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Writing makes an exact man
P-SAT 2014 - 15
(PIONEER’S SCHOLARSHIP/ADMISSION TEST)
{8TH CBSE}
General Instructions:The question paper consist of TWO sections (A), (B).
Section A contains 45 objective questions for maths.
Section B contains 45 objective questions for General science.
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:_____________________________________
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Section – A {Mathematics}
1.
If p : q = 1
1
1
2
and q : r = 3 : 4 , express p : q : r in simplest terms.
:3
2
2
3
(a) 27 :63:98
(b) 63:27:98
(c) 98:63:27
(d) None of these
Sol: (a)
p:q =1
1
1 3 7
:3 = : =3:7
2
2 2 2
q:r=3:4
2
14
= 3:
= 9 : 14
3
3
L.C.M. of two values of q i.e. 7 and 9 is 63
[To find p : q : r. make q same in both cases]
Thus p : q = 3 : 7 = 27 : 63
q : r = 9 : 14 = 63 : 98
p : q : r = 27 : 63 : 98, which is in simplest form.
Also
2.
p : q : r = 27 : 98
r : p = 98 : 27, which in also in simplest form.
If 3A = 5B = 6C, find A : B : C
(a) 5:6:10
(b) 10:6:5
(c) 6:10:5
(d) None of these
Sol: (b)
A=
x
x
x
,B= ,C=
3
5
6
A:B:C=
=
x x x 1 1 1
: : = : :
3 5 6 3 5 6
[L.C.M of 3, 5, 6 = 30]
1
1
1
30: 30: :30
3
5
6
= 10 : 6 : 5
3.
Divide Rs 645 into three parts such that the first part is
2
of the second part and the ratio between
5
second and third parts is 4 : 3.
(a) Rs 120, Rs 300, Rs 200
(b) Rs 120, Rs 300 , Rs 225
(c) Rs 220, Rs 300, Rs 225
(d) None of these
Sol: (b)
Since the ratio between second and third parts is 4 : 3, let second and third parts be Rs. 4x and Rs. 3x
respectively
The n first part =
1
8x
of Rs 4x = Rs.
2
5
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8
x 4x 3x 645
5
According to given,
8x 20x 15x
645
5
x
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43
x 645
5
645 5
75
43
First part = Rs.
8
75
5
Rs 120,
Second part = Rs (4 × 75) = Rs 300,
and third part = Rs (3 × 75) = Rs 225
4.
Find the fourth proportional to
(a) 1.8
2 3
, ,2
3 5
(b) 2.2
(c) 1.4
(d) None of these
Sol: (a)
Let the fourth proportional be x, then
2 3
, ,2 x are in proportion
3 5
Using cross product rule, we get
2
3
x
2
3
5
5.
3
3
2
5
2
x
9
1.8
5
Find the third proportional to 2.4 kg, 9.6 kg.
(a) 40 Kg
(b) 35.4Kg
(c) 38.4Kg
(d) None of these
Sol: (c)
Let the third proportional to 2.4 kg, 9.6 kg be x kg
Then 2.4 kg, 9.6 kg and x kg are in continued proportion
2.4
9.6
9.6
x
24
96
9.6
x
1
4
9.6
x
x 9.6 38.4
Hence the third proportional is 38.4 kg.
6.
1
Find the mean proportional between 8 and 27
3
(a) 15
(b) 20
(c) 25
(d) 35
1
Sol: (a) Mean proportional between 8 and 27.
3
=
25
27
3
25 9
225 15
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7.
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A bag contains Rs 59.60 in the form of 50 paise, 25 paise and 20 paise coins in the ratio 3 : 7 : 5. Find the
number of each type of coins.
(a) 42 ,98,70
(b) 50,60,70
(c) 80,20,30
(d) None of these
Sol: (a)
Since 50 paise, 25 paise and 20 paise coins are in the ratio 3 : 7 : 5, let the number of these coins be 3x,
7x, and 5x respectively.
The value of 50 paise coins = Rs
50
3x
100
Rs
3x
,
2
The value of 25 paise coins = Rs
25
7x
100
Rs
7x
, and
4
The value of 20paise coins = Rs
20
5x
100
Rs x
According to given
3x
2
6x 7x 4x
4
x
7x
4
x 59.50
119
4
17
119
x
4
2
119 4
14
2 17
The number of 50 paise coins = 3 × 14 = 42,
The number of 25 paise coins = 7 × 14 = 98 and
The number of 20 paise coins = 5 × 14 = 70
8.
