# CBSE Sample Paper of Mathematics for Class IX

** Mathematics**

**Design of Sample Question Paper**

**SA-I Class IX (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-I I | 3 | 10 | 30 |

LA | 4 | 6 | 24 |

TOTAL | 34 | 80 |

**Blue Print Sample Question Paper-1 SA-1**

**I Term**

Topic / Unit | MCQ | SA(I) | SA(II) | LA | Total |

Number System | 2(2) | 2(4) | 3(9) | - | 7(15) |

Algebra | 2(2) | 1(2) | 2(6) | 3(12) | 8(22) |

Geometry | 6(6) | 4(8) | 3(9) | 3(12) | 16(35 |

Coordinate Geometry | - | 1(2) | 1(3) | - | 2(5) |

Mensuration | - | - | 1(3) | - | 1(3) |

TOTAL | 10(10) | 8(16) | 10(30) | 6(24) | 34(80) |

**Sample Question Paper Mathematics First Term (SA-I) Class IX 2010-2011**

**Time:**3

**to**3Y

_{2}

**hours**

**M.M.: 80**

**General Instructions**i) All questions are compulsory.ii) 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. However, 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 carry 1 mark each.**1. Decimal expresion of a rational number cannot be(a) non-terminating (B) non-terminating and recurring(C) terminating (D) non-terminating and non-recurring2. One of the factors of (9x

^{2}-1) – (1 +3x)

^{2}is(A) 3+x (B) 3-x (C) 3×-1 (D) 3x+13. Which of the following needs a proof?(A) Theorem (B) Axiom (C) Definition (D) Postulate4. An exterior angle of a triangle is 110° and the two interior opposite angles are equal. Each of these angles is(A) 70° (B) 55° (C) 35° (D) 110°5. In APQR, if ZR > ZQ, then(A) QR>PR (B) PQ>PR (C) PQ<PR (D) QR<PR6. Two sides of a triangle are of lengths 7 cm and 3.5 cm. The length of the third side of the triangle cannot be(A) 3.6 cm (B) 4.1cm (C) 3.4 cm (D) 3.8 cm.7. A rational number between 2 and 3 is(A) 2.010010001… (B) ^6 (C) 5/2 (D) 4.^/28. The coefficient of x

^{2}in (2x

^{2}-5) (4+3x

^{2}) is(A) 2 (B) 3 (C) 8 (D) -79. In triangles ABC and DEF, ZA = ZD, ZB = ZE and AB=EF, then are the two triangles congruent? If yes, by which congruency criterion?(A) Yes, byAAS (B) No (C) Yes, by ASA (D) Yes.byRHS10. Two lines are respectively perpendicular to two parallel lines. Then these lines to each other are(A) Perpendicular (B) Parallel(C) Intersecting (D) incllined at some acute angle

**SECTION – B**Question numbers 11 to 18 carry 2 marks each.

- x is an irrational number. What can you say about the number x
^{2}? Support your answer with examples. - Let OA, OB, OC and OD be the rays in the anticlock wise direction starting from OA, such that ZAOB = ZCOD = 100°, ZBOC = 82° and ZAOD = 78°. Is it true that AOC and BOD are straight lines? Justify your answer.

_8_. 75 |

Is |

( 8~ | )^{3} | MY | |

,15, | J | v3j |

**Q**

**R.**

**Fig. 3**

- In an isosceles triangle, prove that the altitude from the vertex bisects the base.
- Write down the co-ordinates of the points A, B, C and D as shown in Fig. 4.

**SECTION C**Question numbers 19 to 28 carry 3 marks each.19. Simplify the following by rationalising the denominators2^6 6^2ORV5+V3

**If ^ =**

*rr=.*^{a_}v

^{1}^b, find the values of a and b.20.If a=9

**-4**>/5, find the value of a-—.aORIf x = 3+2V2, find the value of x

^{2}+21. Represent 73~5

^{on}the number line.122. If(x-3)and

^{x}*~~r*are both factors of ax

^{2}+5x+b, show that a=b.o23. Find the value of x

^{3}+y

^{3}+15xy-125 when x+y=5.ORIf a+b+c=6, find the value of (2-a)

^{3}+(2-b)

^{3}+(2-c)

