
!)
y2dfiiaebf efbdda….(ii)
y16.BE BF CE FE From (i) and (ii)—=— or —=—Let BD=x => CD=3x, In right triangle ADCCA
2=CD
2+AD
2…………….. (i)and AB
2 = AD
2+BD
2AD
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
2V2+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+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+3
yBut 4q and 4q+2 are both even => a is of the form 4q+1 or 4q+3Let—— = x where x is a rational number
2V3 = 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) = x
2-2x-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 x
2-4x-60
LHS can be written assinG(1-sin
2e)(1-cos
26) sin6cos6sine cos6,cos6 = sine cos6 1sin
26 + cos
20 sin
20 cos
26sin6cos6 sine cos61tanO+cote
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 = 66.43
SECTION Dp(x) = 2x
4+7x
3-19x
2-14x+30If 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+7x-15Now 2x
2+7x-15 = 2x
2+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-(sec
29-tan
29)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)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 |
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