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P-SAT 2014 - 15
(PIONEER’S SCHOLARSHIP/ADMISSION TEST)
{10TH 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.
Find the positive value of k, for which the equation x2 + kx +64 =0 and x2 – 8x +k = 0 will
both have real roots.
(a) 5
(b) 16
(c) 10
(d) 15
Sol:(b)
If the equation x2 + kx + 64 = 0 has real root, then D
k2 – 256 0
k2
256
k 16 ( k 0)
0.
k2 (16)2
.........(i)
If the equation x2 – 8x +k =0 has real root,
then D 0
64 4k 0
k 16
From (i) and (ii), we get
2.
4k 64
..........(ii)
k = 16.
In the following determine the set of values of ‘p’ for which the given equation has real
roots:
2x2 + 3x + p =0
(a) p
9
8
(b) p
7
8
(c) p
9
8
(d) p
9
8
Sol:(d)
2x2 + 3x + p = 0
The equation has real roots if D 0
3.
Swati can row her boat at a speed of 5 km/h in still water. If it takes her 1 hour more to
row the boat 5.25 km upstream then to return downstream, find the speed of the stream.
(a) 8km/h
(b) 1km/h
(c) 2km/h
(d) 3km/h
Sol:(c)
Let the speed of the stream x km/h.
Speed of the boat upstream = (5 –x) km/h.
Speed of the boat downstream = (5 +x) km/h.
According to the given condition,
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5.25
5 x
5.25
1
5 x
21
21 1
1
4 5 x 5 x
5 x 5 x
25 x2
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1
4
42x 100 4x2
4x2 42x 100 0
2x2 21x 50 0
(2x 25)(x 2) 0
25
x 2,
2
Rejecting x
25
2
The speed of the stream = 2km/h.
4.
In a flight for 3000 km, an aircraft was slowed down due to bad weather. Its average speed
for the trip was reduced by 100 km/hr and consequently time of flight increased by one
hour. Find the original duration of flight.
(a) 5 hour
(b) 6 hour
(c) 3 hour
(d) 4 hour
Sol:(a)
Let the original speed = x km/hr
Normal time of flight =
3000
hours.
x
……….(i)
New speed = (x –100)km/hr.
Time of flight when the speed is reduced =
3000
hours.
x 100
......(ii)
According to the given condition,
3000 3000
1
x 100
x
…………..[From (i) and (ii)]
3000x 3000x 300000
1
x(x 100)
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x(x 100) 300000
x2 100x 300000 0
(x 600)(x 500) 0
x 600 0
or x 500 0
x 600,
500, reject ing x
500, we get
Original speed of aircraft = 600 km/hr
Original duration of flight =
5.
3000
5 hours.
600
If p, q, r are rational numbers and p q r, then roots of the equation (p2 –q2) x2 –(q2 – r2)
x+(r2 –p2) =0 are
p2 q2
(c) 1, 2 2
r p
p2 r2
(b) 2 , 2
q q
p r
(a) ,
q p
p2 r2
(d) 1, 2 2
p q
Sol:(d)
p2 q2 x2
x
=
=
=
=
=
q2 r2 x
q2 r 2
q2 r 2
r2 p2
q2 r 2
2
0
4 p2 q2 r2 p2
2 p2 q2
q4 r 4 2q2r2 4 p2r2 p4 q2r 2 p2q2
2 p2 q2
q2 r 2
q4 r 4
2p2
2
2q2r2 4p2q2 4p2r 2
2 p2 q2
q2 r2
q2 r2 2p2
2
2 p2 q2
q2 r 2
q2 r2 2p2
2 p2 q2
q2 r2 q2 r2 2p2 q2 r2 q2 r2 2p2
,
2 p2 q2
2 p2 q2
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2
2
2
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2
2q 2p
2p 2r
,
2 p2 q2 2 p2 q2
=
2 q2 p2
=
2 p2 q2
,
p2 r2
p2 q2
p2 r2
= 1, 2 2
p q
6.
For an A.P. if tp + tp+2q =k(tp+q). Then find k.
(a) –1
(b) –2
(c) 1
(d) 2
Sol:(d)
Let a be the first term and d be the common difference of an AP.
t p a (p 1)d,
tp
2q
tp
q
a (p 2q 1)d,
a (p q 1)d
tp tp
2q
2a (2p 2q 2)d
2[a (p q 1)d] 2t p
7.
q
Which term of the progression
1
2
19,18 ,17 ,........ is the first negative term.
5
5
(a) 26
(b) 33
(c) 25
(d) 10
Sol:(c)
1
2
The given expression is 19, 18 , 17 ,........
5
5
Here,
T2 T1
T3 T2
91
19
5
91 95
5
87 91
5
5
4
5
............ i
4
5
Therefore,, (i) is an AP with a = 19, d =
4
5
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Let the nth term of the given AP be the first negative term.
Then, nth term < 0
Tn 0
19 (n 1)
4
5
0
(99 4n) 0
4n 99
3
n 24
4
n = 25, i.e., 25th term is the first negative term in the given AP
8.
If the pth, qth, rth term of an AP be x, y, z respectively then
x(q–r) +y(r–p)+z(p–q) =k. find the value of k.
(a) 0
(b) 1
(c) –1
(d) 2
Sol:(a)
Let a be first and d be the common difference of AP
tp
x
a (p 1)d x
......(i)
tq y
a (q 1)d y
.......(ii)
tr z
a (r 1)d z
...........(iii)
Substituting the values of x, y, z from (i), (ii), (iii), we get
x(q r) y(r p) z(p q)
a (p 1)d (q r) [a (q 1)d](r p) [a (r 1)d](p q)
a[(q r) (r p) (p q)] d[(p 1)(q r) (q 1)(r p) (r 1)(p q)]
a(0) d[p(q r) q(r p) r(p q) (q r r p p q)]
d(0 0) 0
9.