Add the following: 3a – 2b + 5c + 7, 5b – 3c – 4 and – 2a + 3b – 11c
(a) a + 2b +3c +3
(b) a + 4b – 11c + 8
(c) a + 6b -9c +3
(d) None of these
(c) x2yz
(d) None of these
Sol: (c)
Column method
3a – 2b 5c 7
5b – 3c – 4
2a 3b 11c
a 6b 9c 3,
9.
which is the required sum.
Subtract : –3x2yz from –2x2yz
(a) -5 x2yz
(b) -x2yz
Sol: (c)
–2x2yz – (–3x2yz) = –2x2yz + 3x2yz =(–2 + 3) x2yz =x2yz
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10. Find the product of : 7ab,
(a) 4abc
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4 2 2
5 3
a b c and
bc
5
7
(b) 4
(c) 20a2b3c4
(d) None of these
Sol: (d)
4 2 2
abc
5
7ab
5 3
bc
7
7
4
5
5
7
ab a 2b2c bc3
4a3b4c 4
2 3 5
5 4
11. Divide : 24x y z by 6xy z
(a) –4
(b) –4x2z
(c) –4xyz
(d) None of these
Sol: (d)
– 24x2y 3z5 6xy 5z4
24x2y 3z5
= 6xy 5z4
12. Simplify the following expressions: a2
(a) a/b
x 2y 3z5
xy 5z4
24
6
4
x2
1
y
z5
5 3
4
4
xz
y2
a a2 a
b a
(b) b/a
(c) a2
(d) None of these
(c) 20cm
(d) None of these
Sol: (a)
a3
b a
a a2 a a2 ba a a2 a
a2
a2
a
ba
a
13.
1
f
1
u
a
a a
b
a
b
1
. Find u when f = 6 cm and v = 10 cm.
v
(a) 10 cm
(b) 15cm
Sol: (b)
To make u as the subject of the formula, we have
1
f
1
u
1
u
1
v
v f
fv
1
u
1 1
f v
u
fv
, u is the subject of theformula.
v f
Substituting f = 6 cm and v = 10 cm, we get
u
6 10
60
cm =
cm 15cm.
10 6
4
14. Use the laws of exponents to simplify the following : 23
(a) 260
(b)230
4 5
(c) 220
(d) None of these
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Sol: (a)
4 5
23
5
23 4
5
212
212 5 260
2
15. Use the laws of exponents to simplify the following :
3
(a) 11/4
(B) 17/4
0
2
2
3
(c) 13/4
(d) None of these
(c) 1/256
(d) None of these
Sol: (c)
0
2
3
2
2
3
1
1
2
3
2
24
16. Simplify the following :
(a) 1/64
1
22
32
1
2
1
9 13
1
4
4 4
9
1
25
26
(b) 1/32
Sol: (d)
2
24
25
28 25
26
26
2
8
5
6
2
7
0
2
5
3
17. Simplify and write in the exponential form: 2
(a) 23× 52
(b) 24 × 53
2
3
11
2
2
5
2
8
2
5
0
(c) 2×5
(d) None of these
(c) x
(d) None of these
Sol: (a)
3
2
2
3
–11
2
–5
2
2
2
5
8
0
= 8 × 9 + 112 + 2–5–(–8) –1 = 72 + 121 + 23 –1
= 72 + 121 + 8 – 1 = 200
= 2 × 2 × 2 × 5 × 5 = 23 × 52
xm
18. Simplify :
n
xn
l
xl
xm xn xl
(a) 0
m
2
(b) 1
Sol: (b)
xm
x
n
xn
m
n
x
l
zl
x
l
m
2
xm
x
n n l l m
m n l
2
x2m
x
2n 2l
2m n l
= x2m
2n 2l 2m 2n 2l
x0 1
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xm
xn
19. Simplify :
m n
n l
xn
xl
.
(a) 0
xl
xm
.
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l m
(b) x
(c) 1
(d) None of these
Sol: (c)
m n
xm
xn
2
= xm
n2
xn
. m
x
2
.x n
l2
n l
2
m2
. xl
l m
xl
. m
x
20. Find n so that 211
xm
2
xm
25
(a) 5
n
m n
n2 n2 l2 l2 m2
2 3 22n
. xn
l
n l
. xl
m
l m
x0 1
1
(b) 10
(c) 20
(d) None of these
Sol: (a)
Given 211 25 2 3 22n
211
5
23
26 22n
1
2n 1
4
6 2n 4
6 4 2n
2n 10
| a n = am
n =m
n 5
Hence, n =5.
