Pioneer’s
Aspire Scholarship Exam
Solution
10th CBSE
Examination Centre: Pioneer Education, Sector – 40-D
General Instructions:
The question paper contains 90 multiple choice questions.
There are two sections in the question paper
Section: A– MATHEMATICS (1 to 45)
Section: B– SCIENCE (46 to 90)
Each right answer carries 4 marks and − 1 for every wrong answer.
The Maximum marks are 360 and Maximum Time 2.00 hrs.
Give your response in the OMR Sheet provided with the Question Paper.
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Section-A {Mathematics}
1.
If one of the roots of the quadratic equation x2 + mx + 24 = 0 is 1.5, then what is the value of m?
(a) –22.5
(b) 16
(c) –10.5
(d) –17.5
Ans. (d)
Solution.
We know that the product of roots of a quadratic equation ax2 + bx + = 0 is
In the given equation, x2 + mx + 24 = 0, the product of the roots =
c
.
a
24
24 .
1
The question states that one of the root quadratic equation = 1.5.
If x1 and x2 are the roots of the given quadratic equation and let x1 = 1.5.
Therefore, x2 =
24
16
1
In the given equation, m is the co-efficient of the x term.
We know that the sum of the roots of the quadratic equation ax2 + bc + c = 0 is
b
a
m
1
m
Sum of the roots = 16 + 1.5 = 17 = –17.5.
Therefore, the value of m = –17.5.
2.
A railway half ticket costs half the full fare and the reservation charge is the same on half ticket as on
full ticket. One reserved first class ticket from Chennai to Trivandrum costs Rs. 216 and one full and one
half reserved first class tickets cost Rs. 327. What is the basic first class full fare and what is the
reservation charge?
(a) Rs. 105 and Rs. 6
(b) Rs. 216 and Rs12
(c) Rs. 210 and Rs. 12
(d) Rs. 210 and Rs. 6
Ans. (d)
Solution.
Let half of the full basic fare be Rs. X.
Therefore, full basic fare be Rs. 2X.
Let the reservation charge be Rs. Y per ticket.
Now, one full reservation ticket would cost 2X ( basic fare) +Y(reservation charge)
2X + Y = 216
…..(1)
The total basic fare for one half and one full ticket = X + 2X = 3X and the total reservation charge is 2Y.
Hence, 3X + 2Y = 327.
……(2)
Solving (1) and (2) we get,
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X = 105 and Y = 6
Hence, the full basic fare is 2X = Rs. 210 and the reservation charge is Y = Rs. 6.
3.
If 4x +
6
8
=15 and 6x – =14 then the value of ‘p’ will be (given y = px – 2)
y
y
(a) 2
(b)
4
3
2
3
(d) 3
(c) 5
(d) 7
(c)
Ans. (b)
4.
If x2 – 3x + 1 =0, then the value of x2 +
(a) 3
1
is
x2
(b) 0
Ans. (d)
5.
A man wishes to find the height of a flag post which stands on a horizontal plane; at a point on this
plane he finds the angle of elevation of the top of the flag post to be 450. On walking 30 metres towards
the town he finds the corresponding angle of elevation to be 600. Find the height of the flag post?
(a) 69 m
(b) 70 m
(c) 71 m
(d) 74 m
Ans. (c)
6.
A man on the top of a town, standing on the sea shore finds that a boat coming towards him takes 10
minutes for the angle of depression to change from 300 to 600. Find the time taken by the boat to reach
the shore from the position.
(a) 5 minutes
(b) 10 minutes
(c) 8 minutes
(d) 15 minutes
Ans. (a)
7.
From a point in the cricket ground, the angle of elevation of a vertical tower is found to be
at a
distance of 200 m from the tower. On walking 125 m towards the tower the angle of elevation becomes
2 . Find the height of tower.
(a) 300 m
(b) 100 m
(c) 400 m
(d) 200 m
(c) –1
(d) 2
Ans. (b)
8.
The value of tan 10 tan 20 tan 30 …… tan 890 is
(a) 0
(b) 1
Ans. (b)
9.
A man on the top of a vertical observation tower observes a car moving at a uniform speed coming
directly towards it. If it takes 12 minutes for the angle of depression to change from 300 to 450, how
soon after this will the car reach the observation tower. Give your answer correct to nearest seconds.
