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SET A
SECTION A {MATHEMATICS}
1.
In a certain office, 1/3 of the worker are women1/2 of the women are married and 1/3
of the married women have children. If 3/4 of the men are married and 2/3 of the
married men have children, what part of the workers are without children?
(a) 5/18
(b) 4/9
(c) 11/18
(d) 17/36
Solution:
Let the total no. of workers = x
Women =
1
x
3
⇒ Man =
Married women =
2
x
3
1 x x
=
2  3  6
Married women having children =
Married Men =
32
4  3
1 x  x
=
3  6  18
 x
x =
 2
Married Men having children =
2x x
=
3  2  3
Total No. of workers having children
=
x x x + 6x 7x
+ =
=
18 3
18
18
Total No. of workers without children
= x−
7x 18x − 7x 11x
=
=
18
18
18
Wor kers without children
Total no. of workers
=
11x 11
=
18x 18
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2.
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The perimeter of a rectangle is 80 m. If the length of the field is decreased by 2 m and
breadth increased by 2, the area is increased by 36 sq m. find the length and breadth of
the rectangle.
(a) 30m, 10 m
(b) 40m,20 m
(c)50m,30m
(d) 60m,40m
Solution: (a) 30m, 10 m
3.
R and S start walking toward each other at 10 AM at the speeds of 3 km/h and 4 km/h
respectively. They were initially 17.5 km apart. At what time do they meet?
(a) 2:30 PM
(b) 11:30 AM
(c) 1:30 PM
(d) 12: 30 PM
Solution:
Let they meet after t hrs
Let the distance covered by R in t hrs
= x km
⇒ Distance covered by 5 in t hrs
= (17.5 – x) km
Speed =
Dis tance
Time
⇒ Time =
⇒
Dis tance
Speed
x 17.5 − x
=
3
4
⇒ 4x = 3 × 17.5 – 3x
⇒ 7x = 3 × 17.5
⇒ x=
3 × 175
10 × 7
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⇒ x=
3
× 25
10
⇒ x=
3×5
15
⇒ x=
2
2
Speed =
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Dis tance
Dis tance
⇒ Time =
Time
Time
=
x
3
=
15
5
=
2×3 2
⇒ They will meet at 12 : 30 PM =
4.
A man leaves half of the property for this wife, one third for his son and remaining for
his daughter. If daughter’s share is Rs. 60,000, how much money did the man leave?
(a) Rs 3,60,000
(b) Rs 1,80,000
(c) Rs 1,20,000
(d) 1,60,000
Solution:
Let the total property = Rs x
Wife’s share = Rs
Son’s share = Rs
x
2
x
3
∴ Daughter’s share (remaining of the property) = x −
x x x
− =
2 3 6
But Daughter’s share = Rs 60, 000
∴
5.
x
= 60,000 ⇒ x = Rs 3, 60, 000
6
The perimeter of a rectangular plot is 120 m. If the length of the plot is decreased by 5m
and breadth is increased by 5 m; the area in increased by 75 sq m. Find the length (x)
and breadth (y) of the rectangular plot.
(a) x = 60, y = 30
(b) x = 40, y = 20
(c) x = 50, y = 25
(d) x = 80, y = 40
Solution:
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Let the length of rectangle = x m.
and breadth = y m, and area of rectangle = xy.
∴ Perimeter = 2 (x + y) = 120
⇒ x + y = 60
…(i)
After change
New length = (x – 5) m
New breadth = (y + 5) m
Therefore increased area = 75 m2
⇒ (x – 5) (y + 5) – xy = 75
⇒ xy – 5y + 5x – 25 – xy = 75
⇒ 5x – 5y = 100
⇒ x – y = 20
..(ii)
Solving Eq. (i) and Eq. (ii), we get x = 40 m and y = 2
20 m.
6.
A pharmacist needs to strengthen a 15% alcoholic solution to one of 32% alcohol. How
much pure alcohol should be added to 800 ml of 15% solution?
(a) 600 ml
(b) 800 ml
(c) 400ml
(d) 200 ml
Solution:
Quantity of pure alcohol in 800 ml of 15% solution
= 15% of 800
=
15
× 800 = 120 ml.
