2015 Bull CAT 09

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2015 Bull CAT 09

A) 2ndday B) 3rdday C) 4thday D) 5thday

A) 8.6 cm2 B) 10.5 cm2 C) 12 cm2 D) None of these

DIRECTIONS for the question: Solve the following question and mark the best possible option.

Question No. : 1

There is a 100% green lawn whose green area decreases as the square of number of cows in the lawn. Two cows together ate 8m2

area of lawn on the first day and everyday two new cows join them. The growth of grass on the lawn on a particular day is directly

proportional to the cube of the number of days. If at the end of the first day in 4m2area of grass was grown, on which of the

following days the lawn will be 100% green again?

Explanation:-

Let D be decrease in green area

N1be number of cows, G be growth in area and K1, K2be constants.

Also N2be number of days

D = K1N12and G = K2N2

3

When D = 8, N1= 2 K1= 2 D2= 2(4)2= 32

D3= 2(6)2= 72

D4= 2(8)2= 128

Also, when G = 4, N2= 1

K2= 4 G2= 4(2)3= 32

G3= 4(3)3= 108

G4= 4(4)3= 256

Hence, on third day, total lawn growth = 108+32+4 = 144 m2

The total lawn decrease till the third day = 8+32+72 = 112 m2

The total lawn growth is more than the lawn decreased until that day.

Also, the growth rate of lawn is more than the decreasing of lawn from third day onwards.Hence, from 3rd day onwards lawn will be 100% green.

Hence, [2]


Question No. : 2

Two circles with radius 6cm each meet each other orthogonally. A circle with maximum possible area is drawn in the common

region and the remaining area in the common region is shaded. What is the area of the shaded region?

Explanation:-

Since two circles are orthogonal,

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A) 4cm B) 6cm C) 8cm D) 10cm

= (2 1/4 62) - 62= 18 - 36 = 56.5 - 36 = 20.5 cm2

Also, diameter of small circle = 12 - ? (SQ) = 12 - 8.5 = 3.5

Area = 1.75 1.75 10Required Area = 20.5 - 10 = 10.5 cm2.

Hence, [2]


Question No. : 3

On a frisbee if a square paper has to be fixed in such a way that two of its vertices are on the frisbee and the other two vertices are

outside the frisbee but the mid-point of one side is touching the circumference of the frisbee. If the diameter of frisbee is 20 cm,

find the side of the square paper.

Explanation:-

Let side of square paper be 2x.

CD is tangent to the circle at point S and SP is radius

PSC = 900

SQA = PSC = 900(alternate angles)

QB = x ( from centre bisects the chord)

Draw PR BC

QBRP is a rectangle QB = PR = x

Similarly PRCS is a rectangle

PS = RC = 5 ( dia = 10)

Now, BR = BC - RC2x - 5In BPR

PR2+ RB2= BP2

x2+ (2x - 5)2= 25

x2+ 4x2- 20x + 25 = 25

5x2- 20x = 0 i.e., 5x (x-4) = 0

x = 0 or x = 4 (side of square cannot be 0). So, x = 4 2x = 8.


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A) 109 litres B) 99 litres C) 89 litres D) None of these

A) 3 hrs. B) 4 hrs. C) 4 hrs. D) 5 hrs.

A) r3 r1+ r2 B) r1= r2 C) Both (a) and (b) D) None of these

Question No. : 4

There are two drums of petrol of different prices, their volumes being 220 litres and 180 litres. Equal quantities are drawn from

the two drums, and the petrol drawn from the first drum is poured into the second, and the petrol drawn from the second is

poured into the first. Now the price of petrol per litre in both drums becomes the same. How much is drawn from each drum?

Explanation:-

Drum 1 Drum II

Contents 220 litres 180 litres

Cost/litre Rs. a Rs. b

Total Cost Rs. 220a Rs. 180b

Removed c from each drum and its cost

will beRs. ca Rs. cb

Cost of what is left 220a - ca 180b - cb

New contents cost 220a - ca + cb 180b - cb + ca


Question No. : 5

The time t to go over a fixed distance is in inverse proportion to the average speed v of the vehicle. When the average speed ofthe vehicle is 48 km/hr it takes 6 hours to go over a certain distance. Find the time required to cover the same distance at an

average speed of 72 km/hr.

Explanation:-

Now T = (48 6)/72 = 4 hrs.


Question No. : 6

Two circles are drawn with centre C1and C2and radius r1and r2. The distance between the two centres is more than r1+ r2Another circle is drawn with centre, C3, as the intersection point of the common tangents of circles having centre C1and C2, the

circle with centre C3touches the other two circles. If radius of the larger circle is r3, then which of the following is true?

Explanation:-

Point of intersection of common tangents of circles in this case are collinear i.e., C1- C3- C2. C3should be equidistant from the

centres C1and C2to have the circle with centre touch the other two circles. C3will be the mid -point of C1C2only if the smaller

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A) 1500 m2 B) 1000 m2 C) 3200 m2 D) 2000 m2

A) 1/15 B)14/15 C)

2/5 D)4/5

circles are of the same radius.

r1= r2. There are two possible ways in which the configuration can be drawn as shown below:

In case II we can say that r3> r1+ r2but it is not true in case I. Hence, [2]


Question No. : 7

A hill is in the form of a regular cone with a vertical drop of 20m from the cliff to the bottom of the hill. A tower perpendicular to

the slope of the hill is situated midway down the slope. The tower is 40m tall and the top of the tower is 50m away from the cliff.

The area of the base of the mountain is:

Explanation:-

Since AB is to the slope

OA2= OB2- AB2OA = 30m

Since A is the midpoint OY = 60m

OX = 20m

Area of the base = XY2= 3200 = 3200 m2.

Hence, [3]


Question No. : 8

Two numbers are selected at random without replacement from amongst the first six natural numbers. What is the probability that

the minimum of the two is less than 4?

Explanation:-

The total number of ways of selecting the two numbers is 6 5 = 30.

We want that the minimum of the two numbers is less than 4.

If the smaller number is 1, then the other number can be any of the remaining 5 numbers from 2 to 6.



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A) 60 B) 70 C) 80 D) None of these

A) 6.25cm2 B) 12.5cm2 C) 10cm2 D) Cannot be determined

These are 12 cases.

Since the numbers can be interchanged, the toal number of favourable outcomes is 2 12 = 24.

Thus the required probability is 24/30 = 4/5


Question No. : 9

In ABC, Seg AD seg CB, also AE is angle bisector ofCAB. FindAED.

Explanation:-

? ADC and ADB are right angled triangle

ACD + CAD = 900

x -y + 600= 900

x -y = 300------(I)

x +y + 400= 900------(II)

(? DAB +DBA = 900

x +y = 50

Solving I and II, we get,

y = 10AED is also right angled triangle,

AED +DAE = 900

AED = 900- 100= 800. Hence, [3]


Question No. : 10

In the given figure, D is midpoint of AB and ABC = 90, C = 30. Also DE is parallel to BC, and BE is angle bisector of ABC

Find the area of ABE. (BC = cm.)

Explanation:-

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A) 1 B) 0.6 C) 0.78 D) 0.875

A) -165 B) C) D)

Also AD = BD

Now since DE || BC

ADE =ABC = 90 = 2x as shown in Figure.

DBE + DEB = 2x .......... (exterior angle = Sum of other two interior angle).

x +DEB = 2x

DEB is isosceles

Now DE = DB = AD

DAE =DEA = y 2x+y+y = 180

2y = 180-2x. So, y = 90-x

AEB = 90-x+x = 900

AEB is right angled with AEB = 900

AEB is 450- 450- 900


Question No. : 11

Explanation:-


Question No. : 12

If |a+3| = 8 and |b-6| = 9. What is the minimum possible value of a b? (in numerical value)

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A) a b B) a < b C) Either b > a or b a D) Cannot be determined

A) B) C) D)

Explanation:-

|a + 3| = 8

a+3 = 8 or a+3 = -8

a = 5 or a = -11

|b - 6| = 9

b - 6 = 9 or b-6 = -3

b = 15 or b = 3

Minimum value of a b = -11 15 = -165.