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(a) 16 mins 24 secs
(b) 18 mins 20 secs
(c) 20 mins 44 secs
(d) 14 mins 38 secs
Ans. (a)
10. A cat saw a rat when it was exactly at the middle of a vertical tower. At that moment the angle of
elevation as observed by the cat was 30°. The cat ran a certain distance towards the tower to chose the
rat, which then an to the top of the tower. If the cat, which is now 20 m away from the foot of the tower,
finds the angle of elevation of the rat to be 60°, what distance cat run?
(b) 10 2
(a) 10
(c) 10 3
(d) None of these
Ans. (b)
Solution.
Let the angle made by the foot of the ladder on the ground be ‘ ’ degree
We know that, ladder = 2
i.e.,
(Using of the Wall)
AC = 2 AB
A
In ABC ,
AB
sin
AC
AB
sin
2AB
1
2
B
C
300
x
11. If x = 2, then the value of x +
(a) 1 or 2
x
x .... is
(b) 2
(c) 1 or 4
(d) 2
Ans. (c)
Let
x+
2
2
2+
y
2
x
x x ....
2 ......
y
y
y =y
2
y 2
2
On solving above equation, we get
x = 1 or 4
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12. The area of the triangle formed by the points (k–1, k + 1), (k + 3, k –3) and (k + 1, k + 2) is
(a) 2
(b) 3
(c) 3k
(d) none of these
Ans. (d)
Solution.
The points are (k–1, k + 1), (k + 3, k –3) and (k + 1, k + 2)
So, area of triangle
=
1
k 1
2
=
1
k 1
2
=
1
5k 5 k 3 4k 4
2
=
1
12 6sq.units
2
k 3
5
k 2
k 3 1
k 3
k 2
k 1
k 1
k 1
k 3
k 1 4
13. If the end points of the diameter of a circle are (4, 6) and (8, 4), the radius of the circle is
(b) 5 units
(a) 2 5 units
(c) 10 units
(d) 2 20 units
Ans. (b)
Solution.
Diameter of the circle = Distance between (4, 6) and (8, 4)
= 42 22
16 4 2 5 = Diameter
So, radius of the circle = 5 units
14. If the 3 consecutive vertices of a square are (–1, 4), (–3, 0) and (1, –2). Which of the following can be its
fourth vertex?
(a) (3, 2)
(b) (2, 3)
(c) (–3, 2)
(d) (1, 2)
Ans. (a)
Solution.
Let the points are A(–1, 4), B(–3, 0), C(1, –2) and D(x, y).
The mid-point of
AC =
1 1 4 2
,
2
2
Mid-point of BD =
A (–1, 4)
B (–3, 0)
D (x, y)
C (1, –2)
0,1
x 3 y 0
,
2
2
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x 3
0
2
x 3
and
y
1
2
y = 2.
Thus,
(x, y) = (3, 2)
15. The triangle formed by joining the vertices (3, 2) (5, –3) and (–5, 4) will be a
(a) acute triangle
(b) obtuse triangle
(c) right triangle
(d) isosceles triangle
Ans. (a)
Solution.
We have, (3, 2) (5, –3) and (–5, 4).
AB=
5 3
2
3 2
5 5
BC =
2
and CA =
5 3
= 64 4
70
2
4 3
2
22 52
2
25 49
29
74
2
4 2
BC2 < AB2 + CA2
So,
ABC is an acute triangle.
16. If O is the centroid of the
(a)
3 5
2
(c) 3 / 2
ABC, what is the length of median AD?
(b) 3 5
(d) none of these
Ans. (a)
Solution.
We know, AO = 12 22
5 2/3 AD
( Centroid divides medians in 2 : 1 ratio)
AD = 3 5 / 2
17. If the sum of 1st 9 terms of an AP is equal to the sum of 1st 13 terms, the sum of 1st 22 terms will be
respectively?
(a) 0
(b) 1
(c) –1
(d) 35
Ans. (a)
Solution.
