blue lotus

Blue Lotus

Aptitude

Numerical Reasoning

Numerical Reasoning

• Problems on Numbers

• Problems on Ages

• Ratio and Proportion

• Alligation or Mixture

• Chain Rule

• Partnership

• Venn Diagram

• Area and Volume

• Probability

• Time and Work (Pipes)

• SI and CI

• Average

• Permutation and Combination

• Percentage

Numerical Reasoning

• Boats and Streams

• Time and Distance (Trains)

• Data Sufficiency

• Profit and Loss

• Calendar

• Clocks

• Data Interpretation

• Cubes

Numerical Reasoning

Problems on Numbers

Division Algorithm:

Dividend = the number to be divided.

Divisor = the number by which it is divided.

Dividend / Divisor = Quotient.

Quotient * Divisor = Dividend.

Quotient * Divisor + Remainder = Dividend.

Arithmetic Progression: The nth term of A.P. is given by Tn = a + (n – 1)d;

Sum of n terms of A.P Sn = n/2 *(a + L) or n/2 *[2a+(n-1)d)]

a = 1st term, n = number of term, d= difference, Tn = nth term

Geometrical Progression: Tn = arn – 1.

Sn = a(rn – 1)/(r-1); Where a = 1st term , r = 1st term / 2nd term

Problems on Numbers

Basic Formulae

1. ( a+b)2 = a2 + b2 + 2ab

2. (a-b)2 = a2 +b2 -2ab

3. ( a+b)2 - (a – b)2 = 4ab

4. (a+b)2 + (a – b)2 = 2 (a2 +b2)

5. (a2 – b2) = (a+b) (a-b)

6. (a+b+c)2 =a2 +b2 +c2 + 2(ab +bc+ca)

7. (a3 +b3) = ( a+b) (a2 +ab +b2)

8. (a3 –b3) = (a-b) (a2 - ab + b2)

Problems on Numbers

Three numbers are in the ratio 3:4:5. the sum of

the largest and the smallest equal to the sum of

the third and 52. Find the smallest number ?

Problems on NumbersSolution:

Let the numbers be 3x, 4x and 5x

Then 5x+3x = 4x +52 8x – 4x = 52 4x = 52 x = 52/4 x = 13

The smallest number = 3x = 3*13 = 39.

Problems on Numbers

What is one half of two third of three fourths of

four fifths of five sixth of six sevenths of seven

eights of eight ninth of nine tenths of thirty?

Problems on Numbers

Solution:

= ½ * 2/3 *3/4 * 4/5 * 5/6*6/7*7/8*8/9*9/10 *30

= 3

Problem on Numbers

If the operation ^ is defined by the

equator x ^ y = 2x + y what is the value

of a in 2 ^ a = a ^ 3? (Sathyam)

Problem on Numbers

Solution:

2(2) = a ^ 3

4 + a = 2a + 3

a = 1

Problem on Numbers

There are 150 weight some are 1 kg weight

and some are 2 kg weights. The sum of the

weights is 260. what is the number of 1 kg

weight. (TCS)

Problem on NumbersSolution:

X + 2Y = 260

X + Y = 150

On Solving Two Equations Y = 110

X + Y = 150

X = 150 – 110 = 40 Kg

Problem on Numbers

The cost of 1 pencil, 2 pens and 4 erasers is Rs.

22, while the cost of five pencils, four pens and

two eraser is 32. how much will 3 pencils, 3

pens and 3 eraser? (TCS)

Problem on NumbersSolution:

Let Pencil be x, Pens be y, Erasers be z

x + 2y + 4z = 22

5x + 4y + 2z = 32

Adding we get 6x+6y+6z = 54

3x + 3y + 3z = 27

3 Pencil, 3 Pens and 3 Eraser is Rs. 27.

Problem on Numbers

If the numerator of a fraction is increased by

25% and denominator decrease by 20%, the

new value is 5/4. what is the original value?

(TCS)

Problem on Numbers

Solution:

( x + 25x/100) / (y – 20y/100) = 5/4

125x / 80y = 5/4

x/y = 5/4 * 80/ 125 = 4/5

Problem on Numbers

The difference between two numbers is 1/7 of

the sum of these two numbers. What is the

ratio of the two numbers? (Wipro)

Problem on Numbers

(x- y ) = 1/7 (x+y)

7( x- y) =( x + y)

7x – 7y = x + y

6x = 8y

x/y = 3 / 4

Problem on Numbers

A fraction has a denominator greater than its

numerator by 4. but if you add 10 to the denominator,

the value of the fraction would then become 1/8, what

is the fraction? (Caritor)

Problem on Numbers

Solution:

x/(x+4+10) = 1/8

x/(x+14) = 1/8

8x =( x + 14)

7x = 14; hence x = 2

x/(x+4 )= 2/(2+4 )= 2/6

The ages of two persons differ by 10 years. If 5

years ago, the elder one be 2 times as old as the

younger one, find their present ages.

Problems on Ages

Solution:

x - y = 10; x = 10 + y

x - 5 = 2(y-5)

y + 10 -5 = 2y -10

y+5 = 2y -10

2y- y = 15

y=15 and x = 25Their present ages are 15 years and 25 years.

Problems on Ages

The present ages of three persons are in the

proportion of 4:7:9. Eight years ago, the sum

of their ages was 56. Find their present ages ?

Problems on Ages

Solution:

Three person’s ratio = 4:7:9 Total = 4+7+9 = 20

Sum of their age = 56,

after 8 years their sum = 56 +24 = 80

A’s age = 4/20 *80 = 16

B’s age = 7/20 *80 = 28

C’s age = 9/20 *80 = 36

Their present ages are 16, 28 and 36.

Problems on Ages

Problems on Ages

Father’s age is 5 times his son's age.4 years back

the father was 9 times older than his son. Find

the father's present age? (TCS)

Problems on Ages

Solution: F = 5SF – 4 = 9(S-4)F – 5s = 0F – 9S = -36 + 4 = -32 4S = 32S = 8Father age = 40 years

Problems on Ages

One year ago Pandit was three times his sister’s

age. Next year he will be only twice her age.

How old will Pandit be after five years?

(TCS)

Problems on Ages

Solution:(P-1) = 3(S -1)P + 1 = 2( s+1)P – 3S = -3 + 2 = -2P – 2S = 2-1 = 1S = 3P – 3(3) = -2P – 9 = -2P = -2 + 9 = 7After 5 years = 12

Problems on Ages

A father is 30 years older than his son, however

he will be only thrice as old as his son after 5

years what is father’s present age?

