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1

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P-SAT 2014 - 15

(PIONEER’S SCHOLARSHIP/ADMISSION TEST) {9TH CBSE} General Instructions:-

The question paper consist of FOUR sections (A), (B), (C), (D).

Section A contains 45 objective multiple choice questions of Mathematics.

Section B {Physics 46 – 60}, Section C {Chemistry 61 – 75} and Section D {Biology 76-90}.

Each right answer carries 4 marks and wrong –1.

Maximum Marks 360.

Maximum Time 180 minutes.

Give your response in the OMR sheet given to you with question paper.

Properly write down your roll no, name, contact number in the OMR sheet.

If there is any inappropriate filling of the circles in OMR then that sheet will be

disqualified.

Name: _______________________________Father Name:______________________________ Mobile: ______________________________School:_____________________________________

Solution Visits: www.pioneermathematics.com/latest_updates.com

http://www.pioneermathematics.com/latest_updates.com



2

Section-A {Mathematics}

1. The graph y = 6 is a line

(a) parallel to x-axis at a distance 6 units from the origin

(b) parallel to y-axis at a distance 6 units from the origin

(c) making an intercept 6 on the x-axis

(d) making an intercept 6 on both axis

Ans. (a)

2. The equation y = 4x – 7 has

(a) no solution (b) unique solution

(c) infinitely many solution (d) exactly two solution

Sol: (c)

A linear equation in two variable has infinitely many solutions.

3. A quadrilateral , whose diagonals bisect at right angles is called

(a) Trapezium (b) rectangle (c) a parallelogram with unequal adjacent sides (d)

rhombus

Sol: (d)

4. In the given figure, ABCD and PQRC are rectangles and Q is the mid-point of AC then

Find the value of k. If PR = k(AC)

(a) 2

4 (b)

1

2 (c)

1

3 (d)

2

3

Sol: (b)



3

Given : ABCD and PQRC ate two rectangles and Q is the mid-point of AC

To Prove:

Proof : DP = PC

QC = PR

QC = 1

AC2

1PR AC

2

5. In a parallelogram ABCD, AB = 10 cm and AD = 6 cm. The bisector of A meets DC in E.

AE and BC produced meet at F. Find the length of CF.

(a) 4 cm (b) 3 cm (c) 6 (d) 5 cm

Ans. (a)

1 2

1 3 then 2 3

i.e., AB = BF = 10 cm

But BC = 6 cm then CF = 4 cm

6. In the given figure, P is the mid-point of side BC of a parallelogram ABCD such that

BAP DAP . Find the value of k , AD = k (CD).



4

(a) 3 (b) 2 (c) 5 (d) 6

Sol: (b)

1 2

But ABCD is ||gm then 2 3

then 1 3 i.e., AB BP

but P is mid-point of BC. Then

BP = PC

Now BP = AB = DC

and AD = BC = BP + PC

= 2BP = 2AB = 2CD

7. E and F are respectively the mid-points of the non-parallel sides AD and BC of a

trapezium ABCD. If EF||AB then find EF= k (AB + CD).

(a) 1

3 (b)

1

4 (c)

2

3 (d)

1

2

Ans. (d)

Construction : Join B to E which meets CD produced at G.



5

Since AB || DC and E, F are mid-points of AD and BC

then in ABE and EDG,

1 2 [Vertically opposite angles]

3 4 [Alternate interior angles]

Hence DE = AE

ABE DGE

BE= EG [ASA congruency]

i.e., AB = GD

Then CG = CD +DG = CD + AB

Now in BCG, BF FC and BE EG

BF BEi.e., 1 then EF||GC

FC EG

and1

EF GC2

1i.e., EF [CD AB]

2

8. A parallelogram and a rectangle have common base and equal areas. Then

(a) the perimeter of the rectangle is smaller than the perimeter of the parallelogram. (b)

the perimeter of the rectangle is greater than the perimeter of the parallelogram. (c) the

perimeter of the rectangle is equal to the perimeter of the parallelogram.

(d) the perimeter of the rectangle is greater than and equal to perimeter of the

parallelogram.

Ans. (a)

9. In the given figure, the sides of PQR are 30 m, 24 m and 18 m and sides of ABC are

25 m, 24 m and 7 m. Find the area between the triangles (shaded region)



6

(a) 140 m2 (b) 123 m2 (c) 132 m2 (d) 136 m2

Sol: (c)

Area of 1

PQR 24 182

= 216 m2

Area of ABC 2124 7 84m

2

Area of shaded region = ar( PQR) ar( ABC)

= 216 – 84 =132 m2.

10. A circle is inscribed in an equilateral triangle of side 12 cm touching the sides of the

triangle. Find the radius of the inscribed circle.

(a) 2 3 cm (b) 2 4 cm (c) 3 4 cm (d) 3 6 cm

Sol: (a)

Let radius of in circle = x cm

Ar of ( ABC) = ar ( BOC) ar ( COA) ar ( AOB)



7

23 1 1 1(12) BC OD AC OE AB OF

4 2 2 2

1 1 1

36 3 12 x 12 x 12 x2 2 2

36 3 18x

x 2 3 cm

11. In ABC, median AD, BE and CF intersects each other at G. Then find the value of k. If

BE +CF >k (BC).