S9 = S13
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i.e.,
9
2a
2
13
2a 13 1)d
2
9 1 d
18a + 72 d = 26a + 156d
i.e.,
8a = – 84d
i.e.,
2a = –21d
Now,
S22 = 1
22
2a
2
22 1 d
11 2a 11d
= 11(2a – 2a) = 0
Shortcut Method
If
Sp = Sq
Sp+q = S13
So, if
S9 = S13
S9+13 = S22 = 0
18. In a AP series of 100 terms a1 + a50 + a51 + a100 = 40, so what is the sum of all the terms of the series ?
(a) 10000
(b) 5000
(c) 1000
(d) none of these
Ans. (a)
Solution.
As AP has 100 terms, we know
a1 + a100 = a2 + a99 = a3 + a98 = … and so on
So,
a1 + a50 + a51 + a100 = 400
a1 + a100 =
Thus,
S100 =
1
(400) = 200
2
a1 a100
100 100 100 10000
2
19. The ratio of the sum of 1st ‘P’ terms of two AP series is (4P +16) : (7P –17), what is the ratio of their 6th
terms?
(a) 1
(b)
15
43
(c)
59
160
(d) none of these
Ans. (a)
Solution.
Sp
Sp'
4p 16
7p 17
p / 2 2a
p 1 d
p / 2 2a'
p 1 d'
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=
2a
p 1d
2a'
p 1 d'
Let
Given,
P = 11
2a 10d
2a' 10d'
a 5d
a' 5d'
4 11 16
7 11 17
60
1
60
a6
1
a6 '
20. If a7 = 13a13, where a7 and a13 are the 7th and the 13th terms respectively and the common difference is
6, then 29th term of the same AP is
(a) 80
(b) 93
(c) –93
(d) none of these
Ans. (b)
Solution.
a7 = 13a13
a + 6d = 13(a + 12d)
12d = –25d
= –25 6 = –150
a = –75
a29 = a + (28)d = –75 + 28 6
= –75 + 168 = 93
21. In the fig. shown, PT and PAB are the tangent and the secant drawn to a circle. If PT = 12 cm and PB = 8
cm, then AB is equal to
(a) 16 cm
(b) 10 cm
(c) 4 5 cm
(d) 18 cm
T
Ans. (b)
B
Solution.
A
P
OA = OD = Radii
So,
PT2 PA PB
(relation between PT tangent and secant PAB)
122 = PA 8
PA = 18
AB = PB – PA = 18 – 8 = 10 cm.
22. In the fig. shown, find the value of
(a) 1000
(b) 1200
(c) 800
(d) 600
DEC.
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Ans. (c)
Solution.
ACD = Exterior angle of
Also,
ACD =
Now, in
AOC = 400 + 200 = 600
ABE = 600
(angle in the same arc)
ABE,
BAE +
ABE +
AEB = 1800
400 + 600 +
Thus,
AEB =
AEB = 1800
DEC = 800 (vertically opposite angles)
23. PQRS is a rectangle with PQ = 16 cm and QR = 12 cm. If ABCD is a rhombus formed by joining the midpoints of rectangle PQRS, what is the radius of the circle inscribed
P
in rhombus ABCD?
A
Q
(a) 2.4 cm
D
(b) 4.8 cm
O
B
(c) 3.6 cm
(d) none of these
S
C
R
Ans. (b)
Solution.
OD = 8 cm and OA = 6 cm
AD = 10 cm
Let us draw OT
Area of
AOD
(using Pythagoras theorem)
AD.
1
1
AO OD =
AD OT
2
2
1
1
6 8=
2
2
10
OT
OT = 4.8 cm
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24. Three identical circles of radius r cm each are placed inside an equilateral triangle. Find the length of
the side of the equilateral triangle in terms of r.
(a) 2r
(b) 3r
(c) 2r ( 2 +1)
(d) 2r ( 3 +1)
Ans. (d)
Solution.
Radius of each circle = r
We want to know the length of AB i.e., side of equilateral triangle in terms of r.
In
AOD, we know OD
and
AD
DAO = 300
(radius on tangent)
(equilateral triangle)
AOD = 600
As we know, in triangle having angles 900, 600 and 300 have opposite sides 2a, a 3 and a, respectively.