Problems on Ages

Solution:

F = S + 30

F + 5 = 3(S+5)

S+30 + 5 = 3S + 15

2S = 20

S= 10

F = 10 + 30 = 40

Problems on Ages

A father is three times as old as his son after 15

years the father will be twice as old as his son’s

age at that time. What is the father’s present

age ? (TCS)

Problems on Ages

Solution:

F = 3S

F + 15 = 2(S +15)

Father’s age = 45, Son’s age = 15

Ratio and Proportion

• Ratio: The Relationship between two

variables is ratio.

• Proportion: The relationship between

two ratios is proportion.


The two ratios are a : b and the sum nos. is x

ax bx -------- and ------- a + b a + b

Similarly for 3 numbers a : b : c


If Rs. 1260 is divided among A, B, C in the

ratio 2 : 3 : 4 what is C’s share?


Solution:

C’s Share = 4/9*1260

C’s share = Rs. 560


To 15 liters of water containing 20% alcohol,

we add 5 liters of pure water. What is the % of

alcohol?


Solution:

15 lit 20 %

20 lit (15+5) x by solving we get

= 15%

15% alcohol


What number should be added or subtracted

from each term of the ratio 17 : 24 so that it

becomes equal to 1 : 2


Solution:

Let the number be x.

17 + x/24 + x = 1/2

Solving the above equation,

The number to be subtracted is 10.


The ratio of white balls and black balls is 1:2. If

9 gray balls are added it becomes 2:4:3. Then

what is the number of black balls ?


Solution: Ratio of all the three balls = 2:4:3Ratio of two balls before adding gray = 1:29 gray ratio =3 3 parts = 9 balls 1 part = 9/3 4 parts =? = 9*4/3 =12 Number of black balls is 12


Rs. 770 was divided among A, B and C such that

A receives 2/ 9th of what B and C together

receive. Find A’s share?


Solution:

A = 2/9 (B+C)

B+C =9A/2

A+B+C = 770

A + 9A/2 = 770

11A = 770*2

A = 140

•(Quantity of cheaper / Quantity of costlier)

(C.P. of costlier) – (Mean price)

= --------------------------------------

(Mean price) – (C.P. of cheaper)

Alligation or Mixture


Cost of Cheaper Cost of costlier c d

Cost of Mixture m

d-m m-c

(Cheaper quantity) : (Costlier quantity) = (d – m) : (m – c)

A merchant has 100 kg of salt, part of which he sells at 7% profit and the rest at 17% profit. He gains 10% on the whole. Find the quantity sold at 17% profit?


Solution: 7 17 10 (17-10) (10-7) 7 : 3The ratio is 7:3The quantity of 2nd kind = 3/10 of 100kg = 30kg


A 3-gallon mixture contains one part of S and

two parts of R. In order to change it to mixture

containing 25% S how much R should be

added?


Answer:

R : S2 : 175% : 25%3 : 1

1 gallon of R should be added.


In an examination out of 480 students 85% of

the girls and 70% of the boys passed. How

many boys appeared in the examination if total

pass percentage was 75%


Solution:70 85 7510 5

Number of Boys = 480 * 10/15Number of Boys = 320


In two varieties of tea, one costing Rs. 25/kg.

and the other costing RS. 30/kg are blended to

produce blended variety of tea in ratio 2:3. find

the cost price of the mixture ?


Solution: 25 30 x

(30- 28) (28-25) 2 : 3 Let mixed price be xIf you subtract 28 from 30 you will get 2 and if you subtract 25 from 28 you will get 3.



A person has Rs. 5000. He invests a part of it

at 3% per annum and the remainder at 8% per

annum simple interest. His total income in 3

years is Rs. 750. Find the sum invested at

different rates of interest.


Solution:

Average rate of interest = 5% per annum

3% 8%

5%

3% 2%

Investment at 3% per annum = 3x5000/5= 3000

Investment at 8% per annum = 2x5000/5=2000

Chain Rule

Direct Proportion :

A B

A B

Indirect Proportion:

A B

A B

Chain Rule

• A Garrison of 500 men had provision for 27 days. After 3 days, a reinforcement of 300 men arrived. The remaining food will now last for how many days?

Chain Rule

Solution:

Men days

500 24

800 x

800X = 500x24

X =(500x24)/800 =15 days

Chain Rule

If 20 men take 15 days to complete a job. In how

many days will 25 men finish the work?

CTS Question

Chain Rule

Solution:

Men Days

20 15

25 x

x/15=20/25

x = (20*15)/25 =12

They will take 12 days

If 11.25m of a uniform iron rod weighs 42.75 kg,

what will be weight of 6m of the same rod?

Chain Rule

Solution:

length ( m ) weight ( kg ) 11.25 42.75

6 x

Since it is a direct proportion, x 6 6 x 42.75 --------- = -------- x = --------------- 42.75 11.25 11.25

The weight of rods x = 22.8 kg

Chain Rule

Chain Rule

A stationary engine has enough fuel to run 12

hours when its tank is 4/5 full. How long will it

run when the tank is 1/3 full?

TCS Question

Chain Rule

Answer:

Tank hours

4/5 12

1/3 x

4/5 x = 12 * 1/3

It will run for 5 hours

Chain Rule

20 men complete one - third of a piece of work

in 20 days. How many more men should be

employed to finish the rest of the work in 25

more days?

Chain RuleSolution: Men days work 20 20 1/3 work done = 1/3 x 25 2/3 remaining = 1-1/3=2/3More work, more men (direct proportion)More days, less men (indirect Proportion)1/3 *X = (2/3 )*20*(20 /25) X = 800/25 = 32More men to be employed = (32-20)12 More people needed to finish the job

Chain Rule

15 men take 21 days of 8 hrs each to do a piece of work. How many days of 6 hrs each would 21 women take, if 3 women do as much work as two men?

Chain Rule

Solution:3 women = 2 men21 women = 14 men

Men Days Hrs15 21 814 x 6x/21=(15/14)/(8/6)x= 30 days

Types:

• A invested Rs.x and B invested Rs.y then

A:B = x : y

• A invested Rs. x and after 3 months B invested

Rs. y then the share is

• A:B = x * 12 : y * 9

Partnership

Sanjiv started a business by investing Rs. 36000. After 3 months Rajiv joined him by investing Rs. 36000. Out of an annual profit of Rs. 37100 find the share of each.

Satyam Question

Partnership

Solution :

36000 * 12 : 36000 * 9

4 : 3Sanjiv’s share of profit = (4*37100)/7 = 21200

profit = 21200

Rajiv’s share of profit = 15900

Partnership

A sum of money is divided among A, B, C such

that for each rupee A gets, B gets 65 paise and c

gets 35 paise if c’s share is Rs. 560. what is the

sum?

