Download Solution Of P-SAT 2014 For Class 9th CBSE

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P-SAT 2014 - 15
(PIONEER’S SCHOLARSHIP/ADMISSION TEST)
{9TH CBSE}
General Instructions:The question paper consist of FOUR sections (A), (B), (C), (D).
Section A contains 45 objective multiple choice questions of Mathematics.
Section B {Physics 46 – 60}, Section C {Chemistry 61 – 75} and Section D {Biology 76-90}.
Each right answer carries 4 marks and wrong –1.
Maximum Marks 360.
Maximum Time 180 minutes.
Give your response in the OMR sheet given to you with question paper.
Properly write down your roll no, name, contact number in the OMR sheet.
If there is any inappropriate filling of the circles in OMR then that sheet will be
disqualified.
Name: _______________________________Father Name:______________________________
Mobile: ______________________________School:_____________________________________
Solution Visits: www.pioneermathematics.com/latest_updates.com
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Section-A {Mathematics}
1.
The graph y = 6 is a line
(a) parallel to x-axis at a distance 6 units from the origin
(b) parallel to y-axis at a distance 6 units from the origin
(c) making an intercept 6 on the x-axis
(d) making an intercept 6 on both axis
Ans. (a)
2.
The equation y = 4x – 7 has
(a) no solution
(b) unique solution
(c) infinitely many solution
(d) exactly two solution
Sol: (c)
A linear equation in two variable has infinitely many solutions.
3.
A quadrilateral , whose diagonals bisect at right angles is called
(a) Trapezium (b) rectangle (c) a parallelogram with unequal adjacent sides (d)
rhombus
Sol: (d)
4.
In the given figure, ABCD and PQRC are rectangles and Q is the mid-point of AC then
Find the value of k. If PR = k(AC)
(a)
2
4
(b)
1
2
(c)
1
3
(d)
2
3
Sol: (b)
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Given : ABCD and PQRC ate two rectangles and Q is the mid-point of AC
To Prove:
Proof : DP = PC
QC = PR
QC =
1
AC
2
PR
5.
1
AC
2
In a parallelogram ABCD, AB = 10 cm and AD = 6 cm. The bisector of
A meets DC in E.
AE and BC produced meet at F. Find the length of CF.
(a) 4 cm
(b) 3 cm
(c) 6
(d) 5 cm
Ans. (a)
1
2
1
3 then
2
3
i.e., AB = BF = 10 cm
But BC = 6 cm then CF = 4 cm
6.
In the given figure, P is the mid-point of side BC of a parallelogram ABCD such that
BAP
DAP . Find the value of k , AD = k (CD).
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(a) 3
(b) 2
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(c) 5
(d) 6
Sol: (b)
1
2
But ABCD is ||gm then
then
1
2
3
3 i.e., AB BP
but P is mid-point of BC. Then
BP = PC
Now BP = AB = DC
and AD
= BC = BP + PC
= 2BP = 2AB = 2CD
7.
E and F are respectively the mid-points of the non-parallel sides AD and BC of a
trapezium ABCD. If EF||AB then find EF= k (AB + CD).
(a)
1
3
(b)
1
4
(c)
2
3
(d)
1
2
Ans. (d)
Construction : Join B to E which meets CD produced at G.
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Since AB || DC and E, F are mid-points of AD and BC
then in ABE and EDG,
1
2
[Vertically opposite angles]
3
4
[Alternate interior angles]
Hence DE = AE
 ABE
DGE
BE= EG
[ASA congruency]
i.e., AB = GD
Then CG = CD +DG = CD + AB
Now in BCG, BF FC and BE EG
i.e.,
8.
BF
BE
1
then EF||GC
FC
EG
and EF
1
GC
2
i.e., EF
1
[CD AB]
2
A parallelogram and a rectangle have common base and equal areas. Then
(a) the perimeter of the rectangle is smaller than the perimeter of the parallelogram. (b)
the perimeter of the rectangle is greater than the perimeter of the parallelogram. (c) the
perimeter of the rectangle is equal to the perimeter of the parallelogram.
(d) the perimeter of the rectangle is greater than and equal to perimeter of the
parallelogram.
Ans. (a)
9.
In the given figure, the sides of
PQR are 30 m, 24 m and 18 m and sides of ABC are
25 m, 24 m and 7 m. Find the area between the triangles (shaded region)
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(a) 140 m2
(b) 123 m2
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(c) 132 m2
(d) 136 m2
Sol: (c)
Area of PQR
1
24 18
2
= 216 m2
Area of ABC
1
24 7 84m2
2
Area of shaded region = ar( PQR) ar( ABC)
= 216 – 84 =132 m2.
10. A circle is inscribed in an equilateral triangle of side 12 cm touching the sides of the
triangle. Find the radius of the inscribed circle.
(a) 2 3 cm
(b) 2 4 cm
(c) 3 4 cm
(d) 3 6 cm
Sol: (a)
Let radius of in circle = x cm
Ar of ( ABC) = ar ( BOC) ar ( COA) ar ( AOB)
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3
(12)2
4
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1
1
1
BC OD
AC OE
AB OF
2
2
2
1
1
1
12 x
12 x
12 x
2
2
2
36 3
36 3 18x
x 2 3 cm
11. In ABC, median AD, BE and CF intersects each other at G. Then find the value of k. If
BE +CF >k (BC).
