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 3Y2 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
© terminating (D) non-terminating and non-recurring
2. One of the factors of (9x2-1) - (1 +3x)2 is
(A) 3+x (B) 3-x © 3x-1 (D) 3x+1
3. Which of the following needs a proof?
(A) Theorem (B) Axiom © Definition (D) Postulate
4. 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° © 35° (D) 110°
5. In APQR, if ZR > ZQ, then
(A) QR>PR (B) PQ>PR © PQ<PR (D) QR<PR
6. 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 © 3.4 cm (D) 3.8 cm.
7. A rational number between 2 and 3 is
(A) 2.010010001… (B) ^6 © 5/2 (D) 4.^/2
8. The coefficient of x2 in (2x2-5) (4+3x2) is
(A) 2 (B) 3 © 8 (D) -7
9. 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 © Yes, by ASA (D) Yes.byRHS
10. Two lines are respectively perpendicular to two parallel lines. Then these lines to each other are
(A) Perpendicular (B) Parallel
© 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 x2? 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.
OR
In 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?
2
15. 16.
-13
27
In Fig. 3, ifABIICDthen find the measure of x.
88°
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 denominators
2^6 6^2
OR
V5+V3 rr=.
If ^ = a_v1 ^b, find the values of a and b.
20.
If a=9 -4 >/5, find the value of a-—.
a
OR
If x = 3+2V2, find the value of x2 +
21. Represent 73~5 on the number line.
1
22. If(x-3)and x ~~r are both factors of ax2+5x+b, show that a=b.
o
23. Find the value of x3+y3+15xy-125 when x+y=5.
OR
If 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. 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 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.
OR
Without 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)
- | | | | --- | --- | | | | | |
Factorize x12-y12.
- In Fig. 9, PS is bisector of ZQPR; PT IRQ and ZQ>ZR. Show that
ZTPS = -(ZQ-ZR). R
OR 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 and
ZBAD = ZCAE. Prove that BC = DE.
34. In Fig. 12, if Zx=Zy and AB = BC, prove that AE = CD.
Answers
Section A
1.
6.
(D) ©
2. 7.
(D) ©
3. 8.
(A) (D)
4. 9.
(B) (B)
5. 10.
(B) (B)
SECTION B
11. 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° OR
ZQ=180o-r70o+30o]=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
y2
y2
y2
1
1
2
| ( 8’ | 3 | r-i> | 3 |
| — | + | + | |
| U5, | v 3 , |
-(i)
y2
8 1 1 8-5-3 . AS 15-3-5 = ^5—= °
y2
••• (1) = 3
_8_ v15y
_8_ 75
RHS
y2
Justification : By the formula: If a+b+c=0, then a3+b3+c3=3abc
y2
| 1 “| | 2 | ||||
| 3 | |||||
| 15. | K27, | v3 J |
1 4 = 9
_1V 1
16. Zx=-70°+88°=18°
(v ZQLM=180°-110o=70° and ABIICD=*ZPML=88°)
17. Let ABC be isosceles a in which AB=AC Draw AD1BC
A ‘sADB and ADC are congruent by RHS .-. BD=DC(cpct)
i.e, Altitude AD bisects the base BC
18. 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-V3
6-3
72 + 73 S + J3~(2)-(3)
= 2718-2712 + 2^-2^ = 672-276
OR
i +y2 i +y2
LHS =
| 5-3 |
75 + 73 _(V5 + y3)(75 + 73 75-73 ~
= ®±|^I = 4+715 = a-7l5b
a=4, b=-1
20.
= 9-475 =>- = —
a 9-475 81-80
= 9 + 475
… a~=9-475-9-475 = -875 a
OR
x=3+2j2 => x2=9+8+12>/2 = 17 +12V2
1= 1 17-12V2=17_^ x2 17 + 12V2 289-288
.-. x2+— =17 + 12V2 +17-12V2 = 34
‘A’ respresents V3 ~5 on the number line
3.5 cm
Let f(x) = ax2+5x+b
f(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)
=> x3+y3+15xy=125
=> x3+y3+15xy-125=0
OR 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)=0
AB=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°
- In right triangle QTR, x=90°-40°=50° Again y is the exterior angle of APSR => y=30o+x=50°+30o=80°
- BD+DE = CE+DE => BE=CD In A’s ABE and ACD
BE=CD,AE=AD, ZADE=ZAED .-. AABE = AACD (SAS)
Vz
Vz
VA
VA 1
28. s= y =21, let a=18cm, b=10cm, c=42-(28)=14cm
Ar(A) = Vs(s-a)(s4D)(&<:) = ^21(3)(11)(7) = 2lVncm2 SECTION-D
29. 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-10
2p+q=6 => 10a-12+4a-10=6 => 14a=28 => a=2
OR
x2-5x+6 = (x-2)(x-3)
P(x) = x4-5x3+8x-10x+12
P(2) = 16-40+32-20+12=0
P(3) = 81-135+72-30+12=0
… (x-2)(x-3) divides P(x) completely
30. 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 +y2
Yz+Vz
1 1 1
Now 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) OR
ZB+ZC=90° => ZB=90°-ZC
ZBAL=90°-ZB=90°-(90° - ZC) = ZC
.-. ZBAL = ZACB
ZBAD+ZDAC = ZCAE+ZCAD => ZBAC=ZDAE
In a’s ABC and ADE
AB=AD, AC=AE and ZBAC=ZDAE .-. a’s are congruent
.-. BC = DE (cpct)
Zx=Zy => ZBDC=ZAEB In a’s ABE and CBD
AB=BC, ZB=ZB, ZBDC=ZAED 1 .-. AABE - ACBD [AAS]
.-. AE=CD
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