If a, b, c, d, e and f are in AP, then e –c is equal to
(a) 2(c –a)
(b) 2( f –d)
(c) 2(d –c)
(d) d –c
Sol:(c)
a, b, c d, e, f are in A.P.
b = a + D, c = a + 2D, d = a + 3D, e = a + 4D, f = a + 5D
e – c = (a + 4D) – (a + 2D)
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= 2D
Also, 2(d – c) = 2 (3D – 2D) = 2D
10. If a1, a2, a3,………is an AP such that a1 + a5 + a10 +a15 + a20 + a24 = 300, then a1 + a2 + a3 +…..a24
is equal to
(a) 909
(b) 75
(c)1200
(d) 900
Sol:(c)
a1, a2, a3, ……is an A.P.
a2 = a1 + d, a3 = a1 + 2d, a4 = a1 + 3d, ……..
Also, a1 + a5 + a10 + a15 + a20 + a24 = 300
a1 + a1 + 4d + a1 + 9d + a1 + 14d + a1 + 19d + a1 + 23d = 300
6a1 + 69d = 300
2a1 + 23d = 100
Now, a1 + a2 + a3 + ……..a24
= a1 + a1 + d + a1 + 2d + a1 + 3d + ……….a1 + 23d
= 24a1 + (d + 2d + 3d + ………..23d)
= 24a1 +
23
[2d + (23 – 1)d] = 24a1 + 23(d + 11d)
2
= 24a1 + 23 × 12d = 12(2a1 + 23d)
=12(100)=1200
11. An aeroplane flying horizontal at a height of 2500m above the ground is observed at an
elevation of 60o, and after 15 seconds, the elevation is observed to be 30o. Find the speed
of the aeroplane in km/hr.
(a) 692.8 km/hr.
(b) 792.8 km/hr.
(c) 612.8 km/hr. (d) 680.8 km/hr.
Sol:(a)
Initially the aeroplane is at A and after 15 seconds its position is at C.
AC is the distance covered in 15 sec.
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Also
AOB 60o ;
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COD 30o ;
AB = CD =2500m
In rt. ABO,
OB
cot 60o
AB
In rt.
2500
3
OB
............(i)
CDO,
OD
cot 30o
CD
OD 2500 3
..................(ii)
From (i) and (ii)
BD OD OB 2500 3
BD 2500
3 1
3
2500
2500
3
2500 2
3
3
3
1
3
5000 1.732 8660
m.
3
3
8660
m. is the distance covered in 15 sec.
3
Speed =
8660
8660
m / sec.
60 60km / hr 692.8km / hr.
3 15
3 15 1000
Speed of the aeroplane is 692.8 km/hr.
12. If a leap year is selected randomly. What is the probability of getting 52 Mondays?
(a) 1/7
(b) 5/7
(c) 2/7
(d) 1
Sol:(b)
Leap year
366 days
366 days
52 weeks + 2 days
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Sun, Mon Mon, Tue ,
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Tue, Wed , Wed, Thursday
Thursday, Friday , Friday, Sat , Sat, Sun
Total cases = 7
favourable cases = 5
Probability =
5
7
13. In the given figure, ABCD is a trapezium in which AB||CD. Line segments RS and LM are
drawn parallel to AB such that AJ = JK =KP. If AB = 0.5m and AP = BQ = 1.5 m. Find the
length of LM.
(a) 2.654m
(b) 3.654m
(c) 1.654m
(d) 5.63m
Sol:(c)
AB||RS||LM||CD.
ARJ
ACP 60o
ALK
AJ = JK = KP = 0.5 m.
In rt. APC,
AC
cosec60o
AP
15 2 3
10
3
AC 1.5
2
3
3 1.73m
AC = BD
Now, RS = RJ +JJ’+J’S=2RJ+0.5
…………(i)
[RJ=J’S]
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In rt AJR,
RJ
cot 60o
AJ
RJ 0.5
1
3
5
10
3
3
RJ = 0.288m
Substituting in (i), we get
RS =0.5 +2×0.288=0.5+0.576=1.076m
Now LM = LK +KK’ +K’M=2LK+0.5
…………(ii)
[LK=K’M]
In rt. AKL,
LK
AK
cot 60o
1
0.577m
3
LK 1.0
Substituting in (ii), we get
LM = 2× 0.577+0.5=1.154+0.5=1.654m.
14. If the angle of elevation of the cloud from a point h m above a lake is
depression of its reflection in the lake is
and the angle of
, find the height of the cloud is
(a)
h(tan
tan
tan )
.
tan
(b)
h(tan
tan
tan )
.
tan
(c)
h(tan
tan
tan )
.
tan
(d)
h(tan
tan
tan )
.
tan
Sol:(a)
In AOB
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y
x
tan
y xtan
y
tan
x
In AOB
tan =
2h y
x
x tan
tan
y
tan
2h y
2h y
y
tan
tan
y
2h tan
tan
tan
1
2h
height of the cloud= h + y
=h
2h tan
tan tan
h tan
tan
h tan
tan
15. The angle of elevation of a cliff from a fixed point is
meters towards the top of the cliff at an angle of
. After going up a distance of k
, it is found that the angle of elevation is
, Then the height of the cliff is
(a)
k(cos sin cot )
cot
cot
(b)
k(cos sin cot )
cot
cot
(c)
k(cos sin cot )
cot cot
(d)
k(cos sin cot )
cot
cot
Sol:(a)
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CE
OE
CD
BD
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k sin
AD k sin
k cos , OA hcot
OA OE hcot
k cos
AB CE h k sin
BD
CD
tan
h k sin
hcot
k cos
and simplify.
16. If the point (x, y) be equidistant from the points (a +b, b –a) and (a –b, a +b), then x = k y.
find k.
(a)
b
a
(b)
a
b
(c)
a
b
(d)
b
a
Sol:(b)
Let P(x, y), A(a+b, b–a) and B(a-b, a+b) be the given points.
Since AP = BP,
AP2 BP2
(x a b)2 (y b a)2
(x a b)2 (y a b)2
(x a b)2 (x a b)2
(y a b)2 (y b a)2
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(x a b x a b)(x a b x a b)
(y a b y b a)(y a b y b a)
(2x 2a)( 2b) (2y 2b)( 2a)
4bx 4ab 4ay 4ab
4bx 4ay
bx ay
17. Find the ratio in which the line segment joining (5, 6) and (2, –3) is divided by x-axis.
(a) 1:3
(b) 2:5
(c) 2:1
(d) 1:1
Sol:(c)
Let the required ratio be k:1.