100
Suppose that x ml of pure alcohol is added to the 15% solution to get 32% alcoholic
solution.
∴ New volume of the solution
ion = (800 + x) ml.
∴ Quantity of pure alcohol in (800 + x) ml of 32% solution = (120 + x) ml,
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∴ 32% of (800 + x) = (120 + x)
⇒
32
( 800 + x ) = (120 + x )
100
⇒ 32(800 + x) = 100(120 + x)
⇒ 25600 + 32x = 12000 + 100x
⇒ 68x = 13600 ⇒ x = 200 ml
∴ 200 ml of pure alcohol should be added.
7.
A number consists of two digits. The digit at ten’s place is three time the digit at unit’s
place. The number formed by reversing the digits is 54 less than the original number.
Find the number?
(a) 91
(b) 92
(c) 93
(d) 94
Solution:
Let the digit at units place be = x
Therefore digit at ten’s place = 3x
∴ Number = 10(3x) + x = 31 x
New number by reversing the digits = 10x + 3x = 13x
According to given condition we get
31x – 13x = 54
⇒
18x = 54 ⇒ x = 3
∴ Digit at unit place = 3 digit at ten’s place = 3 × 3 = 9
∴ Number = 93
8.
The angles of a triangle are 5(x –1), 3(2x –5) and 9x. Name the triangle.
(a) isosceles triangle
(b) equilateral triangle
(c) right angled triangle
(d) isosceles right angled triangle
Solution:
We know that sum of three angles of a triangle is 1800. Three angles are
5(x – 1), 3(2x – 5) and 9x.
∴ 5(x – 1) + 3(2x – 5) + 9x = 1800
⇒ 5x – 5 + 6x – 15 + 9x = 1800
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⇒ 20x – 20 =
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1800
⇒ 20x = 180 + 20 = 200
∴ x = 10
Three angles are
5(10 – 1)0, 3(2. 10 – 5)0, (9.10)0
or 450, 450, 900
∴ It is a isosceles right angled triangle.
9.
If
x +5 x +3 x +1
−
=
, find y when y = –7x.
6
4
9
(a) 1
(b)2
(c) 3
(d) 4
Solution:
x +5 x +3 x +1
−
=
6
4
9
⇒
6 ( x + 5) 9 ( x + 3) 4 ( x + 1 )
−
=
36
36
36
⇒ 6x + 30 – 9x – 27 = 4x + 4
⇒ – 3x + 3 = 4x + 4
⇒ 7x = – 1
 1
⇒ x = − 
 7
 1
Now y = – 7x = – 7.  −  = 1
 7
10. If A scooterist drives at the rate of 24 km/h, He reach his destination 5 minutes too late.
If he drives at the rate of 30 km/h, he reached his destination 4 minutes earlier. How far
his destination?
(a) 16 km
(b) 17 km
(c) 18 km
(d) 19 km
Solution:
Let the distance be x km
Time taken in Ist case =
x
× 60
24
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5x
mintues
2
Time taken in IInd case =
x
× 60 minutes = 2x minutes
30
Difference of time in two cases = 9 minutes
∴
5x
− 2x = 9
2
⇒ 5x – 4x = 18
⇒ x = 18 km
11. A diagonal of a rectangle is inclined to one side of the rectangle at 350. The acute angle
between the diagonals is
(a) 900
(b) 950
(c) 70o
(d)1300
Solution:(c)
Ext. angle = sum of two opposite angles ∴ x = 350 + 350 = 700
12. If angles A, B, C and D of the quadrilateral ABCD, taken in order, are in the ratio
3 : 7 : 6 : 4, then ABCD is a
(a) rhombus
(b) trapezium
(c) kite
(d) parallelogram
Solution:
Angles are 3k, 7k, 6k and 4k and sum of ∠ A + ∠ B = ∠ C + ∠ D = 7k + 3k
= 6k + 4k = 10k = 1800
Hence BC AD
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Hence ABCD is a trapezium i.e. C.
13. The figure formed by joining the mid
mid-points
points of the sides quadrilateral PQRS taken in
order, is a square only if,
(a) PQRS is a rhombus
(b) diagonals of PQRS are perpendicular
(c) diagonals of PQRS are equal and perpendicular
(d) diagonals of PQRS are equal
Solution : (c)
14. Which of the following is not true for a parallelogram?
(a) opposite sides are equal
(b) opposite angles are equal
(c) diagonals bisect each other
(d) opposite
osite angles are bisected by the diagonals
Solution:(c)
Angles will be bisected only when all sides are equal.