Partnership

Solution

A : B : C

100 : 65 : 35

20 : 13 : 7 Total = 20+13+7 = 40

C’ share = 560

7/40 *X =560

X= 3200

Partnership

A starts business with Rs.3500 and 5 months

after B joins A as his partner. After a year the

profits are divided in the ratio of 2:3. How much

did B contribute ?

Partnership

Solution:

A :B =3500*12 : 7X

42000 : 7X = 2: 3

7X * 2 = 42000 *3

X = 42000 * 3/14

X = 9000

B’s contribution is Rs.9000

Partnership

A and B invest in a business in the ratio 3 : 2. If

5% of the total profit goes to charity and A’s

share is Rs. 855 what is the total profit?

Partnership

Solution:

A : B = 3:2

Let Profit be X

X – 5% of X

X- 5X/100 = 95X/100

3/5 * 95X/100 = 285

19X= 28500

X = 26500/19 = 1500

Total profit is Rs.1500

Partnership

Partnership

A and B enter into partnership for a year. A contributes Rs.1500 and B Rs. 2000. After 4 months, they admit C who contributes Rs. 2250. If B withdraws his contribution after 9 months, find their profit share ratio at the end of the year?

Partnership

Solution:

A: B: C = 1500*12: 2000*9: 2250*8

= 18000: 18000: 18000

= 1: 1: 1

Profit share at the end of the year,

1: 1: 1

• If A can do a piece of work in n days,

• then A’s 1 day’s work = 1/n

• If A is thrice as B, then:

Ratio of work done by A and B = 3:1

Time and Work

Pipes and Cisterns

• P1 fills in x hrs. Then part filled in 1 hr is 1/x

• P2 empties in y hrs. Then part emptied in 1 hr

is 1/y

• P1 and P2 both working simultaneously which fills in x

hrs and empties in y hrs resp ( y>x) then net part filled is 1/x – 1/y

• P1 can fill a tank in X hours and P2 can empty the full

tank in y hours( where x>y), then on opening both pipes, the net part empties in hour 1/y -1/x

Pipes and Cisterns

One fast typist types same matter in 2 hours and

another slow typist types the same matter in 3

hours. If both do combine in how much time will

they finish?

TCS Question

Time and Work

Solution:

Fast typist = 1/2 ; slow typist = 1/3 ;

Together:

= 1/2 + 1/3 = 5/6 so 6/5 hrs

The work will be completed in 6/5 Hrs.

Time and Work

A and B can finish a piece of work in 30 days, B and C

in 40 days, while C and A in 60 days .In how many days

A, B and C together can finish the work ?

Time and Work

Solution:A + B = 30 days = 1/30B + C = 40 days = 1/40C +A = 60 days = 1/60All work together A+B+C+B+C+A = 1/30 +1/40 +1/602(A+B+C) = 1/30+1/40+1/60 = (4+3+2) /120 = 9/120*2 = 9/240 = 3/80 = 26 2/3 A, B and C can finish the work in 26 2/3 days

Time and Work

10 men can complete a piece of work in 15 days

and 15 women can complete the same work in 12

days. If all the 10 men and 15 women work

together, In how many days will the work get

completed ?

Time and Work

Solution:10 men = 15 days means 1day work = 1/1515 men = 12 days means 1 day work = 1/1210 men + 15 women = 1/15 + 1/12 = 4+5/60 = 9/60 = 3/20 20/3 days = 6 2/3 days

The work will be completed in 6 2/3 days.

Time and Work

Time and Work

A work done by two people in 24 minutes. One

of them can do this work alone in 40 minutes.

How much time is required to do the same work

by the second person?

TCS Question

Time and Work

Solution :

A and B together = 1/24; A = 1/40; B = ?

= 1/24 – 1/40 = 2/120

= 1/60

The second person will complete in 60 minutes.

A cistern has two taps which fill it in 12 minutes

and 15 minutes respectively. There is also a

waste pipe in the cistern. When all the pipes are

opened, the empty cistern is filled in 20 minutes.

How long will a waste pipe take to empty a full

cistern ?

Time and Work (Pipes)

Solution:This problem is based on 2nd method.All the tap work together = 1/12 + 1/15 - 1/20 = 5/60 + 4/60 – 3/60 = 6/60 = 1/10 The waste pipe can empty the cistern in 10

minutes.


A tap can fill a cistern in 8 hours and another

can empty it in 16 hours. If both the taps are

opened simultaneously, Find the time ( in

hours) to fill the cistern


Solution:

Tap 1 = 1/8 (fill); Tap 2 = 1/16 (empty)

= 1/8 – 1/16

= 1 / 16

Total time taken to fill the cistern = 16 hours



A water tank is 2/5th full. Pipe A can fill the tank

in 10 minutes and the pipe B can empty it in 6

minutes. If both the pipes are open, how long

will it take to empty or fill the tank completely?


Answer :A = 1/10; B = 1/6 = 1/10 -1/6 = - 1/15Empty in 15 minutesTo empty 2/5 of the tank 2/5 * 15 = 6 Time taken (empty)= 6 minutes

Area and Volume

Cube:

• Let each edge of the cube be of length a. then,

• Volume = a3cubic units

• Surface area= 6a2 sq.units.

• Diagonal = √3 a units.

Cylinder:

• Let each of base = r and height ( or length) = h.

• Volume = πr2h

• Surface area = 2 πr h sq. units

• Total Surface Area = 2 πr ( h+ r) units.

Area and Volume

Cone:

• Let radius of base = r and height=h, then

• Slant height, l = √h2 +r2 units

• Volume = 1/3 πr2h cubic units

• Curved surface area = πrl sq.units

• Total surface area = πr (l +r)

Area and Volume

Sphere:

• Let the radius of the sphere be r. then,

• Volume = 4/3 πr3

• Surface area = 4 π r2sq.units

Area and Volume

Circle: A= π r 2

Circumference = 2 π r

Square: A= a 2

Perimeter = 4a

Rectangle: A= l x b

Perimeter= 2( l + b)

Area and Volume

Triangle:

A = ½*base*height

Equilateral = √3/4*(side)2

Area of the Scalene Triangle

S = (a+b+c)/ 2

A = √ s*(s-a) * (s-b)* (s-c)

Area and Volume

What is the cost of planting the field in the form

of the triangle whose base is 2.8 m and height

3.2 m at the rate of Rs.100 / m2

Area and Volume

Solution:

Area of triangular field = 1/2 * 3.2 * 2.8 m2

= 4.48 m2

Cost = Rs.100 * 4.48

= Rs.448.