OD = Radius = r
AD = r 3
Now,
in
(opposite to 600)
BO’E, we get
EB = r 3
Thus,
AB = AD + DE + EB
= r 3 +2r+ r 3 = 2r( 3 1 )
25. If TA is a tangent to the circle at B, what is the value of x + y + z ?
(a) 560
(b) 1120
(c) 280
(d) 650
Ans. (a)
Solution.
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x=
PQB = 280
Also,
OB TB
We get,
PBO = 900 – x = 900 – 280 = 620
We get,
OBP =
Now, in
OPB = 620
BPQ,
BPQ +
PBQ +
PQB = 1800
(620 + y) + (620 + z) + x = 1800
x + y + z = 1800–1240=560
26. A student cuts off two circles of radii r1 and r2 from a sheet of thick paper. He keeps these two in a
vertical position in contact with each other at a point P, Now, he keeps a rod of moderate thickness on
the top points A, B of the two circles. If r1
(a) 30°
r2, then for any values of r1 and r2 the angle APB is equal to
(b) 60°
(c) 90°
(d) 120°.
Ans. (c)
Solution:
AT = TB also TB = TP
TAP =
ATP =
TPA also
TPB
A
T
B
BTP = 900
TAP +
TBP +
TAP +
27. If a and
P
TPB = 90°
Similarly,
also,
TBP =
TPB= 90°
TPB +
APB = 1800
APB=90°
are roots of the equation x2+2x+ 1 = 0, then the equation whose roots are
1
1
(a) x2+ x + = 0
2
2
(b) x2+x +
1
=0
2
(c) x2+
1
x+ 1 = 0
2
1
and
1
is a
(d) x2 + 2x +1 = 0.
Ans.(d)
Solution:
2
1
1
28. If a = sin
(a)
3 2
2
1
2
2
1
4
, b = cos
1
4
1
equation is x2 2x 1 0
and c = –cosec
(b) 1
4
, then the value of a3 + b3 + c3 is
(c) 0
(d)
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Ans.(d)
Solution:
a sin
1
4
2
a3 b3 c3
1
Similarly b
1
1
2 2
2 2
2
and c
2 2
2
3 2
2
29. In the figure, O is the centre of circle;
BCO = m°,
BAC = n°. Then
(a) m + n = 90°
A
(b) m + n = 180°
n
O
(c) 2m + n = 180°
(d) m + 2n = 180°.
Ans. (a)
B
m
Q
Solution.
Since OB = OC
SCO =
OBC = m
SOC = 180° - 2m also 180° - 2m = 2n
2m + 2n=180°
m + n = 90°.
30. If 2x–1 + 2x+1 = 320, then x is equal to
(a) 6
(b) 8
(c) 5
(d) 7
(c) 2
(d) –2
Ans. (d)
Solution:
2x .2 1 2x .2 2x
1
2
2
Thus 2x–1 . 5 = 320
2x
1 4
2
2x .5
2
2x 1.5
x=7.
31. The solution of (25)x–2 = (125)2x –4 is
(a) 3/4
(b) 0
Ans. (c)
Solution:
(25)x–2 = (125)2x–4
(5)2x–4 = (5)6x–12
2x – 4 = 6x – 12 =
x= 2.
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32. In this figure the centre of the circle is O. AB
BC, ADOE is a straight line, AP
length twice the radius. Then:
AD , and AB has a
C
D
A
O
E
P
B
(a) AP2 = PB.AB
(b) AP . DO
PB . AD
(c) AB2
(d) AB. AD
OB . AO
AD . DE
Ans. (d)
Solution.
Since AB is tangent to the circle, we have AD/ AB AB/ AE, AB2
But AE
AD
AB2
AD AD
AD2
2r
AD
AB2 – AD. AB
Since AP
AD. AP2
AB ;
AD2
AB
AD. AE.
AD . AB.
AB AB.AD .
AB AB – AP
AB . PB.
33. Applied to a bill for Rs. 10,000 the difference between a discount of 40% and two successive discounts
of 36% and 4%, expressed in rupees, is:
(a) 0
(b) 144
(c) 256
(d) 400
Ans. (b)
Solution:
40% of Rs.10,000 is Rs. 4,000; 36% of Rs. 10,000 is Rs. 3,600; 4% of (Rs. 10,000 - Rs. 3,600) is Rs. 256.