^{3}-3(2-a)(2-b)(2-c)24.25.26.27.28.

a | ||

b/ | c | |

( (-3,0) | ^{1} 0 | (3,0) |

Fig. 5 |

**rig. o**Find the area of a triangle, two sides of which are 18 cm and 10 cm and the perimeter is 42 cm.

**SECTION D**

**Question numbers 29 to 34 carry 4 marks each.**29. Let p and q be the remainders, when the polynomials x

^{3}+2x

^{2}-5ax-7 and x

^{3}+ax

^{2}-12x+6 are divided by (x+1) and (x-2) respectively. If 2p+q=6, find the value of a.ORWithout actual division prove that x

^{4}-5x

^{3}+8x

^{2}-10x+12 is divisible by x

^{2}-5x+6.30. Prove that:(x+y)

^{3}+ (y+z)

^{3}+ (z+x)

^{3}– 3(x+y) (y+z) (z+x) = 2(x

^{3}+y

^{3}+z

^{3}-3xyz)

Factorize x^{12}-y^{12}.- In Fig. 9, PS is bisector of ZQPR; PT IRQ and ZQ>ZR. Show that

### Answers

**Section A**1.6.(D) (C)2. 7.(D) (C)3. 8.(A) (D)4. 9.(B) (B)5. 10.(B) (B)

**SECTION B**11. x

^{2}may be irrational or may not be.For example ; if

_{X}=V3 , x

^{2}=3 rational; if

_{x}=2+>/3, x

^{2}=7+4^ -> irrational

^{o}-r70

^{o}+30

^{o}]=80° which is largest.-. Longest side is PR

- By Euclid’s I Axiom, which states. [“Things which are equal to the same thing are equal to one another”]
- The LHS can be written as

_{2}y

_{2}y

_{2}112

(^{8}‘ | 3 | r-i^{>} | 3 |

— | + | + | |

U5, | v 3 , |

*y*8 1 1 8-5-3 .

_{2}^{AS}15-3-5

^{=}^5—

^{=}°

*y*••• (1) = 3_8_

_{2}_{v}15

_{y}_8_ 75RHS

*y*Justification : By the formula: If a+b+c=0, then a

_{2}^{3}+b

^{3}+c

^{3}=3abcy

_{2}

1 “| | 2 | ||||

3 | |||||

15. | K27, | v3 J |

^{1}**4**= 9_1V 116. Zx=-70°+88°=18°(v ZQLM=180°-110

^{o}=70° and ABIICD=*ZPML=88°)17. Let ABC be isosceles a in which AB=AC Draw AD1BCA ‘sADB and ADC are congruent by RHS .-. BD=DC(cpct)i.e, Altitude AD bisects the base BC18. The coordinates of the points are: A(2,4), B(0, -3), C(-3, -5) and D(5, 0)

**y**

**y**

^{1}**/2+**

^{1}/2+^{1}/2+^{1}/2**SECTON-C**19.27*5 6>/2 2V6(V2-V3~)

_{+}6V2(V6-V36-372 + 73

*S*

*+*

*J3~*

*(2)-(3)= 2718-2712 + 2^-2^ = 672-276OR*

**i**

**+y**

_{2}**i**

**+y**LHS =

_{2}5-3 |

**= ®±|^I =**4+715 = a-7l5ba=4, b=-120.= 9-475 =>- = —a 9-475 81-80= 9 + 475… a~=9-475-9-475 = -875 aOR

**=> x**

*x=3+2j2*^{2}=9+8+12>/2 = 17 +12V2

**1=**

^{1}17-12V2

_{=17}_^ x

^{2}17 + 12V2 289-288.-. x

^{2}+— =17 + 12V2 +17-12V2 = 34‘A’ respresents

**V3**~5 on the number line3.5

**cm**Let f(x) = ax

^{2}+5x+bf(3) = 0 => 9a+15+b=0 9a+b=-15

**0)**

**(i) = (ii) => a=b**If x+y=5 => x+y+(-5)=0., (x)

^{3}+(y)

^{3}+(-5)

^{3}= 3(x)(y)(-5)=> x

^{3}+y

^{3}+15xy=125=> x

^{3}+y

^{3}+15xy-125=0OR a+b+c=6 => (2-a)+(2-b)+(2-c)=0.-. (2-a)

