# 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**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.

**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<b2.In Fig. 1, the graph of a polynomial

(D) |

(A) |

(C) |

L p(_{X}) | |

Fig. 1 |

^{2}0-Sec

^{2}0 .Given that tanG^, the value of

_{Cosec2+Sec2Q}«(A) -1(B) 1(O (D) -x

(A) (C) |

(B) (D) |

3 443 |

_5_ 1212 5147 |

will terminate after how many places of decimal?(D) will not terminate |

8. |

9. |

^{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

**SECTION-B**

**Question numbers 11 to 18 carry 2 marks each.**

#-c |

- 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-y**

**12 Fig. 4**14. If 7sin

^{2}9+3Cos

^{2}9 = 4, show that tan6=1OR„ ’15 (2 + 2sin9)(1-sin9)If Cot9=—, evaluate

*)*

*-f±*

**f**8 (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}

18. |

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 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}-2×-15, then form a quadratic polynomial whose zeroes are

*(2a)*and (2/3).

24. 25. |

26. |

Prove that |

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 |

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) = 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*

Evaluate |

^{2}55° + sin

^{2}35

^{c}tan 10° tan 20° tan 60° tan 70° tan 80°

p^{2}+1 |

^{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 |

### 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 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

^{2}9+3cos

^{2}9=4 or 3(sin

^{2}9 + cos

^{2}9) + 4sin

^{2}6 = 41 1=> sin

^{2}9=- sin9=- ^9 = 30° 4 2.-. tan9 = tan30

**°=-^L**V3OR15cot 9=— (given) 82(1 + sin6)(1 -sine)Given expression =————————— = cot

^{2}9

’15^{x:} |

**v**8,64_

_{a}- be bd

^{15}

^{deiiac}^ec

**=5**a <

^{!})

*y*dfiiaebf efbdda….(ii)

_{2}**y**16.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

^{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 |

^{32}

^{12}x10 = 30+6.25 = 36.25 64-32

**i**

**+y**

_{2}**SECTION C**19. 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

**y**But 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 “**

^{w}5xAs 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}-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

- sinG |

^{2}e)(1-cos

^{2}6) sin6cos6sine cos6,cos6 = sine cos6 1sin

^{2}6 + cos

^{2}0 sin

^{2}0 cos

^{2}6sin6cos6 sine cos61tanO+cote

si |

_{+}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- ^{x}i”^{25} di = -i—10 | -2 | -1 | 0 | 1 | 2 |

fidi | -14 | -12 | 0 | 10 | 16 |

^{1}/

_{2}x = 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 |

**SECTION D**p(x) = 2x

^{4}+7x

^{3}-19x

^{2}-14x+30If two zeroes of p(x) are

**and -V2.-. (x+V2~)(x-^) or x**

*-J2*^{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×-15Now 2x

^{2}+7×-15 = 2x

^{2}+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-(sec

^{2}9-tan

^{2}9)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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