21. Determine 8x
x
If 9x
(a) 1
2
240 9x .
(b) 2
(c) 0
(d) None of these
Sol: (b)
Given 9x + 2 = 240+ 9x
9x × 92 – 9x = 240
80×9x = 240
(32)x = 3
x=
9x (81 –1) = 240
9x=
240
3
80
32x = 31
2x = 1
1
2
(8x)x
1
8
2
1
2
41/2
22
1/2
2
2
1
2
21 2.
22. Using special expansions, find the value of : (1003)2
(a) 100600
(b) 1006009
(c) 100006
(d) None of these
Sol: (b)
(1003)2 = (1000 + 3)2
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=
(1000)2
+
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2×1000×3+(3)2
=1000000 + 6000+9 = 1006009
1
3 , evaluate: x2
x
23. If x
(a) 11
1
x2
(b) 13
(c) 9
(d) 15
(c) 29
(d) 30
Sol: (a)
Given
1
x
3
x
x
2
1
2 x
x
x2 2
x2
1
x2
2
1
x
x
x2
32
1
x
2
9
9
9 2 11
24. If a + b = 7 and ab = 10, find the value of a2 + b2.
(a) 39
(b) 49
Sol: (c)
We know that (a +b)2 = a2 + b2 + 2ab
72 a2 b2 2 10
49 20 a2 b2
a2 b2 29
25. If 3p – 4q = 5 and pq = 3, find the value of 27p3 – 64q3.
(a) 680
(b) 665
(c) 775
(d) 780
(c) (x–5)(3x–4)
(d) (x–2) (3x–4)
Sol: (b)
Given 3p – 4q = 5
(3p –4q)3 = 53
(3p)3 (4q)3 3 3p 4q (3p 4q) 125
27p3 64q3 36pq(3p 4q) 125
27p3 64q3 36 3 5 125
27p3 64q3 125 540 665
26. Factorise the following trinomial:3x2 – 10x + 8
(a) (x–2) (x–6)
(b) (x–4) (3x–4)
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Sol: (d)
We want to find two integers whose sum is –10 and product is 3× 8 i.e. 24 By trial, we see that
(–6) + (–4) = –10 and (–6) (–4) = 24
3x2 – 10x + 8 = 3x2 – 6x – 4x + 8
= 3x( x – 2) –4(x –2)
= (x – 2)(3x –4)
27. Reduce the following algebraic fraction to lowest terms:
(a) 1/x
(b) 1/x2
3x
6x 9x2
(c) 1/(2-3x)
(d) 1/(2 +3x)
Sol: (c)
3x
1
3x(2 3x) (2 3x)
28. Find the area of the given figure ABCD. A(4,1) B(-2,1) C(-3,-2) D(3,-2)
(a) 18
(b) 20
(c) 30
(d) 10
Sol: (a)
Points A, B, C, D are marked in the adjoining figure. It is easy to see that it is parallelogram.
Area = base × height
= 6 units × 3units == 18 square units.
29. If the base of a right-angled triangle is 6 units and the hypotenuse is 10 units, find its area.
(a) 30 sq. units
(b) 20 sq. units
(c) 24 sq. units
(d) 48 sq. units
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Sol: (c)
Let ABC be a right – angled triangle at i.e.
From
B 90o and its base BC = 6 units and Hypotenuse AC = 10 units
ABC, by Pythagoras theorem, we get
AB2 = 102 –62 = 100 –36 = 64
64 units = 8 units
AB =
Area of
=
ABC, =
1
base height
2
1
6 8 sq. units = 24 sq. units
2
30. Calculate the area of a triangle whose sides are 29 cm, 20 cm and 21 cm. Hence find the length of the
altitude corresponding to the longest side.
(a) 210 cm2, 420/29 (b) 420 cm2,420/29
(c) 320 cm2,420/29
(d) 110 cm2,12
Sol: (a)
Since the sides of the triangle are 29cm, 20cm and 21cm
s = semiperimeter =
Area
29 20 21
2
s(s a)(s b)(s c)
Area of triangle = 35 35 29 35 20 35 21 cm3
= 35 6 15 14 cm3
= 7 5 2 3 3 5 7 2 cm3
= 5 7 2 3 cm3 210 cm3
The longest side of the triangle is 29 cm, let h cm be the length of the corresponding altitude, then
area of triangle =
210 =
1
base height
2
1
29 h
2
h
420
14
14
29
29
Hence, the length of the altitude corresponding to the longest side = 14
14
cm.