Area and Volume

Area of a rhombus is 850 cm2. If one of its

diagonal is 34 cm. Find the length of the other

diagonal?

Area and Volume

Solution:

850 = ½ * d1 * d2

= ½ * 34 * d2

= 17 d2

d2 = 850 / 17

= 50 cm

Second diagonal = 50cm

Area and Volume

A grocer is storing small cereal boxes in

large cartons that measure 25 inches by 42 inches

by 60 inches. If the measurement of each small

cereal box is 7 inches by 6 inches by 5 inches then

what is maximum number of small cereal boxes

that can be placed in each large carton ?

Area and Volume

Solution:

No. of small boxes = (25*42*60 ) / ( 7*6*5 )

= 300

300 boxes of cereal box can be placed.

Area and Volume

Area and Volume

If the radius of a circle is diminished by 10%,

what is the change in the area in percentage?

Area and Volume

Solution:

= D1 + D2 – D1*D2 /100

= 10 + 10 – 10*10/100

= 20 -1

= 19%

New area changed = 19%.

Area and Volume

A circular wire of radius 42 cm is bent in the

form of a rectangle whose sides are in the ratio

of 6:5. Find the smaller side of the rectangle?

Area and Volume

Solution:length of wire = 2 πr = (22/7*14*14)cm = 264cmPerimeter of Rectangle = 2(6x+5x) cm = 22xcm 22x =264 x = 12 cmSmaller side = (5*12) cm = 60 cm

Area and Volume

A man is running around a rectangle. It takes

time, 2 times in traveling length than traveling

width and the total perimeter is 300 m. Find the

Area?.

Area and Volume

Solution:Breadth = XLength = 2XArea = 2X*X = 2X2

Perimeter 6X = 300m X = 50mArea = 2x2 =2*50*50Area = 5000 sq.m.

• Probability:

P(E) = n(E) / n(S)

• Addition theorem on probability:

n(AUB) = n(A) + n(B) - n(AB)

• Mutually Exclusive:

P(AUB) = P(A) + P(B)

• Independent Events:

P(AB) = P(A) * P(B)

Probability

There are 19 red balls and One black ball. Ten balls are placed in one jar and remaining in one jar. What is probability of getting black ball in right jar ?

(Infosys -2008)

Probability

Answer:

Probability is 1/2.

Probability

There are 5 red shoes 4 green shoes. If one

draws randomly a shoe what is the probability of

getting a red shoe?

CTS Question

Probability

Answer:

The probability is 5/9

Probability

A bag contains 2 red, 3 green and 2 blue balls

are to be drawn randomly. Two balls are drawn

at random. What is the probability that the balls

drawn contain only blue balls ?

Probability

Answer :

The probability is 1/21

Probability

Probability

Sam and Jessica are invited to a dance party. If

there are 7 men and 7 women in total at the dance

and 1 woman and 1 man are chosen to lead the

dance, what is the probability that Sam and Jessica

will not chosen to lead the dance ?

Probability

Answer:

The Probability of Selecting = 1/7*7 =

1/49

The Probability of not Selecting = 1-1/49

= 48/49

Probability

The letters of the word SOCIETY are placed

in a row. What is the probability that the three

vowels come together?

Probability

Answer:

Required Probability = (5!*3! )/7!

= 1/7

Simple Interest = PNR / 100

Amount A = P + PNR / 100

When Interest is Compound annually:

Amount = P (1 + R / 100)n

Compound Interest = A - P

Simple / Compound

Interest

• Half-yearly C.I.:

Amount = P (1+(R/2)/100)2n

• Quarterly C.I. :

Amount = P (1+(R/4)/100)4n

Simple / Compound Interest

Simple/compound interest

Difference between C.I and S.I for 2 years = P*(R/100)2.

Difference between C.I and S.I for 3 years = P{(R/100)3+3 (R/100)2 }

What is the S.I. on Rs. 3000 at 18% per annum

for the period from 4th Feb 1995 to 18th April 1995

Sathyam Question


Answer:

Time = 24+ 31+17 = 73 days = 73/365 = 1/5

P = 3000; R = 18%;

= PNR/100 = 3000*1*18/100*5

The simple interest is Rs. 108


A sum of money doubles itself at C.I. in 15 years.

In how many years will it become eight times?

Satyam Question


Solution:

A = P(1+R/100)15

2P =P(1+R/100) 15 ; 2 = (1+R/100)15

If A = 8P

8P = P(1+R/100)n

23 = ( 1+R/100)

[(1+R/100) 15]3 = (1+R/100)n

n = 3*15

It will take a period of 45 years.


Raja borrowed a certain money at a certain rate of S.I.

After 5 years, he had to pay back twice the amount

that he had borrowed. What was the rate of interest?

TCS Question


Solution:

A = 2P

A = P + PNR/100

2P = P(1+NR/100)

2 = (1+5*R/100)

1 = R/20

The rate of interest is 20%


Simple/compound interest

In simple interest what sum amounts to Rs. 1120

in 4 years and Rs. 1200 in 5 years?

CTS Question

Simple/Compound interest

Answer : Interest for 1 year = 1200 – 1120 = 80Interest for 4 year = 80*4 = 320A = 1120P = A – P = 1120 – 320The Principal is Rs. 800


A simple interest amount for Rs. 5000 for 6 months is Rs. 200. What is the annual rate of interest?

CTS Question


Solution:

P = 5000; N = 6/12 = ½

I = 200

R = I *100 / P*N

=200*100*2/5000*1

= 40/5 = 8%

The annual rate of interest is 8%


A man earns Rs. 450 as an interest in 2 years on a certain sum invested with a company at the rate of 12% per annum. Find the sum invested.


Solution:

P = I*100/R*N

= 450*100/12*2

Principal = Rs. 1875


If Rs. 85 amounts to Rs. 95 in 3 years, what

Rs. 102 will amount in 5 years at the same rate percent?


Solution:

Let P = Rs. 85; A = Rs. 95; I = 10/3 in 1 year

Rate = I*100/P*N =4% ( app)

Amount = P+PNR/100 = 102+20 =122

Hence the amount in 5 years = Rs. 122


What will be the difference between S.I and C.I on a sum of Rs. 4500 put for 2 years at 5% per annum?


Solution:

C.I – S.I = P (R/100)2

= 4500(5/100)2 = 11.25

Difference = Rs. 11.25


What will be the C.I on Rs. 15625 for 2½ years at 4% per annum?