Rs. 3,600 + Rs. 256 = Rs. 3,856; the difference is Rs. 4,000 - Rs. 3,856 = Rs. 144; or two successive
discounts of 36% and 4% are equivalent to one discount of 38.56%.
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34. Triangle PAB is formed by three tangents to circle O and
APB = 40°; then angle AOB equals:
R
B
P
(a) 45°
40o
O
S
A
(b) 50°
T
(c) 55°
(d) 70°
Ans. (d)
Solution:
P = 40°;
PAB +
TAS=180°–
TAS +
PBA = 180° – 40° = 140°.
PAB;
RBS= 180° –
PBA
RBS = 360° – 140° = 220°.
Since OA and OS bisect angles TAS and RBS, respectively
OAS+
1
OBS= (220°)= 1 1 0
2
AOB= 180° – 1 1 0 ° = 70°.
The number of degrees in
35. If x = 1 1
(a) x =1
1
AOB is independent of tangent ASB.
1 ......... , then :
(b) 0 < x <1
(c) 1 < x < 2
(d) x is infinite
Ans. (c)
Solution:
x=
x 1 , x2 = 1 + x, x2 – x – 1=0, and x 1.62
1 < x < 2.
36. The area of the largest triangle that can be inscribed in a semi-circle whose radius r is:
(a) r2
(b) r3
(c) 2r2
(d)2r3
Ans. (a)
Solution:
Of all triangles inscribable in a semi-circle, with the diameter as base, the one with the greatest area is
the one with the largest altitude (the radius); that is, the isosceles triangle. Thus the area is
1
. 2r . r = r2
2
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37. John ordered 4 pairs of black socks and some additional pairs of blue socks. The price of the black socks
per pair was twice that of the blue. When the order was filled, it was found that the number of pairs of
the two colours had been interchanged. This increased the bill by 50% .The ratio of the number of pairs
of black socks to the number of pairs of blue socks in the original order was:
(a) 4 : 1
(b) 2 : 1
(c) 1 : 4
(d) 1 : 2
Ans. (c)
Solution:
Let x be the number of pairs of blue socks, and let y be the price of a pair of blue socks. Then 1y is the
price of a pair of black socks and
x × 2y + 4 × y =
x= 16;
3
(4 –2y + x–y). Divide by y and solve for x.
2
4 : x = 4 : 16 = 1 : 4.
38. If the radius of a circle is increased 100%, the area is increased:
(a) 100%
(b) 200%
(c) 300%
(d) 400%
Ans. (c)
Solution:
Let r be the original radius; then 2r = new radius, r2 =original area, 4 r2 = new area.
Percent increase in area=
4 r2
r2
r2
×100=300.
39. A point is selected at random inside an equilateral triangle. From this point perpendiculars are dropped
to each side. The sum of these perpendiculars is:
(a) least when the point is the centre of gravity of the triangle
(b) greater than the altitude of the triangle
(c) equal to the altitude of the triangle
(d) one-half the sum of the sides of the triangle
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Ans. (c)
Solution.
Let P be an arbitrary point in the equilateral triangle ABC with sides of length s, and denote the
perpendicular segments by pa, pb, pc. Then Area ABC = Area APB + Area BPC + Area CPA =
=
1
1
(spa + spb + spc )= s(pa + pb + pc)
2
2
A
Pa P
Pb
Pc
B
Also, Area ABC =
s
C
1
s h, where h is the length of the altitude of ABC. Therefore, h = pa + pb + pc and this
2
sum does not depend on the location of P.
40. Tom, Dick and Harry started out on a 100 km/hr journey. Tom and Harry went by automobile at the
rate of 25 km/hr, while Dick walked at the rate of 5 km/hr. After a certain distance, Harry got off and
walked on at 5 km/hr, while Tom went back for Dick and got him to the destination at the same time
that Harry arrived. The number of hours required for the trip was:
(a) 5
(b) 6
(c) 7
(d) 8
Ans. (d)
Solution:
Let t1, t2, t3 be the number of hours, respectively, that the car travels forward, back to pick up Dick, then
forward to the destination. Then we may write
25 . t1 – 25 . t2 + 25 . t3 = 100
for car
5 . t1 + 5 . t2 + 25 . t3 = 100
for Dick
25 . t1 + 5 . t2 + 5 . t3 = 100
for Harry.