Question No. : 13

Compare a and b, if a > 0 and b > 0 and 8a2 4ab 2a + b = ba2 4ab + b2a = 0.

Explanation:-

From first equation

8a2- 4ab -2a + b = 0

4a(2a-b) - (2a-b) = 0

(4a-1)(2a-b) = 0

from both cases b > a.

Hence, [2]


Question No. : 14

One person each is to be selected from teams A, B and C. Team A consists of 2 boys and 1 girl, team B consists of 2 girls and 1

boy and team C consists of 2 girls and 2 boys. Find the probability that the new team consists of 1 girl and 2 boys.

Explanation:-

Team A Team B Team C

2B, 1G 2G,1B 2G, 2B

No. of ways of selection

Girl from 'A' = 1 1 2C1= 2

Girl from 'B' =2C12C1

2C1= 8

Girl from 'C' =2C1 1 2C1= 4

Total ways = 14

Total number of ways of selecting the team = 3C13C1

4C1= 3 3 4

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A) 1.6m/s B) 5m/s C) 0.8m/s D) Data insufficient

A) 10s B) 20s C) 10-20s D) Data insufficient

Hence, [4]


Question No. : 15

Chetan and Ketan decide to play a game. The arena for the game are two circular lawns with centres at C1and C2 respectively

Chetan says that he shall run back and forth between C1and C2. Ketan meanwhile is to run around the circumference of circle with

centre C1. Ketan starts running from the point A where Chetan touches circle with centre C1on his way out of circle with centre C1Chetan starts from point A, and Chetan finds that whenever he touches the circumference of circle with centre C1, Ketan is also

there and they have to avoid crashing against each other. Also, whenever Ketan touches point A he find Chetan there.

If Ketans speed is 5m/s, and he meets Chetan in multiple of 10 seconds only, what is Chetans approximate speed? Assume that

Chetan meets Ketan 10seconds after the start of the race.

Explanation:-

Distance travelled by Chetan = speed 10

Distance travelled by Ketan = 510 = 50 metres

Hence, [1]

Note: 10 seconds because they meet for the first time when Chetan has travelled distance C1A and Ketan has completed one circle


Question No. : 16

Chetan and Ketan decide to play a game. The arena for the game are two circular lawns with centres at C1and C2 respectivelyChetan says that he shall run back and forth between C1and C2. Ketan meanwhile is to run around the circumference of circle with

centre C1. Ketan starts running from the point A where Chetan touches circle with centre C1on his way out of circle with centre C1Chetan starts from point A, and Chetan finds that whenever he touches the circumference of circle with centre C1, Ketan is also

there and they have to avoid crashing against each other. Also, whenever Ketan touches point A he find Chetan there.

Ketan and Chetan are joined by another friend Vetan who moves between the two points C 2 and where that Chetans line of

motion cuts circle with centre C2. If Vetan is at C2when Chetan is at C1and Vetan starts off in a direction opposite to Chetan

initially, then when will Vetan and Chetan have to try and avoid a crash for the first time? [Given: radius of circle C 2= 3m, Vetans

speed = 0.5m/s. Use data from previous questions if necessary.]

Explanation:-

All the data except the distance between the two circles has been given. Hence, data in insufficient.

Hence, [4]


Question No. : 17

Two circles intersect each other at C and D. Line AB is their common tangent. What is the sum of measures of ACB andADB?

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A) 180 B) 90 C) 270 D) Indeterminate

A) 1, 1 B) a1, a2 C) a2, a3 D) a1+ a3, a2 a3

Explanation:-

BAC =ADC (as both intercept same arc)

ABC = 1/2m (arc BC) =BDC

BAC +ABC =ADC + BDC

Add ACB to both sides,

BAC +ABC +ACB

=ADC + BDC +ACB

ADB +

ACB = 1800

[ the sum of all angles of a triangle = 1800].

Hence, [1]


Question No. : 18

Consider a quadratic equation (x + a1) (x + a2) + a3= 0. If the roots of this equation are b1and b2, then find the value of c1, c2i.e.,

the roots of the quadratic equation (x + b1) (x + b2) a3= 0?

Explanation:-The quadratic equation:

(x+a1) (x+a2) + a3= 0

x2+ a1x + a2x + a1a2+ a3= 0

x2+ (a1+ a2) x + (a1a2+ a3) = 0

The roots of this equation are b1, b2.

b1+ b2= -(a1+ a2)

b1b2= (a1a2+ a3)

The second quadratic equation is

(x+b1) (x+b2) - a3= 0

x2+ (b1+ b2)x+ (b1b2-a3) = 0

x2+ [-(a1+a2)x] + [(a1a2+ a3) - a3] = 0

x2- (a1+ a2)x +a1a2= 0

The roots of this equation are c1,c2

c1+ c2= (a1+a2)

c1c2= a1a2

The roots of this equation are a1,a2.

Hence, [2]


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A) 18 B) C) D)

A) 1000 B) 2500 C) 2000 D) Cannot be determined

Question No. : 19

Suppose there is a video cassette. If it is played from the beginning, the time required to complete half the cassette is 72 minutes

at a constant rate. What is the time required when the winding radius of the tape is further reduced by half of the unwinding side?

(in numerical value)

Explanation:-

Let the initial winding radius be 4r i.e. when the casette is viewed from beginning to the end.

Decrease in winding area a result reduction of the radius by half i.e. when half the casette is completed = (4r)2- (2r)2 = [16r2

- 4r2] = 12 r2

This is equal to the product of the length of wound tape ?1and thickness d.

The length, in turn is equal to the product of the rate at which the casette is played and the time t1required to complete half the

casette. But rate at which the casette is played is constant (given). The thickness of the tape will also be constant throughout.

= rate t1

12r2= d

12r2= (72d) rate ----(i)

Decrease in winding area as a result of further reduction of radius by half from 2r to r

= (2r)2- (r)2

= [4r2- r2] = 3r2

3r2=3r2= rate t2 d

3r2= (t2d) rate ---- (ii)

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


Question No. : 20

The population of the planet Andromeda at any time t can be calculated by the following relation: M W + MW/1000, where M is

the number of males on the planet at time t and W is the number of females on the planet at time t. Right now the probability of a

person being a female on this planet is 0.6 and tomorrow this probability decreases by 10%. (Assume there are only male and

female on this planet.)

The number of males tomorrow will be:

Explanation:-

x = 2000 0.54 + 0.54x x = 2348

the number females tomorrow will be = 2348

Now, all the question can be answered.

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A) 1548 B) 2348 C) 3248 D) 2148

A) 25% B) 33.33% C) 40% D) None of these


Question No. : 21





The number of females tomorrow will be:

Explanation:-

x = 2000 0.54 + 0.54x x = 2348the number females tomorrow will be = 2348



Question No. : 22





The increase in the number of males is by:

Explanation:-

x = 2000 0.54 + 0.54x x = 2348

the number females tomorrow will be = 2348


DIRECTIONS for the question: The question below is followed by two statements marked I and II. Mark as your answer.

Question No. : 23

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A) if the question can be answered by using only one of the statements

B) if the question can be answered by using either of the statements alone

C) if the question can be answered by using both the statements together D) if the question cannot be answered

A) 310 214 B) 133 77 210 C) 77 216 D) 33 77 210

A) 2 x B) -2 x 2 C) - x D) 0 x

The figure below is a right angled prism. What is the surface area of BAED, given that area on the face ABC = x?

I. Area of EFCA = z II. Area of DBCF = y

Explanation:-

Statement I and II alone are insufficient to answer the question,

Combining both the statement, we get,

AC.BC = x ...... (i);

BC.FC = y ....(ii)FC.AC = z ...... (iii)

Similarly we can find, BC2and FC2. Given that AB2 = AC2+ BC2, we know AB2

Area of DEAB = AB.EA = AB.FC ( EA = FC)

Area of DEAB can be found. Hence, [3]

DIRECTIONS for the question:Answer the question independently of any other question.