Solution:

A = P(1+R/100)n

A = 15625 [(1+4/100)2 ( 1+(4*1/2) / 100)]

= 17238

C.I = A – P

C.I = 17238 - 15625

Compound interest = Rs. 1613

Average

• Average is a simple way of representing an

entire group in a single value.

• “Average” of a group is defined as:

X = (Sum of items) / (No of items)

Total temperature for the month of

September is 840C. If the average

temperature of that month is 28C. Find out

the days in the month of September?

Average

Solution

Number of days= 840/28=30 days

Average

The painter is paid x rupees for painting every 10m of a wall and y rupees for painting every extra meter. During one week, he painted 10m on Monday, 13m on Tuesday, 12m on Wednesday, 11m on Thursday and 12m on Friday. What is his average daily earning in rupees for the five day week?

Average

Solution:

Day 1 = x, Day 2 = x+3y, Day 3 = x+2y,

Day 4 = x+y, Day 5 = x+2y

Average = (x+ x+3y+ x+2y+ x+y+ x+2y) / 5

= 5x+8y / 5 = 5x/5 + 8y / 5

Average for 5 days is x+ (8y/5)

Average

Average

The average of 11 observations is 60. If the

average of 1st five observations is 58 and that

of last five is 56, find sixth observation?

Average

Solution:

5 observations average = 58

Sum = 58*5 = 290

Last 5 observation average = 56

Sum = 56*5 = 280

Total sum of 10 numbers = 570 (290 + 280)

Total sum of 11 numbers = 660 (11*60)

6th number = 90 (660 -570)

Average

The average of age of 30 students is 9 years. If

the age of their teacher included, it becomes 10

years. Find the age of the teacher?

Average

Solution:

Total age of 30 students = 30*9 = 270

Including the teacher’s age = 31*10 = 310

Difference is = 310-270 = 40 years

Permutations and Combinations

• Factorial Notation: n! = n(n-1)(n-2)….3.2.1 • Number of Permutations: n!/(n-r)!

• Combinations: n!/r!(n –r)!

A foot race will be held on Saturday. How many

different arrangements of medal winners are

possible if medals will be for first, second and

third place, if there are 10 runners in the race …


Solution:

n = 10

r = 3

n P r = n!/(n-r)!

= 10! / (10-3)!

= 10! / 7!

= 8*9*10

= 720

Number of ways is 720.


To fill a number of vacancies, an Employer must

hire 3 programmers from 6 applicants, and two

managers from 4 applicants. What is total

number of ways in which she can make her

selection ?


Solution:

It is selection so use combination formula

Programmers and managers = 6C3 * 4C2

= 20 * 6 = 120

Total number of ways = 120 ways.


In how many ways can the letters of the word

BALLOON be arranged so that two Ls do not

come together?


Solution:

Total arrangement = 7! / 2!*2! (L and O occurred twice) =1260Ls come together (BAOON) (LL) = 6! / 2! = 3* 4* 5*6* = 360Ls not come together1260 – 360 = 900Number of ways = 900.



A man has 7 friends. In how many ways can

he invite one or more of them to a party?


Solution:

In this problem, the person is going to select his friends for party, he can select one or more person, so

= 7C1 + 7C2+7C3 +7C4 +7C5 +7C6 +7C7

= 127

Number of ways is 127

Percentage

• By a certain Percent, we mean that many

hundredths.

• Thus, x Percent means x hundredths, written

as x%

Percentage

Two numbers are respectively 30% and 40%

less than a third number. What is the second

number as a percentage of first?

Percentage

Solution:Let 3rd number be x.1st number = x – 30% of x = x – 30x/100 = 70x/ 100 = 7x/102nd number = x – 40% of x = x – 40x/100 = 60x/ 100 = 6x/10Suppose 2nd number = y% of 1st number6x / 10 = (y/100 )* 7x /10 y = 600 / 7 y = 85 5/7 Hence it is 85 5/7%

Percentage

After having spent 35% of the money on

machinery, 40% on raw material, 10% on staff,

a person is left with Rs.60,000. What is the total

amount of money spent on machinery and the

raw materials?

Percentage

Solution:Let total salary =100%Spending:Machinery + Raw material + staff = 35%+40%+10% = 85%Remaining percentage = 100 %– 85% = 15%15 % of X = 60000X = 4, 00,000In this 4, 00,000 75% for machinery and raw material = 4, 00,000* 75/100 = 3, 00,000

Percentage

If the number is 20% more than the other, how much percent is the second number less

than the first?

Percentage

Solution:

Let X =20

= X / (100+X) *100%

= 20 /120 *100%

=16 2/3%

The percentage is 16 2/3%

Percentage

An empty fuel tank of a car was filled with A type petrol. When the tank was half empty, it was filled with B type petrol. Again when the tank was half empty, it was filled with A type petrol. When the tank was half – empty again, it was filled with B type petrol. What is the percentage of A type petrol at present in the tank?

Percentage

Solution:Let capacity of the tank be 100 liters. Then,Initially: A type petrol = 100 litersAfter 1st operation:A = 100/2 = 50 liters, B = 50 litersAfter 2nd operation:A = 50 / 2+50 = 75 liters, B = 50/2 = 25 litersAfter 3rd operation:A = 75/2= 37.5 liters, B = 25/2 +50 = 62.5 litersRequired Percentage of Type A is = 37.5%

Percentage

Find the percentage increase in the area of a Rectangle whose length is increased by 20% and breadth is increased by 10%

Percentage

Answer:

Percentage of Area Change=( X +Y+ XY/100)%

=20+10+20*10/100

=32%

Percentage

If A’s income is 40% less than B’s income, then how much percent is B’s income more than A’s income?

Percentage

Answer:

Percentage = R*100%/(100-R) = (40*100)/ (100-40)

=66 2/3%

Percentage

One side of a square is increased by 30%. To maintain the same area by how much percentage the other side will have to be decreased?

Percentage

Answer:

Percentage = r*100%/(100+r)

= (30*100) / 130

= 23 1/3%

Boats and streams

•Up stream – against the stream

•Down stream – along the stream

•u = speed of the boat in still water

•v = speed of stream

•Down stream speed (a)= u+v km / hr

•Up stream speed (b)=u-v km / hr

•u = ½(a+b) km/hr

•V = ½(a-b) km / hr

A man can row Upstream at 12 kmph and

Downstream at 16 kmph. Find the man’s rate in

still water and rate of the current?

Boats and streams

Solution:

Rate in still water = 1/2 (16 + 12) = 14 Kmph

Rate of Current = 1/2 (16 – 12 ) = 2 Kmph

Boats and streams

A Boat is rowed down a river 40 km in 5 hr and

up a river 21 km in 7 hr. Find the speed of the

boat and the river?

