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7/29/2019 Mathmatics Question Type Odisha Board Exam
1/24
SAMP
LE
Roll No. of the Candidate
2013AH
Grand Total Marks Grand Total
in words .......................................................................................................
[No. of Qns. : 14] Time : 2 Hours]
[No.of Printed Pages: 24 ] [Full Marks : 100
Signature of Examiner &
Regd. No.
Signature of the Scrutiniser
& Regd. No.Signature of the Chief Evaluator &
Regd. No.
Signature of the Deputy Chief &
Regd. No.
Date............................
Q. No. 1 2 3 4 5 6 7 8 9 10 11 12 13 14
Marks
REGULAR SET : A
AR 15
_el[u c iP^ :
K. _g_Z ifM Ce LZU _Aa _e Gj C_e c\Z [a _g iL I _ iL_g_Z ifM Ce LZe _ZK_e _g iL I _ iL ijZ ckA @ G[eiU iKZ _Z _e VK bae fL jAQ ^ ^jc ckA @
L. ~\ KQ ZU _eflZ jG, Za ZU~q _g_Z ifM Ce LZU _el Mj \dZe[a ^elKu `eA @C MUG VK _g_Z cM @
FOR USE AT THE EVALUATION CENTRE
(Varified and found correct)
Full signature of the Invigilator
MTH
MATHEMATICS
QUESTION - CUM - ANSWER BOOKLET
7/29/2019 Mathmatics Question Type Odisha Board Exam
2/24
SET : A
(2)
: i]eY iP^ :: GENERAL INSTRUCTIONS :
l Gj _g_Z ifM CeLZe Cbd \NI il CeckK, Ga a^ _g ej@Q The question-cum-Answer Booklet consists of objective and subjective types of questions.
l _ZK a^ _g c Ce fLa _A ce eZ i^ \@~AQ _el[c^ iji e Ce fLa @agK Required space has been provided against each objective type question. The candidate has
to choose the correct choice and write the complete answer along with its bit number in the
space provided.
l \NCeckK _g c ce Ce fLa c eZ i^ \@~AQ _el[c^ iji^ ce Ce fLa @agK For subjective type questions required space for each question has been provided. The
candidate has to answer the questions in the space provided.
l ^g @ ~d _ZK _ge Ce \a @agK Follow the instructions given against both the objective and subjective types of questions.
l Ce Kk Ga ^k af _G _^e fLa @agK Candidate should write the answer with blue / black ball point pen only .
l a^ _gMWKe Cee KYi KUKU K fbA Ce fLf Zje cfu^ Ke~a^j In case of Multiple Choice, Fill in the blanks and Matching questions, scratching / rewriting /
marking /over writing is not permitted, thereby rendering to disqualification for evaluation.
l _g_Z ifM Ce LZe [a _g _Xa c _el[c^u15 c U@]K icd \@~a Extra 15 minutes shall be given for reading the question paper.
l _g_Z ifM Ce LZe ghe ifM [a Lf _, bf Ke[a ~KYi _ge Ce fLa^c aaje Ke~a Extra two rulling pages given at the end of this booklet may be utilised for writing the wrong
answers or extra answers if necessary.
: INSTRUCTIONS TO THE CANDIDATES :A : Check up the number of questions and number of printed pages mentioned on the top left-
hand and right-hand corner of this page with the number of questions and number of printed
pages of this question paper soon after the receipt of the question paper. Verify whether the
Set Symbol in this page is printed in all other pages and write this Set Symbol on the space
provided for writing the Set Symbol on the cover page of the answer paper.
B : In case of discrepancy, collect a fresh correct question paper in exchange of the defective
one from the invigilator of your hall / room after receipt of the question paper.
7/29/2019 Mathmatics Question Type Odisha Board Exam
3/24
SET : A
(3)
x~ Q{ AB$# QSL Z Dd >
The figures in the right-hand margin indicate marks.