Then the coordinates of the point of division are
2k 5 3k 6
,
.
k 1 k 1
This point lies on the x-axis whose equation is y = 0.
3k 6
0
k 1
3k 6,
or k 2
Line segments joining the two points is divided in the ratio of 2 :1 internally by x-axis.
18. If ‘a’ is the length of one of the sides of an equilateral triangle ABC, base BC lie on x-axis
and vertex B is at the origin, find the y-coordinates of the vertex A of triangle ABC.
(a)
3a
2
(b)
a
2
(c)
a
3
(d) None of these
Sol:(a)
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Let C (a, 0) and A(x1, y1)
Now, AB =
x12 y 12
BC = a
AC = (x1 a)2 y 12
ABC is an equilateral
AC = BC
(x1 a)2 y 12
a
(x1 a)2 y 12 a2
x12 a2 2x1a y 12 a2
...........(i)
x12 y 12 2x1a
Also AB = BC
x12 y 12
a
x12 y 12 a2
..................(ii)
From (i) and (ii)
2x1a a2
x1
a
and proceed
2
19. If P and Q are two points whose coordinates are (at2, 2at) and
S is the point (a, 0). Then
(a)
t2
(b)
a
2a
,
respectively and
2
t
t
1
1
is
SP SQ
t
(c) independent of t
(d) none of these
Sol:(C)
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(at 2 a)2 (2at)2
SP
a2t 4 a2 2a2t 2 4a2t 2
(at 2 a)2
(at 2 a) a(t 2 1)
SQ
a
a
t2
2
a
a
t2
2
2a
t
2
a
a
t2
a
1 t2
t2
20. If the mid-point of the line joining (3, 4) and (k, 7) is (x, y) and 2x+2y +1 =0. Find the value
of k.
(a) –1
(b) 2
(c) 1
(d) -15
Sol:(d)
x
3 k
,y
2
4 7
2
2x +2y +1 =0
2
3 k
2
2
11
1 0
2
Solve and find the value of k.
21. In figure, PT is a tangent and PAB is a secant. If PT = 6 cm, AB = 5 cm, find the length of PA.
(a) 5 cm
(b) 2 cm
(c) 4 cm
(d) 6 cm
Sol:(c)
Join OT, OA and OP. Draw OC AB.
Let radius of circle = r.
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OT PT
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[Radius from the point of contact tangent]
In right angled
OTP
OP2 = PT2 +OT2
OP2 = 62 +r2
OP2 –r2 = 36
OP2 –OA2 =36
………..(i)
Also in right angled OCA ,
OA2 = OC2+AC2
OP2 –(OC2 +AC2)=36
[from (i)]
PC2 –AC2=36
[From OP2 –OC2=PC2]
(PC –AC) (PC +AC) =36
AP(PC+BC) =36
AP(PB)=36
[ from centre of circle bisects the chord]
AP(AP+5) =36
(AP+9) (AP–4)=0
AP2 5AP–36=0
AP=4 or –9
AP = 4 cm
22. Two tangent PA and PB are drawn to the circle with centre O, such that APB 120o . If
OP = k (AP). Then find k.
(a) 1
(b) 2
(c) 3
(d) 4
Sol: (b)
A circle C(O, r) , PA and PB are tangents to the circle from P, outside the circle such that
APB 120o . OP is joined.
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To prove : OP = 2PA
Construction: Join OA and OB
Proof: Consider s PAO and PBO
PA =PB
[Tangent to circle, from a point outside it, are equal]
OP = OP
[common]
OAP
OAP
OBP 90o
OBP
OPA
OPB
[RHS]
1
1
APB
120o 60o
2
2
In right angles OAP,
AP
cos600
OP
1
2
OP 2AP
23. In the figure, AB is diameter of a circle with centre O and QC is a tangent to the circle at C.
If
CAB 30o , Then find the value of
CBA ?
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(a) 20o
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(b) 50o
(c) 60o
(d) 80o
Sol: (c)
In
AOC, OA = OC
CAO 300
ACO
s of equal sides are equal]
[Opp.
ACB 90o
Also
[Radius of the same circle]
[Also in semicircle]
OCB 90o 300 600
In
COB, OC = OB
[Radius of the same circle]
OBC 60o
OCB
[Opp.
s of equal sides]
CBA 60o
24. AB is diameter and AC is a chord of a circle such that BAC 30o . If then tangent at C
intersects AB produced in D, then BC = k (BD). Then k is
(a) 1
(b) 2
(c)
1
2
(d) 3
Sol: (a)
Join OC
Now
ACB 90o
ABC 60o
Also OB = OC
OB = OC
OBC
Now
OCB 60o
CBD 180o 60o 120o
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and
o
o
BCD 90
60
BDC 30o
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o
30
BC BD
25. In fig., a circle is inscribed in a quadrilateral ABCD in which Bo 90o . If AD = 23 cm, AB =
29 cm and DS = 5 cm, find the radius (r) of this circle.
(a) 11cm
(b) 2 cm
(c) 10 cm
(d) 8 cm
Sol:(a)
DS=RD=5cm(Length of the tangent drawn from an external point to a circle are equal)
AR=18cm
AQ=18cm
BQ=11cm
r=11cm
26. Find the radius of circle if an arc of angle 40o has length of 4 cm. Hence, find area of the
sector formulate by this arc.
(a) 25 cm2
(b) 36 cm2
(c) 38 cm2
(d) 30 cm2
Sol:(b)
Length of arc (l) = 4 cm
Angle ( ) = 40o
Let radius of the circle be r cm.
Length of the sector (l) =
or r
l 180o
4
180o
40o
r
180o
18cm
Radius of the circle = 18 cm.
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2
Area of the sector =
18 18 40o
360o
r
360o
2
Medical and Non - Medical Classes
o
(18) 40
360o
36 cm2
27. The minute hand of a clock is 10 cm long. Find the area swept by the minute hand between
9:00 a.m. and 9:35 a.m.