29
31. Find the area of an equilateral triangle of side 8 m correct to three significant figures.
(a) 27 m2
(b) 27.0 m2
(c) 27.3 m2
(d) 27.7 m2
Sol : (d)
Given the side of an equilateral triangle = a = 8 m
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Area of the equilateral triangle =
=
Medical and Non - Medical Classes
3 2
a
4
3
8 8 m2 16 3 m3
4
=(16 × 1.732) m3 = 27.712 m2
= 27.7 m2, correct to 3 significant figures.
32. A bicycle wheel has a diameter (including the tyre) of 70 cm. How many times would the travel rotate to
cover a distance of 4.4 km ?
(a) 1000
(b) 3000
(c) 4000
(d) 2000
Sol: (d)
Given,
diameter of the bicycle = 70 cm
radius r
70
cm 35 cm
2
Circumference of the wheel = 2 r
2
22
35 cm
7
220 cm
The distance covered by the wheel in one revolution = 220 cm
Since the distance to be covered by the wheel = 4.4 km
= (4.4×1000×100) cm = (44×10000) cm,
the number of times the wheel would rotate =
44 10000
2000 .
220
33. A copper wire when bent in the form of a square encloses an area of 121 cm2. If the same wire is bent
into the form of a circle, find the area of the circle.
(a) 49cm2
(b) 154 cm2
(c) 98 cm2
(d) 216 cm2
Sol: (b)
Let the side of the square be a cm.
Area of square = a2 cm2
According to given, a2 = 121
a = 121 11
The length of the wire = perimeter of the square
= 4a cm = (4×11) cm = 44 cm.
Let r cm be the radius of the circle
Circumference of the circle = 2 r
As the same wire is to bent into the form of circle,
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2 r = 44
2
22
r 44
7
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r 7
22
7 7 cm2 154cm2
7
The area of the circle = r2
34. The area of three faces of a box are 120 cm2, 72 cm2 and 60 cm2. What is the volume of the box ?
(a) 720 cm3
(b) 360 cm3
(c) 180 cm3
(d) 90 cm3
Sol: (a)
Let the dimensions of the box be l cm, b cm and h cm.
According to given,
l ×b = 120
l × h = 72
b × h = 60
Multiplying these equations, we get
l2 × b2 × h2 = 120 × 72 × 60
l × b × h = 120 72 60
602 122
l × b × h= 60× 12 = 720
Hence the volume of the box = 720 cm3
35. The external dimensions of an open rectangular wooden box are 98 cm by 84 cm by
77 cm. If the wood is 2 cm thick all around, find the internal dimensions of the box.
(a) l=90cm , b=80cm,h=40cm
(b) l=94cm , b=80cm,h=75cm
(c) l=90cm , b=80cm,h=20cm
(d) l=20cm , b=80cm,h=40cm
Sol: (b)
The internal dimensions of the box are :
length = 98 cm – 4 cm = 94 cm,
breadth = 84 cm –4 cm = 80 cm,
height = 77 cm –2 cm = 75 cm
( Since the box is open at the top, so only the thickness of bottom is to be reduced from the length .)
36. 3
(a) 18
6
9
15
(b) 24
?
39
(c) 20
(d) 30
Sol: (b)
24 Each term is the addition of two previous number.
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37. 11
13
17
(a) 2
19
?
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29
(b) 23
(c) 27
(d) 22
Sol: (b)
23 all number are prime number in series.
38. Study the diagram given here carefully and answer the questions that follow :
Youths in villages
Unemployed youths
Educated
Uneducated unemployed youths in villages are represented by
(a) 1
(b) 4
(c) 6
(d) 7
Ans. (b)
39. Nidhi travels 10 km to the north, turns left and travels 5 km, and then again turns right and covers
another 5 km, and then turns right, and travels another 5 km. How far is she from the starting point ?