D A >Answerallquestions.
aRMYZ / Algebra1. _ZK^ _Yke ic]^ Ke : 5
Solve by the method of substitution
2x + 3y 2 = 0
10x 6y 3 = 0
L/ (OR)
_aMe _eYZ Ke ic]^ Ke :Solve by completing the squares :
2x2 9x + 4 = 0_________________________________________________________________
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7/29/2019 Mathmatics Question Type Odisha Board Exam
4/24
SET : A
(4)
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2. a = 2
1
3 2
1
3
jf, _cY Ke ~ 2a3 + 6a = 3 5
If a = 2
1
3 2
1
3
, then prove that 2a3 + 6a = 3
L/ (OR)
^c ieYe \@~A[a Z[ake cc^ ^d Ke :
Find the mean of the data given in the following table :
ibM 0-10 10-20 20-30 30-40 40-50Class Interval
aeeZFrequency 3 8 25 35 19
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7/29/2019 Mathmatics Question Type Odisha Board Exam
5/24
SET : A
(5)
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3. DABC icKY ZbRe mB = 900, a = 4 i.c. I c = 3 i.c. Gje Ke \NI lZ`k ^e_Y Kea _A GK _aj PZ _Z Ke 5
In the right angled tri angle, DABC, mB = 900, a = 4 cm and c = 3 cm. Preparea flow-chart to find the length of its hypotenuse and area.
L/ (OR)
RY aq GK _Ke95 Uu@ 10% KZK @g 98 Uu aK cfe KY i[ecU190 Uu ahK @d Kf Za i KZ Uu a^~M Ke[f, ^d Ke
A man, purchasing some 95 rupees 10% share selling at market value of Rs.
98.00 earned Rs. 190.00 per annum. Find the amount of money he invested.
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7/29/2019 Mathmatics Question Type Odisha Board Exam
6/24
SET : A
(6)
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4. ^cfLZ _gc^u ce ~ KYi PeUe Ce \@ 4 x 4Answer any four of the following questions :
(i) _Zu adie \AMY I _Ze adi cg 90 cZ _Zu adi I _Z adie \AMY cg60 _Z I _Zue adi ^d Ke Sum of the twice fathr's age and son's age is 90. But the sum of father's age and
twice the son's age is 60. Find father's and son's age.
(ii) ax2 + bx + c = 0 icKeYe MUG ck @_eUe Z^MY jf, _cY Ke ~3b2 = 16ac (a 0)If one of the roots of the equation, ax2 + bx + c = 0 is three times that of hte
other, then prove that 3b2 = 16ac.
(iii) iek Ke :
Simplify :
3 loga2 + 2 log
a3 + log
a5 log
a9 (a > 0, a 1)
(iv) ^c ieYe \@~A[a Z[ake cc ^d Ke :Find the median of the data given in the following table :
fu 15 16 17 18 19 20ScoresaeeZFrequency 5 9 15 20 14 2
7/29/2019 Mathmatics Question Type Odisha Board Exam
7/24
SET : A
(7)
(v) (x)2
K \K e_e _eYZ Ke Give the value of (x)
2in binary form.
(1101)2 + (x)2 = (11101)2 + (11)2
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7/29/2019 Mathmatics Question Type Odisha Board Exam
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SET : A
(8)
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5. ic _ge Ce \@ : 2 x 6Answer all questions :
(i) px 4y = 0 I 16x py = 0 ijicKeY\de @iL ic]^ [f, p ec^ ^d Ke
7/29/2019 Mathmatics Question Type Odisha Board Exam
9/24
SET : A
(9)
If the simultaneous equations, px 4y = 0 and 16x py = 0 have infinite
number of solutions, then find the value of p.
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(ii) 2x2 + 5x c = 0 icKeYe MUG ck 2 jf, c e c ^d Ke If one of the rootsof the equation, 2x2 + 5x c = 0 is 2, then find the
value of c._________________________________________________________________
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(iii) log 0.01986 e KeUeK I ci ^d Ke Find the characteristic and mantesa of log 0.01986
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7/29/2019 Mathmatics Question Type Odisha Board Exam
10/24
SET : A
(10)
(iv) 15 U fa]ue cc^ 52.5 I ic^ue cc 53.2 jf, ic^ue MeK^d Ke
The mean of 15 scores is 52.5 and their median is 53.2. Find their mode.
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(v) (64)10
K \K e_e _Kg Ke :Express (64)
10in binary form.
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(vi) ic]^ Ke :
Solve :
1
61 2 x = 216
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7/29/2019 Mathmatics Question Type Odisha Board Exam
11/24
SET : A
(11)
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6. ^cfLZ _gc^ue Kak Ce fL : 1 x 5Write the answer only of the following questions.
(i)x y
3 40+ = I
y x
4 30 = icKeY \de ic]^ KZ ?
What is the solution of the two equations,x y
3 40+ = and
y x
4 30 = ?
Ans : _________________________________________________________
(ii) 3x2 + 9x = 2 icKeYe ck\de MY`k KZ ?What is product of the roots of the equation, 3x2 + 9x = 2 ?
Ans : _________________________________________________________
(iii) 161
1
2
e c^ KZ ?