(a)
3
4
(b)
3
2
(c)
2
3
(d)
4
3
Sol: (b)
BE + EC > BC
BE
1
AC BC
2
Similarly, CF
BE CF
1
AB BC
2
1
(AC AB) 2BC
2
BE CF 2BC
1
(AC AB)
2
BE CF 2BC
1
BC
2
BE CF
 AC AB BC
3
BC
2
12. ABCD is a parallelogram X and Y are the mid-points of BC and CD respectively. Then find
the value of k. If ar(AXY) k ar (|| gm ABCD) .
(a)
2
8
(b)
4
8
(c)
3
8
(d)
3
9
Sol: (c)
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1
ar ( DBC)
4
ar( CYX)
ar(DBC)
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1
ar ( gm ABCD)
2
ar ( CYX)
1
ar (|| gm ABCD)
8
Similarly,
ar ( AYD)
1
ar (|| gm ABCD)
4
Now, use
ar ( AXY) ar (|| gm ABCD)
ar ( ABX) ar( AYD) ar( CYX).
13. PQRS is a rectangle inscribed in a quadrant of a circle of radius 13 cm. A is any point on
PQ. If PS = 5 cm, then find the value of ar (PAS) = ?
(a) 30 cm2.
(b) 40 cm2.
(c) 50 cm2.
(d) 60 cm2.
Sol: (a)
 A is any point on PQ
PA < PQ
ar ( PQR)
1
PQ QR
2
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=
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1
12 5 30 cm2
2
PS = 5 cm
PA < PQ
ar( PAS) ar PQR
ar( PAS) 30 cm2
14. ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Then find
the value of k. If ar (BDE) = k ar(ABC).
(a)
1
6
(b)
1
7
(c)
1
4
(d)
1
3
Sol: (c)
ar ( ABC)
3
(side)2
4
Let
side =a cm
ar ( ABC)
3 2
a
4
BD
a
2
ar( BDE)
3 a
4 2
2
3 a2
4 4
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2
3a
4 4
3 2
a
4
ar( BDE)
ar( ABC)
ar( BDE)
1
4
1
ar ( ABC)
4
15. ABCD is a parallelogram in which BC is produced to E such that CE = BC(in figure). AE
Intersects CD at F. If ar (DFB) = 3 cm2, find the area of the parallelogram ABCD.
(a) 13 cm2
(b) 12 cm2
(c) 15 cm2
(d) 10 cm2
Sol: (b)
In
ADF and ECF
ADF
ECF
Alternate int erior angles
CEF
Alternate int erior angle
AD = CF
DAF
ADF
ECF
DF = CF
(CPCT)
BF is median of BCD
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ar ( BDF)
1
ar( BCD)
2
ar ( BCD)
1
ar(||gm ABCD)
2
ar ( BDF)
1 1
ar(||gm ABCD)
2 2
1
ar(||gm ABCD)
4
ar ( BDF)
1
ar(||gm ABCD)
4
3
3 4 ar(||gm ABCD)
ar(||gm ABCD) 12cm2
16. ABCD is a trapezium in which AB ||DC, DC = 30 cm and AB = 50cm. IF X and Y are,
respectively the mid-points of AD and BC, then find the value of k.
If ar (DCYX) = k ar (XYBA).
(a)
7
9
(b)
5
9
(c)
5
8
(d)
4
8
Sol: (a)
In DCY and PBY
CY = BY
C
B
1
2
DCY
(Vertically apposite angles)
PBY
(ASA congruency)
DC =PB
[CPCT]
PB = 30 cm
AP = 50 cm + 30 cm = 80 cm
Using mid-point theorem
XY =
1
1
AP
80cm 40cm
2
2
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Let distance between AB, XY and DC is h cm.
DCYX
In
ar(XYBA)
1
h(30cm 40cm) 35hcm
2
1
h(40cm 50cm) 45hcm
2
ar( DCYX) 35 7
ar( XYBA) 45 9
ar ( DCYX)
7
( XYBA)
9
17. If O is centre of circle as shown in figure, find
(a) 40o
(b) 60o
RTQ
(C) 45o
(d) 90o
Sol: (c)
In the given figure,
5
5
6 360o
140o 360o
5 360o 140o 220o
 6 140o
...(i)
5 2 2 [The angle subtended by a chord is twice the angle subtended by it on
the circumference in the alternate segment]
220o
2 2
[from(i)]
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o
2 110
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.......(ii)
4 is the exterior angle.
In cyclic quadrilateral PQTS,
2 [Exterior angle of cyclic quadrilateral is equal to interior opposite angle]
4
4 110o
[From (ii)]
RQT 110o
3
1
[Exterior angle of cyclic quadrilateral is equal to interior
opposite angle]
3 45o
[ 1 45o ]
RTQ 45o
18. In the given figure, AB is the chord of a circle with centre O. AB is produced to C such
that BC = OB, CO is joined and produced to meet the circle in D. If
ACD y and
AOD x , find the value of k. If x = ky.