15. In the figure, T, V and S are the mid
mid-points
points of the sides PQ, QR and RP, respectively of an
equilateral triangle PQR. Than ∆TVS is an
(a) equilateral triangle
(b) right angled triangle
(c) isosceles triangle
(d) isosceles right angled triangle
Solution:
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Given : Equilateral ∆ PQR, in which T, V and S are the mid-points of the sides PQ, QR and
RP.
To Prove : ∆ STV is an equilateral triangle.
Proof : T is the mid-point of PQ and S is the mid-point of PR.
∴
1
TS = QR
2
(1)
Similarly
1
TV = PR
2
(2)
and
1
SV = PQ
2
(3)
But
PQ = QR = RP
1
1
1
PQ = QR = RP
2
2
2
∴
SV = TR = TV
∴ ∆ STV is an equilateral triangle.
16. In the ∆ABC , AD is the median through A and E is the mid-point of AD. BE produced
meets AC at F. Then, AF = k × AC find k.
(a)
1
2
(b)
1
3
(c)
1
4
(d)
1
8
Solution:
Given : ∆ ABC, in which E is the mid-point of the median AD.
To Prove :
AF =
1
AC
3
Construction : Draw DC
BF, cutting AC at G.
Proof : In the ∆ ADG, EF
DC and E is the mid-point of AD.
∴
F is the mid-point of AG (mid-point theorem).
∴
AF = FG
(1)
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Now in the ∆BCF, DG BF and D is the mid
mid-point of BC.
∴ G is the mid-point
point of CF (mid
(mid-point theorem)
∴ FG = GC
(2)
From Eqs. (1) and (2), we get
AF = FG = GC
Since AC = AF + FG + GC
∴
AC = 3 AF
∴
AF =
1
AC
3
17. ABC is triangle, right angled at B and P is mid-point of AC. Then PB = PA =AC= k. find the
value of k.
(a)
1
2
(b)
1
3
(c)
1
4
(d)
1
8
Solution:
Given : In the ∆ABC , P is the mid
mid-point of AC and ∠ B = 900.
To Prove : PB = AP =
1
AC.
2
Construction: Draw PK BC
Proof : ∠2 + ∠ABC = 1800
(the sum of the interior angles of
sides)
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But
∠ABC = 900
∴
∠2 = 900
But
∠1 + ∠2 = 1800
∴
∠1 = 900
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Now in the ∆PAK and the ∆PBK ,
∠1 = ∠2
(each 900)
PK = PK
(common)
AK = BK (P is the mid-point
point of AC and PK
BC)
∴
∠PAK ≅ ∆PKB
∴
PA = PB
But
PA = PC
Now
AC = PA + PC ⇒ AC = PA + PA = 2PA
⇒
PA =
∴
1
PB = PA = AC
2
1
AC
2
18. ABCD is a parallelogram as shown in fig in which ∠DAB = 750 and ∠DBC = 600 .
Calculate ∠BDC .
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(a) 300
(b)
600
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(c) 450
(d) 900
Solution:
In the In the gm ABCD, BC AD and BD is a transversal.
∴
∠CBD = ∠BDA
600 = ∠BDA
∴
∠BDA = 600
Now in the ∆ABD ,
∠A + ∠ABD + ∠BDA = 1800
⇒ 750 + ∠ABD + 600 = 1800
⇒ ∠ABD = 450
∴ ∠BDC = ∠ABD
∴ ∠BDC = 450
19. In the figure ABCD is a square and DCE is an equilateral triangle. Then ∠DAE is.
(a) 300
(b) 450
(c) 600
(d) 150
Solution:
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Given : ABCD is a square and CDE is a quadrilateral triang
triangle.
To Prove :
∠ DAE = 150 and AE = BE.