What is he value of 161
1
2 ?
Ans : _________________________________________________________
(iv)log 8
log 1
2
2 6e iekKZ c^ KZ ?
What is he simplified value oflog 8
log 1
2
2 6?
Ans : _________________________________________________________
(v) _aj PZe Z[_ag _A aajZ iKZe MUG PZ Ke Draw the figure of the symobol used for "Impute" in a flow-chart.
Ans : _________________________________________________________
7/29/2019 Mathmatics Question Type Odisha Board Exam
12/24
SET : A
(12)
RcZ I ZKYcZGeometry and Trigonometry
= 227d i
7. _cYKe ~, MUG ae K Ve ic\ea R c^ue \N ic^ 5Prove that chords of a circle equidistant from the centre are of equal length.
L/ (OR)
_cY Ke ~ ae GK aji aP c \A MUG Q\K aKA I B ae Q\Kf Ga PT ae GK gKL jf, PA . PB = PT2
Prove that if a secant to a circle passing through an external point P meets the
cirlce at A and B and PT is a tangent segment then PA . PB = PT2
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7/29/2019 Mathmatics Question Type Odisha Board Exam
13/24
SET : A
(13)
8. A I B K ag \AU a _eeKP I Q ae Q\ Ke P c \A AB ijice iek eL a \dK ~[Kc M I N ae Q\ Ke
_cY Ke ~ MN = 2AB 5Two circles having centres at A and B intersect each other at points P and Q.
A straight line through P parallel to AB meets the circles at points M and N
respectively. Prove that MN = 2AB
L/ (OR)
GK ae ai MN M Ve @uZ gK C_e A I B G_e \AU a ~_e
A M B Ga AN I BN aK ~[Kc P I Q ae Q\ Ke _cY Ke ~PN
BN
QN
AN=
MN is the diameter of a circle. A and B are points on the tangent drawn at M
such that A M B. AN and BN meet the circle at P and Q respectively.
prove thatPN
BN
QN
AN=
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7/29/2019 Mathmatics Question Type Odisha Board Exam
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SET : A
(14)
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9. (A + B + C) = 1800 jf, _cY Ke ~tan A + tan B + tan C = tanA . tan B . tan C 5
If (A + B + C) = 1800, then Provl that
tan A + tan B + tan C = tanA . tan B . tan C
L/ (OR)
i~e KYK C^Z 600 e450 K ji _Aae GK e QA 24 cUe a _Af eCyZ d Ke ( 3 = 1.732)
When the sun's angle of elevation decreased from 600 to 450, the shadow of a
pillar increased by 24 metres. Find the height of the pillar. ( 3 = 1.732)
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7/29/2019 Mathmatics Question Type Odisha Board Exam
15/24
SET : A
(15)
10. D ABC e AB = 7 i.c., m C = 600 I CX cc = 5 i.c.
ZbRU @u^ Ke 5
In D ABC, AB = 7 cm, m C = 600 and median CX = 5 cm. Construct the
triangle.
L/ (OR)
ABCD @dZ PZe BC = 6 i.c., I AB = 5 i.c. @dZ PZU @u^ Ke Gje _ea@u^ Ke In rectangle ABCD, BC = 6 cm and AB = 5 cm. Construct the rectangle and
then construct its circum-circle.
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7/29/2019 Mathmatics Question Type Odisha Board Exam
16/24
SET : A
(16)
11. 4.5 i.c. ai ag GK a @u^ Ke Gj ce GK ic\aj ZbR
@fL^ Ke ~je bc = 7 i.c. 2Construct a circle of radius 4.5 cm and inscribe in it an isosceles triangle whose
base is 7 cm.
L/ (OR)
4.5 i.c. ai ag GK a @u^ Ke Gj C_e P GK a @ P Ve a_Z GK gK @u^ Ke Construct a circle of radius 4.5 cm. Take a point P on it. At P construct a
tangent to the circle.
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12. ^cfLZ _gc^u ce ~KYi Z^Ue Ce \@ : 4 x 3
Answer any three of the following questions.
(i) MUG @aKe lZe Pe_U aW \ae cUeK1
3Uu jiae 288 Uu Ly
jf @ _e]e \N ^d Ke
Rs. 288.00 was spent for fencing around a semi-circular field at the rate of
Rs.1
3per metre. Find the length of the semi circumference.
(ii) \AU ae aie @^_Z 3:5 _[c ae lZ`k 198 a.i.c. jf, \Zd aelZ`k ^d Ke The ratio of the diameters of two circles is 3:5. If the area of the first circle is 198
sq.cm. then find the area of the second circle.