(a) 3

4 (b)

3

2 (c)

2

3 (d)

4

3

Sol: (b)

BE + EC > BC

1

BE AC BC2

Similarly, 1

CF AB BC2

1BE CF (AC AB) 2BC

2

1

BE CF 2BC (AC AB)2

1

BE CF 2BC BC AC AB BC2

3

BE CF BC2

12. ABCD is a parallelogram X and Y are the mid-points of BC and CD respectively. Then find

the value of k. If ar(AXY) k ar (|| gm ABCD) .

(a) 2

8 (b)

4

8 (c)

3

8 (d)

3

9

Sol: (c)



8

1ar( CYX) ar ( DBC)

4

1ar(DBC) ar ( gm ABCD)

2

1

ar ( CYX) ar (|| gm ABCD)8

Similarly,

1ar ( AYD) ar (|| gm ABCD)

4

Now, use

ar ( AXY) ar (|| gm ABCD) ar ( ABX) ar( AYD) ar( CYX).

13. PQRS is a rectangle inscribed in a quadrant of a circle of radius 13 cm. A is any point on

PQ. If PS = 5 cm, then find the value of ar (PAS) = ?

(a) 30 cm2. (b) 40 cm2. (c) 50 cm2. (d) 60 cm2.

Sol: (a)

A is any point on PQ

PA < PQ

1ar ( PQR) PQ QR

2



9

= 2112 5 30 cm

2

PS = 5 cm

PA < PQ

ar( PAS) ar PQR

2ar( PAS) 30 cm

14. ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Then find

the value of k. If ar (BDE) = k ar(ABC).

(a) 1

6 (b)

1

7 (c)

1

4 (d)

1

3

Sol: (c)

23ar ( ABC) (side)

4

Let side =a cm

23ar ( ABC) a

4

aBD

2

2 23 a 3 aar( BDE)

4 2 4 4



10

2

2

3 aar( BDE) 14 4ar( ABC) 43

a4

1ar( BDE) ar ( ABC)

4

15. ABCD is a parallelogram in which BC is produced to E such that CE = BC(in figure). AE

Intersects CD at F. If ar (DFB) = 3 cm2, find the area of the parallelogram ABCD.

(a) 13 cm2 (b) 12 cm2 (c) 15 cm2 (d) 10 cm2

Sol: (b)

In ADF and ECF

ADF ECF Alternate int erior angles

AD = CF

DAF CEF Alternate int erior angle

ADF ECF

DF = CF (CPCT)

BF is median of BCD



11

1

ar ( BDF) ar( BCD)2

but gm1ar ( BCD) ar(|| ABCD)

2

gm1 1ar ( BDF) ar(|| ABCD)

2 2

gm1ar ( BDF) ar(|| ABCD)

4

gm13 ar(|| ABCD)

4

gm3 4 ar(|| ABCD)

gm 2ar(|| ABCD) 12cm

16. ABCD is a trapezium in which AB ||DC, DC = 30 cm and AB = 50cm. IF X and Y are,

respectively the mid-points of AD and BC, then find the value of k.

If ar (DCYX) = k ar (XYBA).

(a) 7

9 (b)

5

9 (c)

5

8 (d)

4

8

Sol: (a)

In DCY and PBY

CY = BY

C B

1 2 (Vertically apposite angles)

DCY PBY (ASA congruency)

DC =PB [CPCT]

PB = 30 cm

AP = 50 cm + 30 cm = 80 cm

Using mid-point theorem

XY = 1 1

AP 80cm 40cm2 2



12

Let distance between AB, XY and DC is h cm.

In 1

DCYX h(30cm 40cm) 35hcm2

1ar(XYBA) h(40cm 50cm) 45hcm

2

ar( DCYX) 35 7

ar( XYBA) 45 9

7ar ( DCYX) ( XYBA)

9

17. If O is centre of circle as shown in figure, find RTQ

(a) 40o (b) 60o (C) 45o (d) 90o

Sol: (c)

In the given figure, o5 6 360

o o o5 140 360 6 140

o o o5 360 140 220 ...(i)

5 2 2 [The angle subtended by a chord is twice the angle subtended by it on

the circumference in the alternate segment]

o220 2 2 [from(i)]



13

o2 110 .......(ii)

In cyclic quadrilateral PQTS, 4 is the exterior angle.

4 2[Exterior angle of cyclic quadrilateral is equal to interior opposite angle]

o4 110 [From (ii)]

oRQT 110

3 1 [Exterior angle of cyclic quadrilateral is equal to interior

opposite angle]

o o3 45 [ 1 45 ]

oRTQ 45

18. In the given figure, AB is the chord of a circle with centre O. AB is produced to C such

that BC = OB, CO is joined and produced to meet the circle in D. If ACD y and

AOD x , find the value of k. If x = ky.