(a) 190.23 cm2 (b) 180.02 cm2
(c) 183.33 cm2
(d) 186.22 cm2
Sol:(c)
The minute hand describe a circle of radius 10 cm In an hour, i.e., 60 minutes it describes
angle of 360o.
So, in 35 minutes, the minute hand describes an angle of
360o 35
=210o
60
Thus, the minute hand describes a sector of angle 210o and radius 10 cm in 35 minutes.
Required area swept by the minute hand
=Area of the sector
r2
=
360o
=
22 (10)2 210
7 360o
22 10 10 210
183.33 cm2
o
7 360
28. A wire bent into the form of a square encloses an area of 121 sq. cm. If the wire is bent in
the form of a circle, find the area of circle. Use
(a) 154 cm2
(b) 150 cm2
22
7
(c) 155 cm2
(d) 140 cm2
Sol:(a)
Let each side of square be ‘a’ and ‘r’ be the radius of the circle
Area of the square = a2
But area of the square = 121 cm2
a2 = 121
a = 121 11cm.
Perimeter of the square = 4×side
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4×11 = 44 cm
But circumference of the circle = perimeter of the square.
2 r 44
r
2
22
r 44
7
44 7
7cm.
2 22
area of the circle = r2
=
22
(7)2
7
22
7 7 154cm2
7
29. ABCD is a square of side 4 cm. At each corner of the square a quarter circle of radius 1 cm
and the centre of square, a circle of radius 1 cm are drawn, as shown in fig., Find the area
of the shaded region. (Use
(a) 8.72 cm2
3.14 ).
(b) 6.55 cm2
(c) 9.72 cm2
(d) 10.52 cm2
Sol:(c)
Each side of the square = 4cm
Area of the square ABCD = (side)2
Radius of the inner circle = 1 cm
Area of inner circle = r2 3.14(1)2 3.14cm2
r2
Area of the segment APQ =
360o
3.14 (1)2 90o
=
360o
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= 3.14
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1 3.14 2
cm
4
4
Total area of 4 such segments
4
1
3.14cm2 3.14cm2
4
Total area of the circle and segments (unshaded portion)
=3.14+3.14=6.28 cm2
Area of the shaded region = Area of the square –Area of the unshaded portion
= 16 – 6.28=9.72 cm2
30. ABCDEF is any hexagon with different vertices A, B, C, D, E and F as the centres of circles
with same radius r are drawn. The area of the shaded portion is
(a) r2
(b) 2 r2
(c) 3 r2
(d) 4 r2
Sol:(b)
Sum of interior angle in polygon = (n –2) 180o
In hexagon;
= (6 – 2) 180o
= 4×180o
=720o
720
120o
one interior angle =
6
Area of shaded portion
120o
r2
=6
o
360
= 2 r2
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31. How many spherical lead shots each 4.21 cm in diameter can be obtained from a
rectangular solid lead with dimensions 66 cm, 42 cm, and 21cm.?
(a) 1600
(b) 1500
(c) 1520
(d) 1250
Sol:(b)
Dimensions of the rectangular solid are 66 cm, 42 cm, 21 cm.
……….(i)
Volume of the solid = 66 × 42 × 21 cm3
Diameter of a spherical lead shot = 4.2 cm
radius = 2.1 cm
Volume of a spherical lead shot
4 22
(2.1)3
3 7
………….(ii)
Number of lead shots
Volume of the rec tan gular solid
Volume of one spherical solid
=
66 42 21 21
88 (2.1)3
=
66 2 1000
1500
88
[From (i) and (ii)]
32. A solid cylinder has a total surface area 462 sq. cm. Its curved surface area is one-third of
the total surface area. Find the volume of the cylinder.
(a) 539 cm3
(b) 639 cm3
(c) 400 cm3
(d) 300 cm3
Sol:(a)
Let r be the radius of the base and h be the height of the cylinder
Total surface area = 462 cm2
………(i)
and curved surface area
1
462 154cm2
3
……….(ii)
2 rh 2 r2 462cm2
[From (i)]
154 + 2 r2 =462
[From (ii)]
2 r2 462 154 308
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r2
308 7
49
2 22
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r 7 cm
From (ii) we have, the curved surface area.
2 rh 2
22
7 h 154
7
h
7
cm
2
Volume of the cylinder
= r2h
22
7
7 7
539cm3
7
2
33. A building is in the form of a cylinder surmounted by a hemispherical vaulted dome which
contains 17.7 m3 of air and its internal diameter is equal to the height of the crown of the
vault above the floor. Find the height of the building. [Use=
(a) 3m
(b) 2m
22/7 ].
(c) 4m
(d) 5m
Sol:(a)
Let r be the radius of the base of the building. It is given that the height of the building is
equal to the diameter of the base. Therefore, the height of the building = 2r
Also height of the vaulted dome = r
Therefore, the height of the cylindrical portion of the building = 2r –r = r
Volume of the building = Volume of the cylinder + Volume of the hemisphere
=
.r2 .r
2 3
r
3
5 3
r
3
5 3
r 17.7
3
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17.7 3 7 17.7 21
5 22
110
r3
r
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17.7 21
110
1/3
Height of the building
17.7 21
110
= 2r 2
1/3
3m
34. In the given figure, a cone of radius 10 cm is divided into two parts by drawing a plane
through the mid-point of its axis, parallel to its base. Compare the volume of the two parts.
(a) 1:8
(b) 1:2
(c) 1:7
(d) 2:3
Sol:(c)
Let h be the height of the given cone. On dividing it into two parts, we get
(i) Frustum of the cone with R = 10 cm
and height =
h
cm
2
(ii) A similar cone with r = 5 cm and height =
h
cm
2
Volume of the smaller cone
Volume of the frustum cone
=
=
1 2 h
r
3
2
h
[R 2 r2 Rr]
2
1
3
(5 5)
25 1
2
[(10) (5) 10 5] 175 7
2
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35. Three cubes of metal whose edges are in the ratio 3 : 4 : 5 are ,melted down into a single
cube whose diagonal is 12 3 cm. Find the edges of the three cubes.
(a) 6 cm, 8 cm, 10 cm.