^{3}+(2-b)

^{3}+(2-c)

^{3}= 3(2-a)(2-b)(2-c) .-. (2-a)

^{3}+(2-b)

^{3}+(2-c)

^{3}-3(2-a)(2-b)(2-c)=0AB=BC=AC=6 units as AABC is equilateral AO bisects base BC => OB=3 units.-. OA

^{2}=AB

^{2}-OB

^{2}=6

^{2}-3

^{2}= 27

*OA=3y/3*^{C}=> x=30°

- In right triangle QTR, x=90°-40°=50° Again y is the exterior angle of APSR => y=30
^{o}+x=50°+30^{o}=80° - BD+DE = CE+DE => BE=CD In A’s ABE and ACD

*Vz*

*Vz*

*VA***128. s=**

*VA***y**=21, let a=18cm, b=10cm, c=42-(28)=14cmAr(A) = Vs(s-a)(s4D)(&<:) = ^21(3)(11)(7) = 2lVncm

^{2}

**SECTION-D**29. Let P(x) = x

^{3}+2x

^{2}-5ax-7 and Q(x) = x

^{3}+ax

^{2}-12x+6 P(-1) = pandQ(2) = q… p=-l+2+5a-7 => p=5a-6 q=8+4a-24+6 q=4a-102p+q=6 => 10a-12+4a-10=6 => 14a=28 => a=2ORx

^{2}-5x+6 = (x-2)(x-3)P(x) = x

^{4}-5x

^{3}+8x-10x+12P(2) = 16-40+32-20+12=0P(3) = 81-135+72-30+12=0… (x-2)(x-3) divides P(x) completely30. Let x+y=p, y+z=q, z+x=r … LHS = p

^{3}+q

^{3}+r

^{3}-3pqr= (P+q+r) (p

^{2}+q

^{2}+r

^{2}-pq-qr-rp)

**i**

**+y**

_{2}**i**

**+y**

_{2}*Yz+Vz*1 1 1Now p+q+r=2(x+y+z)p

^{2}+q

^{2}+r

^{2}-pq-qr-rp = (x+y)

^{2}+(y+z)

^{2}+(z+x)

^{2}-(x+y)(y+z)-(y+z)(z+x)-(z+x)(x+y)x

^{2}+y

^{2}+2xy+z

^{2}+2yz+2zx x

^{2}+y

^{2}-xy +z

^{2}-yz -xz-y

^{2}-xy -z

^{2}-yz -xz x

^{2}-xy -yz -xz x

^{2}+y

^{2}+z

^{2}-xy-yz-zx.-. (p+q+r) (p

^{2}+q

^{2}+r

^{2}-pq-qr-rp) = 2(x+y+z) (x

^{2}+y

^{2}+z

^{2}-xy-yz-zx)= 2(x

^{3}+y

^{3}+z

^{3}-3xyz)

_{X}12.

_{y}12 – (

_{X}6-y6)(x

^{6}+y

^{6})= (x

^{3}-y

^{3})(x

^{3}+y

^{3})(x

^{2}+y

^{2})(x

^{4}+y

^{4}-x

^{2}y

^{2})= (x-y)(x

^{2}+y

^{2}+xy)(x+y)(x

^{2}+y

^{2}-xy)(x

^{2}+y

^{2})(x

^{4}+y

^{4}-x

^{2}y

^{2})ZQ+ZR=180°-2ZQPS=180°-2 [ZQPT+ZTPS]

^{0}-2[90°-Z1 + ZTPS]=>Z1+Z2 = 2Z1-2ZTPS =>ZTPS

**=1**(Z1-Z2) =

**1**(ZQ-ZR) ORZB+ZC=90° => ZB=90°-ZCZBAL=90°-ZB=90°-(90° – ZC) = ZC.-. ZBAL = ZACBZBAD+ZDAC = ZCAE+ZCAD => ZBAC=ZDAEIn a’s ABC and ADEAB=AD, AC=AE and ZBAC=ZDAE .-. a’s are congruent.-. BC = DE (cpct)Zx=Zy => ZBDC=ZAEB In a’s ABE and CBDAB=BC, ZB=ZB, ZBDC=ZAED 1 .-. AABE – ACBD [AAS].-. AE=CD

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