Proof: In the ∆ ADE and the ∆ BCE
AD = BC
DE = CE
and
(sides of a square)
(sides of the equilateral triangle)
∠ ADE = ∠ BCE = 900 + 600 = 1500
(since ∠ ADC = ∠ BCE = 900 and ∠ EDC = ∠ ECD = 600)
∴
∆ADE ≅ ∆BCE
∴
AE = BE
(c.p.c.t.e.)
Now in the ∆ADE ,
∠ ADE = ∠ ADC + ∠ CDE
∴
∠ADE = 900 + 600 = 1500
∴
∠DAE + ∠DEA = 1800 − 1500 = 300
But
DA = DE
∴
∠ DAE = ∠ DEA
∴
2 ∠ DAE = 300
∴
∠ DAE = 150
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0
20. In the following figure find the value of x .
(a) 300
(b) 600
(c) 450
(d) 900
Solution: (d) 900
2y + 2z +80 +100 = 360
y + z = 90
Also x + y + z = 1800
∴
x = 900
21. AD, the median of ∆ABC divides it into two
(a) Right triangles
(b) Congruent triangles
(c) Triangles of equal area
(d) Two equilateral triangles
Solution:(c)
Triangles of equal area
22. The area of the figure formed by joining the mid
mid-points
points of the adjacent sides of a
parallelogram with diagonals 10 cm and 20 cm is
(a) 200 cm2
(b) 100 cm2
(c) 50 cm2
(d) 25 cm2
Solution:(b)
Area of triangle is half of the parallelogram which are on the same base and between
same parallels.
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There is no option in the given question (in order to minimize fluke and guessing
probability )
mid-points
points of the adjacent sides of a rectangle of
23. The figure obtained by joining the mid
sides 18 cm and 8 cm is :
(a) a rectangle of area 72 cm2
(b) a square of area 72 cm2
(c) a trapezium of area 72 cm2
(d) a rhombus of area 72 cm2
Solution:(d)
Area of rectangles = 18 × 8 = 144 cm2
Area of rhombus PQRS =
=
1
area ABCD
2
1
× 144 = 72 cm2
2
24. If a triangle and a parallelogram are on the same base and between same parallels, then
the ratio of the area of the triangle to the area of parallelogram is
(a) 1: 4
(b) 1 : 3
(c) 1 : 2
(d) 1 : 1
Solution:(c)
Area of a triangle is half the area of parallelogram which are on the same base and
between same parallels.
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area of ABCD = DC × EF
area of ∆ DCE =
1
DC × EF
2
25. In a parallelogram ABCD, AB = 10 cm. The altitudes corresponding to AB and AD are
respectively 14 cm and 16 cm. Find AD.
(a) 8.75 cm
(b) 8.70 cm
(c) 6.75 cm
(d) 7.75 cm
Solution:
AB = 10 cm
AF = 16 cm and AE = 14 cm
area of a parallelogram = base × altitude
⇒ AD × AE = AB × AE
⇒ AD × 16 = 10 × 14
∴ AD =
10 × 14
= 8.75 cm
16
26. BD is one of the diagonals of a quadrilateral ABCD. AM and CN are the perpendicular
from A and C, respectively, on B
BD. Then area (Quad. ABCD) = k × BD (AM + CN). Find the
value of k.
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(a)
1
2
(b)
1
3
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(c)
1
4
(d)
1
8
Solution:
ABCD is a quadrilateral
AM ⊥ BD and CN ⊥ BD.
Area of quad. ABCD = ar ( ∆ABD ) + ar ( ∆BCD)
1
1
Area ( ABCD) = BD × AM + × BD × CN
2
2
=
1
BD ( AM + CN )
2
27. Compute the area of trapezium ABCD in the figure
(a) 14 cm2
(b) 16 cm2
(c) 8 cm2
(d) 20 cm2
Solution:
Join BE
BE2 = BC2 – EC2 = 52 – 32 = 16
⇒ BE = 4 cm
∴ Area of trapezium =
1
( a + b ) × height
2
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=
1
[2 + ( 2 + 3)] × 4
2
=
1
× 7 × 4 = 14 cm2
2
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28. In the given figure G is the centroid of the ∆ ABC, such that GD = 3cm and BC = 4cm. find
the area of the ∆ ABC.