(a) 5
(b) 3
(c) 6
(d)4
Sol: (b)
OB = BC
[Given]
BOC
BCO
BOC y
OBA
BOC
OCB
[Exterior angle of a
is equal to sum of the opposite interior
angles]
OBA y y 2y
Also,
OA =OB
OAB
[Radii of the same circle]
OBA 2y
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Now,
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AOD is exterior angle if AOC
AOD
OAC
OCA
x 2y y
x 3y
19. In the given figure, determine b
(a)9o
(b)10o
(c)12o
(d)13o
Sol: (d)
In CAE, C
43o
A
62o
AEC 180o
AEC 180o
AEC 75o
In cyclic quadrilateral AEDB, a + AEC 180o
a =105o
and
A
BDE 180o
BDE 1180 ,also BDE c 1180
c 620
Similarly in AFB,
a + 62o +b = 180o
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105o
+
62o
+b
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=180o
b = 13o
20. In the given figure, O is the centre of the circle.
(a) x +y=z
(b)y +z =x
(c)x +z=y
(d) none of these
Sol: (a)
ABC x p
[Exterior angle of a ]
y = x + p +p = x+ 2p
z = 2 ABC 2x 2p
=(x +2p) + x = y + x
21. The sum of angles formed in the four segments exterior to a cyclic quadrilateral by the
sides is = x (900). Find the value of x?
(a)6
(b)
7
(c)
8
(d)4
Sol: (a)
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E
F
G
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H 6 right angles
In cyclic quadrilateral
similarly
and
E 180o
1
HAEB,
3
2
.....................(i)
F 180o
G 180o
.....................(ii)
.....................(iii)
Adding eq.(i), (ii) and (iii), we get
1
E
2
F
3
G 3 180o
2
3 540o
E
F
G
1
E
F
G
H 6 90o
22. In the given figure, O is the centre and AE is the diameter of the semicircle ABCDE. If AB
= BC and
(a)30o
AEC 50o , then find
(b) 40o
CBE .
(c) 45o
(d) 500
Sol: (b)
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Join OC,
2 AEC
In
AOC 100o
AOC
AOB and BOC, AB = BC, AO =OC
and
OB = OB
BOA
BOC
BOA 50o and
BOA
BOC 50o
CEO
[Each 50o]
BO || CE [If corresponding
AOC
Now,
COE 180o
s are equal, lines are ||]
[ AOC 180o ]
COE 80o
COE 2 CBE
CBE 40o
23. In the given figure,
(a) 40o
ADC 130o and chord BC = chord BE. Find
(b) 100o
(c) 90o
CBE
(d) 80o
Sol: (b)
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Join OC
AOC 260o
AOC 100o
Reflex
COB 180o 100o 80o
1
COB
2
1
CEB
80o
2
CEB
CEB
ECB 40o
CBE 180o (40o 40o ) 100o
24. AB and AC are two chords of a circle of radius r such that AB = 2AC. If p and q are the
distances of AB and AC from the centre. Find the value of k. If k(q2) = p2 + 3r2.
(a) 4
(b) 3
(c) 5
(d) 2
Sol: (a)
AM2 = r2 –p2
1
AB
2
2
r2 p2
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AB2
=
4(r2
–
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p2)
(2AC)2 = 4(r2 –p2) ( AB 2AC)
4AC2 =4(r2 –p2)
But
………(i)
AN2 = r2 – p2
AC
2
2
r 2 q2
AC2 = 4(r2 –q2)
….(ii)
From (i) and (ii)
r2 –p2 = 4r2 –4q2
4q2 = 4r2 –r2 +p2
25. In figure, O is the centre of the circle,
(a) 45o, 12o
(b) 30o, 15o
BCO 30o Find x and y.
(c) 16o, 40o
(d) 18o, 60o
Sol: (b)
In OCP,
OPC 90o
[AP BC]
OCP 30o
POC 90o 30o 60o
DO AP,
DOP 90o
COD 90o 60o 30o
CBD
1
COD
2
[Angle subtended theorem]
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1
30o 15o
2
y 15o
y
AOD 90o
ABD
1
AOD
2
[Angle subtended theorem]
In ABP,
x 45o y 90o 180o
x 45o 15o 90o 180o
x 180o 150o 30o
26. A metallic sheet is of the rectangular shape with dimensions 48cm×36 cm. From each of
its corners, a square of 8 cm is cut off and an open box is made of the remaining sheet,
then the volume of the box is
(a) 5102 cm3
(b) 5012 cm3
(c) 5000 cm3
(d) 5120 cm3
Sol: (d)
When squares of 8 cm is cut off , then
length of the box = (48 – 16) cm = 32 cm
breadth of the box = (36 – 16) cm = 20 cm
height of the box = 8 cm
volume of the box = 32×20×8 = 5120 cm3
27. How many spherical lead shots each 4.2 cm in diameter can be obtained from a
rectangular solid lead with dimensions 66cm, 42 cm and 21 cm?
(a) 1560
(b) 1500
(c) 1400
(d) 1630
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Sol: (b)
Dimensions of the rectangular solid are 66cm, 42 cm21 cm.
Volume of the solid = 66×42×21 cm3
……..(i)
Diameter of a spherical lead shot = 4.2 cm
radius = 2.1 cm
Volume of spherical lead shot
=
4 22
(2.1)3
3 7
........(ii)
Number of lead shots
Volume of the rec tangular solid
Volume of one spherical lead shot
=
66 42 21 21
88 (2.1)3
from (i) and (ii)
66 2 1000
1500
88
28. A sector of a circle of radius 12 cm has the angle 120o. It is rolled up so that two
bounding radii are joined together to form a cone. Find the volume of the cone.