(a) 5 (b) 3 (c) 6 (d)4

Sol: (b)

OB = BC [Given]

BOC BCO

BOC y

OBA BOC OCB [Exterior angle of a is equal to sum of the opposite interior

angles]

OBA y y 2y

Also, OA =OB [Radii of the same circle]

OAB OBA 2y



14

Now, AOD is exterior angle if AOC

AOD OAC OCA

x 2y y

x 3y

19. In the given figure, determine b

(a)9o (b)10o (c)12o (d)13o

Sol: (d)

In oCAE, C A AEC 180

o o o43 62 AEC 180

oAEC 75

In cyclic quadrilateral AEDB, a + oAEC 180

a =105o

o

0 0 0

and A BDE 180

BDE 118 ,also BDE c 118 c 62

Similarly in AFB,

a + 62o +b = 180o



15

105o + 62o +b =180o

b = 13o

20. In the given figure, O is the centre of the circle.

(a) x +y=z (b)y +z =x (c)x +z=y (d) none of these

Sol: (a)

ABC x p [Exterior angle of a ]

y = x + p +p = x+ 2p

z = 2 ABC 2x 2p

=(x +2p) + x = y + x

21. The sum of angles formed in the four segments exterior to a cyclic quadrilateral by the

sides is = x (900). Find the value of x?

(a)6 (b) 7 (c) 8 (d)4

Sol: (a)



16

E F G H 6 right angles

In cyclic quadrilateral

HAEB, o1 E 180 .....................(i)

similarly o2 F 180 .....................(ii)

and o3 G 180 .....................(iii)

Adding eq.(i), (ii) and (iii), we get

o1 E 2 F 3 G 3 180

oE F G 1 2 3 540

oE F G H 6 90

22. In the given figure, O is the centre and AE is the diameter of the semicircle ABCDE. If AB

= BC and oAEC 50 , then find CBE.

(a)30o (b) 40o (c) 45o (d) 500

Sol: (b)



17

Join OC,

o2 AEC AOC AOC 100

In AOB and BOC, AB = BC, AO =OC

and OB = OB

BOA BOC

o oBOA 50 and BOC 50

BOA CEO [Each 50o]

BO || CE [If corresponding s are equal, lines are ||]

Now, oAOC COE 180 [ oAOC 180 ]

o

o

COE 80

COE 2 CBE

CBE 40

23. In the given figure, oADC 130 and chord BC = chord BE. Find CBE

(a) 40o (b) 100o (c) 90o (d) 80o

Sol: (b)



18

Join OC

Reflex

o

o

o o o

AOC 260

AOC 100

COB 180 100 80

o

1CEB COB

2

1CEB 80

2

oCEB ECB 40

o o o oCBE 180 (40 40 ) 100

24. AB and AC are two chords of a circle of radius r such that AB = 2AC. If p and q are the

distances of AB and AC from the centre. Find the value of k. If k(q2) = p2 + 3r2.

(a) 4 (b) 3 (c) 5 (d) 2

Sol: (a)

AM2 = r2 –p2

2

2 21AB r p

2



19

AB2 = 4(r2 – p2)

(2AC)2 = 4(r2 –p2) ( AB 2AC)

4AC2 =4(r2 –p2) ………(i)

But AN2 = r2 – p2

2

2 2ACr q

2

AC2 = 4(r2 –q2) ….(ii)

From (i) and (ii)

r2 –p2 = 4r2 –4q2

4q2 = 4r2 –r2 +p2

25. In figure, O is the centre of the circle, oBCO 30 Find x and y.

(a) 45o, 12o (b) 30o, 15o (c) 16o, 40o (d) 18o, 60o

Sol: (b)

In OCP,

o

o

o o o

OPC 90 [AP BC]

OCP 30

POC 90 30 60

DO AP,

o

o o o

DOP 90

COD 90 60 30

1CBD COD

2 [Angle subtended theorem]



20

o o

o

o

1y 30 15

2

y 15

AOD 90

1ABD AOD

2 [Angle subtended theorem]

In ABP,

o o o

o o o o

o o o

x 45 y 90 180

x 45 15 90 180

x 180 150 30

26. A metallic sheet is of the rectangular shape with dimensions 48cm×36 cm. From each of

its corners, a square of 8 cm is cut off and an open box is made of the remaining sheet,

then the volume of the box is

(a) 5102 cm3 (b) 5012 cm3

(c) 5000 cm3 (d) 5120 cm3

Sol: (d)

When squares of 8 cm is cut off , then

length of the box = (48 – 16) cm = 32 cm

breadth of the box = (36 – 16) cm = 20 cm

height of the box = 8 cm

volume of the box = 32×20×8 = 5120 cm3

27. How many spherical lead shots each 4.2 cm in diameter can be obtained from a

rectangular solid lead with dimensions 66cm, 42 cm and 21 cm?

(a) 1560 (b) 1500 (c) 1400 (d) 1630



21

Sol: (b)

Dimensions of the rectangular solid are 66cm, 42 cm21 cm.

Volume of the solid = 66×42×21 cm3 ……..(i)

Diameter of a spherical lead shot = 4.2 cm

radius = 2.1 cm

Volume of spherical lead shot

= 34 22(2.1) ........(ii)

3 7

Number of lead shots

Volume of the rectangular solid

Volume of one spherical lead shot

=3

66 42 21 21from (i) and (ii)

88 (2.1)

66 2 10001500

88

28. A sector of a circle of radius 12 cm has the angle 120o. It is rolled up so that two

bounding radii are joined together to form a cone. Find the volume of the cone.