Question No. : 24

If - 5 < x < 5, what is the greatest value of (13 - x)7 (7 + x)3?

Explanation:-

The maximum value of xa yboccurs when a/x = b/y.

So, the maximum value of (13 - x)7 (7 + x)3occurs when 7 / (13 - x) = 3 / (7 + x). Solving this equation yields x = - 1.

So the maximum value of os (13 - x)7 (7 + x)3is 147 63 = 210 33 77


Question No. : 25

The solution set of the inequality |x3- 6x2+ 12x 6| (x-2)3is

Explanation:-

Given |x3- 6x2+ 12x 6| > (x-2)3

|x3- 6x2+ 12x 6| > x3- 6x2+ 12x - 8

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A) 5 B) C) D)

A) -1,3 B) 7, 8 C) 2, 7 D) -2, 7

If a = x3-6x2+ 12x - 8, then the given inequality is |a+2| > a,

Now for a < 0, as |a+2| > 0, |a+2| > a and for a > 0, as |a+2|= a + 2 and a+2 > a.

hence the inequality is true for all real values.


Question No. : 26

C is a circle with centre C0. Let PQ be a diameter of C. C1is the midpoint of C0Q, C2is the midpoint of C1Q and so on. A certain

number of circles are constructed with diameters as C0Q,C1Q, C2Q, and so on. If the sum of the areas of the constructed circles is

341 cm2and the area of the circle with centre C0in 1024 cm2, how many circles are constructed? (in numerical value)

Explanation:-

Let the radius of the circle with centre C1be r and let it area be A. The area of the biggest circle (with centre C0) is 4A.

The sum of the areas of n circles, each with radius half on the preceding one, beginning one, beginning with the one with centre

C1 is

S the areas form a GP with the common ratio of .

The sum of the areas of an circles is

Alternative method:

The radii of successive circles are half these of the preceding one. Therefore the areas are in a G.P with common ratio 1/4. The

areas of the first few circles that are constructed area 256, 64, 16, 4, 1. The sum of these is 341. Therefore there are 5 circles.


Question No. : 27

The roots of the equation (x + 1) (x + 8) + 10 = 0 are given as p and q. Find the roots of (x + p) (x + q) - 4 = 0

Explanation:-

(x + 1) (x + 8) + 10 = 0 can be expanded as x2+ 9x + 8 + 10 = 0

x2+ 9x + 18 = 0

Sum of roots p + q = -9

Product of roots pq = 18

(x + p) (x + q)-4 = 0 can be written as x2+ (p + q) x + pq - 4 = 0

x2- 9x + 18 - 4 = 0

X2- 9x + 14 = 0

Roots are 7 and 2

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A) Real and positive B) Real and negative C) Are rational numbers D) Cannot be determined

A) B) C) D) None of these

A) B) Rs. 1.33 and Rs. 1.66 C) Rs. 2.57 and Rs. 0.43 D) Data insufficient


Question No. : 28

If m > 0, n > 0 and p > 0 & m, n, p are all real, then both the roots of the equation mx 2+ nx + p = 0 are

Explanation:-The sum of the roots given is -ve and the product is +ve, hence the roots have to be -ve only. If a and b are the roots of given

equation. i.e a + b is negative, and a b is positive, hence a and b have to be -ve.


Question No. : 29

Ram can row a boat in still water at 10 kmph. He decides to go boating in a river. To row upstream he takes 2 hours and to row

downstream he takes 1 1/2hours. Find the speed of the river (in kmph).

Explanation:-

Suppose the speed of the river is 'y' kmph.

While rowing upstream he takes 2 hrs and while rowing downstream he takes 1 1/2hours. As the Distance covered is constant the

ratio of the net Speeds of the boat while going upstream and downstream will be the inverse of the ratio of the Time taken.


Question No. : 30

Rajesh and Rohit have some money with which they go shopping. The amount with Rajesh and Rohit comprise the same two

digits but in reverse order. Each boy spends all his money on 33 apples and oranges. The number of oranges bought by Rohit is

equal to number of apples bought by Rajesh. The number of oranges purchased by Rohit is equal to the amount with Rajesh,which is 4.5 times the number of apples, Rohit has purchased. Find the unit price of an apple and an orange.

Explanation:-

Let x be the number of oranges purchased by Rohit and hence the number of apples purchased by him is 33 x. Also, we have, x =

4.5 (33 x).

Hence, 5.5x = 148.5x = 27

Hence, number of oranges purchased by Rohit = 27 and apples = 6

Amount with Rajesh = 27 = number of oranges purchased by Rohit. Hence, amount with Rohit = 72

Let a be cost price of orange and b be the cost price of apple, then we have,

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A) 1001 B) C) D)

A) The rightmost digit B) The leftmost digit C) Both [a] and [b] D) None of these

A) 4, 6, 8, 5, 10 gallons B) 6, 9, 10, 12, 5 gallons C) 4, 6.5, 7.5, 8, 9 gallons D) 4, 7, 7.5, 8, 6 gallons

72 = 27a + 6b .(i)

27 = 6a + 27b

Hence, 33a + 33b = 99

a + b = 3 .(ii)

{(i) 6 (ii)}21a = 54 a = 18/7 = 2.57

Hence, b = 3/7. Hence 3 rdoption


Question No. : 31

Two person A and B, are counting number from 1 to 2000, but A is counting in ascending order i.e., 1, 2, 3, ., 2000 while B is

counting in descending order i.e., 2000, 1999, 1998 ..... 1. The counting speed of A is 3 times faster than counting speed of B. Both

start counting at the same time till they reach 2000 and 1 respectively and then repeat the cycle but in reverse order, this process

goes on till both count the same number at the same time. The process will stop at which number if counting speed of A is 30

number per minute? Both can count for maximum of 10 hrs. (in numerical value)

Explanation:-

A B Time

1 - 1500 2000 1501 50 min

1501 2000, 2000

10011500 1001 50 min

Thus at the end of 100 min, both A and B will count 1001.


Question No. : 32

A ten digit number is such that all its digits are distinct. If the number is added to its reverse (ten digit number) then also we get aten digit number. Then the number 9 cannot be:

Explanation:-

9 cannot be the leftmost or rightmost digit. (As the sum of the even digit number and its reverse will become an eleven digit

number).


Question No. : 33

A bottler had 5 empty casks, numbered 1 to 5, and he tried the following experiment:

First he filled up cask 1 and cask 2 with beer, and found that together they held 10.5 gallons. Then he poured the beer from cask 1

into cask 3, and to fill up cask 3 he had to add another 3.5 gallons. He then poured cask 2 into cask 4, and had to add another 1.5

gallons to completely fill cask 4. Then he poured cask 3 into cask 5, and to fill cask 5 he was obliged to add another 1.5 gallons.

Then he filled cask 3 from cask 4, but as he had some left over in cask 4 he put it into cask 1. He then found that to fill 1 he had to

add another 3.5 gallons. What was the capacity of each Cask?

Explanation:-

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A) 7 B) 6 C) 8 D) 9

A) Satyam B) HCL Tech C) Wipro D) Infosys

Let a, b, c, d and e represents the quantity of cask 1, 2, 3, 4 and 5 respectively.

a + b = 10.5; a + 3.5 = c; b + 1.5 = d;

c + 1.5 = e

d c + 3.5 = a, a + c d = 3.5

a + a + 3.5 b 1.5 = 3.5

2a b = 1 .5, a + b = 10.5

3a = 12, a = 4, b = 6.5, c = 7.5, d = 8, e = 9.


Question No. : 34

In a month, Rohan goes to market for n times, all these days are decided by a rule that he will not go on same day again unless he

covers all other days of a week. Also the difference between the days of his any two outings is never same. If 1st of July is Sunday,

then what is the maximum value of n?

Explanation:-

For n to be maximum we have to arrange days will minimum difference between them. Now total day in July are 31.

Let | denotes the days on which Rohan goes to market then | 0 | 1 | 2 | 3 | 4 | 5 | 6 | = 29 days.The number between | represents the difference in days between two visits.