(a) 189.57 cm3
(b) 190.57 cm3
(c) 185.57 cm3
(d) 170.57 cm3
Sol: (a)
 Angle of sector = 120o =
Length of are =
=
1
of central angle
3
1
1
of circumference = 2 r
3
3
1
22
2
12cm
3
7
Let radius of cone = x cm
Now circumference of base of cone = length of arc
1
22
22
2
12 2
x
3
7
7
x
4 cm
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radius = 4 cm, slant height = 12 cm,
h
l2 r 2
h
144 16
128 8 2cm
Volume of cone =
1 2
rh
3
29. Two solid spheres made of the same metal have weights 5920 g and 740 g, respectively.
Determine the radius of the larger sphere, if the diameter of the smaller one is 5cm.
(a) 10 cm
(b) 5 cm
(c) 6 cm
(d) 8 cm
Sol: (b)
m1 = 5920 g
m2 =740 g
Density (D) =
Volume =
Mass
Density
V1 =
5920 3
cm
D
V2 =
740 3
cm
D
V1
V2
5920
D
740
D
4 3
r
3 1
4 3
r
3 2
r13
r23
Mass
Volume
5920
740
5920
740
r2 = 5/2 cm
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3
1
r
5
2
r13
3
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5920
740
5920 125
740
8
r13 125
r13 5cm
Radius of longer sphere = 5 cm
30. The water for a factory is stored in a hemispherical tank whose internal diameter is 14
m. The tank contains 50 kilolitres of water. Water is pumped into the tank to fill to its
capacity. Calculate the volume of water pumped into the tank.
(a) 840.66m3
(b) 670.66m3
(c) 668.66m3
(d) 650.66m3
Sol: (c)
Water in the tank = 50,000 litres = 50 m3
volume of tank =
2 3
r
3
2 22 14.7 14.7 14.7
3 7
2
2
2
3
718.66m
Volume of water pumped in the tank
= 718.66 –50 =668.66 m3
31. A sphere and a right circular cylinder of the same radius have equal volume. By what
percentage does the diameter of the cylinder exceed its height ?
(a) 45%
(b) 40%
(c) 30%
(d) 50%
Sol: (d)
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Let radius of sphere be r
4 3
r
3
r 2h
h
4
r
3
Diameter =2r
Increased diameter = 2r
%
4
2r
r
3
3
2r
100
3
4
r
3
% increased =
2 3
3 4
100 50
32. If each edge of a cube is increased by 25%, then find the percentage increase in its
surface area
(a) 50.01%
(b) 52.23 %
(c) 56.25 %
(d) 26.25%
Sol: (c)
Let edge of cube = x units
Surface area – 6x2 square units
New edge = x +
25
5x
x
units
100
4
New surface area = 6
25 2
x square units
16
75 2
x square units
8
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Increase in surface area =
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75 2
x 6x2
8
27 2
x
8
27 2
x
8
% increase =
100 56.25%
6x2
33. Find the mean of the following distribution
Frequency
Variable
4
4
8
6
14
8
11
10
3
12
(a)8.05
(b) 10
(c) 15 (d) 43
Sol: (a)
Frequency
Variable
fixi
4
4
16
8
6
48
14
8
112
11
10
110
3
12
36
Total = 332
thus mean = 332/40 = 8.05
34. The mean marks (out of 100) of boys and girls in an examination are 70 and 73,
respectively. If the mean marks of all the students in that examination is 71. Find the
ratio of number of boys to the number of girls.
(a) 2:1
(b) 2:3
(c) 5: 4
(d) 1: 5
Sol: (a)
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Let the boys be x
Number of girls be (100 –x)
Mean of boys =
70=
Total marks
No. of boys
Total marks
x
Total marks = 70x
Mean of girls =
73=
Total marks
No. of girls
Total marks
(100 x)
Total marks = 73(100–x)
Total marks = 70x + 7300 –73x
= (7300–3x)
71 =
7300 3x
100
7300–3x = 7100
–3x = 7100-7300
–3x = –200
x=
200
3
200
3
Number of boys =
Number of girls =
200
Ratio = 3
100
3
100
3
2:1
Ratio of boys to girls = 2 :1
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35. In a class there are x boys and y girls, A student is selected at random, then probability
of selecting a girl is
(a)
x
y
(b)
x
(c)
x y
y
(d)
x y
y
x
Sol: (c)
36. ABC is an equilateral triangle. P is a point on BC such that BP : PC = 2 : 1. Find the value
of k. If k(AP2) = 7AB2.
(a) 8
(b)9
(c) 3
(d) 5
Sol: (b)
Draw AD BC
As ABC is an equilateral triangle, D is mid-point of BC
Given BP : PC = 2 : 1
Let PC = x, then BP = 2x
BC = BP + PC = 2x + x=3x
AB = 3x
….(i) (
AB = BC)
As D is mid-point of BC,
BD = DC =
1
1
3
BC = 3x = x
2
2
2
3
1
DP = DC –PC= x –x= x
2
2
In ABD, D 90o ,
AB2 = AD2 +BD2
AD2
=
AB2
–
BD2=(3x)2
3
– x
2
2
27 2
x
4
…..(ii)
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o
In
ADP, D 90
AP2 = AD2 + DP2
27
= x2
4
= 7x
2
1
x
2
2
[Using (ii)]
1
7. AB
3
2
[Using (i)]
9AP2 = 7AB2
37. ABC is an isosceles triangle. AB = AC=10 cm, BC = 12 cm. PQRS is a rectangle drawn
inside the isosceles triangle. Give PQ = SR = y cm and PS = QR =2x cm. Then x =
(a) 6
3y
4
(b) 6
4y
3
(c) 3
2y
4
(d) 5
6y
4
Sol: (a)
In ABC, AB AC
Draw AD
BC, then, D is mid-point of BC.