This system of simultaneous equations is equivalent to the system
t1 – t2 + t4 = 4
t1 + t2 + 5t3 = 20
5t1 + t2 + t3 = 20
whose solution is t1 = 3, t2 = 2, t3 = 3.
Hence t1 + t2 + t3 = 8 = total number of hours.
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41. Two pipes can fill up a tank in 30 minutes and 40 minutes respectively and another pipe can empty the
full tank in 20 minutes. If all the pipes are opened when the tank is half full, then the time required to fill
the tank will be
(a) 1hr.
(b) 2 hrs.
(c) 3 hrs.
(d) 2 hrs. 30 mins.
Ans. (a)
Solution:
Part emptied by pipe C in 1 min =
1
20
1
1
1
1
30 40 20 120
In 1min Part of tank fill =
120 min required to fill the tank. Since the tank is half filled, then the time required = 1 hour.
42. Which one of the following is correct?
(a) ( 3 7
7
3)( 3 7
7
3 2)
24 ( 3 7
7
3 4)( 3 7
7
3 6)
(b) ( 3 7
7
3)( 3 7
7
3 2)
24 ( 3 7
7
3 4)( 3 7
7
3 6)
(c)
(3 7
7
3)( 3 7
7
3 2)
24 ( 3 7
7
3 4)( 3 7
7
3 6)
(d)
(3 7
7
3)( 3 7
7
3 2)
24 ( 3 7
7
3 4)( 3 7
7
3 6)
Ans.(a)
Solution:
3
=
=
7
3
7
7
3
3 4
7
7
3
7
2
3
3
7
6
3
7
7
3
4
6
3
7
7
3
2
7
3 6
3
7
7
3
24
24
43. If a clock is started at noon, then at 16 minutes past 5 p.m. the hour hand turns through
(a) 155°
(b) 158°
(c) 160°
(d) 163°.
Ans. (b)
Solution:
Started from noon to 5.00 pm, hour hand travels
150° = (30 × 5)
=
30
×16 = 8°
60
in 60 min. it travels = 30°
Total
in 1 min, it travels =
30o
in 16 min, it travels
60o
s = 158° 60
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44. The hypotenuse of an isosceles right-angled triangle is q. If we describe equilateral triangles (outwards)
on all its 3 sides, then the total area of the re-entrant hexagon thus obtained is
(a)
3 +2)
q2(
(b)
q2 2 3 1
4
(b)
q2 4 3 1
(d)
4
q2 5 3 2
4
Ans. (b)
Solution:
A
AB2 + BC2 = q2
2
AB2
=
q2
AB3
q2
=
2
AB
2
B
Total area
3
=
4
=
q
2
3 2
q
4
q
q
2
3
4
3 2
q
4
q
2
2
q2
4
3 2
q
4
C
1 q q
. .
2 2 2
q2
2 3 1
4
45. An insect climbs up a 10 metre high pole. It climbs up 3 metres in the first minute but slips down 2
metres in the second minute, again climbs up 3 metres in the third minute and slips down 2 metres in
the fourth minute and so on. It reaches the top of the pole in
(a) 15 minutes
(b) 17 minutes
(c) 19 minutes
(d) 20 minutes.
Ans. (c)
Solution:
In 2 min insect climb = 1 m
in 18th min he had covered = 9 m .
in 19th min. he will be on the top.