(b) 2 cm, 5cm, 20 cm.
(c) 8 cm, 3cm, 21 cm.
(d) 9 cm, 5cm, 20 cm.
Sol:(a)
3 a, a3 (3x)3 (4x)3 (5x)3 . calculate and proceed.
12 3
36. A die is thrown once. Find the probability of getting a prime number
(a)
1
3
(b)
1
2
(c)
3
2
(d)
3
4
Sol:(b)
Event E : getting a prime number (i.e., getting a natural number p>1 and having 1 and p as
its only divisors)
E = {2, 3, 5}
Number of favourable cases = n(E) =3
P(E) =
n(E) 3 1
n(S) 6 2
37. Three unbiased coins are tossed. What is the probability of getting at most two heads
(a) 7/8
(b) 1
(c)1/8
(d) 1/4
Sol:(a)
Let E be the event of getting at most two heads i.e., number of heads is
2, i.e.,
2 or 1, 0. By 0 heads, we mean all tails.
E = {HHT, HTH, THH, HTT, THT, TTH, TTT}
n(E) = 7
38. A bag contains 12 balls out of which x are white.
(i) If one ball is drawn at random, what is the probability that it will be a white ball?
(ii) If 6 more white balls are put in the bag, the probability of drawing a white ball will be
double than that in (i) .Find x.
(a) (i)
x
12
(ii) 3.
(b) (i)
x
(ii) 5.
12
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(c) (i)
x
10
(ii) 6.
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(d) (i)
x
8
(ii) 2
Sol:(a)
n(s) = 12
(i) A = ball down is white
n(A) = x
p(A) =
x
12
(ii) When 6 more white balls are put in the bag, total no. of balls in bag= 12 + 6 =18
No. of white balls = x +6
n(s) = 18,
B = ball drawn is white
n(B) = x +6
p(B) =
x 6
18
A.T.Q.,
p(B) = 2× p(A)
x 6
18
2 x
12
39. Solve for x:
x 3
1
1
1
1
(x 1)(x 2) (x 2)(x 3) (x 3)(x 4) 6
(a) x=2, x =–3 (b) x =7, x =–2
(c) x =10, x =–5
(d) x =6, x =–8
Ans.(b)
1
1
1
1
(x 1)(x 2) (x 2)(x 3) (x 3)(x 4) 6
1
1
1
1
(x 1)(x 2) (x 3)(x 2) (x 3)(x 4) 6
1
1
1
1
(x 1)(x 2) (x 3) x 2 x 4
1
6
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1
1
x 4 x 2
(x 1)(x 2) (x 3) (x 2)(x 4)
1
6
1
1
2x 6
(x 1)(x 2) (x 3) (x 2)(x 4)
1
6
1
2
1
(x 1)(x 2) (x 2)(x 4) 6
1
1
2
(x 2) x 1 x 4
1
6
1
x 4 2x 2
(x 2) (x 1)(x 4)
1
3x 6
x 2 (x 1)(x 4)
1
6
1
6
3
1
(x 1)(x 4) 6
18 = x2 – 5x + 4
(x–7) (x+ 2) = 0
x = 7, x = –2.
40. A man is starting on the desk of a ship which is 10 m above the water level. He observed
the angle of elevation of the top of a hill as 60o and the angle of depression of the base of
the hill as 30o. Find the distance of the hill from the ship.
(a) 10 3 m
(b) 5 3 m
(c) 8 3 m
(d) 8 10 m
Sol:(a)
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In AEC
CE
AE
tan60o
x
AE
x
AE
3
3
.....(i)
In AED
ED
AE
tan30o
1
3
10
AE
AE 10 3
.........(ii)
From (i) and (ii)
10 3
x
3
x = 10 3
3
x = 30 m.
Height of hill = 10+30 = 40 m.
Distance of the hill = 10 3 m .
41. A two digit number is such that product of its digits is 18.
When 63 is subtracted from the number, the digits interchanging their places. Find the
number.
(a) 56
(b) 98
(c) 92
(d) 50
Sol:(c)
Let the required number be 10x +y, where x is ten’s digit and y is one’s digit.
xy = 18
10x + y – 63=10y + x
9x – 9y = 63
x–y= 7
x=7+y
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y(7+y)= 18
y2 +7y – 18 =0
(y+9) (y – 2) =0
y = 2, –9.
y = 2 (as y > 0)
x=9
Number is 92.
42. What is the probability that date of birth of a person is in the month of January?
(a) 1/12
(b) 31/365
(c) 1/365
(d) 30/365
Sol:(b)
Total no. of days in Jan = 31
Total no. of days in a year = 365
Probability =
31
365
43. Determine the ratio in which the line 2x + y = 4 divides the line segment joining the points
(2, - 2) and (3, 7).
(a) 2:7
(b) 7:2
(c) 2:9
(d) 9:2
Ans (c)
P=
3k 2 7k 2
,
k 1 k 1
point P lies on 2x + y = 4
3k 2 7k 2
2
4
k 1
k 1
6k+4 + 7k-2=4k+4
9k +2 = 4
9k = 2
k=
2
9
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3 1
1
44. The 2nd, 31st and the last term of an AP are 7 , and 6 , respectively. Find the number
4 2
2
of terms.
(a) 55
(b) 59
(c) 60
(d) 44
Sol:(b)
Let a be the first term and d the common difference of the AP
Given, T2 7
3
4
a d
31
4
1
2
T31
and,
a + 30d =
.........(i)
1
2
…..(ii)
Subtracting (i) from(ii), we get
29d
1 31
2 4
29
4
d
1
4
Putting the value of d in (i), we get
Tn
13
2
i.e.,
a+(n–1)d =
8 (n 1)
8
n 1
4 4
1
4
13
2
13
2
13
2
32 – n+1 = –26
n = 59
Hence, first term = 8 and number of terms = 59.
45. A thief runs away from a police station with a uniform speed of 100m/minutes. After one
minute a policeman runs behind the thief to catch him. He goes at speed of 100m/minute
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in first minute and increases his 10m each succeeding minute. After how many minutes
the policeman will catch the thief?