But BC = 12 cm (given)
Also PBQ
SRC
BD = 6 cm
BQ RC
QD = DR = x cm
BQ = BD –QD = (6 –x) cm.
In
ABD, D 90o
AB2 = AD2 +BD2
AD2 = AB2 –BD2
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AD2
=
(10)2
–(6)2
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= 100 –36 = 64
AD = 8cm
In ABD, PQ|| AD,
PQ
AD
BQ
BD
6y
8
6 x
y
8
6 x
6
x 6
3
y as required
4
38. P is any point in the interior of a triangle ABC.
(a) PA +PB < AC +BC
(b) PA +PB > AC +BC(c) PA +PB =AC +BC (d) None of these
Sol: (a)
According to the given statements, the figure will be as shown below.
Produce, BP to meet AC at point M.
Since, the sum of any two sides of a triangle is greater than its third side.
In BCM,BC CM
and, In
BM..................I
APM, AM PM AP............II
Adding I and II, we get
BC + CM + AM + PM > BM +AP
BC + (CM +AM) > BM –PM +AP
BC +AC > PB +PA
i.e. PB + PA < BC+AC
39. The diagram shows a right pyramid that has an isosceles triangular base. If the volume
of the pyramid is 330 cm3, calculate its height , h
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(a) 16.5 cm
(b) 15.2 cm
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(c) 12.30 cm
(d) 10.23 cm
Sol: (a)
AD
132 52
144
12cm
Area of base =
1
BC AD
2
1
10 12
2
60 cm2
Volume of pyramid =
330
1
(Area of base) × Height
3
1
60 h
3
20h 330
h
330
16.5 cm
20
40. The diagram shows a cylinder with a diameter of 10 cm and of height 15 cm. The
shaded portion in the form of cone, with base diameter 10 cm and height 6 cm, is
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hollowed out. Find the volume of the remaining solid, in cm3
(a) 235
cm3
(b) 325
cm3
(c) 255
cm3
(d) 366
cm3
Sol: (b)
Volume of cylinder = r2h
52 15
=
= 375 cm3
Volume of cone =
=
1
3
1 2
rh
3
52 6
= 50 cm3
Volume of remaining solid = 375
50
325 cm3
41. If a leap year is selected randomly. Find the probability of getting 53 Sundays?
(a) 1/7
(b) 2/7
(c) 1/365
(d) 53/365
Sol: (b) In a leap year there are 366 days
thus the number of Sunday that can be completed in this leap year =52
thus were are left with two days and they can be
Monday and Tuesday
Tuesday and Wednesday
Wednesday and Thursday
Thursday and Friday
Friday and Saturday
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Saturday and Sunday
thus probability of having 53 Sunday is
2
7
42. 19 cards numbered 1, 2, 3, 4, …………, 16, 17, 18,19 are put in a box and mixed
thoroughly. One person draws a cards from the box at random. Find the probability that
the number on the card is (i) a prime number,
(a)
8
19
(b)
7
19
(c)
7
13
(d)
4
19
Sol: (a)
Let E2 = event of getting a prime number. Then, E2 = {2, 3, 5, 7, 11, 13, 17, 19}
n(E2) =
8.
P(getting a prime number) = P(E2) =
n(E2 ) 8
n(S) 19
43. A bag contains 5 black balls, 4 blue balls, 4 white balls and 6 red balls. One ball is drawn
at random from the bag. Find the probability that the ball drawn is neither black nor
blue.
(a) 10/19
(b)
5
19
(c)
8
19
(d)
8
11
Sol: (a)
Let E4 = event of getting a ball which is neither black nor blue.
Then, n(E4) = (4 + 6) = 10
P (getting a ball which is neither black nor blue)=P(E4) =
n(E4 ) 10
n(S) 19
44. In an examination, the mean of marks scored by a class of 40 students was calculated
as 72.5. Later on, it was detected that the marks of one student were wrongly copied as
48 instead of 84. Find the correct mean.
(a) 71.2
(b) 73.4
(c) 70.6
(d) 75.6
Sol: (b)
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Incorrect sum of marksof 40students
40
Mean of marks=
Incorrect sum of marksof 40students
40
72.5=
Incorrect sum of marks of 40 students = 72.540 = 2900.
Since the marks of one student were wrongly copied as 48 instead of 84, correct sum of
the marks of 40 students
= 2900 – 48 + 84 = 2936.
2936
73.4
Correct mean = 40
45. Mean temperature of a city of a certain week was 250C. If the mean temperature of
Monday, Tuesday, Wednesday and Thursday was 230C and that of Thursday, Friday,
Saturday and Sunday was 280, find the temperature of Thursday.
(a) 270C
(b)280C
(c) 290C
(d) 300C
Sol: (c)
Mean temperature of week = 25oC,
The sum of temperatures of 7 days of the week = 7 × 25oC = 175oC.
……(i)
Sum of temperatures of Monday, Tuesdays, Wednesdays and Thursday.
=4 ×23oC = 92oC.