(a) 15 km
(b) 4 km
(c) 5 km
(d) 10 Km
Ans. (a)
40. Sit and Mona are Narain’s wives, Bindu is Mona’s step-daughter. How is Sita related to Bindu ?
(a) Sister
(b) Mother-in-law
(c) Mother
(d) None of these
Ans. (c)
41. Problem Figures
Answer Figures
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Ans. (b)
42. Problem Figures
Answer Figures
Ans. (d)
43. Express the following algebraic fractions in lowest terms:
(a) x
12x
18
(b) 4x/9
(c)2x/3
(d) 12x/18
(b) 20
(c) 200
(d) 2000
(c)
(d)168
Sol: (c)
12x 2 2 3 x
=
2 3 3
18
2x
3
44. Evaluate : 1052 – 952
(a) 2
Sol: (d)
1052 – 952 = (105+95)(105–95)
= 200 × 10 = 2000
45. Simplify : 1620 × 1614
(a) 163
1628
(b)
162
166
Sol: (c)
1620 1614 1628
= 1620+14
= 1634
1628
1628=1634–28 = 166
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Section – B {Science}
46. The hottest planet in the solar system is
(a) Mars.
(b) Sun.
(c) Mercury.
(d) Venus.
Ans. (d)
47. The large number of rocks that lie between the orbits of Mars and Jupiter are called
(a) comets.
(b) asteroids.
(c) meteors.
(d) meteorites.
Ans. (b)
48. Astronomers gave the Big Bang theory based on the observation that
(a) the universe is expanding.
(b) the universe is contracting.
(c) the size of the universe is not changing. (d) a few years ago people saw a big explosion in the sky.
Ans. (a)
49. The distance of the star Proxima Centauri from the earth is
(a) 4.3 million kilometers.
(b) 81/4 light minutes.
(c) 2 million light years.
(d) 4.3 light years.
Ans. (a)
50. The pole Star is in the constellation of
(a) Ursa Major.
(b) Ursa Minor.
(c) Orion.
(d) Scorpius.
Ans. (b)
51. Artificial satellites are sent in space by scientists because
(a) they look beautiful in the sky.
(b) every country wants to break the record of the largest number of satellites sent.
(c) they serve as stopovers for people going to the moon.
(d) they are useful in communication, forecasting weather, locating minerals and in studying outer
space.
Ans. (d)
52. A spherical mirror with its reflecting surface on the outside is a
(a) plane mirror.
(b) concave mirror.
(c) convex mirror.
(d) either concave or convex, depending on which way you look at it.
Ans. (c)
53. Which of these is not a pollutant unless present in excess?
(a) Sulphur dioxide
(b) Carbon dioxide
(c) Carbon monoxide
(d) Nitrogen dioxide
Ans. (b)
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54. Development and movement of cyclones is studied by the
(a) Indian Postal Department
(b) Indian Forensic Department
(c) Indian Meteorological Department
(d) Indian Geographical Department
Ans. (c)
55. Band of which of these colours is not seen in a spectrum?
(a) Green
(b) Yellow
(c) Purple
(d) Red
Ans. (c)
56. Suppose a swimming pool is filled with an imaginary liquid which is optically rarer than air. For a
person looking at the pool from outside, the apparent depth of the pool will be ……..
(a) less than the real depth
(b) more than real depth
(c) same as the real depth
(d) cannot be predicted
Ans. (b)
57. Which friction is self adjusting friction
(a) static friction
(b) limiting friction
(c) kinetic friction
(d) none of these.
Ans. (b)
58. If the body is moving in the inclined plane with constant velocity then
(a) force due to weight is equal to force due to friction
(b) force due to weight is greater than force due to friction
(c) force due to weight is less than force due to friction
(d) none of these
Ans. (a)
59. Friction
(1) is a tangential force
(2) is a normal force
(3) is a tangential force and act between the two surface of contact
(4) it does not depend upon the area of contact.
(a) only (1) is correct
(b) only (3) is correct
(c) (3) and (4) is correct
(d) (2) and (4) is correct.
Ans. (c)
60. The splitting of white light into its constituent colours is called
(a) refraction
(b) dispersion
(c) deviation
(d) displacement
Ans. (b)
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61. The pointed teeth in your mouth are
(a) premolars
(b) incisors
(c) molars
(d) canines
(c) Steel
(d) Vaccum
Ans. (d)
62. In which medium does sound travel fastest ?
(a) Air
(b) Water
Ans. (c)
63. Friction is
(a) always a disadvantages.
(b) always an advantage.
(c) sometimes a disadvantage and sometimes an advantages.
(d) neither a disadvantage, nor an advantage.