(a) 189.57 cm3 (b) 190.57 cm3 (c) 185.57 cm3 (d) 170.57 cm3

Sol: (a)

Angle of sector = 120o = 1

3of central angle

Length of are = 1

3of circumference =

12 r

3

=1 22

2 12cm3 7

Let radius of cone = x cm

Now circumference of base of cone = length of arc

1 22 222 12 2 x

3 7 7

x 4 cm



22

radius = 4 cm, slant height = 12 cm,

2 2h l r

h 144 16 128 8 2cm

Volume of cone = 21r h

3

29. Two solid spheres made of the same metal have weights 5920 g and 740 g, respectively.

Determine the radius of the larger sphere, if the diameter of the smaller one is 5cm.

(a) 10 cm (b) 5 cm (c) 6 cm (d) 8 cm

Sol: (b)

m1 = 5920 g

m2 =740 g

Density (D) = Mass

Volume

Volume = Mass

Density

V1 = 35920cm

D

V2 = 3740cm

D

1

2

5920V D

740VD

31

32

4r

592034 740r3

3132

r 5920

r 740

r2 = 5/2 cm



23

31

3

r 5920

7405

2

31

5920 125r

740 8

31r 125

31r 5cm

Radius of longer sphere = 5 cm

30. The water for a factory is stored in a hemispherical tank whose internal diameter is 14

m. The tank contains 50 kilolitres of water. Water is pumped into the tank to fill to its

capacity. Calculate the volume of water pumped into the tank.

(a) 840.66m3 (b) 670.66m3 (c) 668.66m3 (d) 650.66m3

Sol: (c)

Water in the tank = 50,000 litres = 50 m3

volume of tank = 32r

3

3

2 22 14.7 14.7 14.7

3 7 2 2 2

718.66m

Volume of water pumped in the tank

= 718.66 –50 =668.66 m3

31. A sphere and a right circular cylinder of the same radius have equal volume. By what

percentage does the diameter of the cylinder exceed its height ?

(a) 45% (b) 40% (c) 30% (d) 50%

Sol: (d)



24

Let radius of sphere be r

2 34r h r

3

4h r

3

Diameter =2r

Increased diameter = 4 2r

2r r3 3

2r100

3%4

r3

% increased = 2 3

100 503 4

32. If each edge of a cube is increased by 25%, then find the percentage increase in its

surface area

(a) 50.01% (b) 52.23 % (c) 56.25 % (d) 26.25%

Sol: (c)

Let edge of cube = x units

Surface area – 6x2 square units

New edge = x +25 5x

x units100 4

New surface area = 2256 x

16 square units 275

x8

square units



25

Increase in surface area = 2 2 275 27x 6x x

8 8

% increase =

2

2

27x

8 100 56.25%6x

33. Find the mean of the following distribution

Frequency Variable

4

8

14

11

3

4

6

8

10

12

(a)8.05 (b) 10 (c) 15 (d) 43

Sol: (a)

Frequency Variable fixi

4

8

14

11

3

4

6

8

10

12

16

48

112

110

36

Total = 332

thus mean = 332/40 = 8.05

34. The mean marks (out of 100) of boys and girls in an examination are 70 and 73,

respectively. If the mean marks of all the students in that examination is 71. Find the

ratio of number of boys to the number of girls.

(a) 2:1 (b) 2:3 (c) 5: 4 (d) 1: 5

Sol: (a)



26

Let the boys be x

Number of girls be (100 –x)

Mean of boys = Total marks

No. of boys

70= Total marks

x

Total marks = 70x

Mean of girls = Total marks

No. of girls

73= Total marks

(100 x)

Total marks = 73(100–x)

Total marks = 70x + 7300 –73x

= (7300–3x)

71 = 7300 3x

100

7300–3x = 7100

–3x = 7100-7300

–3x = –200

x =200

3

Number of boys = 200

3

Number of girls =100

3

Ratio =

200

3 2:1100

3

Ratio of boys to girls = 2 :1



27

35. In a class there are x boys and y girls, A student is selected at random, then probability

of selecting a girl is

(a) x

y (b)

x

x y (c)

y

x y (d)

y

x

Sol: (c)

36. ABC is an equilateral triangle. P is a point on BC such that BP : PC = 2 : 1. Find the value

of k. If k(AP2) = 7AB2.

(a) 8 (b)9 (c) 3 (d) 5

Sol: (b)

Draw AD BC

As ABC is an equilateral triangle, D is mid-point of BC

Given BP : PC = 2 : 1

Let PC = x, then BP = 2x

BC = BP + PC = 2x + x=3x

AB = 3x ….(i) ( AB = BC)

As D is mid-point of BC,

BD = DC = 1

2BC =

1

23x =

3x

2

DP = DC –PC=3

x2

–x=1

x2

In oABD, D 90 ,

AB2 = AD2 +BD2

AD2 = AB2 – BD2=(3x)2 –2

23 27x x

2 4 …..(ii)



28

In oADP, D 90

AP2 = AD2 + DP2

=2

227 1x x

4 2 [Using (ii)]

=2

2 17x 7. AB

3 [Using (i)]

9AP2 = 7AB2

37. ABC is an isosceles triangle. AB = AC=10 cm, BC = 12 cm. PQRS is a rectangle drawn

inside the isosceles triangle. Give PQ = SR = y cm and PS = QR =2x cm. Then x =

(a) 3y

64

(b) 4y

63

(c) 2y

34

(d) 6y

54

Sol: (a)

In ABC, AB AC

Draw AD BC, then, D is mid-point of BC.