(a) 14 cm2
(b) 16 cm2
(c) 18 cm2
(d) 20 cm2
Solution:
In the ∆GBC, BC = 4 cm
and
GD = 3 cm
∴
Area of the ∆ GBC =
1
1
× base × altitude = × 4 × 3 = 6 cm2
2
2
Now area of the ∆ABC = 3 × area of ∆GBC
= 3 × 6 = 18 cm2.
29. The area of a trapezium iss ____________the product of its height and the sum of the parallel
sides.
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(a)
1
2
(b)
1
3
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(c)
1
4
(d)
1
8
Solution:
Given : Trapezium ABCD in which AB DC. Let AB = a, DC = b and distance between the
parallel sides is DL = BN = h.
To Prove : ar(ABCD) =
1
(a + b) h
2
Construction : Join BD.
Proof : Since BD is the diagonal of ABCD.
∴ ar ( ABCD) = ar ( ∆ABD) + ar ( ∆BCD )
=
1
1
AB × DL + DC × NB
2
2
=
1
1
a×h + b×h
2
2
=
1
h ( a + b)
2
30. The medians of a ∆ABC intersect at G. then ∆AGB = s ( ∆ABC ) . Find the value of s.
(a)
1
2
(b)
1
3
(c)
1
4
(d)
1
8
Solution:
Given : A triangle ABC in which the medians AD, BE and CF intersect at the point G.
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To Prove : ar ( ∆ABC ) = 3 ar ( ∆AGB )
Proof : In the DABC,
ar ( ∆ADB ) = ar ( ∆ADC )
(1)
ar ( ∆BAE) = ar ( ∆BCE)
(2)
ar ( ∆BCF ) = ar ( ∆ACF )
(3)
(A median divides the triangles into two parts of equal area).
In the DBCG, DC is a median.
∴
ar ( ∆GBD) = ar ( ∆GDC )
(4)
Subtracting Eq. (4) from Eq. (1)
ar ( ∆ABD) − ar ( ∆GBD) = ar ( ∆ADC ) − ar ( ∆GDC )
∴
ar ( ∆ABG ) = ar ( ∆AGC )
Similarly ar ( ∆ABG ) = ar ( ∆BGC )
(5)
… (6)
Comparing Eq. (5) and Eq. (6)
ar ( ∆AGC ) = ar ( ∆BGC ) = ( ∆AGB )
As
ar ( ∆ABC ) = ar ( ∆ABG ) + ar ( ∆ACG ) + ar ( ∆BGC )
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= 3 ar ( ∆AGB )
or
ar ( ∆AGB) = 1 /3 ar ( ∆ABC )
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SECTION – B {PHYSICS}
31. A cylindrical object floats in water such that
3
th of its volume is immersed in water.
4
Its density in kg m–3 is.
(a) 250
(b) 0.75
(c) 0.25
(d) 750
Solution:(d)
According to the principle of floatation,
Weight of the substance = Weight of the liquid displaced
volume × Density × g =
3
× 1000 × g = 750
4
∴ Density = 750
32. The apparent weight of an object on the surface of the moon, if the mass of the object
and liquid displaced are X and Y respectively, on earth is.
(a)
X−Y
g
(b)
g
(X − Y)
6
(c) 6g ( X − Y )
(d)
6
(X − Y)
g
Solution:
The weight of the body on the surface of earth = X g
The weight of liquid displaced is = Yg
The apparent weight is on earth is Xg – Yg
Now the value of g on the surface of earth is
surface of the moon is
1
of g therefore apparent weight on the
6
g
(X − Y)
6
33. The roof of a house blows out during a storm because
(a) the wind blows with high velocity under the roof.
(b) the weight of the roof is less than the weight of an equal volume of air.
(c) the wind blows with high velocity over the roof, causing an upthrust on the bottom
side.
(d) the wind blows with high velocity over the roof, causing low pressure under the
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roof.
Solution:(c)
34. The sudden fall in pressure due to a rise in humidity indicates
(a) a cyclone
(b) rainfall
(c) a dust storm
(d) dry weather
Solution:(a)
35. The density of a substance is 0.6 g cm–3. If it floats with
1
rd of its volume in a liquid find
3
the density of the liquid.
