Mathematics
Design of Sample Question Paper
SAI Class IX (20102011)
Type of Question 
Marks per question 
Total No. of Questions 
Total Marks 
M.C.Q. 
1 
10 
10 
SAI 
2 
8 
16 
SAI I 
3 
10 
30 
LA 
4 
6 
24 
TOTAL 

34 
80 
Blue Print Sample Question Paper1 SA1
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 (SAI) Class IX 20102011
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.
SectionA
Question numbers 1 to 10 carry 1 mark each.
1. Decimal expresion of a rational number cannot be
(a) nonterminating (B) nonterminating and recurring
(C) terminating (D) nonterminating and nonrecurring
2. One of the factors of (9x
^{2}1) – (1 +3x)
^{2} is
(A) 3+x (B) 3x (C) 3×1 (D) 3x+1
3. Which of the following needs a proof?
(A) Theorem (B) Axiom (C) 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° (C) 35° (D) 110°
5. In APQR, if ZR > ZQ, then
(A) QR>PR (B) PQ>PR (C) 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 (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.^/2
8. The coefficient of x
^{2} in (2x
^{2}5) (4+3x
^{2}) is
(A) 2 (B) 3 (C) 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 (C) 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
(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.
OR
In APQR, ZP=70°, ZR=30°. Which side of this triangle is the longest? Give reasons for your answer.
13.
14.
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.
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 coordinates 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_}v
^{1} ^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 x
^{2} +
21. Represent 73~5
^{on} the number line.
1
22. If(x3)and
^{x}~~r are both factors of ax
^{2}+5x+b, show that a=b.
o
23. Find the value of x
^{3}+y
^{3}+15xy125 when x+y=5.
OR
If a+b+c=6, find the value of (2a)
^{3}+(2b)
^{3}+(2c)
^{3}3(2a)(2b)(2c)
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 x
^{3}+2x
^{2}5ax7 and x
^{3}+ax
^{2}12x+6 are
divided by (x+1) and (x2) respectively. If 2p+q=6, find the value of a.
OR
Without 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
ZTPS = (ZQZR). 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) (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
12. No, AOC and BOD are not straight lines
v i) ZAOC = 182° *180°
ii) ZBOD = 178° *180° OR
ZQ=180
^{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
y
_{2}
y
_{2}
y
_{2}
1
1
2
(^{8}‘ 
3 
ri^{>} 
3 
— 
+ 

+ 
U5, 

v 3 , 

(i)
y_{2}
8 1 1 853 .
^{AS} 1535
^{=} ^5—
^{=} °
y_{2}
••• (1) = 3
_8_
_{v}15
_{y}
_8_ 75
RHS
y_{2}
Justification : By the formula: If a+b+c=0, then a
^{3}+b
^{3}+c
^{3}=3abc
y
_{2}
^{1} 4 = 9
_1V 1
16. 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 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
SECTONC
19.
27*5 6>/2 2V6(V2V3~)
_{+} 6V2(V6V3
63
72 + 73
S + J3~ (2)(3)
= 27182712 + 2^2^ = 672276
OR
i+y_{2} i+y_{2}
LHS =
75 + 73 _(V5 + y3)(75 + 73 7573 ~
= ®±^I = 4+715 = a7l5b
a=4, b=1
20.
= 9475 => = —
a 9475 8180
= 9 + 475
… a~=94759475 = 875 a
OR
x=3+2j2 => x
^{2}=9+8+12>/2 = 17 +12V2
1= ^{1} 1712V2
_{=17}_^ x
^{2} 17 + 12V2 289288
.. x
^{2}+— =17 + 12V2 +1712V2 = 34
‘A’ respresents
V3~5 on the number line
3.5
cm
Let f(x) = ax
^{2}+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)
=> x
^{3}+y
^{3}+15xy=125
=> x
^{3}+y
^{3}+15xy125=0
OR a+b+c=6 => (2a)+(2b)+(2c)=0
.. (2a)
^{3}+(2b)
^{3}+(2c)
^{3} = 3(2a)(2b)(2c) .. (2a)
^{3}+(2b)
^{3}+(2c)
^{3}3(2a)(2b)(2c)=0
AB=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
25. DrawADIIPQ, BEIILMIIPQ
=> ZPAD=15° => ZDAB=20°
=> ZDAB=ZABE=20° and ZEBM=ZBML=10
^{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
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(sa)(s4D)(&<:) = ^21(3)(11)(7) = 2lVncm
^{2}
SECTIOND
29. Let P(x) = x
^{3}+2x
^{2}5ax7 and Q(x) = x
^{3}+ax
^{2}12x+6
P(1) = pandQ(2) = q
… p=l+2+5a7 => p=5a6 q=8+4a24+6 q=4a10
2p+q=6 => 10a12+4a10=6 => 14a=28 => a=2
OR
x
^{2}5x+6 = (x2)(x3)
P(x) = x
^{4}5x
^{3}+8x10x+12
P(2) = 1640+3220+12=0
P(3) = 81135+7230+12=0
… (x2)(x3) divides P(x) completely
30. 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}pqqrrp)
i+y_{2} i+y_{2}
Yz+Vz
1 1 1
Now p+q+r=2(x+y+z)
p
^{2}+q
^{2}+r
^{2}pqqrrp = (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}xyyzzx
.. (p+q+r) (p
^{2}+q
^{2}+r
^{2}pqqrrp) = 2(x+y+z) (x
^{2}+y
^{2}+z
^{2}xyyzzx)
= 2(x
^{3}+y
^{3}+z
^{3}3xyz)
_{X}12.
_{y}12 – (
_{X}6y6)(x
^{6}+y
^{6})
= (x
^{3}y
^{3})(x
^{3}+y
^{3})(x
^{2}+y
^{2})(x
^{4}+y
^{4}x
^{2}y
^{2})
= (xy)(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]
= 180
^{0}2[90°Z1 + ZTPS]
=>Z1+Z2 = 2Z12ZTPS =>ZTPS
=1(Z1Z2) =
1(ZQZR) 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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