But BC = 12 cm (given) BD = 6 cm

Also PBQ SRC BQ RC

QD = DR = x cm

BQ = BD –QD = (6 –x) cm.

In oABD, D 90

AB2 = AD2 +BD2 AD2 = AB2 –BD2



29

AD2 = (10)2 –(6)2 = 100 –36 = 64

AD = 8cm

In ABD, PQ|| AD,

PQ BQ y 6 x

AD BD 8 6

6y 3

6 x x 6 y as required8 4

38. P is any point in the interior of a triangle ABC.

(a) PA +PB < AC +BC (b) PA +PB > AC +BC(c) PA +PB =AC +BC (d) None of these

Sol: (a)

According to the given statements, the figure will be as shown below.

Produce, BP to meet AC at point M.

Since, the sum of any two sides of a triangle is greater than its third side.

In BCM,BC CM BM..................I

and, In APM,AM PM AP............II

Adding I and II, we get

BC + CM + AM + PM > BM +AP

BC + (CM +AM) > BM –PM +AP

BC +AC > PB +PA

i.e. PB + PA < BC+AC

39. The diagram shows a right pyramid that has an isosceles triangular base. If the volume

of the pyramid is 330 cm3, calculate its height , h



30

(a) 16.5 cm (b) 15.2 cm (c) 12.30 cm (d) 10.23 cm

Sol: (a)

2 2AD 13 5

144

12cm

2

1Area of base = BC AD

2

110 12

2

60 cm

Volume of pyramid = 1

3(Area of base) × Height

1330 60 h

3

20h 330

330h 16.5 cm

20

40. The diagram shows a cylinder with a diameter of 10 cm and of height 15 cm. The

shaded portion in the form of cone, with base diameter 10 cm and height 6 cm, is



31

hollowed out. Find the volume of the remaining solid, in cm3

(a) 235 cm3 (b) 325 cm3 (c) 255 cm3 (d) 366 cm3

Sol: (b)

Volume of cylinder = 2r h

= 25 15

= 3375 cm

Volume of cone = 21r h

3

= 215 6

3

= 350 cm

Volume of remaining solid = 3375 50 325 cm

41. If a leap year is selected randomly. Find the probability of getting 53 Sundays?

(a) 1/7 (b) 2/7 (c) 1/365 (d) 53/365

Sol: (b) In a leap year there are 366 days

thus the number of Sunday that can be completed in this leap year =52

thus were are left with two days and they can be

Monday and Tuesday

Tuesday and Wednesday

Wednesday and Thursday

Thursday and Friday

Friday and Saturday



32

Saturday and Sunday

thus probability of having 53 Sunday is 2

7

42. 19 cards numbered 1, 2, 3, 4, …………, 16, 17, 18,19 are put in a box and mixed

thoroughly. One person draws a cards from the box at random. Find the probability that

the number on the card is (i) a prime number,

(a)8

19 (b)

7

19 (c)

7

13 (d)

4

19

Sol: (a)

Let E2 = event of getting a prime number. Then, E2 = {2, 3, 5, 7, 11, 13, 17, 19} n(E2) =

8.

P(getting a prime number) = P(E2) = 2n(E ) 8

n(S) 19

43. A bag contains 5 black balls, 4 blue balls, 4 white balls and 6 red balls. One ball is drawn

at random from the bag. Find the probability that the ball drawn is neither black nor

blue.

(a) 10/19 (b)5

19 (c)

8

19 (d)

8

11

Sol: (a)

Let E4 = event of getting a ball which is neither black nor blue.

Then, n(E4) = (4 + 6) = 10

P (getting a ball which is neither black nor blue)=P(E4) = 4n(E ) 10

n(S) 19

44. In an examination, the mean of marks scored by a class of 40 students was calculated

as 72.5. Later on, it was detected that the marks of one student were wrongly copied as

48 instead of 84. Find the correct mean.

(a) 71.2 (b) 73.4 (c) 70.6 (d) 75.6

Sol: (b)



33

Mean of marks=

Incorrect sum of marksof 40students

40

72.5=

Incorrect sum of marksof 40students

40

Incorrect sum of marks of 40 students = 72.540 = 2900.

Since the marks of one student were wrongly copied as 48 instead of 84, correct sum of

the marks of 40 students

= 2900 – 48 + 84 = 2936.

Correct mean =

293673.4

40

45. Mean temperature of a city of a certain week was 250C. If the mean temperature of

Monday, Tuesday, Wednesday and Thursday was 230C and that of Thursday, Friday,

Saturday and Sunday was 280, find the temperature of Thursday.

(a) 270C (b)280C (c) 290C (d) 300C

Sol: (c)

Mean temperature of week = 25oC,

The sum of temperatures of 7 days of the week = 7 × 25oC = 175oC. ……(i)

Sum of temperatures of Monday, Tuesdays, Wednesdays and Thursday.