(a) 4minutes
(b) 5 minutes
(c) 6 minutes
(d) 3
minutes
Sol:(b)
Let the policeman catches the thief in n minutes.
Since the thief ran one minute before the police,
therefore the time taken by thief before being caught = (n+1) minutes
Distance travelled by the thief in (n +1) minutes
= 100(n+1) metres
In first minute, speed of policeman
=100 m/minute.
In second minute, speed of policeman
=110 m/minute.
In third minute, speed of policeman
=120 m/minute and so on.
Speeds 100, 110, 120,……form an AP
Total distance travelled by the policeman in n
minute =
n
2 100 (n 1)10
2
On catching the thief by policeman, distance travelled by the thief = Distance travelled by
policeman.
100(n+1) =
n
[1×100+(n–1)10]
2
100n + 100 =100n +
n
(n –1)10
2
100 = n(n–1) 5
n2 – n–20 = 0
(n –5)(n+4)=0
n –5=0
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n=5
n+4=0
n = –4 is not possible
Time taken by the policeman to catch the thief = 5 minutes.
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Section-B {Physics}
46. A beam of monochromatic light is refracted from vacuum into a medium of refractive
index 1.5. The wavelength of refracted will be(a) dependent on intensity of reflected light
(b) same
(c) smaller
(d) larger
Ans.(c)
During reflection frequency remain same as it is a property of source, in denser medium
speed of light decreases hence wavelength decreases (v = f )
47. Focal length of a convex lens will be maximum for(a) blue light
(b) yellow light
(c) green light
(d) red light
Sol:(d)
f
1
as
1
∴
R
v
, fR
fv
Hence focal of lens is maximum for red colour and minimum for violet colour.
48. A ray of light is incident on the surface of separation of a medium with the velocity of light
at an angle 45o and is refracted in the medium at an angle 30o. What will be the velocity of
light in the medium(a) 1.96×108m/s
(b) 2.12×108m/s (c) 3.18×108m/s (d) 3.33×108m/s
Sol:(b)
sin i
sin r
2
1
o
c1
c2
3 108
c2
sin45
sin30o
c2
3 108
2
2.12 108 m / s
49. A ray of light strikes a transparent surface from air at an angle . If the angle between the
reflected and refracted ray is a right angle, the refractive index of the other surface is
given by(a)
1/ tan
(b)
tan2
(c)
sin
(d)
tan
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Sol:(d) i+r =90o
sini
sinr
;
sin
sin 90
tan
50. A concave mirror and a convex mirror are placed co-axially, their reflecting surfaces facing
each other. Their focal length are15 and 12 cm. respectively. An object placed between
them is 20 cm. from the concave mirror. The image formed by it is at the object itself.
Calculate the distance of the concave mirror from the object.
(a) 60 cm
(b) 36 cm
(c) 12 cm
(d) none of these
Sol: (d)
Because the distance between object and concave mirror is already given i.e. 20cm.
Note: If the distance between convex mirror and concave mirror were asked then:
“In a concave mirror for the image to be formed at the object itself, the latter must be at
the centre of curvature. But here, it is at a distance of 20 cm. from mirror whose distance
is not the radius of curvature. The radius of curvature is twice the focal length or 2× 15 or
30 cm. The fact of the matter is the reflected rays from the concave mirror are reflected
back along the same path forming the image at the object itself. This can happen only if the
reflected rays falling on the mirror are normal to the mirror.
O is object OP = 20 cm, f = 15 cm. (of concave mirror)
The reflected rays from concave mirror are reflected back by convex mirror.
C must be centre of curvature of the latter. So its radius curvature = twice its focal
length i.e. 12×2 or 24 cm.
Now, it there no convex mirror, the concave mirror would form an image of O at C. Then
we would have for concave mirror
u= +20cm. f= +15 cm v = PC =?
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Now,
1
v
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1 1
1
1
1
i.e.,
u f
v
20
15
1 1 1
1
or v PC 60
v 15 20 60
But we want OP’
Now PO’ = PC–P’C=60–2×12=36cm.
(P’C= radius of curvature of convex mirror i.e. twice its focal length)
Direction:
(A) Statement –I is true, Statement-II is true; Statement-II is a correct explanation for
statement-I.
(B) Statement-I is true, Statement-II is true; statement-II is not correct explanation for
statement-I.
(C) Statement-I is true, Statement-II is false.
(D) Statement-I is false, Statement-II is true.
51. Statement-I : When a concave mirror is held under water, its focal length will increase.
Statement- II : The focal length of a concave mirror is independent of the medium in
which it is placed.
Sol:(d)
52. Statement-I : Keeping a point object fixed, if a plane mirror is moved, the image will also
move.
Statement-II : In case of a plane mirror, distance of object and its image is equal from any
point on the mirror.
Sol:(d)
53. Light waves projected on oil surface show seven colours due to (a) Polarization
(b) Diffraction
(c) Reflection
(d) Interference
Sol:(d)
54. A fish looking up through the water sees the outside world contained in a circular horizon.
If the refractive index of water is 4/3 and fish is 12 cm. below the surface, the radius of
this circle in cm. is (a) 36 5
(b) 4 5
(c) 36 7
(d) 36 / 7
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Sol:(d)
1
sin ic
r
r2 h2
Using h = 12 cm ,
4/3 , We get
36
cm.
7
55. What will be the refractive index of glass for total internal reflection –
(a)
3 1
2
(b)
5 1
2
(c)
2 1
2
(d)
3
2
Sol:(d)
56. A light ray is incident perpendicularly to one face to a 90o prism and is totally internally
reflected at the glass-air interface. If the angle of reflection is 45o, we conclude that the
refractive index n–
(a) n
1
2
(b) n
2
(c) n
1
2
(d) n
2
Sol:(b)
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57. A person wishes to distinguish between two pillars located at a distance of 11 km. What
should be the minimum distance between the pillars?
(a) 3.2 m
(b) 2.3 m
(c) 2.5 m
(d) 2.6 m
Sol:(a)
As limit of resolution of eye is 1 minute, the pillars will be seen distinctly if
i.e.,
0o1'
d
as 1'
rad
D 180 60
180 60
i.e., d
11 103
m
180 60
i.e., d 3.2m
So the minimum distance between the pillars at which they are just resolved (i.e., distinct)
from 11 km is 3.2m.