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Section-B {Science}
46. Focal length of a convex lens will be maximum for –
(a) blue light
(b) yellow light
(c) green light
(d) red light
Ans. (d)
47. If a convex lens of focal length 80 cm and a concave lens of focal length 50 cm are combined together,
what will be their resulting power –
(a) +6.5 D
(b) –6.55 D
(c) +7.5 D
(d) –0.75 D
Ans. (d)
48. When a mirror is rotated through an angle θ , the refracted ray from it turns through an angle of –
(a) θ
(b) θ /2
(c) 2 θ
(d) 0
Ans. (c)
49. A ray of light strikes a transparent surface from air at angle θ . If the angle between the reflected and
refracted ray is a right angle, the refractive index of the other surface is given by –
(a) μ 1/ tanθ
(b) μ tan2 θ
(c) μ sin θ
(d) μ tan θ
Ans. (d)
50. A point source of light B, placed at a distance L in front of the centre of a mirror of width d, hangs
vertically on a wall. A man walks in front of the mirror along a line parallel to the mirror at a distance
2L from it as shown. The greatest distance over which he can see the image of the light source in the
mirror is –
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(a) d/2
(b) d
(c) 2d
(d) 3d
Ans. (d)
51. A ray of light passes through four transparent media with refractive indices μ1 , μ 2 , μ3 , and μ 4 as
shown in the figure. The surfaces of all media are parallel. If the emergent ray CD is parallel to the
incident ray AB, we must have:
(a) μ1 μ 2
(b) μ 2 μ3
(c) μ3 μ 4
(d) μ 4 μ1
Ans. (d)
52. An observer can see through a pin-hole the top end of a thin rod of height h, placed as shown in the
figure. The beaker height is 3h and its radius h. When the beaker is filled with a liquid up to a height
2h, he can see the lower end of the rod. Then the refractive index of the liquid is –
(A) 5/2
(b) 5 / 2
(c) 3 / 2
(d) 3/2
Ans. (b)
53. A ray of light is incident at the glass-water interface at angle i, it emerges finally parallel to the surface
of water, then the value of μ g would be :
(a) (4/3) sin i
(b) 1/sin i
(c) 4/3
(d) i
Ans. (b)
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54. A ray of light is incident on the surface of separation of a medium with the velocity of light at an angle
450 and is refracted in the medium at an angle 300. What will be the velocity of light in the medium –
(a) 1.96
108 m/s
(b) 2.12
108 m/s
(c) 3.18
108 m/s
(d) 3.33
108 m/s
Ans. (b)
55. 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 the fish is 12 cm. Below the surface, the radius of this circle in
cm is –
(a) 36 5
(c) 36 7
(b) 4 5
(d) 36 / 7
Ans. (d)
56. What will be refractive index of glass for total internal reflection –
(a)
3 1
2
(b)
5 1
2
(c)
2 1
2
(d)
3
2
Ans. (d)
57. A light ray is incident perpendicularly to one face to a 900 prism and is totally internally reflected at
the glass-air interface. If the angle of reflection is 450, we conclude that the refractive index n –
(a) n
1
2
(b) n
2
(c) n
1
2
(d) n
2
Ans. (b)
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58. Which of the prism is used to see infrared spectrum of light –
(a) Rock salt
(b) Nicol
(c) Flint
(d) Crown
Ans. (a)
59. Myopia is due to –
(a) elongation of eye ball
(b) irregular change in focal length
(c) shortening of eye ball.
(D) older age
Ans. (a)
60. Rainbow is formed due to a combination of –
(a) Refraction and absorption
(b) Dispersion and focusing
(c) Refraction and scattering
(d) Dispersion and total internal reflection
Ans. (d)
61. A person using a lens as a simple microscope sees an –
(a) inverted virtual image
(b) inverted real magnified image
(c) upright virtual image
(d) upright real magnified image
Ans. (c)
62. Blue colour of sky is due to phenomenon of –
(a) Reflection
(b) Refraction
(c) Scattering
(d) Dispersion
Ans. (c)