Boats and streams

Solution:

Speed of the Boat Downstream = 40/7 = 8 (a)

Speed of the Boat Upstream = 21/7 = 3 (b)

Speed of the Boat = 1/2 ( a + b ) = 1/2 ( 8+3 )

= 5.5 Kmph

Speed of the River = 1/2 ( a – b ) = 1/2 (8 – 3)

= 2.5 kmph

Boats and streams

A boat’s crew rowed down a stream from A to B

and up again in 7 ½ hours. If the stream flows at

3km/hr and speed of boat in still water is 5

km/hr. find the distance from A to B ?

Boats and streams

Solution: Down Stream = Sp. of the boat + Sp. of the stream = 5 +3 =8Up Stream = Sp. of the boat – Sp. of the stream = 5-3 = 2Let distance be XDistance/Speed = Time X/8 + X/2 = 7 ½ X/8 +4X/8 = 15/2 5X / 8 = 15/2 5X = 15/2 * 8 5X = 60 X =12

Boats and streams

Boats and Streams

A boat goes 40 km upstream in 8 hours and 36 km downstream in 6 hours. Find the speed of the boat in still water in km/hr?

Boats and Streams

Solution:

Speed of the boat in upstream = 40/8 = 5 km/hr

Speed of the boat in downstream= 36/6 =6 km/hr

Speed of the boat in still water = (5+6 ) / 2

= 5.5 km/hr

Boats and Streams

A man rows to place 48 km distant and back in 14 hours. He finds that he can row 4 kmph with the stream in the same time as 3 Kmph against the stream. Find the rate of the stream?

Boats and Streams

Solution:Down stream 4 km in x hours. Then,Speed Downstream = 4/x km/hr, Speed Upstream = 3/x km/hr48/ (4/x) + 48/(3/x) = 14x = 1/2Speed of Downstream = 8,Speed of upstream = 6Rate of the stream =1/2 (8-6) km/hr = 1 km/hr

Time and Distance

•Speed:-

• Distance covered per unit time is

called speed.

Speed = distance/time (or)

•Distance = speed*time (or)

•Time = distance/speed

• Distance covered α Time (direct variation).

• Distance covered α speed (direct variation).

• Time α 1/speed (inverse variation).

Time and Distance

• Speed from km/hr to m/sec - (Multi by 5/18).

• Speed from m/sec to km/h, - (Multi by 18/5).

• Average Speed:-

Average speed = Total distance traveled Total time taken

Time and Distance

From height of 8 m a ball fell down and each

time it bounces half the distance back. What

will be the distance traveled?

Sathyam Question

Time and Distance

Solution:

= 8 + 4 + 4+2+2+1+1+0.5+0.5 +….etc.

The total distance traveled is 24 m

Time and Distance

Two cars are 15 km apart. One is running at a

speed of 50 kmph and the other at 40 kmph. How

much time will it take for the two cars to meet?

Sathyam Question

Time and Distance

Solution:

Time taken

=Distance / (S1 – S2)

= 15 / (50 – 40)

= 15 / 10

= 1.5

It will take 1½ hours.

Time and Distance

The center of a storm shifts 22.5 miles in 1 hour. At

the same rate what time will it take to move 60

miles?

TCS Question

Time and Distance

Answer:

For 22.5 miles it takes 1 hour

It means for 60 miles T = D / S

Time taken = 60 / 22.5

It will take 2 2/3 hours.

Time and Distance

Time and Distance

By walking at ¾ of his usual speed, a man

reaches office 20 minutes later than usual.

Find his usual time?

Time and Distance

Solution:

Usual time = Numerator * late time

= 3*20

= 60

Time and Distance

A man on motorcycle rides 110 miles in 330 minutes. What is his average speed in miles per hour?

TCS Question

Time and Distance

Answer:

Speed = D / T =110*60 /330

The average speed = 20 miles/hour

Time and Distance (Trains)

A train starts from Delhi to Madurai and at the

same time another train starts from Madurai to

Delhi after passing each other they complete

their journeys in 9 and 16 hours, respectively.

At what speed does second train travels if first

train travels at 160 km/hr ?


Solution:

Let x be the speed of the second train

S1 / S2 = √T2/T1

160/x = √16/9

160/x = 4/3

x = 120

The speed of second train is 120km/hr.


Two hours after a freight train leaves Delhi a

passenger train leaves the same station traveling in

the same direction at an average speed of 16 km/hr.

After traveling 4 hours the passenger train overtakes

the freight train. What was the average speed of the

freight train? Wipro Question


Solution :

Speed of Passenger train = 16 kmph

Distance = 16*4 = 64

Speed of freight train = Distance / ( S1 + S2 )

= 64 / (4+2)

= 64/6

= 10.6 km/hr

The average speed = 10.6 km/hr


There are 20 poles with a constant distance

between each pole. A train takes 24 sec to

reach the 12 pole. How much time will it take

to reach the last pole ?


Solution:

To cross 11 poles it is taking 24 sec

To cross 19 poles it will take x time

Poles time

11 24

19 x

11x = 19 * 24

x = 19* 24 /11

x = 41.45 sec

It reaches the last pole in 41.45sec


120 m long train crosses the pole after

2½ sec. Find how much time it takes to cross

140 m long platform?

Caritor Question


Solution:

To cross 120 m it is taking 2 ½ sec. (5/2sec)

To cross (120 +140)=260 m it will take x sec

120x = 260*5/2 (apply chain rule)

= 5 5/12

It takes 5 5/12 seconds.


A train X speeding with 120 kmph crossed another train Y, running in the same direction, in 2 minutes. If the lengths of the trains X and Y be 100 m and 200m respectively, what is the speed of train Y?


Solution:Let the speed of train Y be x km/hrRelative Speed of X to Y = (120 –x) km/hr = [(120 –x)*5/18] m/sec =( 600 – 5x) / 18 m/secT = D / Rel. Speed300 / (600 – 5x /18) = 120 ( 2 Minutes ) 5400 = 120 (600 -5x) x = 111 m/sec.

Profit and Loss

• Cost Price. - CP

• Selling Price. - SP

• Profit or Gain. - P = SP – CP

• Loss. - L = CP - SP

.

• Gain% = [(Gain*100)/C.P.]

• Loss% = [(Loss*100)/C.P.]

• S.P. = ((100 + Gain%)/100)C.P.

• S.P. = ((100 – Loss%)/100)C.P.

Profit and Loss

Anu bought a necklace for Rs. 750 and sold it

for Rs. 675. Find her Loss percentage?.