=4 ×23oC = 92oC. ………..(ii)

Sum of temperatures of Thursday, Friday, Saturday and Sunday.

= 4× 28oC = 112oC …………..(iii)

Sum of temperatures of Monday to Sunday and, Thursday

= 92oC + 112oC [Using (ii) and (iii)]

= 204oC …………..(iv)

Temperature of Thursday = 204oC – 175oC [using (iv) and (i)]

= 29oC



34

Section–B {Physics}

46. A U tube is partially filled with mercury. If water is added in one arm and an oil is added

in the other arm, find the ratio of water and oil columns so that the mercury level is

same in the two arms of U tube. Given : density of water = 103 kg m–3, density of oil =

900 kg m–3.

(a) 9

10 (b)

8

10 (c)

5

10 (d)

4

10

Sol: (a)

Since the level of mercury is same in the two arms of the U tube, therefore

Pressure of water column on the surface of mercury in one arms = Pressure of oil

column on the surface of mercury in the other arm. i.e.,

1 1 2 2h g h g

where h1 = height of water column,

1 = density of water = 103 kg m–3.

h2 = height of oil column , and

2 =density of oil = 900 kg m–3.

1 13

2 2

h 900 9

h 10 10

47. A solid density has weight W. then find its apparent weight

when it is completely submerged in a liquid of density L .

(a) LW 1 (b) LW 1 (c) L

W 1 (d) L

W 1

Sol: (b)

Given, weight of the solid = W

Mass of the solid = W/g

Volume of the solid = Mass W / h

Density

Volume of liquid displaced = Volume of the solid



35

W / g

Upthrust on the solid = Volume of liquid displaced × density of the liquid ×acceleration

due to gravity

= L L

W / g Wg

Apartment weight = True weight – upthrust

= LL

WW W 1

48. A bock of wood floats in water with 2/5th of its volume above the surface. The density of

wood is

(a) 0.12g/cm3 (b) 1.2 g/cm3 (c) 0.6g/cm3 (d) 4.2 g/cm3

Sol: (c)

49. A solid weights 30 gf in air and 26 gf when completely immersed in a liquid of relative

density 0.8. Find the relative density of the solid.

(a) 6 (b)5 (c) 8 (d) 4

Sol: (a)

Given weight of solid in air = 30 gf and in liquid = 26 gf., R. D. of liquid = 0.8

Density of liquid = 0.8 g cm–3

Given, weight of solid = 30 gf

Mass of solid = 30 g

Density of solid = 3Mass 306 g cm

Volume 5

Hence, relative density of solid = 6.

50. A body is initially at rest. It undergoes one dimensional motion with constant

acceleration. The power delivered to it at time t is proportional to

(a) t1/2 (b) t (c) t3/2 (d) t2

Sol: (b)



36

2

u at at

P F ma

ma at ma t

Since ma2 is constant

P t

51. The blades of wind mill sweep out a circle of area A = 30 m2. If the wind flows at a

velocity 36km / h perpendicular to the circle, then the mass of air passing through

in it time t = 1 minute is (take density of air = 1.2 kg m–3)

(a) 216 kg (b) 2160 kg (c) 21600 kg (d) 216000 kg

Sol: (c)

Mass = volume × density

= A t t

= 30 × 10 × 60 × 1.2 ( =30km/h= 10 ms–1)

= 21600 kg

52. A mass of 10 g moving horizontally with a velocity of 100 cm s–1 strikes a pendulum bob

of mass 10 g. The two masses stick together (see in the figure). The maximum height

reached by the system is (g = 10 ms–2).

(a) zero (b) 5 cm (c) 2.5 cm (d) 1.25 cm

Sol: (c)

K. E. mass of 10 g is converted into O.E. of the system.

21 1 1 2

21 1

1 2

1i.e., m (m m )gh

2

mh

2(m m )g

10 10000

2 20 1000

=10

2.5cm.4

53. A body is moving uni-directionally under the influence of a source of constant powder.

Its displacement in time t is proportional to



37

(a) t2 (b) t3/2 (c) t (d) t1/2

Sol: (b)

W FS m aSP

t t t

=q2

2 3

m distance mSS

t t t

or 2 3PS t

m

or 2 3S t P

constt.m

or 3/2S t

54. If the momentum of a body increase by 30%, then its kinetic energy will increase by

(a) 30 % (b) 60 % (c) 69% (d) 90%

Sol: (c)

2pK.E., E

2m

30 13pNow, p' p 30%p p p

100 10

22 2p' 13 pNew K.E. , E'

2m 10 2m

169E

100

E' E%increse in K.E. 100

E

=

169 1691 100 100

100 100

69%

55. A motor can pump up water to fill a tank of volume 500 m3 in 25 minutes, which is

placed at a height of 20 m. If efficiency of the motor is 40%, calculate the power of the

motor.



38

(a) 6.53 W (b) 7.52 W (c) 5.36 W (d) 9.82 W

Sol: (a)

mass of water to fill the tank

m = Volume × density of water

m = 50 m3 × 1000 kg m–3 = 5×104 kg

Height of tank, h = 20 m

Work done by the pump to fill the tank

= mgh = 5×104 × 9.8×40

= 9.8×106 J

Efficiency = 40% = 40 2

100 5

Useful work,

6 62W 9.8 10 3.93 10 J

5

Time, t = 215 minutes = 15 × 60 = 900s

Power, P = 6W 3.92 10

6.53kWt 900

56. A man makes a short and loud sound in front of a hill and the echo is heard after 3

seconds. On moving closer to the hill by 165m, the echo is heard after 2 seconds.