58. A beam of light consisting of red, green and blue colours is incident on a right-angled
prism. The refractive indices of the material of the prism for the above red, green and blue
wavelengths are 1.39, 1.44 and 1.47 respectively. The prism will
(a) separate the red colour from the green and blue colours
(b) separate the blue colour from the red and green colours
(c) separate all the three colours from one another
(d) not separate even partially any colour from the other two colours.
Sol:(a) The colours for which i > θC, will get total internal reflection : i > θC or sin i > sin θC
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or sin 45° >
1
or
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1
2
1
or for which μ > 2 or μ > 1.414
Hence, the rays for which μ > 1.414 will get TIR.
For green and blue μ > 1.414, so they will suffer TIR on face AC. Only red comes out from this
face.
59. A light beam is travelling from Region I to Region IV (Refer Figure). The refractive
index in Regions I, II, III and IV are n0,
n0 n0
n
, and 0 and respectively. The angle of
2 6
8
incidence θ for which the beam just misses entering Region IV is
(a) sin–1
3
4
(b) sin–1
1
8
(c) sin–1
1
4
(d) sin–1
1
3
Sol:(b)
Critical angle from region III to region IV
n0 sin θC =
n0 / 8
n0 /6
3
4
Now applying Snell’s law in region I and region III
n0 sin θ =
or sin θ =
n0
sin θC
6
1
sin
6
θ = sin–1
c
1 3
6 4
1
8
1
8
60. A rectangular glass slab ABCD of refractive index n1 is immersed in water of refractive
index n2
(n1 > n2). A ray of light is incident at the surface AB of the slab as shown. The maximum
value of the angle of incidence αmax, such that the ray comes out only from the other
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surface CD, is given by
(a) sin–1
n1
n
cos sin 1 2
n2
n1
(b) sin–1 n1 cos sin
(c) sin–1
n1
n2
(d) sin–1
1
1
n2
n2
n1
Sol:(a)
Rays come out only from CD, means rays after refraction from AB get total internally
reflected at AD. From the figure
r1 + r2 = 90o
r1 = 90o – r2
(r1)max = 90o – (r2)min and (r2)min = θC
where sin θC =
(for total internal reflection at AD)
n2
n
or θ = sin–1 2
n1
n1
(r1)max = 90o – θC Now applying Snell’s law at face AB
sin max
sin max
sin max
n1
n2 sin r1 max sin 900 C
cos C
or sin
max
n1
cos
n2
C
max
sin
1
n1
cos
n2
C
sin
1
n1
cos sin
n2
1
n2
n1
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Section-C {Chemistry}
Direction:
(A) Statement –I is true, Statement-II is true; Statement-II is a correct explanation for
statement-I.
(B) Statement-I is true, Statement-II is true; statement-II is not correct explanation for
statement-I.
(C) Statement-I is true, Statement-II is false.
(D) Statement-I is false, Statement-II is true.
61. Statement-I : Soap are formed by saponification reaction.
Statement-II : In a saponification reaction, an organic acid combine with an alkali like NaOH
or KOH.
Sol:(c)
In a saponification reaction, an organic ester (triglyceride in nature) combine with NaOH or
KOH to form soap and glycerol.
62. Sodium hydrogen carbonate is used to distinguish :
(a) Ethanol and methanol
(b) Ethanol and ethene
(c) Ethene and ethyne
(d) Ethanol and ethanoic acid
Sol:(d)
Sodium hydrogen carbonate distinguishes ethanol from ethanoic acid. The acid gives brisk
effervescence of carbon dioxide gas with sodium hydrogen carbonate while ethyl alcohol
does not.
CH3COOH aq.
NaHCO3
CH3COONa(aq.) H2O (aq.) CO2(g)
C2H5OH(aq.) NaHCO3(aq.)
No action
63. Acetic acid was added to a solid X kept in a test tube. A colourless and odourless gas Y was
evolved. The gas was passed through lime water which turned milky. It was concluded that:
(a) solid X is sodium hydroxide and the Y is CO2
(b) solid is sodium carbonate and the gas Y is CO2
(c) solid X is sodium acetate and the gas Y is CO2
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(d) solid X is sodium hydrogen carbonate and the gas Y is SO2
Sol:(b)
Only CO2 gas turns lime water milky.
64. When sodium hydrogen carbonate powder is added to acetic acid, a gas evolves. Which one of
the following statements is not true for this gas? It
(a) turns lime water milky
(b) extinguishes a burning splinter
(c) dissolves in a solution of sodium hydroxide
(d) turns acidified potassium dichromate solution green.
Sol:(d)
It is wrong statement. Acetic acid cannot turn acidified potassium dichromate solution green
since no chemical reaction is possible.
65. Four students observed the colour and odour of acetic acid in its reaction with sodium
hydrogen carbonate. They tabulated their observations as below:
Student
Colour of acetic
acid
Odour of acetic
acid
1.
Blue
Fruity
Action with sodium
hydrogen
carbonate
Gas evolved bubbles
2.
Colourless
Smell of vinegar
Effervescence
3.
Light green
Odourless
Gas evolved without
bubbles
4.
Light brown
Rotten egg
Effervescence
The correct set of observations is that of student.
(a) 1
(b) 2
(c) 3
(d) 4
Sol: (b)
This student (2) has made the correct observation. Acetic acid is colourless with vinegar
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smell. It gives effervescence of carbon dioxide on reacting with solution hydrogen
carbonate.
66. According to IUPAC system, the correct name of the organic compound
CH3
Br
O
|
||
CH CH2 C OH
(a) 2-Bromobutanoic acid
(b) 2-Bromobutyric acid
(c) 3-Bromobutanoic acid
(d) 3-Bromo-2-hydroxybutan-2-one.
Sol:(c)
67. Alcohol can be produced by the hydration of :
(a) alkenes
(b) alkynes
(c) alkanes
(d) acids
Sol:(a)
H2C CH2 H2O
Ethene
CH3CH2OH
Ethanol or
Ethyl alcohol
Alkenes upon hydration in the presence of dilute acid form alcohols.