Profit and Loss

Solution:

CP = 750, SP = 675 L = 750 – 675 = 75

Loss% = loss /CP *100

= 75*100/750

= 10%

Loss 10%

Profit and Loss

A shopkeeper bought a watch for Rs. 400 and

sold it for Rs. 500. What is his profit percentage?

TCS Question

Profit and Loss

Solution:

CP = 400; SP = 500 P = 500 – 400 = 100

Profit % = Profit /CP *100

= 100*100/400

= 25%

Profit 25%

Profit and Loss

By Selling 15 Mangoes , a Fruit vendor recovers

the cost price of 20 Mangoes. Find the profit

percentage?

Profit and Loss

Solution:

The expenditure and the revenue are equated,

Percentage of profit =Goods left*100

Goods sold

= ( 5*100 ) /15 = 33.3%

Profit and Loss

Profit and Loss

A shopkeeper loses 7% by selling a cricket ball

for Rs. 31. for how much should he sell the ball

so as to gain 5%

Profit and Loss

Solution:

First case S.P = Rs. 31 and Loss% = 7%

C.P = [100/(100 – loss%)]*S.P = (100*31) / (100-7)

= 100 / 3

Second case, C.P = Rs 100 / 3 and gain% = 5%

S.P = [(100+gain 5%) / 100]*C.P

= [ (100+5 ) / 100] * 100/3

= Rs. 35

Profit and Loss

What is the selling price of a Toy car if the cost

of the car is Rs. 60 and a profit of 10% over

selling price is earned?

CTS Question

Profit and Loss

Solution:

Profit = 60*10/100

= 6

Selling Price = C.P + Profit

= 60 + 6

Selling Price = Rs. 66

Profit and Loss

Find the single discount to a series discount

20%, 10% and 5%.

Profit and Loss

Answer:

SP = [( 80*90*95 )/ 100*100*100 ]* CP

= 0.684 CP

Discount = (1 – 0.684) * 100%

= 0.316 *100 %

Discount = 31.6%

CalendarOdd days:

0 = Sunday

1 = Monday

2 = Tuesday

3 = Wednesday

4 = Thursday

5 = Friday

6 = Saturday

CalendarMonth code: Ordinary year

J = 0 F = 3

M = 3 A = 6

M = 1 J = 4

J = 6 A = 2

S = 5 O = 0

N = 3 D = 5

Month code for leap year after Feb. add 1.

Calendar

Ordinary year = (A + B + C + D )-2

-----------------------take remainder

7

Leap year = (A + B + C + D) – 3

------------------------- take remainder

7

What is the day of the week on 30/09/2007?

Calendar

Solution:

A = 2007 / 7 = 5

B = 2007 / 4 = 501 / 7 = 4

C = 30 / 7 = 2

D = 5

( A + B + C + D )-2

= -----------------------

7

= ( 5 + 4 + 2 + 5) -2

----------------------- = 14/7 = 0 = Sunday

7

Calendar

What was the day of the week on 13th May,

1984?

Calendar

Solution:

A = 1984 / 7 = 3

B = 1984 / 4 = 496 / 7 = 6

C = 13 / 7 = 6

D = 2

( A + B + C + D) -3

= -----------------------

7

= 14/7= 0, Sunday.

Calendar

Calendar

On what dates of April 2005 did Sunday fall?

CalendarSolution: You should find for 1st April 2005 and then you find the Sundays

date.A = 2005 / 7 = 3B = 2005 / 4 = 501 / 7 = 4C = 1 / 7 = 1D = 6 (A + B + C + D) -2 = ----------------------- 7 3 + 4 + 1 + 6 -2 ----------------------- = 12 / 7 = 5 = Friday. 7 1st is Friday means Sunday falls on 3, 10, 17, 24

Calendar

What was the day on 5th January 1986?

CalendarSolution:

A = 1986 / 7 = 5B = 1986 / 4 = 496/7 = 6C = 5 / 7 = 5D = 0 (A + B + C + D) -2 = ----------------------- 7 5 + 6 + 5 + 0-2 = ----------------------- = 14 / 7 = Sunday 7

Clock:

•In every Hour, the minute hand gains 55 minutes on the hour hand

•In every hour both the hands coincide once. = (11m/2) – 30h (hour hand to min hand)

= 30h – (11m/2) (min hand to hour hand)

•If you get answer in minus, you have to subtract your

answer with 360 o

Clocks

Clocks

Find the angle between the minute hand and

hour hand of a clock when the time is 7:20.

Solution:

= 30h – (11m/2)

= 30 (7) – 11 20/2

= 210 – 110

= 100

Angle between 7: 20 is 100o

Clocks

Clocks

How many times in a day, the hands of a

clock are straight?

Clocks

Solution:

In 12 hours, the hands coincide or are in opposite direction 22 times a day.


Clocks

How many times do the hands of a clock

coincide in a day?

ClocksSolution:



Clocks

At what time between 7 and 8 o’clock will the

hands of a clock be in the same straight line but,

not together?

Clocks

Solution: h = 7

= 30h – 11m/2

180 = 30 * 7 – 11 m/2

On simplifying we get ,

5 5/11 min past 7

Clocks

At what time between 5 and 6 o’clock will the

hands of a clock be at right angles?

Clocks

Solution: h = 5

90 = 30 * 5 – 11m/2

Solving

10 10/11 minutes past 5

Clocks

Find the angle between the two hands of a clock

at 15 minutes past 4 o’clock

Clocks

Solution:

Angle = 30h – 11m/2

= 30*4 – 11*15 / 2

The angle is 37.5o

Clocks

At what time between 5 and 6 o’clock are the

hands of a clock together?

Clocks

Solution: h = 5

O = 30 * 5 – 11m/2

m = 27 3/11

Solving

27 3/11 minutes past 5

In interpretation of data, a chart or a graph is

given. Some questions are given below this chart

or graph with some probable answers. The

candidate has to choose the correct answer from

the given probable answers.

Data Interpretation

• 1. The following table gives the distribution of students according to

professional courses:

__________________________________________________________________

Courses Faculty

___________________________________

Commerce Science Total

Boys girls Boys girls

___________________________________________________________

• Part time management 30 10 50 10 100

• C. A. only 150 8 16 6 180

• Costing only 90 10 37 3 140

• C. A. and Costing 70 2 7 1 80

__________________________________________________________________

• On the basis of the above table, answer the following questions:

The percentage of all science students over

Commerce students in all courses is

approximately:

(a) 20.5 (b) 49.4

(c) 61.3 (d) 35.1

Data Interpretation

Answer:

Percentage of science students over commerce

students in all courses = 35.1%

Data Interpretation

What is the average number of girls in all

courses ?