(a) 0.9 g/cm–3
(b) 1.8 g/cm–3
(c) 0.6 g/cm–3
(d) 0.2 g/cm–3
Solution:
Let, volume of the substance = V cm3
∴ Mass of the substance = Volume × Density
= V × 0.6 g
And
weight of the substance = V × 0.6 gf
Volume of the liquid displaced =
1
V
3
∴ Mass of the liquid displaced = Volume × Density
=
1
V×ρ
3
And weight of the liquid displaced =
1
V × ρ gf
3
According to the principle of floatation,
Weight of the substance = Weight of the liquid displaced
or
V × 0.6 =
or
ρ=
2
V×ρ
3
0.6 × 3
g cm−3 = 0.9 g cm−3 .
2
36. A bullet moving with a velocity of 300 m/s is just able to pierce a block of wood 2.0 cm
thick. What should be its velocity if it is required to pierce a block of wood (same type)
18 cm thick ?
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(a) 300 m/s
(b) 600 m/s
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(c) 900 m/s
(d) 1200 m/sec
Solution:
Given that :
v1 = 300 ms–1
s1 = 2.0 cm = 0.02 m
s2 = 18 cm = 0.18 m
We know that ,
Energy of the bullet = Work done by the bullet
or
1
mv 2 = F s
2
In first case,
1 2
mv1 = F s1
2
In second case,
1 2
mv 2 = F s2
2
v 22 s2
Dividing the two, 2 =
v 1 s1
or
=
v 22 =
s2
× v12
s1
0.18
2
2
× (300) = 9 × (300)
0.02
or
2
9 × (300) ms −1 = 900 ms −1
37. If the kinetic energy of a body increases by 300%, Calculate the percentage increase in
momentum
(a) 100%
(b) 150%
(c) 200%
(d) 50%
Solution:(a)
Let kinetic energy be E1
Therefore,
E2 = ( E1 + 300% of E1 )
E2 = 4E1
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Now,
Medical
edical and Non - Medical Classes
E × 2m
P1
= 1
P2
E2 × 2m
P1 1
=
P2 2
P2 = 2P1
Now, increase in linear momentum given by
2P 1 − P1
× 100 =100%
P1
38. A car and a truck are moving on the road with equal kinetic energy. Breaks are applied
to both of them which produce equal retarding forces in them. Both will cover distance
S1 and S2 respectively so which of the following is correct?
(a) S1 > S2
(b) S1 = S2
(c) S1 < S2
(d) S1 ≥ S2
Solution: (b)
39. In the given figure initially a body of mass ‘m’ is placed at point A and finally it is placed
at point B by moving it along the inclined plane making an angle θ with the horizontal
surface. Then the potential energy at point B wil
will be
(a) mgh
(b) mg h sin θ
(c) mg h cos θ
(d) mgh / 2
Solution:(a)
40. What is the power of pump
ump which takes 10 seconds to lift 100 kg of water to a water
tank situated at a height of 10 m? Take g = 10m/s2
(a) 1 kw
(b) 2 kw
(c) 3 kw
(d) 4 kw
Solution:
Power =
=
Work done
time
mgh 100 × 10 ×10
=
time
10
= 1000 w = 1 kw
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SECTION C {CHEMISTRY}
41. From 200 mg of CO2, 1021 molecules are removed. How many moles of CO2 are left?
(a) 2.88×10–3
(b) 1.88×10–3
(c) 288×10–3
(d) 2.88×10–5
Solution:
Gram-molecular mass of CO2 = 44 g
Mass of 1021 molecules of CO2 =
44
× 1021 = 0.073g
23
6.02 × 10
Mass of CO2 left = (0.2 – 0.073) = 0.127 g
Number of moles of CO2 left =
0.127
= 2.88 × 10–3
44
42. How many years in it would take to spend one avogadro’s number of rupees at a rate 10
lakh of rupees in one second?
(a) 1.9089×1025 year
(b) 1.9089×109 year
(c) 1.9089×1010 year
(d) 1.9089×1012 year
Solution:
Number of rupees spent in one second = 106
Number of rupees spent in one year
= 106 × 60 × 60 × 24 × 365
Avogadro’s number of rupees will be spent in
6.02 × 1023
= 6
10 × 60 × 60 × 24 × 365
= 19.089 × 109 year
= 1.9089 × 1010 year
43. One atom of an element weight 6.644×10–23 g. Calculate the number of gram atom in 40
kg of it.