Calculate the velocity of sound and the distance of hill from the first position.

(a) 330 ms–1 (b) 356 ms–1 (c) 536 ms–1 (d) 226 ms–1

Sol: (a)

Let, velocity of sound = V

And distance of hill from first position = x

Distance of hill from second position = (x – 165) m

Now, distance travelled by the sound in going and coming back from the first position

=2x

Time = 3s

Hence, Distance = Velocity × time



39

or 2x = 3V ….(i)

Similarly, for the second position,

=2(x–165) = 2V ……..(ii)

Dividing equation (i) from (ii), we get

x 3

(x 165) 2

or 2x =3x –495

or x =495 m

Putting this value of x in equation (i) , we get

2×495=3V

or 1 12 495V ms 330ms

3

57. A source of sound produces waves of wavelength 0.80 m in air. The same source of

sound produces waves of wavelength 4.0m in water. If the velocity of sound in air = 332

ms–2, find the velocity of sound in water.

(a) 1860 ms–1 (b) 1660 ms–1 (c) 1760 ms–1 (d) 1540 ms–1

Sol: (b)

We know that,

Velocity = V = v

for water, V1 = 1 1v

And for air, V2 = 2 2v

1 1 1

2 2 2

V v

V v

Here, v1 = v2 ) Since source of sound is the same)

V2 = 332 ms–1

2 0.80m

1 4.0m



40

1V 4.0

332 0.80

or 1 11

332 4.0V ms 1660ms

0.80

58. The following graph shows the displacement vs distance of a pulse on a role at two

different times. Find the speed of the pulse.

(a) 5.5 cm s–1 (b) 4.9 cm s–1 (c) 5.9 cm s–1 (d) 4.5 cm s–1

Sol: (d)

From the graphs

Distance between the two maxima = 37.5–15=22.5 cm

Time interval = 10–5 = 5s

Speed of the pulse = distance travelled

Time taken

= 122.54.5cm s

5

59. A sonar emits pulses on the surface of water which are detected after reflection from

the bottom. If the time interval between the emission and detection of the pulse is 2 s,

find the depth of water. (Take velocity of sound in water as 1531 ms–1)

(a) 1531 m (b) 1630 m (c) 1450 m (d) 1860 m

Sol: (a)

Given time = 2 s, velocity of sound in water = 1531 ms–1

Let the depth of water = d

Therefore, total distance travelled by sound before it is detected by the Sonar= 2d

Using the expression Distance = velocity × Time



41

We have 2d = 1531× 2

Therefore, 1531 2

d 1531m2

60. A body is vibrating 6000 times in one minute. If velocity of sound in air is 360 m/s, find

the frequency of the vibration in Hz

(a) 400 Hz (b) 100 Hz (c) 200 Hz (d) 300 Hz

Sol: (b) Given Number of vibrations = 6000.

time (t) = 1 min = 60s, velocity (v) = 360 ms–1

number of waves 6000v 100Hz

time taken 60



42

Section–C {Chemistry}

61. Rearrange the following (I to IV) in the order of increasing masses and choose the

correct answer from (a), (b), (c) and (d) [Atomic masses: N = 14, O = 16, Cu = 63]

I. 1 molecule of oxygen

II. 1 atom of nitrogen

III. 1×10–10 g-molecule of oxygen

IV. 1×10–10 g-atom of copper

(a) II < I< III < IV (b) IV < III< II <I

(c) II<III<I<IV (d) III < IV < I < II

Sol: (a)

1 molecule of O2 weights = A A

32 14g; 1N atm g

N N

1×10–10 g=molecule of oxygen = 32×10–10g;

1×10–10 g-atom of copper = 63.5×10–10g.

62. Simplest formula of the compound containing nitrogen (14g) for each 40 g of oxygen is

(A) NO (b) N2O3 (c) N2O4 (d) N2O5

Sol: (d)

40 g oxygen = 40 5

g atom;16 2

14 g of nitrogen = 1 g-atom

The formula is NO5/2 or N2O5.

63. 2.0 g of oxygen contains number of atoms same as in

(a) 4g of S (b) 7 g of nitrogen (c) 0.5 g of H2 (d) 12.3 g of Na

Sol: (a)

Both 2 g of oxygen and 4 g of sulphur represent 0.125 g-atom of element.

64. Number of water molecules in the drop of water, if 1 mL of water has 20 drops and A is

Avogadro’s number , is

(a) 0.5 A/18 (b) 0.05 A (c) 0.5 A (d) 0.05 A/18

Sol: (d)



43

1 drop of water =2

1H O

1 1mL g (d 1g mL )

20 20

18 g of water = A molecules

1 A 1 0.05Ag water molecules

20 18 20 10

65. 30 3114 15Si and P are

(a) Isotopes (b) isobars (c) isomorphs (d) isotones

Sol: (d)

These contain same number of neutrons and hence are isotones.