Hence maximum days are 8. Note since we have to follow the first rule, the difference has to be arranged in different way and we

can use 7 and 8 instead of 5 and 6 difference or any such combination as the total days are 31

One solution of this is 1, 9, 14, 17, 18, 20, 26, 29 or 30. Note multiple solutions are possible but in all case you cannot arrange

them in such way that n > 8, as then numbers of days he does not go to market + number of days he goes to market > 31.

DIRECTIONS for the question: Study the table/s given below and answer the question that follows.

Question No. : 35

B.S.E

Name Open High Low CloseMkt. Cap

(Rs. Crores)Volume (number of shares traded today)

Infosys 4006 4050 3580 3650 24988 4,92,925

Wipro 1750 1843 1703 1822 39567 3,90,503

Satyam 268 283 262 274 8870 62,23,894

NIIT 210 223 206 218 887 6,69,324

HCL Tech 280 298 269 293 8223 18,71,348

N.S.E

Name Open High Low Close Mkt. Cap(Rs. Crores) Volume (number of shares traded today)

Infosys 4018 4062 3610 3672 25138 8,96,638

Wipro 1748 1835 1710 1835 39850 6,57,782

Satyam 266 285 261 276 8935 95,20,124

NIIT 213 225 207 221 900 12,32,426

HCL Tech 278 299 270 294 8251 25,32,523

Which share traded the highest total volume in money terms (i.e., volume number closing price) on BSE and NSE combined?

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A) Satyam on NSE B) Infosys on BSE C) HCL Tech on NSE D) Infosys on NSE

Explanation:-

Volume in money terms

Name BSE NSE Total ( 106Rs.)

Infosys4,92,925

36508,96,638 3672 5091

Wipro3,90,503

18226,57,782 1835 1918

Satyam62,23,894

27495,20,124 276 4333

NIIT 6,69,324 218 12,32,426 221 418

HCL Tech18,71,348

29325,32,523 294 1293

From table Infosys has the maximum volume in money terms. Hence, [4]


Question No. : 36

B.S.E



Infosys 4006 4050 3580 3650 24988 4,92,925

Wipro 1750 1843 1703 1822 39567 3,90,503

Satyam 268 283 262 274 8870 62,23,894

NIIT 210 223 206 218 887 6,69,324

HCL Tech 280 298 269 293 8223 18,71,348

N.S.E



Infosys 4018 4062 3610 3672 25138 8,96,638

Wipro 1748 1835 1710 1835 39850 6,57,782

Satyam 266 285 261 276 8935 95,20,124

NIIT 213 225 207 221 900 12,32,426

HCL Tech 278 299 270 294 8251 25,32,523

If variance is defined as difference between highest and lowest value of a scrip expressed as percentage of lowest value, then

which scrip showed the highest variance on which stock exchange?

Explanation:-

Variance

Name BSE (%) NSE (%)

Infosys 13.12 12.5

Wipro 8.22 7.3

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A) I, II, IV B) II, III C) III, IV D) II, III, IV

Satyam 8 9.2

NIIT 8.25 8.7

HCL Tech 10.8 10.7

From the table, it can be seen that highest variance is shown by Infosys on BSE. (In fact, you just need to calculate variance for

given 4 options). Hence. [2]


Question No. : 37

B.S.E



Infosys 4006 4050 3580 3650 24988 4,92,925

Wipro 1750 1843 1703 1822 39567 3,90,503

Satyam 268 283 262 274 8870 62,23,894

NIIT 210 223 206 218 887 6,69,324

HCL Tech 280 298 269 293 8223 18,71,348

N.S.E



Infosys 4018 4062 3610 3672 25138 8,96,638

Wipro 1748 1835 1710 1835 39850 6,57,782

Satyam 266 285 261 276 8935 95,20,124

NIIT 213 225 207 221 900 12,32,426

HCL Tech 278 299 270 294 8251 25,32,523

Market capital of any company at a given point of time, is given by the product of number of shares of the company and the share

price at that time. The market capital given in the tables are calculated on closing price. Based on this, which of the following are

true?

I. The market capital of Infosys increased marginally during the day

II. Wipro's market capital growth was the highest during the day

III. At NSE, market capital of Satyam and NIIT grew at the same rate.

IV. NIIT's market cap growth was higher at BSE compared to that at NSE

Explanation:-

Market Capital = Number of shares share price

As, no. of shares are constant, all changes in market capital are corresponding to share price.

Market capital change during the day is equivalent to change in stock price i.e., change in closing price over the opening price

of any stock.

I. Infosys' price is coming down during the day implying a fall in market capital. Hence I is False.

Market capital growth (change of closing price over opening

Name BSE (%) NSE (%)

Infosys -8.88% -8.6%

Wipro 4.1% 4.9%

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A) Satyam B) HCL Tech C) Wipro D) Infosys

Satyam 2.2% 3.75%

NIIT 3.8% 3.75%

HCL Tech 4.6% 5.75%

From table, growth of HCL Tech at NSE is the highest. II is not true.

III. From above table, III is true

IV. From above table, NIIT's growth at BSE is higher than that at NSE. Hence, IV is true

Hence, [3]


Question No. : 38

B.S.E



Infosys 4006 4050 3580 3650 24988 4,92,925

Wipro 1750 1843 1703 1822 39567 3,90,503

Satyam 268 283 262 274 8870 62,23,894

NIIT 210 223 206 218 887 6,69,324

HCL Tech 280 298 269 293 8223 18,71,348

N.S.E



Infosys 4018 4062 3610 3672 25138 8,96,638

Wipro 1748 1835 1710 1835 39850 6,57,782

Satyam 266 285 261 276 8935 95,20,124

NIIT 213 225 207 221 900 12,32,426

HCL Tech 278 299 270 294 8251 25,32,523

Which company has the highest number of shares?

Explanation:-

Market Capital = Number of shares Closing price

Name Number of shares

Infosys 6.846 107

Wipro 21.72 107

Satyam 32.37 107

NIIT 4 107

HCL Tech 28.1 107

From the table, Satyam has the highest number of share. Hence, [1]

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A) 85 B) C) D)

A) - 3.8 10-3 B) - 3.8 10-2 C) - 2.4 10-3 D) - 11.2 10-3

DIRECTIONS for the question: Read the information given below and answer the question that follows.

Question No. : 39

In an experiment 100 readings are taken. The average of the readings should be 76. But one reading is misread. Due to this,

average of first 50 reading and the misread reading is coming to be 74 and that of last 50 reading and misread is coming to be 75.

If the misread reading is read as 28. Then,

Find the original reading.

Explanation:-

x + x2 + x3 + -----x50 + x = 3774

x1 + x2 ----- x50 = 3746

x51 + x2 + ---- x100 + x (misread) = 3825

x51 + x52 + ---- x100 = 3797x1 +x2 + ----- x100 = 7543

x1 to x100 include the misread reading

Sum of 99 correct reading = 7543-28 = 7515

Original reading that was misread = 7600-7515 = 85.


Question No. : 40


average of first 50 reading and the misread reading is coming to be 74 and that of last 50 reading and misread is coming to be 75.If the misread reading is read as 28. Then,

Find the approximate relative error in the average.

Explanation:-

x1+ x2 + ----- + x100 = 7515 after misread

Hence, [4]


Question No. : 41




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A) 57 B) 67 C) 48 D) 51

A) 20 B) -20 C) -30 D) 40

Find the approximate percentage error in the reading itself (i.e. misreading).

Explanation:-

Approximate percentage error = 100 = 67.1%. Hence, [2]


Question No. : 42




In an experiment 100 reading are taken. The average of these reading should be 28. But 5 readings are misread. Due to this the

average is coming to be 26. If the relative error in the average of that 5 readings is -2, find the average of the 5 misread readings.

Explanation:-

- 40 = Average due to misread - 20

Average due to misread = -20.

Hence, [2]


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A) 48 B) 56 C) 68 D) 72

Question No. : 43

MnD Ltd. Is a software company that has developed three solutions viz. Edusoft, Financo and Elevate. During testing the company

found a number of bugs in these solutions.

