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# CBSE Sample Paper of Mathematics for Class IX

## Mathematics

### 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-1I 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 2010-2011Time: 3 to 3Y2 hours M.M.: 80General Instructionsi)             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-AQuestion 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 (9x2-1) – (1 +3x)2 is(A)    3+x                           (B)    3-x                            (C)        3x-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 x2 in (2x2-5) (4+3x2) 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 angleSECTION – BQuestion numbers 11 to 18 carry 2 marks each.
1. x is an irrational number. What can you say about the number x2? Support your answer with examples.
2. 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.
ORIn APQR, ZP=70°, ZR=30°. Which side of this triangle is the longest? Give reasons for your answer. 13.14.
 _8_. 75
 Is
In Fig. 2, it is given that Z1 =Z4 and Z3=Z2. By which Euclid’s axiom, it can be shown that if Z2 = Z4 then Z1 = Z3.
 ( 8~ )3 MY ,15, J v3j
How will you justify your answer, without actually calculating the cubes?215. 16.-1327In Fig. 3, ifABIICDthen find the measure of x.88°QR.Fig. 3
1. In an isosceles triangle, prove that the altitude from the vertex bisects the base.
2. Write down the co-ordinates of the points A, B, C and D as shown in Fig. 4.
SECTION CQuestion numbers 19 to 28 carry 3 marks each.19.    Simplify the following by rationalising the denominators2^6        6^2ORV5+V3               rr=.If      ^ = a_v1 ^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 x2 +21.     Represent 73~5 on the number line.122.         If(x-3)and x~~r are both factors of ax2+5x+b, show that a=b.o23.         Find the value of x3+y3+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
In Fig. 8, D and E are points on the base BC of a AABC such that BD=CE and AD=AE.Prove that AABC = AACD.                                                   „rig. oFind the area of a triangle, two sides of which are 18 cm and 10 cm and the perimeter is 42 cm.SECTION DQuestion numbers 29 to 34 carry 4 marks each.29.         Let p and q be the remainders, when the polynomials x3+2x2-5ax-7 and x3+ax2-12x+6 are divided by (x+1) and (x-2) respectively. If 2p+q=6, find the value of a.ORWithout actual division prove that x4-5x3+8x2-10x+12 is divisible by x2-5x+6.30.         Prove that:(x+y)3 + (y+z)3 + (z+x)3 – 3(x+y) (y+z) (z+x) = 2(x3+y3+z3-3xyz)
1. Factorize x12-y12.
2. In Fig. 9, PS is bisector of ZQPR; PT IRQ and ZQ>ZR. Show that
ZTPS = -(ZQ-ZR).                                                                                             ROR                                                                                                     A In AABC, right angled at A, (Fig. 10), AL is drawn perpendicular to BC. Prove that ZBAL = ZACB.33.    In Fig. 11, AB=AD, AC=AE andZBAD = ZCAE. Prove that BC = DE. 34.     In Fig. 12, if Zx=Zy and AB = BC, prove that AE = CD.

Section A1.6.(D) (C)2. 7.(D) (C)3. 8.(A) (D)4. 9.(B) (B)5. 10.(B) (B) SECTION B11.         x2 may be irrational or may not be.For example ; if X=V3 , x2=3    rational; if x=2+>/3, x2=7+4^ -> irrational 12.         No, AOC and BOD are not straight lines v i) ZAOC = 182° *180°ii) ZBOD = 178° *180° ORZQ=180o-r70o+30o]=80° which is largest.-. Longest side is PR
1. By Euclid’s I Axiom, which states. ["Things which are equal to the same thing are equal to one another"]
2. The LHS can be written as
y2y2y2112
 (8‘ 3 r-i> 3 — + + U5, v 3 ,
-(i)y28 1 1 8-5-3 . AS 15-3-5 = ^5—= °y2••• (1) = 3_8_ v15y_8_ 75RHSy2Justification : By the formula: If a+b+c=0, then a3+b3+c3=3abcy2
 1 “| 2 3 15. K27, v3 J
1 4 = 9_1V   116.         Zx=-70°+88°=18°(v ZQLM=180°-110o=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)yy1/2+1/2+1/2+1/2SECTON-C19.27*5        6>/2      2V6(V2-V3~) + 6V2(V6-V36-372 + 73   S + J3~ (2)-(3)= 2718-2712 + 2^-2^ = 672-276ORi+y2 i+y2LHS =
 5-3
75 + 73 _(V5 + y3)(75 + 73 75-73 ~= ®±|^I = 4+715 = a-7l5ba=4, b=-120.= 9-475 =>- = —a   9-475    81-80= 9 + 475… a~=9-475-9-475 = -875 aORx=3+2j2 => x2=9+8+12>/2 = 17 +12V21= 1 17-12V2=17_^ x2 17 + 12V2    289-288.-. x2+— =17 + 12V2 +17-12V2 = 34‘A’ respresents V3~5 on the number line3.5 cmLet f(x) = ax2+5x+bf(3) = 0 => 9a+15+b=0    9a+b=-150) (i) = (ii) => a=bIf x+y=5 => x+y+(-5)=0., (x)3+(y)3+(-5)3 = 3(x)(y)(-5)=>    x3+y3+15xy=125=>    x3+y3+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.-. OA2=AB2-OB2=62-32 = 27     OA=3y/3 25.          DrawADIIPQ, BEIILMIIPQ=> ZPAD=15° => ZDAB=20°=> ZDAB=ZABE=20° and ZEBM=ZBML=10C=> x=30°
1. In right triangle QTR, x=90°-40°=50° Again y is the exterior angle of APSR => y=30o+x=50°+30o=80°
2. BD+DE = CE+DE => BE=CD In A’s ABE and ACD
BE=CD,AE=AD, ZADE=ZAED .-. AABE = AACD (SAS)VzVzVAVA 128.     s=y=21, let a=18cm, b=10cm, c=42-(28)=14cmAr(A) = Vs(s-a)(s4D)(&<:) = ^21(3)(11)(7) = 2lVncm2SECTION-D29.          Let P(x) = x3+2x2-5ax-7 and Q(x) = x3+ax2-12x+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=2ORx2-5x+6 = (x-2)(x-3)P(x) = x4-5x3+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 = p3+q3+r3-3pqr= (P+q+r) (p2+q2+r2-pq-qr-rp)i+y2 i+y2Yz+Vz1 1 1Now p+q+r=2(x+y+z)p2+q2+r2-pq-qr-rp = (x+y)2+(y+z)2+(z+x)2-(x+y)(y+z)-(y+z)(z+x)-(z+x)(x+y)x2+y2+2xy+z2+2yz+2zx x2+y2 -xy +z2 -yz -xz-y2 -xy -z2 -yz -xz x2 -xy       -yz -xz x2+y2+z2-xy-yz-zx.-. (p+q+r) (p2+q2+r2-pq-qr-rp) = 2(x+y+z) (x2+y2+z2-xy-yz-zx)= 2(x3+y3+z3-3xyz)X12.y12 – (X6-y6)(x6+y6)= (x3-y3)(x3+y3)(x2+y2)(x4+y4-x2y2)= (x-y)(x2+y2+xy)(x+y)(x2+y2-xy)(x2+y2)(x4+y4-x2y2)ZQ+ZR=180°-2ZQPS=180°-2 [ZQPT+ZTPS] = 1800-2[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