………..(ii)
Sum of temperatures of Thursday, Friday, Saturday and Sunday.
= 4× 28oC = 112oC
…………..(iii)
Sum of temperatures of Monday to Sunday and, Thursday
= 92oC + 112oC
= 204oC
[Using (ii) and (iii)]
…………..(iv)
Temperature of Thursday = 204oC – 175oC
[using (iv) and (i)]
= 29oC
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Section–B {Physics}
46. A U tube is partially filled with mercury. If water is added in one arm and an oil is added
in the other arm, find the ratio of water and oil columns so that the mercury level is
same in the two arms of U tube. Given : density of water = 103 kg m–3, density of oil =
900 kg m–3.
(a)
9
10
(b)
8
10
(c)
5
10
(d)
4
10
Sol: (a)
Since the level of mercury is same in the two arms of the U tube, therefore
Pressure of water column on the surface of mercury in one arms = Pressure of oil
column on the surface of mercury in the other arm. i.e.,
h1 1g h2 2g
where h1 = height of water column,
1
= density of water = 103 kg m–3.
h2 = height of oil column , and
2
=density of oil = 900 kg m–3.
h1
h2
1
2
900 9
103 10
47. A solid density
has weight W. then find its apparent weight
when it is completely submerged in a liquid of density
(a) W 1
(b) W 1
L
L
L
(c) W 1
.
(d) W 1
L
L
Sol: (b)
Given, weight of the solid = W
Mass of the solid = W/g
Volume of the solid =
Mass
Density
W/h
Volume of liquid displaced = Volume of the solid
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W/g
Upthrust on the solid = Volume of liquid displaced × density of the liquid ×acceleration
due to gravity
=
W/g
L
g
W
L
Apartment weight = True weight – upthrust
=W
W
L
W 1
L
48. A bock of wood floats in water with 2/5th of its volume above the surface. The density of
wood is
(a) 0.12g/cm3
(b) 1.2 g/cm3
(c) 0.6g/cm3
(d) 4.2 g/cm3
Sol: (c)
49. A solid weights 30 gf in air and 26 gf when completely immersed in a liquid of relative
density 0.8. Find the relative density of the solid.
(a) 6
(b)5
(c) 8
(d) 4
Sol: (a)
Given weight of solid in air = 30 gf and in liquid = 26 gf., R. D. of liquid = 0.8
Density of liquid = 0.8 g cm–3
Given, weight of solid = 30 gf
Mass of solid = 30 g
Density of solid =
Mass
Volume
30
6 g cm
5
3
Hence, relative density of solid = 6.
50. A body is initially at rest. It undergoes one dimensional motion with constant
acceleration. The power delivered to it at time t is proportional to
(a) t1/2
(b) t
(c) t3/2
(d) t2
Sol: (b)
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u at at
P F
ma
ma at ma2t
Since ma2 is constant
P
t
51. The blades of wind mill sweep out a circle of area A = 30 m2. If the wind flows at a
velocity
36km / h perpendicular to the circle, then the mass of air passing through
in it time t = 1 minute is (take density of air = 1.2 kg m–3)
(a) 216 kg
(b) 2160 kg
(c) 21600 kg
(d) 216000 kg
Sol: (c)
Mass = volume × density
=A
t t
= 30 × 10 × 60 × 1.2
(
=30km/h= 10 ms–1)
= 21600 kg
52. A mass of 10 g moving horizontally with a velocity of 100 cm s–1 strikes a pendulum bob
of mass 10 g. The two masses stick together (see in the figure). The maximum height
reached by the system is (g = 10 ms–2).
(a) zero
(b) 5 cm
(c) 2.5 cm
(d) 1.25 cm
Sol: (c)
K. E. mass of 10 g is converted into O.E. of the system.
1
i.e., m1 12 (m1 m2 )gh
2
m 1 12
h
2(m1 m2 )g
10 10000
2 20 1000
=
10
2.5cm.
4
53. A body is moving uni-directionally under the influence of a source of constant powder.
Its displacement in time t is proportional to
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(a)
t2
(b)
t3/2

P
m
Medical and Non - Medical Classes
(c) t
(d) t1/2
Sol: (b)
W
t
P
FS
t
m aS
t
m distance
=
t
t2
or
S2
P 3
t
m
or
S2
t3
or
S t 3/2
mSq2
S
t3
constt.
54. If the momentum of a body increase by 30%, then its kinetic energy will increase by
(a) 30 %
(b) 60 %
(c) 69%
(d) 90%
Sol: (c)
K.E., E
p2
2m
Now, p' p 30%p p
p'2
New K.E., E'
2m
30
13p
p
100
10
13
10
2
p2
2m
169
E
100
%increse in K.E.
E' E
100
E
169
169
1 100
100
= 100
100
69%
55. A motor can pump up water to fill a tank of volume 500 m3 in 25 minutes, which is
placed at a height of 20 m. If efficiency of the motor is 40%, calculate the power of the
motor.
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(a) 6.53 W
(b) 7.52 W
Medical and Non - Medical Classes
(c) 5.36 W
(d) 9.82 W
Sol: (a)
mass of water to fill the tank
m = Volume × density of water
m = 50 m3 × 1000 kg m–3 = 5×104 kg
Height of tank, h = 20 m
Work done by the pump to fill the tank
= mgh = 5×104 × 9.8×40
= 9.8×106 J
Efficiency = 40% =
40
100
2
5
Useful work,
W
2
9.8 106 3.93 106 J
5
Time, t = 215 minutes = 15 × 60 = 900s
W
Power, P =
t
3.92 106
900
6.53kW
56. A man makes a short and loud sound in front of a hill and the echo is heard after 3
seconds. On moving closer to the hill by 165m, the echo is heard after 2 seconds.