63. An endoscope is employed by a physician to view the internal parts of a body organ.
It is based on the principle of –
(a) refraction
(b) reflection
(c) total internal reflection
(d) dispersion
Ans. (c)
64. The asexual process replaced by the sexual method is known as –
(a) Semigany
(b) Amphimixis
(c) Apospory
(d) Apomixis
Ans. (d)
65. Gemmule formation in sponges is helpful in –
(a) Parthenogenesis
(b) Sexual reproduction
(c) Only dissemination
(d) Asexual reproduction
Ans. (d)
66. Ovulation in mammals is caused by –
(a) FSH and TSH
(b) FSH and LH
(c) FSH and LTH
(d) LTH and LH
Ans. (b)
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67. If both ovaries are removed from a rat, then which hormone is decreased in blood –
(a) Oxytocin
(b) Estrogen
(c) Prolactin
(d) Gonadotrophic
Ans. (b)
68. Which part of the ovary in mammals acts as an endocrine gland after ovulation –
(a) Vitelline membrane
(b) Graffian follicles
(c) Stroma
(d) Germinal epithelium
Ans. (b)
69. Tunica albuginea is the covering around
(a) ovary
(b) testis
(c) kidney
(d) heart
(c) golgi bodies
(d) ribosomes
(c) dog and goat
(d) goat and fish
Ans. (b)
70. Acrosome is made up of –
(a) mitochondria
(b) centrioles
Ans. (c)
71. Example of External fertilization is (a) fish and frog
(b) frog and monkey
Ans. (a)
72. Development of embryo in human body takes place in –
(a) ovary
(b) oviduct
(c) uterus
(d) ostium
Ans. (c)
73. Which of the following statements is not true for ethane –
(a) It can be chlorinated with chlorine
(b) It can be catalytically hydrogenated
(c) When oxidised produces CO2 and H2O (d) It is a homologous of isobutane
Ans. (b)
74. Which gives CH4 when treated with water –
(a) Silicon carbide
(b) Calcium carbide
(c) Aluminium carbide
(d) Iron carbide
(b) CO + N2
(c) CO + H2
(d) CO + N2 + H2
Ans. (c)
75. Water gas is –
(a) CO + CO2
Ans. (c)
76. Which one gives only one monosubstitution product on chlorination –
(a) n-pentane
(b) Neopentane
(c) Isopentane
(d) n-butane
Ans. (b)
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77. Main constituent of marsh gas is –
(a) C2H2
(b) CH4
(c) H2S
(d) CO
Ans. (b)
78. Ethyl bromide gives ethylene when reacted with –
(a) Ethyl alcohol
(b) Dilute H2SO4
(c) Aqueous KOH
(d) Alcoholic KOH
Ans. (d)
79. Which is the most reactive hydrocarbon in the following –
(a) Ethane
(b) Ethyne
(c) Ethene
(d) Methane
Ans. (b)
80. Which of the following gases is used for welding –
(a) Methane
(b) Ethane
(c) Acetylene
(d) Ethene
Ans. (c)
81. The IUPAC name of the compound having the formula (CH3)3CCH=CH2 is –
(a) 3, 3, 3-trimethyl-1-propane
(b) 1, 1, 1-trimethyl-1-1-butene
(c) 3,3-dimethyl-1-butene
(d) 1,1-dimethyl-1,3-butene
Ans. (c)
82. On moving from left to right across a period in the table the metallic character –
(a) Increases
(b) Decreases
(c) Remains constant
(d) first increases and then decreases
Ans. (b)
83. The size of the following species increases in the order –
(a) Mg2+ < Na+ < F– < Al
(b) F– < Al < Na+ > Mg2+
(c) Al < Mg < F– < Na+
(d) Na+ < Al < F– < Mg2+
Ans. (a)
84. The set representing the correct order of first ionisation potential is –
(a) K > Na > Li
(b) Be > Mg > Ca
(c) B > C > N
(d) Ge > Si > C
(c) ZnO
(d) SiO2
Ans. (b)
85. Which one of the following oxides is neutral –
(a) CO
(b) SnO2
Ans. (a)
86. Arrange S, O and Se in ascending order of electron affinity –
(a) Se < S < O
(b) Se < O < S
(c) S < O < Se
(d) S < Se < O
Ans. (a)
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87. One important merit of modern periodic table is –
(a) it explains why element in the same group have the same chemical properties
(b) Hydrogen has been placed accurately
(c) Isobars have not been placed separately
(d) It is based on classifying elements according to their atomic masses
Ans. (a)
88. Elements belonging to the same group have similar properties because –
(a) they have similar electronic configuration of the outermost shell
(b) their atomic numbers go on increasing as we move down the group
(c) all of them are metallic elements.
(d) None of the above
89. The elements with atomic numbers 3, 11, 19, 37 and 55 belong to –
(a) alkali metals
(b) alkaline earth metals
(c) halogens
(d) noble gases
Ans. (a)
90. The elements belonging to IA to VII A and 0 groups are known as –
(a) alkali metals
(b) representative elements
(c) transition elements
(d) inner-transition elements
Ans. (b)
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