(a) 15 (b) 12.5

(c) 16 (d) 11

Data Interpretation

Answer:

Average number of girls in all courses = 50 / 4

= 12.5

Data Interpretation

What is the percentage of boys in all courses

over the total students?

(a) 90 (b) 80

(c) 70 (d) 76

Data Interpretation

Answer:

Percentage of boys over all students

= (450 x 100) / 500

= 90%

Data Interpretation

Data Sufficiency

Find given data is sufficient to solve the problem or not.

A.If statement I alone is sufficient but statement II alone is not sufficient

B.If statement II alone is sufficient but statement I alone is not sufficient

C.If both statements together are sufficient but neither of statement alone is sufficient.

D.If both together are not sufficient

Data Sufficiency

What is John’s age?

I. In 15 years will be twice as old as Dias would be

II. Dias was born 5 years ago. (Wipro)

Data Sufficiency

Answer:

c) If both statements together are sufficient but neither of statement alone is sufficient.

Data Sufficiency

What is the distance from city A to city C in kms?

I. City A is 90 kms from city B.

II.City B is 30 kms from city C

Data Sufficiency

Answer:

d) If both together are not sufficient

Data Sufficiency

If A, B, C are real numbers, Is A = C?

I. A – B = B – C

II. A – 2C = C – 2B

Data Sufficiency

Answer:

D . If both together are not sufficient

Data Sufficiency

What is the 30th term of a given sequence?

I. The first two term of the sequence are 1, ½

II. The common difference is -1/2

Data Sufficiency

Answer:

A. If statement I alone is sufficient but statement II alone is not sufficient

Data Sufficiency

Was Avinash early, on time or late for work?

I. He thought his watch was 10 minute fast.

II. Actually his watch was 5 minutes slow.

Data Sufficiency

Answer:

D. If both together are not sufficient

Data Sufficiency

What is the value of A if A is an integer?

I. A4 = 1

II. A3 + 1 = 0

Data Sufficiency

Answer:

B. If statement II alone is sufficient but statement I alone is not sufficient

Cubes

A cube object 3”*3”*3” is painted with green

in all the outer surfaces. If the cube is cut into

cubes of 1”*1”*1”, how many 1” cubes will

have at least one surface painted?

Cubes

Answer:

3*3*3 = 27

All the outer surface are painted with colour.

26 One inch cubes are painted at least one surface.

Cubes

A cube of 12 mm is painted on all its sides. If

it is made up of small cubes of size 3mm, and if

the big cube is split into those small cubes, the

number of cubes that remain unpainted is

Cubes

Answer:

= 8

Cubes

A cube of side 5 cm is divided into 125 cubes of equal size. It is painted on all 6 sides.

1. How many cubes are coloured on only one side?

2. How many cubes are coloured on only two side?

3. How many cubes are coloured on only three side?

4. How many cubes are not coloured?

Cubes

Answer:

1. 54

2. 36

3. 8

4. 27

Cubes

A cube of 4 cm is divided into 64 cubes of

equal size. One side and its opposite side is

coloured painted with orange. A side adjacent

to this and opposite side is coloured red. A side

adjacent to this and opposite side is coloured

green? Cont..

Cubes

1. How many cubes are coloured with Red alone?

2. How many cubes are coloured orange and Red alone?

3. How many cubes are coloured with three different colours?

4. How many cubes are not coloured?

5. How many cubes are coloured green and Red alone?

Cubes

Answer:

1. 8

2. 8

3. 8

4. 8

5. 8

CubesA 10*10*10 cube is split into small cubes of equal

size 2*2*2 each. A side and adjacent to it is coloured Pink. A side adjacent to Pink and opposite side is coloured Blue. The remaining sides are coloured yellow.

1. Find the no. of cubes not coloured?2. Find the no. of cubes coloured blue alone?3. Find the no. of cubes coloured blue & pink &

yellow?4. Find the no. of cubes coloured blue & pink ?5. Find the no. of cubes coloured yellow & pink ?

Cubes

Answer:

1. 27

2. 18

3. 4

4. 12

5. 12

Venn Diagram

If X and Y are two sets such that X u Y has 18

elements, X has 8 elements, and Y has 15

elements, how many element does X n Y have?

Venn Diagram

Solution:

We are given n (X uY) = 18, n (X) = 8, n (Y) =15. using the formula.

n( X n Y) = n (X) + n (Y) - n ( X u Y)

n( X n Y) = 8 + 15 – 18

n( X n Y) = 5

Venn Diagram

If S and T are two sets such that S has

21elemnets, T has 32 elements, and S n T has

11 elements, how many element elements does

S u T have?

Venn Diagram

Answer:

n (s) = 21, n (T) = 32, n ( S n T) = 11,

n (S u T) = ?

n (S u T) = n (S) + n( T) – n (S n T)

= 21 + 32 – 11 = 42

Venn Diagram

If A and B are two sets such that A has 40

elements, A u B has 60 elements and A n B

has 10 elements, how many element elements

does B have?

Venn Diagram

Answer:

n ( A) = 40, n ( n B) = 60 and n ( A n B) = 10,

n ( A u B) = n ( A) + n (B) – n ( A n B)

60 = 40 + n (B) – 10

n (B) = 30

Venn Diagram

In a group of 1000 people, there are 750 people

who can speak Hindi and 400 who can speak

English. How many can Speak Hindi only?

Answer:

n( H u E) = 1000, n (H) = 750, n (E) = 400,

n( H u E) = n (H) + n (E) – n( H n E)

1000 = 750 +400 – n ( H n E)

n ( H n E) = 1150 – 100 = 150

No. of people can speak Hindi only

_

= n ( H n E) = n ( H) – n( H n E)

= 750 – 150 = 600

Venn Diagram

In a class of 100 students, the number of students passed in English only is 46, in maths only is 46, in commerce only is 58. the number who passes in English and Maths is 16, Maths and commerce is 24 and English and commerce is 26, and the number who passed in all the subject is 7. find the number of the students who failed in all the subjects.

Venn Diagram

Solution:

No. of students who passed in one or more subjects

= 11+ 9 + 13 + 17 + 15 + 19 + 7 = 91

No of students who failed in all the subjects

= 100 -91 = 9

Venn Diagram

In a group of 15, 7 have studied Latin, 8 have

studied Greek, and 3 have not studied either.

How many of these studied both Latin and

Greek?

Venn Diagram

Answer:

3 of them studied both Latin and Greek.

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