Calculate the velocity of sound and the distance of hill from the first position.
(a) 330 ms–1 (b) 356 ms–1
(c) 536 ms–1
(d) 226 ms–1
Sol: (a)
Let, velocity of sound = V
And distance of hill from first position = x
Distance of hill from second position = (x – 165) m
Now, distance travelled by the sound in going and coming back from the first position
=2x
Time = 3s
Hence, Distance = Velocity × time
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or 2x = 3V
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….(i)
Similarly, for the second position,
=2(x–165) = 2V
……..(ii)
Dividing equation (i) from (ii), we get
x
3
(x 165) 2
or
2x =3x –495
or
x =495 m
Putting this value of x in equation (i) , we get
2×495=3V
2 495
ms
3
V
or
1
330ms
1
57. A source of sound produces waves of wavelength 0.80 m in air. The same source of
sound produces waves of wavelength 4.0m in water. If the velocity of sound in air = 332
ms–2, find the velocity of sound in water.
(a) 1860 ms–1
(b) 1660 ms–1
(c) 1760 ms–1
(d) 1540 ms–1
Sol: (b)
We know that,
Velocity = V = v
for water, V1 = v1
And for air, V2 = v2
V1
V2
v1
v2
1
2
1
2
Here,
v1 = v2 ) Since source of sound is the same)
V2 = 332 ms–1
2
0.80m
1
4.0m
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V1
4.0
332 0.80
V1
or
332 4.0
ms
0.80
1
1660ms
1
58. The following graph shows the displacement vs distance of a pulse on a role at two
different times. Find the speed of the pulse.
(a) 5.5 cm s–1
(b) 4.9 cm s–1
(c) 5.9 cm s–1
(d) 4.5 cm s–1
Sol: (d)
From the graphs
Distance between the two maxima = 37.5–15=22.5 cm
Time interval = 10–5 = 5s
Speed of the pulse =
=
22.5
4.5cm s
5
dis tance travelled
Time taken
1
59. A sonar emits pulses on the surface of water which are detected after reflection from
the bottom. If the time interval between the emission and detection of the pulse is 2 s,
find the depth of water. (Take velocity of sound in water as 1531 ms–1)
(a) 1531 m
(b) 1630 m
(c) 1450 m
(d) 1860 m
Sol: (a)
Given time = 2 s, velocity of sound in water = 1531 ms–1
Let the depth of water = d
Therefore, total distance travelled by sound before it is detected by the Sonar= 2d
Using the expression
Distance = velocity × Time
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We have 2d = 1531× 2
Therefore, d
1531 2
1531m
2
60. A body is vibrating 6000 times in one minute. If velocity of sound in air is 360 m/s, find
the frequency of the vibration in Hz
(a) 400 Hz
(b) 100 Hz
(c) 200 Hz
(d) 300 Hz
Sol: (b) Given Number of vibrations = 6000.
time (t) = 1 min = 60s, velocity (v) = 360 ms–1
v
number of waves
time taken
6000
100Hz
60
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Section–C {Chemistry}
61. Rearrange the following (I to IV) in the order of increasing masses and choose the
correct answer from (a), (b), (c) and (d) [Atomic masses: N = 14, O = 16, Cu = 63]
I. 1 molecule of oxygen
II. 1 atom of nitrogen
III. 1×10–10 g-molecule of oxygen
IV. 1×10–10 g-atom of copper
(a) II < I< III < IV
(b) IV < III< II <I
(c) II<III<I<IV
(d) III < IV < I < II
Sol: (a)
1 molecule of O2 weights =
32
14
g; 1N atm
g
NA
NA
1×10–10 g=molecule of oxygen = 32×10–10g;
1×10–10 g-atom of copper = 63.5×10–10g.
62. Simplest formula of the compound containing nitrogen (14g) for each 40 g of oxygen is
(A) NO
(b) N2O3
(c) N2O4
(d) N2O5
Sol: (d)
40 g oxygen =
40
16
5
g atom;
2
14 g of nitrogen = 1 g-atom
The formula is NO5/2 or N2O5.
63. 2.0 g of oxygen contains number of atoms same as in
(a) 4g of S
(b) 7 g of nitrogen
(c) 0.5 g of H2
(d) 12.3 g of Na
Sol: (a)
Both 2 g of oxygen and 4 g of sulphur represent 0.125 g-atom of element.
64. Number of water molecules in the drop of water, if 1 mL of water has 20 drops and A is
Avogadro’s number , is
(a) 0.5 A/18
(b) 0.05 A
(c) 0.5 A
(d) 0.05 A/18
Sol: (d)
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1 drop of water =
1
1
mL
g
20
20
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(dH2O 1g mL 1 )
18 g of water = A molecules
1
g water
20
65.
30
14
A 1
18 20
0.05A
molecules
10
Si and 31
15P are
(a) Isotopes
(b) isobars
(c) isomorphs
(d) isotones
Sol: (d)
These contain same number of neutrons and hence are isotones.