The number of bugs found in the Financo solution alone is twice the number of bugs found in all the three solutions.

The number of bugs found in the Edusoft soluton alone is four less than three times the number ofbugs found in all the three

solutions.

The number of bugs found in the Elevate solution alone is four more than the number of bugs found in all the three solutions.

24 bugs are found in the Edusoft solution.

Some of the bugs are found in exactly 2 solutions and these numbers for the various possible combinations, are in the ratio 1: 2: 3(in some order).

Ten bugs found in the Edusoft solution are also found in at least one more solution.

Based on the information given above, the total number of bugs found by MnD ltd. is equal to:

Explanation:-

Group solution:

Let the number of bugs developed in all the three solutions be x and the number of bugs involved in exactly two solutions be a, b

and c; then, we get the Venn diagram as follows:

Ten bugs involved in the Edusoft solution are also involved in at least one more solution.

a + x +b = 10 and 3x - 4 = 24 - 10 = 14

x = 6 and a + b = 4

As, a, b and c are in the ratio of 1 : 2 : 3 and a + b = 4, either a = 1 and b = 3 or a = 3 and b = 1

c = 2.

Total number of bugs = 14 + 12 + 10 + 6 + 2 + (a + b) = 44 + 4 = 48.


Question No. : 44

MnD Ltd. Is a software company that has developed three solutions viz. Edusoft, Financo and Elevate. During testing the company

found a number of bugs in these solutions.

The number of bugs found in the Financo solution alone is twice the number of bugs found in all the three solutions.

The number of bugs found in the Edusoft soluton alone is four less than three times the number ofbugs found in all the three

solutions.

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A) I or II B) I or III C) II or III D) None of these

The number of bugs found in the Elevate solution alone is four more than the number of bugs found in all the three solutions.

24 bugs are found in the Edusoft solution.

Some of the bugs are found in exactly 2 solutions and these numbers for the various possible combinations, are in the ratio 1: 2: 3

(in some order).

Ten bugs found in the Edusoft solution are also found in at least one more solution.

Which of the following additional information would enable us to find the exact number of bugs found in each solution?

I. Twenty-one bugs are found in the Financo solution.

II. Six bugs are found in all the three solutions.

III. Twenty-one bugs are found in the Elevate solution.

Explanation:-

Group solution:

Let the number of bugs developed in all the three solutions be x and the number of bugs involved in exactly two solutions be a, b

and c; then, we get the Venn diagram as follows:

Ten bugs involved in the Edusoft solution are also involved in at least one more solution.

a + x +b = 10 and 3x - 4 = 24 - 10 = 14

x = 6 and a + b = 4

As, a, b and c are in the ratio of 1 : 2 : 3 and a + b = 4, either a = 1 and b = 3 or a = 3 and b = 1

c = 2.

We need not check for statement II, as we have already deduced it. We need to calculate the values of a and b to find the exact

number of bugs involved in each solution.

12 + 6 + 2 + a = 21a = 1b = 3.

From statement III:

10 + 6 + 2 + b = 21

b = 3

a = 1.

Thus, we can calculate the exact number of bugs involved in each solution using either statement I or statement III. Hence, 2.

DIRECTIONS for the question:Analyse the graph/s given below and answer the question that follows.

Question No. : 45

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A) Semata P 2001 B) PJB 2001 C) Kongress 2000 D) Kongress 2001

Given below are two graphs giving information about the number of rallies held by some political in India from 1995-2001 and

the money spent by them on these rallies over the period.

Which of the following indicates the most amount of money spent per rally by a party in any particular year?

Explanation:-

Hence option 4

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A) PJB 1998 B) Kongress 1998 C) Kongress 1997 D) PJB 1997


Question No. : 46



If the values of Kongress is interchanged with value of PJB in the year 1998 with respect to the number of rallies organized by

them and compared with the original values of Kongress and PJB in 1997, then who would have spent more money per rally? (The

amount spend by Kongress and PJB is not interchanged in year 1998)

Explanation:-

Number of ralliesTotal amount spent on rallies

(crs.)

Kongress 74 5

PJB 68 4.4

The above figures are the original figures before interchanging for the year 1998.

After Interchanging the figures (Amount spend is not to be interchanged)

Number of rallies Total amount Average spent on rallies

Kongress 68 5 735294

PJB 74 4.4 594595

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The figures for the year 1997

Number of rallies Total amount Average spent on rallies

Kongress 59 3 508475

PJB 65 4 615385

Thus, by comparing the above figures we get the answer as Kongress 1998. Hence, [2]


Question No. : 47



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A) 435600 B) 40000 C) 449821 D) 450000

Assume that the money spent per rally by Semata P goes on decreasing and increasing in alternate years from the year 1997 to

2001 by 10%, keeping 1997 as the base year then the money spent by Semata P per rally in 2001 would be:

Explanation:-

The money per rally spent by Semata P in the year 1997

The money spent by Semata P in 2001. 444444 0.9 1.1 0.9 1.1 435600.

Hence [1]


Question No. : 48



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A) Kongress-108, PJB-100, Semata P-88 B) Kongress-110, PJB-97, Semata P-86 C) Kongress-105, PJB-100, Semata P-88

D) Kongress-114, PJB-94, Semata P-8

The Indian government has imposed restriction that from the year 2002 there would not be more than 650 rallies in totality by all

the parties. There are 4 parties other than Kongress, PJB and Semata P. These three have been allotted a share of maximum 63%,

based on their seniority. If, after their allotment and respective acceptance, the remaining rallies can be distributed amongst otherparties, the rallies held by each one of them in 2002 based on this distribution would be: (Assume each decides to hold -rallies

based on average growth numbers per year from 1995-2001 only.)

Explanation:-

63% share of 650 rallies 410.

Thus, 410 rallies are to be distributed amongst the three based on their average growth for the period gained by each one of

them.

Average growth of Kongress

Thus rallies held by Kongress in 2002 = 97 + 11.2 108 ..............(1)Average growth of PJB

Thus rallies held by PJB in 2002 = 90 + 10.3 = 100 .....................(2)

Average growth of Semata Party

Thus, rallies held by Semata in 2002 = 78 + 9.7 88 ......................(3)

Thus, Kongress - 108, PJB - 100, Semata - 88.

Hence, [1].


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A) Rs.1703675 B) Rs.1675678 C) Rs.1790167 D) Rs.1703703

Question No. : 49



In the year 2002, 88 rallies will be held by Semata P. The party has to distribute the funds for holding these rallies in over 27 states

evenly. How much fund should be allocated per state for rallies? (Assume each state will be holding approximately the same

number of rallies and money spent on rallies was in the year 2002 was 15% more than that of the year 2000.)

Explanation:-

Amount spend on rallies in 2000 = Rs. 4 crores

Amount to be spend on rallies in 2002 = (4 1.15) crores = Rs. 4.6 crores

Amount to be allocated to each state = = Rs. 1703703. Hence [4]

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A) 61000 B) C) D)

DIRECTIONS for the question: Go through the pie chart/s given below and answer the question that follows.

Question No. : 50

In the Trankanian presidential election, if no single candidate secures a simple majority of more than 50% in the first round, then

the top five candidates in terms of votes polled proceed to the next round. If no one still gets a simple majority, then the top three

of that round go into the next round. If the decision is still not decisive in favour of one candidate, only the top two proceed to the

(hopefully) last round. Shown below are the results of Elections '72.

How many more votes were cast in favour of candidate A in Round 2, as that compared to Round 1? (in numerical value)

Explanation:-

Votes for candidate A in Round 1 = 10% of 1850000 = 185000

Votes for candidate A in Round 2 = 12% of 2050000 = 246000

Difference = 61000.


Question No. : 51



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A) 74% B) 90% C) 55% D) 65%



Candidate B made gains only at the expense of 'C and 'Others'. If all the voters in favour of candidate C voted for candidate B in

Round 2, then how much percentage of the 'Others' category votes did candidate B get?