(a) 103
(b) 10–3
(c) 102
(d) 10–3
Solution:
Atomic mass of the element
= Mass of one atom × 6.02 × 1023
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= 6.6444 ×
10–23
× 6.02 ×
Medical and Non - Medical Classes
1023
= 40 g
40 kg = 40, 000 g
No. of gram atoms =
=
Mass of the element in grams
Atomic mass in grams
4000
= 103
40
44. Insulin contains 3.4% sulphur. The minimum molecular weight of insulin is :
(a) 941.176
(b) 942.276
(c) 945.27
(d) None
Solution:(a)
3.4 g S = 100 g insulin
∴ 32 g S =
100 × 32
= 941.176
3.4
Insulin must contain at least one atom of S in its one molecule.
45. 5.6 litre of oxygen at NTP is equivalent to :
(a) 1 mole
(b) ½ mole
(c) ¼ mole
(d) 1/8 mole
Solution:(c)
22.4 litre O2 at STP = 1 mole.
5.6 litre O2 at STP =
5.6 1
= mole
22.4 4
46. 22.4 litre of water vapour at NTP, when condensed to water, occupies an approximate
volume of :
(a) 18 litre
(b) 1 litre
(c) 1 mL
(d) 18 ml
Solution:(d)
22.4 litre water vapour = 1 molar H2O
= 18 g H2O liquid = 18 mL H2O
47. Which has the highest mass:
(a) 1 g atom of C
(b) ½ mole of CH4
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(c) 10 mL of water
(d)
3.011×1023
Medical and Non - Medical Classes
atom of oxygen
Solution:(a)
48. The oxygen obtained by the decomposition of 72 g H2O is :
(a) 72 g
(b) 46 g
(c) 50 g
(d) 64 g
Solution:
2H2O( g ) 
→ 2H2( g ) + O2
36g
32 g
36g H2O give 32 g O2 on decomposition
72 g H2O give =
72
× 32 = 64 g
36
49. The formula of calcium dihydrogen phosphate is :
(a) CaH2PO4
(b) Ca(HPO4)2
(c) Ca(H2PO4)2
(d) Ca3(PO4)2
Solution:(c)
Ca(H2PO4)2
50. The number of oxygen atoms present in 0.25 moles of magnesium per chlorate.
(a) 4 N
(b) 2 N
(c) 6 N
(d) 8 N
Solution:(b)
1 mole of magnesium per chlorate ( Mg(ClO4 )2 ) = 8 mol of oxygen atom
0.25 mol of magnesium per chlorate = 0.25 × 8 = 2
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SECTION D {BIOLOGY}
51. The ability of nitrogen fixation is found in
(a) monerans only
(b) both monerans and protistans
(c) protistans only
(d) fungi only
Solution: (a)
52. A plant body not differentiated into root, stem and leaves is termed as
(a) thallus
(b) Mycelium
(c) hyphae
(d) herb
Solution: (a)
53. The compound plants composed of algae and fungi are called
(a) algae
(b) bryophytes
(c) pteridophytes
(d) lichens
Solution: (d)
54. Which of the following group of plants is called vascular cryptogam?
(a) thallophyta
(b) bryophyta
(c) pteridophyta
(d) angiospermae
Solution: (c)
55. Which of the following plant group bears naked seeds?
(a) pteridophyta
(b) bryophyta
(c)gymnospermae
(d) angiospermae
(c) ferns
(d) Pines
(c) Jelly fish
(d) Sponges
(c) Flat worms
(d) Glow worms
Solution: (c)
56. Most primitive vascular plant are :
(a) Bacteria
(b) Mosses
Solution: (c)
57. Radial symmetry is best seen in:
(a) Fished
(b) Star fish
Solution: (b)
58. Platyhelminthes are commonly known as:
(a) Blind worms
(b) Round worms
Solution: (c)
59. Presence of chitinous exoskeleton is an identifying feature of :
(a) Corals
(b) Molluscs
(c) Coelenterates
(d) Arthropods
Solution: (d)
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60. Which group out of the following includes all marine forms only:
(a) porifera
(b) coelenterate
(c) Mollusca
(d) echinpdermata
Solution: (d)
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