66. Isotopes of an element have
(a) similar chemical properties but different physical properties
(b) similar chemical and physical properties
(c) similar physical properties but different chemical properties
(d) different chemical and physical properties.
Ans.(a)
67. Which of the following triads represents isotones?
(a)
12
6
C,
13
6
(c)
40
18
Ar,
C, 14
6 C
(b)
40
18
Ar,
42
20
(d)
14
7
N,
41
Ca, 21
Sc
42
20
43
Ca, 21
Sc
16
8
O, 18
9 F
Sol: (b)
40
18
Ar,
42
20
43
Ca, 21
Sc have 22 nuetrons each
68. Rutherford experiment which established the nuclear model of the atom used a beam of
(a)
-particles which impinged on a metal foil and got absorbed.
(b) -rays which impinged on a metal foil and ejected electrons.
(c) helium atom, impinged on a metal foil and got scattered.
(d) helium nuclei, impinged on a metal foil and got scattered.
So: (d)
-particles are same as helium nuclei.
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69. For an ement with atomic number 19, the
(a) L shell
(b) M shell
Medical and Non - Medical Classes
19th
electron will occupy
(c) N shell
(d) K shell
Ans: (c)
70. If the nuclide of actinium 89Ac228 emits beta particle , he daughter nuclide will be:
(a)
88Ra228
(b) 90Th228
(c) 87Fr224
(d)
90Th229
Ans: (b)
71. A sample of an element X contain two isotopes
17X35
and
17X37
. if the average atomic
mass of the sample is of the element be 35.5 then the percentage of the these two
isotopes in the sample is:
(a) 25%, 75%
(b) 10%, 90 %
(c) 75%, 25%
(d) 90%, 10%
Ans: (c)
72. The atomic number of an element Y is 20 the electronic configuration of its ion having
inert gas configuration is
(a) 2,8,10
(b) 2, 18
(c) 2,10,8
(d) 2,8,8
(c) 20Ca40, 18Ar40
(d) None of these
Ans: (d)
73. Which of these is a pair of isobars
(a) 6C12, 8O16
(b) 6C13, 6C14
Ans: (c)
74. Alpha particle is emitted by 92X238 during radioactivity the new species Y should be
(a) 90Y234
(b) 90Y238
(c) 92Y234
(d) 94Y234
Ans: (a)
75. Which among the following is true?
(a) Alpha rays are cathode rays.
(b) Electrons make up the cathode ray.
(c) Protons make up the cathode
(d) electromagnetic radiation make up the cathode ray.
Ans: (b)
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Section –D {Biology}
76. Which is correct about earthworm
(a) it has brain but no head
(b) it secrets cocoon around unfertilized egg|f
(c) it has no locomotory organs
(d) it can crawl on smooth surface concerned
Ans.(a)
77. Clitellum of pheretima is primary concerned with
(a) copulation
(b) production of cocoons
(c) excretion
(d) burrowing
Ans.(b)
78. Which of the following statements is true for pheretima?
(a) it is a dioecious animal with distinct sexual dimorphism
(b) in it copulation occurs at night in burrow during rainy season
(c) it can copulate throughout the year when ever it rains
(d) it cannot travel both backwards and forwards
Ans.(b)
79. A skeleton like function during locomotion of pheretima posthuma is performed by
(a) blood
(b) elementary canal laden with mud
(c) coelomic fluid
(d) ventral nerve cord
Ans.(b)
80. Milluscs are
(a) diploblastic and coelomate
(b) triploblastic and acoelomate
(c) triploblastic and pseudocoelomate
(d) triploblastic and coelomate
Ans.(d)
81. Vitamin that is destroyed on heating
(a) vitamin D
(b) vitamin C
(c) vitamin A
(d) vitamin K
Ans.(b)
82. Which of the following sets includes the bacterial diseases?
(a) diphtheria, leprosy, plague
(b) malaria, leprosy, plague
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(c) tetanus, tuberculosis, measles
Medical and Non - Medical Classes
(d) malaria, mumps, poliomyelitis
Ans.(a)
83. Escherichia coli in human intestine help to synthesize
(a) vitamin B and D
(b) vitamin B and C
(c) vitamin A and K
(d) vitamin B and K
Ans.(d)
84. Region of human body where poliomyelitis virus multiplies.
(a) muscle cells
(b) epithelial cells
(c) intestinal cells
(d) nerve cells
Ans.(a)
85. Who discovered the small pox vaccine?
(a) Louis Pasteur
(b) Edward Jenner
(c) Alexander Fleming
(d) Anton van Leeuwenhock
Ans.(b)
86. Which of the following atmospheric pollutants does the exhaust of motor vehicles in
Delhi NOT produce?
(a) SO2
(b) fly ash
(c) hydrocarbons gas
(d) CO
Ans.(b)
87. An American plants that had become a troublesome waterweed in India is
(a) Trapa bispinosa
(b) Cyperus rotundus
(c) Eichhornia crassipes
(d) Trypha latifolia
Ans.(c)
88. The species, which are in danger of extinction, are referred to as
(a) endangered species
(b) vulnerable species
(c) threatened species
(d) rare species
Ans.(a)
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89. Which of the following is an algae
(a) apple moss
(b) irish moss
(c) club moss
(d) reindoor moss
(c) very rare
(d) most extensive soils
Ans: (b)
90. Alluvial soils in India are
(a) found in desert (b) least fertile
Ans.(d)
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