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17/24
SET : A
(17)
(iii) MUG _Rce bc GK icaj ZbR I CyZ 11 i.c. Gje @dZ^ 44 3 N^i.c. jf, Gje _g_Zke lZ`k d Ke
The base of a prism is an equilateral triangle and its height is 11 cm. If its volume
is 44 3 cubic cm. then find its lateral surface area.
(iv) 33 i.c. CyZ ag GK ifee aK_Zke lZ`k 11 i.c. \N ag MUGicN^e icM _Zke lZ`k ic^ ifee bce ai ^d Ke
The curved surface area of a cylinder of height 33 cm is equal to the whole
surface area of a cube of side 11 cm. Find the diameter of the base of the
cylinder.
(v) MUG @dZN^e @dZ^ 2430 N^cUe Gje \\, _i I CyZe @^_Z 6:5:3jf, Gje _g_Zke lZ`k ^d Ke
The volume of a cuboid is 2430 cubic meter and its length, breadth and height
are in the ratio 6:5:3. Find its lateral surface area.
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SET : A
(18)
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13. ic _ge Ce \@ : 2 x 7 = 14Answer all questions :
(i) \ ae AB CDII _cY Ke ~ AC = BD
In the given fiture, AB CDII . Prove that AC = BD
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(ii) \ PZe O, ae K
_cY Ke ~ mOBC = 900 mBACIn the given figure, O is the centre of the circle.
C
B
D
A
O
C
A
B
7/29/2019 Mathmatics Question Type Odisha Board Exam
19/24
SET : A
(19)
Prove that mOBC = 900 mBAC
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(iii) MUG aeK O Ga P GK aji a PQ I PR \AU gK L _cY Ke ~ DOPQ DOPR
O is the centre of a circle and P is an external point. PQ and PR are two
tangent-segments. Prove that DOPQ DOPR
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(iv) MUG a C_eiP ae @uZ gK RS R ij ice jf, _cY Ke ~ PaMc aiRPS K ic\L Ke
P is a point on a circle. At P a tangent is drawn parallel to the chord RS . Provethat the radius through P bisects RPS.
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7/29/2019 Mathmatics Question Type Odisha Board Exam
20/24
SET : A
(20)
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(v) PQ MUG ae R P I Q Ve Cq R _Z f\d ~[Kc aKR I S aeQ\ Ke _cY Ke ~ PR = QS
PQ is a chord of a circle. At P and Q perpendiculars drawn to PQR meet the
cirlce at R and S respectively. Prove that PR = QS._________________________________________________________________
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(vi) \ PZe PT gK L Q\K m AX T = 700I
m BY T = 1500
jf APTe _ecY ^d Ke In the given figure, PT is a tangent segment.
If m AX T = 700 and m BYA = 1500
then find the measure ofAPT._________________________________________________________________
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OA
TP
BY
X
Q
7/29/2019 Mathmatics Question Type Odisha Board Exam
21/24
SET : A
(21)
(vii) _cY Ke :
Prove : cos (450 + q) =1
2
(sin q cos q)
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14. a^ ce \@~A [a ia Cec^u ce _ZK _g _A VK CeU aQ ZcCe LZe fL 1 x 4Choose the correct answer from the possible answers given in the bracket below
each question and write it in your answer book.
(i) \ ae m AXB : m BY C : m C ZA : : 2:3:4 jf,
ACB e _ecY KZ ?
If in the given circle, m AX B : m BY C : m C ZA : : 2 : 3 : 4, thenwhat is the measure ofACB ? (400, 600, 800, 1000)
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(ii) GK ae 8 i.c. \N ag GK R KVe6 i.c. \ee @aiZ aeai KZ i.c.
In a circle a chord of length 8 cm is at a distance of 6 cm from the
centre. What is the length of the radius in cm ? (10, 12, 14, 16)
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B
A
C
X Z
Y
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SET : A
(22)
(iii) MUG ae Ga R Ke 900KY C_^ Ke Re \N 7 i.c.
jf, ai KZ i.c. ?In a circle a chord subtends an angle of 900 at the centre. If the length
of the chord is 7 cm, then what is the length of the radius in cm ?
2
77 2 2 7
7
2, , ,
FHG
IKJ
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(iv) tan 1350e c^ KZ ?
What is the value of tan 1350 (1, 0, 1, 3 )
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ADDITIONAL PAGES
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SET : A
(23)
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SET : A
(24)
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