Explanation:-

In Round 1, B got 22% votes

22% of 1850000 = 407000; C got 5% votes = 5% of 1850000 = 92500.

Others got 35% votes35% of 1850000 = 647500.

In Round 2, B got 53% votes53% of 2050000 = 1086500

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A) 94% B) 86.8% C) 72.4% D) Data insufficient


Question No. : 52





In the second round, at some additional constituencies (total votes = 140000), the ballot-boxes were sent late to the counting

centre. If only 10000 votes were not in favour of B or F, then what minimum percentage of the new votes must go to candidate F

in order to call for a third round?

Explanation:-

In order for Round 3 to be called, candidate B must get 50% or less of the new total of (2050000 + 140000) = 2190000.

50% of 2190000 = 1095000

B already has 1086500 votes (53%), thus if it gets more than 8500 votes out of the new ballot, it gets a simple majority. Thus, inorder to call for Round 3, F must get (140000 - 10000 - 8500) = 121500 votes.

121500/140000 @ 86.8%. Hence, [2]

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A) 5.5% B) 6% C) 6.2% D) 6.5%


Question No. : 53





Because of pending criminal cases against them, candidates E and D are disqualified in the first round of voting and all votes cast

in their favour are nullified. What is candidate C's new share of votes (in percentage)?

Explanation:-

Candidates C got 5% votes = 0.05 1.85 = 0.0925 mn votesCandidate D and E together got 10 + 6 = 16%

These were nullified16% of 1850000 = 296000 .......................(i)

Hence, [2]

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A) 1.85 mn kroner B) 46250 kroner C) 740000 kroner D) Insufficient data


Question No. : 54





The law states that every candidate other than the top five candidates in Round 1, must pay 2.5 kroners for every vote less than

the votes of the lowest of top five. How much will the government earn through this source? [Round off the votes to the nearest

500.]

Explanation:-

While candidate C's fined amount can be calculated. We do not know how many candidate comprise the category 'Others' and

what are their number of votes, thus their individual fines cannot be computed.

Hence, [4]

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A) 5 B) C) D)


Question No. : 55

Out of 1000 teams participating in a tournament (atleast one medal is won by each team), twice as many win all gold, silver and

bronze medals as those winning only bronze and silver medals. 725 teams do not win any gold medal and 620 teams do not win

any silver medal. Number of teams winning only silver medals is 6 times the number of teams winning only gold medals. Out of

the bronze medal winners the difference between the number of teams winning gold medals and those not winning any gold

medal is 180. 50 teams win only gold and silver medals. 700 teams win at least one of the two types of medals, silver or bronze.

How many teams win only gold medal? (in numerical value)

Explanation:-

725 team do not win any gold medal

1000 - 725 = 275 teams win one or more gold medals.

Similarly 1000 - 620 = 380 teams win one or more silver medals.(z + x) - (225 - 2x - y + 2x) = 180

z = 405 - (x + y)

50 + 6y + 2x + x = 380

6y + 3x = 330

2y + x = 110

Also 50 + 6y + 2x + x + 225 - 2x - y + z = 700

275 + x + 5y + 405 - (x + y) = 700

680 + 4y = 700

y = 5 x = 110 - 2y = 100

5 teams win only gold medal.


Question No. : 56

Out of 1000 teams participating in a tournament, twice as many win all gold, silver and bronze medals as those winning only

bronze and silver medals. 725 teams do not win any gold medal and 620 teams do not win any silver medal. Number of teams

winning only silver medals is 6 times the number of teams winning only gold medals. Out of the bronze medal winners the

difference between the number of teams winning gold medals and those not winning any gold medal is 180. 50 teams win only

gold and silver medals. 700 teams win at least one of the two types of medals, silver or bronze.

How many teams do not win any bronze medal?

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A) 20 B) 80 C) 180 D) 220

Explanation:-



Similarly 1000 - 620 = 380 teams win one or more silver medals.

(z + x) - (225 - 2x - y + 2x) = 180

z = 405 - (x + y)

50 + 6y + 2x + x = 380

6y + 3x = 330

2y + x = 110Also 50 + 6y + 2x + x + 225 - 2x - y + z = 700

275 + x + 5y + 405 - (x + y) = 700

680 + 4y = 700

y = 5 x = 110 - 2y = 100

Number of teams winning bronze medals

= 225 2x y + 2x + x + z

= 225 y + x + 405 x y = 630 2y = 620

Number of teams not winning any bronze medal = 1000 620 = 380.


Question No. : 57






How many teams win both gold and bronze medals?

Explanation:-

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A) 265 B) 335 C) 365 D) 405




(z + x) - (225 - 2x - y + 2x) = 180

z = 405 - (x + y)

50 + 6y + 2x + x = 380

6y + 3x = 330

2y + x = 110

Also 50 + 6y + 2x + x + 225 - 2x - y + z = 700

275 + x + 5y + 405 - (x + y) = 700

680 + 4y = 700

y = 5 x = 110 - 2y = 100

225 2x y + 2x = 225 y = 220 teams win both gold and bronze medals.


Question No. : 58


bronze and silver medals. 725 teams do not win any gold medal and 620 teams do not win any silver medal. Number of teamswinning only silver medals is 6 times the number of teams winning only gold medals. Out of the bronze medal winners the



How many teams win only one type of medal?

Explanation:-




(z + x) - (225 - 2x - y + 2x) = 180

z = 405 - (x + y)

50 + 6y + 2x + x = 380

6y + 3x = 330

2y + x = 110

Also 50 + 6y + 2x + x + 225 - 2x - y + z = 700

275 + x + 5y + 405 - (x + y) = 700

680 + 4y = 700

y = 5 x = 110 - 2y = 100

y + 6y + z = 7y + 405 x y

= 405 x + 6y = 405 100 + 30

= 335 teams win only one type of medal.

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A) 300 B) 320 C) 400 D) 420


Question No. : 59






How many teams win bronze but not silver medals?

Explanation:-




(z + x) - (225 - 2x - y + 2x) = 180

z = 405 - (x + y)

50 + 6y + 2x + x = 380

6y + 3x = 330

2y + x = 110

Also 50 + 6y + 2x + x + 225 - 2x - y + z = 700

275 + x + 5y + 405 - (x + y) = 700

680 + 4y = 700y = 5 x = 110 - 2y = 100

225 2x y + z

= 225 2x y + 405 x y

= 630 3x 2y

= 630 300 10 = 320 teams win bronze but not silver medals.


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A) T B) P C) Q D) S

A) No. 2 B) No. 3 C) No. 4 D) Cannot be determined

Question No. : 60

P, Q, R, S, T, U and V are seven students whose pet dogs are standing in a row. The pets are numbered 1 to 7 from left to right.

Neither P's pet nor U's pet are at the ends of the row.

R's pet is to the right of S's pet.

T and Q's pets are adjacent to each other.

V's pet is among the three middle pets in the row.

Q's pet is not adjacent to R's pet but it is one of the two pets between R's and V's pets.

Whose among the following can be pet no. 2?

Explanation:-

From (iii) we get that T's pet and Q's pet are immediately next to each other from (iv), we get V' pet is no. 3 or 4 or 5. From (V), we

know that there are two pets between R's and V's pets and that Q's pet is adjacent to V. Further, since Q's pet is adjacent to V.

Further, since Q's pet and T's pet should be together, the orders should be RTQV or VQTR, if it is RTQV, the right most position V's

pet can occupy is no. 5 and hence, R will be the second pet. From condition (ii), S's pet should be in no. 1 position, 6th pet will be

P's, which will not satisfy condition (i) So, RTQV is not possible.

We have only VQTR. Since S's pet is on the left side of R's pet, only P's and U's pet can then come to the right. But then condition (i)

will not be satisfied. So we should ensure R's pet is in the right most position which means VQTR will be in 4th, 5th, 6th, 7th houses.

To satisfy conditions i) and (ii), we need to have S's pet in the first place. Thus, P's and U's pets will be in 2nd and 3rd position inany order.

Thus the two possible arrangements are SPUVQTR or SUPVQTR.

P's or U's pet can be in position NO. 2.