Ans. (c)
64. Any rock containing metallic mineral is called
(a) Alloy
(b) Ore
(c) Metalloid
(d) Compound
Ans. (b)
65. A metal X can replace another metal Y from its metal salt. Is X above or below Y in the reactivity series ?
(a) X is above Y in the reactivity series
(b) X is below Y in the reactivity series
(c) X is at the top in the reactivity series
(d) None of the above
Ans. (a)
66. Which of these metals cannot displace hydrogen from a dilute acid ?
(a) Iron
(b) Zinc
(c) Silver
(d) Calcium.
Ans. (c)
67. Which of the following represents gold of 100% purity ?
(a) 100 carats.
(b) 1 carat
(c) 18 carats
(d) 24 carats.
Ans. (d)
68. Which one of the following alloys is light and strong ?
(a) Brass
(b) Stainless steel
(c) Duralium
(d) Bronze
(b) a compound
(c) a mixture
(d) none of these
Ans. (c)
69. Pure water is
(a) an element
Ans. (b)
70. Which of these is the male reproductive organ in humans ?
(a) Sperm
(b) Ovum
(c) Testes
(d) Ovaries
Ans. (c)
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71. Which of the following is a hermaphrodite animal ?
(a) Elephant
(b) Cow
(c) Dog
(d) Earthworm
(c) Paramecium
(d) Human
(c) ovary
(d) vagina
Ans. (d)
72. In …., the offsprings grow out of the parent’s body.
(a) Hydra
(b) Amoeba
Ans. (a)
73. In humans, fertilization occurs in
(a) oviduct
(b) uterus
Ans. (a)
74. Which of the following methods of reproduction is sexual ?
(a) vegetative reproduction
(b) regeneration.
(c) binary fission.
(d) none of these.
Ans. (d)
75. Zygote is related to which method of reproduction ?
(a) asexual reproduction
(b) sexual reproduction
(c) spore formation.
(d) vegetative reproduction.
Ans. (b)
76. Which of these is the male reproductive organ in plants ?
(a) pistil
(b) pollen grain
(c) stamen
(d) ovule
Ans. (a,c)
77. In the human urinary system, which of these, in your opinion, is the most important organ ?
(a) kidneys
(b) ureters
(c) urinary bladder
(d) urethra
Ans. (a)
78. ……………are the tiny filtering units of the kidney.
(a) glomerulus
(b) nephrons
(c) tubules
(d) bowmen’s capsules
Ans. (b)
79. Main constituent of acid rain is water with
(a)CO2
(b) NO2
(c) SO2
(d) Both (b) & (c)
Ans. (d)
80. Which compound is added to petrol to prevent engines from knocking?
(a) Lead
(b) Carbon
(c) Hydrogen
(d) Ethanol
Ans. (a)
81. Which of these methods does not result in conservation of water ?
(a) Using drip irrigation
(b) Recycling of water
(c) Cutting vegetation so that less water is consumed
(d) Storing water by dams
Ans. (c)
82. Which of the following is not a pollutant unless present in excess ?
(a) Sulphur dioxide
(b) Carbon dioxide
(c) Carbon monoxide
(d) Nitrogen dioxide
Ans. (b)
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83. Which of the following gases mixes with the blood and prevents it from carrying oxygen ?
(a) Carbon monoxide (b) Carbon dioxide
(c)Nitrogen oxide
(d) Nitrogen trioxide
Ans. (a)
84. Which hormone is responsible for secondary sexual character in females ?
(a) Testosterone
(b) Oestrogen
(c) Thyroxine
(d) Pituitary
Ans. (b)
85. If an organism is a diploid (or 2n) with 16 chromosomes, then how many chromosomes its sperm cells
or egg cells with contain ?
(a) 8
(b) 16
(c) 32
(d) 64
Ans. (a)
86. Which of the following is not a gland ?
(a) Pancreas
(b) Adrenal
(c) Pituitary
(d) Kidney
Ans. (d)
87. Which of the following flows directly into blood ?
(a) Enzyme
(b) Hormone
(c) Minerals
(d) Proteins
Ans. (b)
88. Which hormone is responsible for ovulation ?
(a) LH
(b) Testosterone
(c) Estrogen
(d) FSH
Ans. (c)
89. Endocrine glands
(a) Do not possess ducts
(b) Sometime have duct
(c) Always have duct
(d) Pour their secretion through ducts.
Ans. (a)
90. What type of reproduction occurs in Spirogyra ?
(a) Budding
(b) Fragmentation
(c) Vegetative reproduction
(d) Fission
Ans. (b)
ALL THE BEST
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