68. The heteroatoms present in
CH3 O CH2 CH2 (Br) are
(i) oxygen
(ii) carbon
(iii) hydrogen
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (iii) and (iv)
(iv) bromine
(d) (i) and (iv)
Sol:(d) Both oxygen (O) and bromine (Br) are heteroatoms. Please remember that apart from
C and H atoms, all other atoms present in an organic compound are hetero atoms.
69. CH3CH2OH
Alkaline KMnO4
Heat
CH3COOH
In the above reaction, alkaline KMnO4 acts as
(a) reducing agent
(b) oxidizing agent
(c) catalyst
(d) dehydrating agent
Sol:(b) Alkaline KMnO4 is also known as Baeyer’s reagent. It acts as an oxidizing agent.
Direction:
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(A) Statement –I is true, Statement-II is true; Statement-II is a correct explanation for
statement-I.
(B) Statement-I is true, Statement-II is true; statement-II is not correct explanation for
statement-I.
(C) Statement-I is true, Statement-II is false.
(D) Statement-I is false, Statement-II is true.
70. Statement-I : Chlorine is the most reactive member of the halogen family.
Statement-II: Size of chlorine is more than that of fluorine.
Sol:(d)
Fluorine is the most reactive element belonging to halogen family and not chlorine.
Fluorine is placed above chlorine in the periodic table.
71. Statement-I: In a triad, the three elements present have same gaps of atomic number.
Statement-II: Elements in a triad have similar properties.
Sol:(d)
In triad, the atomic mass of the middle element is the mean of the atomic masses of the
first and third element. And does not have any concern with their atomic numbers.
72. Which of the following sets does not belong to a group?
(a) Li, Na, K
(b) B, C, N
(c) B, Al, Ga
(d) O, S, Se.
Sol:(b)
The element B, C and N belong to second period. They do not belong to the same group.
73. Which of the following gives the correct increasing order of the atomic radii of B, O and F?
(a) F, O, B
(b) N, F, O
(c) O, F, B
(d) B, O, F
Sol:(a)
74. Which of the following elements will form a basic oxide?
(a) An element with atomic number 7
(b) An element with atomic number 17
(c) An element with atomic number 14
(d) An element with atomic number 11
Sol:(d)
The element is Na. It will form basic oxide (Na2O).
75. Which of the following elements would lose an electron most easily?
(a) Li
(b) Na
(c) K
(d) Rb
Sol:(d) The element (Rb) will lose electron most easily.
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Section-D {Biology}
76. A student soaked 5 raisins each of equal weight in three beakers, a, b and c, each
containing 100 ml of distilled water kept at room temperature. He removed raisins from
beaker a after 10 minutes, beaker b after 20 minutes and c after one hour. He calculated
the percentage of water absorption as Pa, Pb, Pc. What is true
(a) Pa = Pb = Pc
(b) Pa < Pb > Pc
(c) Pa > Pb > Pc
(d) Pa < Pb < Pc
Sol:(d)
77. Bishnoi community people lost their lives while protecting
(a) Khejri tree
(b) Sal tree
(c) Pinus tree
(d) Teak tree
(c) Karnataka
(d) Andhra Pradesh
Sol:(a)
78. Indra Gandhi Canal has brought greenery to
(a) Haryana
(b) Rajasthan
Sol:(b)
79. Animals get extinct mainly due to
(a) Predation
(b) Habitat destruction (c) Afforestation
(d) Pollution
Sol:(b)
80. Which of the following bacteria is found in Ganga water?
(a) Coliform bacteria
(b) Streptococcus bacteria
(c) Staphylococcus bacteria
(d) Diplococcus bacteria
Sol:(a)
81. Form the list given below, select the character which can be acquired but not inherited
(a) Colour of eye
(b) Colour of skin (c) Size of body
(d) Nature of hair
Sol:(c)
82. Select the statement that describe characteristics of genes
(i) Genes are specific sequence of bases in a DNA molecule
(ii) A gene does not code for proteins
(iii) In individuals of a given species, a specific gene is located on a particular chromosome
(iv) Each chromosome has only one gene.
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(a) (i) and (ii)
(b) (i) and (iii)
Medical and Non - Medical Classes
(c) (i) and (iv)
(d) (ii) and (iv)
Sol:(b)
83. Some dinosaurs had feathers although they could not fly but birds have feathers that help
them to fly. In the context of evolution this means that
(a) Reptiles have evolved from birds
(b) There is no evolutionary connection between reptiles and birds
(c) Feathers are homologous structures in both the organisms
(d) Birds have evolved from reptiles
Sol:(d)
84. A basket of vegetable contains carrot, potato, radish and tomato. Which of them represent
the correct homologous structures?
(a) Carrot and potato
(b) Carrot and tomato
(c) Radish and carrot
(d) Radish and potato
Sol:(c)
85. ‘Descent with modification ‘ is the central theme of
(a) Fossils
(b) Bones
(c) Birds
(d) Embryo
Sol:(c)
86. Name the parts labelled A in given figure
(a) Fallopian tube
(b) Ovary
(c) Uterus
(d) Cervix
Sol:(a)
87. The correct sequence of organs in the male reproductive system for transport of sperms is
(a) Testis
vas deferens
(c) Testis
urethra
urethra
ureter
(b) Testis
ureter
urethra
(d) Testis
vas deferens
ureter
Sol:(a)
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88. Which among the following disease is not sexually transmitted?
(a) Syphilis
(b) Hepatitis
(c) HIV-AIDS
(d) Gonorrhoea
Sol:(b)
89. The correct sequences of reproductive stages seen in flowering plants is
(a) Gametes, zygote, embryo, seedling
(b) Zygote, gametes, embryo, seedling
(c) Seedling, embryo, zygote, gametes
(d) Gametes, embryo, zygote, seedling
Sol:(a)
90. In human females, an event that reflects onset of reproductive phase is
(a) Growth of body
(b) Changes in hair pattern
(c) Change in voice
(d) Menstruation
Sol:(d)
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