Question No. : 61

P, Q, R, S, T, U and V are seven students whose pet dogs are standing in a row. The pets are numbered 1 to 7 from left to right.

Neither P's pet nor U's pet are at the ends of the row.

R's pet is to the right of S's pet.

T and Q's pets are adjacent to each other.

V's pet is among the three middle pets in the row.

Q's pet is not adjacent to R's pet but it is one of the two pets between R's and V's pets.

Which pet belongs to U?

Explanation:-

From (iii) we get that T's pet and Q's pet are immediately next to each other from (iv), we get V' pet is no. 3 or 4 or 5. From (V), weknow that there are two pets between R's and V's pets and that Q's pet is adjacent to V. Further, since Q's pet is adjacent to V.

Further, since Q's pet and T's pet should be together, the orders should be RTQV or VQTR, if it is RTQV, the right most position V's

pet can occupy is no. 5 and hence, R will be the second pet. From condition (ii), S's pet should be in no. 1 position, 6th pet will be

P's, which will not satisfy condition (i) So, RTQV is not possible.

We have only VQTR. Since S's pet is on the left side of R's pet, only P's and U's pet can then come to the right. But then condition (i)

will not be satisfied. So we should ensure R's pet is in the right most position which means VQTR will be in 4th, 5th, 6th, 7th houses.

To satisfy conditions i) and (ii), we need to have S's pet in the first place. Thus, P's and U's pets will be in 2nd and 3rd position in

any order.

Thus the two possible arrangements are SPUVQTR or SUPVQTR.

U's pet is either 2nd or 3rd.

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A) Rs. 27000 B) Rs. 35000 C) Rs. 40000 D) Rs. 45000

A) Rs. 42000 B) Rs. 45000 C) Rs. 84000 D) Rs. 51000

DIRECTIONS for the question: Go through the following graph/information and answer the question that follows.

Question No. : 62

In order to repay his debt, A decides to try his luck at betting. He bets with B, C and D and doubles his money. He then repays a

quarter of his loan and bets the remaining money. In all he bets four times, each time doubling his money and paying off a quarter

of his debt. B loses half as much as D. C loses Rs. 7,000 more than 1/4 ththe amount lost by D. C loses Rs. 22,000 less than the

average amount lost by B, C, and D. In the end A is left with no money.

How much money does B lose?

Explanation:-

Let A's debt be Q and be started betting with money X. Therefore, after first round of winning and repaying money left with him =

2X - Q/4

Hence, after 4th round of winning and repaying money left with him is:

From question, we know that this amount is zero

Let B, C, and D lost b, c and d amount of money.

Solving these three simultaneous linear equation we have d = 80000, b = 40000 and c = 27000.

Total money won by A = 80000+40000+27000 =147000 and total loan paid by A = Q = 147000 + X

Q = 192000 X = 45000

B loses Rs. 40000. Hence, [3]


Question No. : 63





How much money does A win in round 2?

Explanation:-

Let A's debt be Q and be started betting with money X. Therefore, after first round of winning and repaying money left with him =2X - Q/4





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A) Rs. 42000 B) Rs. 45000 C) Rs. 48000 D) Rs. 51000

A) Rs. 45000 B) Rs. 48000 C) Rs. 150000 D) Rs. 192000


Q = 192000 X = 45000

A wins Rs. (2X - ) in round 2 i.e., (90000-48000) = Rs. 42000.

Hence, [1]


Question No. : 64





How much money did A start with?

Explanation:-

Let A's debt be Q and be started betting with money X. Therefore, after first round of winning and repaying money left with him =2X - Q/4





Total money won by A = 80000+40000+27000 =147000 and total loan paid by A = Q = 147000 + XQ = 192000 X = 45000

A starts with Rs. X = Rs. 45000. Hence, [2]


Question No. : 65





What is A's debt?

Explanation:-


2X - Q/4



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A) Rs. 24000 B) Rs. 48000 C) Rs. 55000 D) Rs. 96000

A) The party finally broke uplate in the night

B) The markets have finally broken outafter a long spell of staying in a narrow range

C) We succeeded in breaking intotheir cocoon with great difficulty

D) The usually composed Tina broke offunder intense pressure




Q = 192000 X = 45000

A's debt = Q = Rs. 192000. Hence, [4]


Question No. : 66





How much money does A win in the 4th round?

Explanation:-


2X - Q/4






Q = 192000 X = 45000

DIRECTIONS for the question: In each sentence, the highlighted word is used in different ways. Choose the option/ options in

which the usage of the word is incorrect.

Question No. : 67

BREAK

Explanation:-

broke up came to an end

broken out existed a defined territory

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A) We were discussing the plan when suddenly she came off with a unique proposition

B) The new proposal hascome in forgreat public criticism C) I do not know how this whole thing came about

D) The argument comes downto whether you are ready to be subservient to his command or not

A) It is extremely hard to recognize that someone has an alcohol problem

B) The distinctions in levels of alcoholism is not as clear cut as imagined

C) Alcoholism silently becomes a part of one's system D) Alcoholism is hard to detect in one's loved ones

breaking into- entering forcibly

broke off - used for ending talks/ relationships/ negotiations/ agreement

The right phrase here should be broke down, which implies an emotional collapse or crying.

DIRECTIONS for the question: In each sentence, the highlighted word is used in different ways. Choose the option/ options in

which the usage of the word is incorrect.

Question No. : 68

COME

Explanation:-

come in for ..receive, get, be the object of

came about ---happen, take place

comes down ---is finally a question of

came off . take place successfully

The right phrase in this case ought to be came up with.

DIRECTIONS for the question: Identify the most appropriate summary for the paragraph.

Question No. : 69

So heres a question for you: how do you tell the difference between someone whos got a bit of a drink problem and someone

who is a chronic alcoholic? Can you differentiate between the person who wakes up down the park, in piss-wet clothes with an

empty bottle of Famous Grouse as a pillow, wearing no shoes; who would literally rather die in astonishing agony from organ

failure than stop drinking and someone who is just having a bit of robust English fun? Youd think the distinction would be fairly

clear cut; but, terrifyingly, it isnt. I say that as a recovering chronic alcoholic who drank heavily nearly every single day for 23years, despite it nearly killing me several times over. I started drinking when I was 13, was drinking every day by the time I was 15

and was, by any sane standards, a desperate alcoholic albeit relatively high-functioning before I hit my twenties. I then drank

heavily every day until I was 37. The odd thing about this condition was that not only did no one around me realise I was suffering

from it I didnt even realise myself.

Explanation:-

In this case, the primary context of the sentence is set-up in the first two sentences of the paragraph wherein the author introduces

the context of the paragraph. The example that follows is an illustration for the given context itself. The context of the paragraph is

that it not easy to distinguish between different types of alcoholics. Keeping this in mind, we find option 2 to be the apt answerhere.

DIRECTIONS for the question: Identify the most appropriate summary for the paragraph.

Question No. : 70

It doesnt take a genius to see weve reached a tipping point in higher education. Universities in the UK are building bigger, hiking

tuition, adding administrators, and expanding student bodies at a higher rate than ever before. Most universities have simply

doubled down on a strategy of infinite expansion, rather than fundamentally rethinking their future. This masquerades as

innovation, usually by invoking misleading comparisons to gargantuan companies like Apple or Amazon. But unsurprisingly, for

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A) The inherent need for an organization to expand needs to be carefully balanced with the needs of education and a solution

needs to be worked out that retains the best of both worlds

B)

The need for universities to transform into corporate like entities is superfluous and instead a more cost-effective and

better solution can be implemented in its placeC) The future of universities does not need to take the path of uncontrolled expansion and we need to consider alternates that

could actually be better as well as viable

D) The pros of a micro-university far outweigh the cons of traditional universities hell-bent on expansion

universities to build, budget, and think like a corporate giant - especially given their shoestring budgets - is a recipe for disaster.