66. Isotopes of an element have

(a) similar chemical properties but different physical properties

(b) similar chemical and physical properties

(c) similar physical properties but different chemical properties

(d) different chemical and physical properties.

Ans.(a)

67. Which of the following triads represents isotones?

(a) 12 13 146 6 6C, C, C (b) 40 42 43

18 20 21Ar, Ca, Sc

(c) 40 42 4118 20 21Ar, Ca, Sc (d) 14 16 18

7 8 9N, O, F

Sol: (b)

40 42 4318 20 21Ar, Ca, Sc have 22 nuetrons each

68. Rutherford experiment which established the nuclear model of the atom used a beam of

(a) -particles which impinged on a metal foil and got absorbed.

(b) -rays which impinged on a metal foil and ejected electrons.

(c) helium atom, impinged on a metal foil and got scattered.

(d) helium nuclei, impinged on a metal foil and got scattered.

So: (d)

-particles are same as helium nuclei.



44

69. For an ement with atomic number 19, the 19th electron will occupy

(a) L shell (b) M shell (c) N shell (d) K shell

Ans: (c)

70. If the nuclide of actinium 89Ac228 emits beta particle , he daughter nuclide will be:

(a) 88Ra228 (b) 90Th228 (c) 87Fr224 (d) 90Th229

Ans: (b)

71. A sample of an element X contain two isotopes 17X35 and 17X37 . if the average atomic

mass of the sample is of the element be 35.5 then the percentage of the these two

isotopes in the sample is:

(a) 25%, 75% (b) 10%, 90 % (c) 75%, 25% (d) 90%, 10%

Ans: (c)

72. The atomic number of an element Y is 20 the electronic configuration of its ion having

inert gas configuration is

(a) 2,8,10 (b) 2, 18 (c) 2,10,8 (d) 2,8,8

Ans: (d)

73. Which of these is a pair of isobars

(a) 6C12, 8O16 (b) 6C13, 6C14 (c) 20Ca40, 18Ar40 (d) None of these

Ans: (c)

74. Alpha particle is emitted by 92X238 during radioactivity the new species Y should be

(a) 90Y234 (b) 90Y238 (c) 92Y234 (d) 94Y234

Ans: (a)

75. Which among the following is true?

(a) Alpha rays are cathode rays.

(b) Electrons make up the cathode ray.

(c) Protons make up the cathode

(d) electromagnetic radiation make up the cathode ray.

Ans: (b)



45

Section –D {Biology}

76. Which is correct about earthworm

(a) it has brain but no head (b) it secrets cocoon around unfertilized egg|f

(c) it has no locomotory organs (d) it can crawl on smooth surface concerned

Ans.(a)

77. Clitellum of pheretima is primary concerned with

(a) copulation (b) production of cocoons

(c) excretion (d) burrowing

Ans.(b)

78. Which of the following statements is true for pheretima?

(a) it is a dioecious animal with distinct sexual dimorphism

(b) in it copulation occurs at night in burrow during rainy season

(c) it can copulate throughout the year when ever it rains

(d) it cannot travel both backwards and forwards

Ans.(b)

79. A skeleton like function during locomotion of pheretima posthuma is performed by

(a) blood (b) elementary canal laden with mud

(c) coelomic fluid (d) ventral nerve cord

Ans.(b)

80. Milluscs are

(a) diploblastic and coelomate (b) triploblastic and acoelomate

(c) triploblastic and pseudocoelomate (d) triploblastic and coelomate

Ans.(d)

81. Vitamin that is destroyed on heating

(a) vitamin D (b) vitamin C (c) vitamin A (d) vitamin K

Ans.(b)

82. Which of the following sets includes the bacterial diseases?

(a) diphtheria, leprosy, plague (b) malaria, leprosy, plague



46

(c) tetanus, tuberculosis, measles (d) malaria, mumps, poliomyelitis

Ans.(a)

83. Escherichia coli in human intestine help to synthesize

(a) vitamin B and D (b) vitamin B and C

(c) vitamin A and K (d) vitamin B and K

Ans.(d)

84. Region of human body where poliomyelitis virus multiplies.

(a) muscle cells (b) epithelial cells

(c) intestinal cells (d) nerve cells

Ans.(a)

85. Who discovered the small pox vaccine?

(a) Louis Pasteur (b) Edward Jenner

(c) Alexander Fleming (d) Anton van Leeuwenhock

Ans.(b)

86. Which of the following atmospheric pollutants does the exhaust of motor vehicles in

Delhi NOT produce?

(a) SO2 (b) fly ash (c) hydrocarbons gas (d) CO

Ans.(b)

87. An American plants that had become a troublesome waterweed in India is

(a) Trapa bispinosa (b) Cyperus rotundus

(c) Eichhornia crassipes (d) Trypha latifolia

Ans.(c)

88. The species, which are in danger of extinction, are referred to as

(a) endangered species (b) vulnerable species

(c) threatened species (d) rare species

Ans.(a)



47

89. Which of the following is an algae

(a) apple moss (b) irish moss (c) club moss (d) reindoor moss

Ans: (b)

90. Alluvial soils in India are

(a) found in desert (b) least fertile (c) very rare (d) most extensive soils

Ans.(d)

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