Weve given the "bigger is better" approach enough of a chance. What we havent done, in any substantial way, is try the opposite:

going smaller. Dont get me wrong. Im not talking about stop-gap austerity measures, like slashing faculty jobs and cutting

student services. What we need is a major shift in how we think about the size, shape, role, and responsibility of the university.

Lets imagine for a minute what it would look like to have a university which occupied just a couple buildings, stripped away al

but a few basic administrative costs, and offered tiny classes for little or no tuition. I want to call this the micro-university.

Explanation:-

Option 1 is ruled out as the author does not mention how the best of both worlds (traditional university types and new ones) need

to be retained.

Option 3 is ruled out as the author only offers one alternate and does not offer alternates. If you compare this option carefully

with option 2, you will see that option 2 differs from option 3 as it only mentions one solution (micro-universities). This is the best

fit in the given case.

Option 4 is incorrect as no analysis of pros and cons is made in the paragraph.

DIRECTIONS for the question : Read the passage and answer the question based on it.

Question No. : 71

Eighty-five people in the world own more wealth than half of the worlds population. This attention-grabbing figure from Oxfams

recent report Even It Up: Time to End Extreme Inequality made headlines as a neat summary of the burgeoning problem of income

and wealth inequality across the globe. Notably, it came with a written endorsement from Andy Haldane, chief economist at the

Bank of England. Since the 2008 financial crisis, the report explains, the number of billionaires has doubled, swelling to 1,645

people. This isnt a problem restricted to affluent countries: absurd levels of wealth exist alongside desperate poverty around the

world.

The credit crunch and ensuing recession didnt level out income disparity, however the rich grew richer, while the global poor

took the hit. Oxfam argue that global inequality isnt inevitable, and that market fundamentalism and deliberate political and

economic drives from governments towards deregulation, rapid reductions in public spending, privatisation, financial and trade

liberalisation, generous tax cuts for corporations and the wealthy, and a race to the bottom to weaken labour rights have all

hastened and bolstered inequality in the current crisis.

For Oxfam, an explicitly non-political charity, coming out against extreme wealth-hoarding and rising inequality was a bold move.

The report quotes Warren Buffett, the fourth wealthiest man in the world, saying: Theres been class warfare going on for the last

20 years and my class has won. This sort of language is unheard of in charity press releases and reports. But for campaigners

fighting to eradicate poverty, its no longer possible to pretend that wealth accumulation and inequality are either benign or

inevitable.

There is only so much charities and aid can do before accepting that poverty doesnt exist in a vacuum many people have far too

little because a small number have far too much. Its a simple concept, but one thats been politically unpalatable in mainstream

discourse until relatively recently. Shortly after the publication of the report, the Charity Commission reprimanded Oxfam for not

doing enough to avoid accusations of political bias in a tweet highlighting a report on how UK government policies were

exacerbating poverty and hunger. At the same time, groups and commentators on the right were criticising the Oxfam inequality

report for its methodology in calculating wealth disparity.

Others leapt to defend the methods. What is very interesting about the Oxfam figures is how the charity was critiqued in some

quarters for producing them. There was also criticism of the methodology used, Danny Dorling, professor of human geography at

the University of Oxford tells me. In fact the methods used were identical to those which the global bank Credit Suisse use when

they produce their world wealth report each year. When the bank does this there is no criticism of their assumptions. The report

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A) How many billionaires were there before the 2008 financial crises?

B) How to develop political will in order to change the attitude towards poverty?

C) How to eliminate poverty across the world? D) How did Oxfam calculate the income inequality figures?

also highlights the equality instinct the deep global cultural instinct, in religion, literature and philosophy, that wide entrenched

gaps between the rich and the poor are unfair and morally wrong. Perpetuating inequality requires political and economic policies

that run counter to our preference for fairness. Grassroots campaigns calling for the redistribution of wealth, including Occupy

and the Robin Hood Tax Campaign, have swelled in numbers since the crash, and even figures on the right including former chair

of the US Federal Reserve Alan Greenspan now say that income inequality is dangerous.

Oxfams solutions are straightforward: living wages, limiting executive pay, redistributive taxation and promoting womens

economic equality and rights. More radical is the fact that their report heralds a sea change in attitudes towards inequality

poverty isnt inevitable, and a concerted political will can change that.

Which of the following is the most suitable question to ask the author of the passage?

Explanation:-

Option 2 is the apt choice. Refer to the lines: Oxfams solutions are straightforward: living wages, limiting executive pay,

redistributive taxation and promoting womens economic equality and rights. More radical is the fact that their report heralds a

sea change in attitudes towards inequality poverty isnt inevitable, and a concerted political will can change that.

The question in option 2 continues the thread of thought in the last paragraph.

The question in option 1 is already answered. Refer to the lines: Since the 2008 financial crisis, the report explains, the number of

billionaires has doubled, swelling to 1,645 people.

Remember, poverty is not the subject of the passage, income inequality is. This is why option 3 is incorrect.

The answer for the option 4 is provided in an indirect manner in the passage (the method used for the same as the one used by

global bank Credit Suisse). Again, this is not a point that introduces something substantial in the passage.

DIRECTIONS for the question : Read the passage and answer the question based on it.

Question No. : 72

Eighty-five people in the world own more wealth than half of the worlds population. This attention-grabbing figure from Oxfams

recent report Even It Up: Time to End Extreme Inequality made headlines as a neat summary of the burgeoning problem of incomeand wealth inequality across the globe. Notably, it came with a written endorsement from Andy Haldane, chief economist at the

Bank of England. Since the 2008 financial crisis, the report explains, the number of billionaires has doubled, swelling to 1,645

people. This isnt a problem restricted to affluent countries: absurd levels of wealth exist alongside desperate poverty around the

world.

The credit crunch and ensuing recession didnt level out income disparity, however the rich grew richer, while the global poor

took the hit. Oxfam argue that global inequality isnt inevitable, and that market fundamentalism and deliberate political and

economic drives from governments towards deregulation, rapid reductions in public spending, privatisation, financial and trade

liberalisation, generous tax cuts for corporations and the wealthy, and a race to the bottom to weaken labour rights have all

hastened and bolstered inequality in the current crisis.

For Oxfam, an explicitly non-political charity, coming out against extreme wealth-hoarding and rising inequality was a bold move.The report quotes Warren Buffett, the fourth wealthiest man in the world, saying: Theres been class warfare going on for the last

20 years and my class has won. This sort of language is unheard of in charity press releases and reports. But for campaigners

fighting to eradicate poverty, its no longer possible to pretend that wealth accumulation and inequality are either benign or

inevitable.

There is only so much charities and aid can do before accepting that poverty doesnt exist in a vacuum many people have far too

little because a small number have far too much. Its a simple concept, but one thats been politically unpalatable in mainstream

discourse until relatively recently. Shortly after the publication of the report, the Charity Commission reprimanded Oxfam for not

doing enough to avoid accusations of political bias in a tweet highlighting a report on how UK government policies were

exacerbating poverty and hunger. At the same time, groups and commentators on the right were criticising the Oxfam inequality

report for its methodology in calculating wealth disparity.

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A) the global community of the rich is too brazen to accept the truth

B) the poor are too weak to come to terms with the reality and demand change

C) there is an undercurrent in the world that seeks change in status quo of global inequality

D) there is an underground movement that is transforming the global economic scenario by bringing in cultural equality

Others leapt to defend the methods. What is very interesting about the Oxfam figures is how the charity was critiqued in some

quarters for producing them. There was also criticism of the methodology used, Danny Dorling, professor of human geography at

the University of Oxford tells me. In fact the methods used were identical to those which the global bank Credit Suisse use when

they produce their world wealth report each year. When the bank does this there is no criticism of their assumptions. The report

also highlights the equality instinct the deep global cultural instinct, in religion, literature and philosophy, that wide, entrenched

gaps between the rich and the poor are unfair and morally wrong. Perpetuating inequality requires political and economic policies

that run counter to our preference for fairness. Grassroots campaigns calling for the redistribution of wealth, including Occupy

and the Robin Hood Tax Campaign, have swelled in numbers since the crash, and even figures on the right including former chair

of the US federal reserve Alan Gre

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