Blue Lotus Aptitude Numerical Reasoning. Problems on Numbers Problems on Ages Ratio and Proportion Alligation or Mixture Chain Rule Partnership Venn Diagram

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

• Cubes

Numerical Reasoning

• Boats and Streams

• Time and Distance (Trains)

• Data Sufficiency

• Profit and Loss

• Calendar

• Clocks

• Data Interpretation

Numerical Reasoning

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); a = 1st term , r = 1st term / 2nd term

Problems on Numbers

Basic Formulae

1. ( a+b)2 = a2 + b2 + 2ab2. (a-b)2 = a2 +b2 -2ab3. ( a+b)2 - (a – b)2 = 4ab4. (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)9. (a3+b3+c3 -3abc) = (a+b+c) (a2+b2+c2-ab-bc-ca) If a+b+c = 0, then (a3+b3+c3) =3abc

Problems on Numbers

A man ate 100 Bananas in 5 days, each day eating 6 more than the previous day. How many bananas did he eat on the first day?

Problems on Numbers

Solution:

First day be x

Then x+6, x+12, x+18, x+24

5x + 60 = 100

x=8

In first day he ate 8 Bananas.

Problems on Numbers

The traffic light at three different road crossings

change after every 48 seconds, 72 seconds, and 108

seconds respectively. If they all change

simultaneously at 10:20:00 hours then at what time

they again change simultaneously?

Problems on Numbers

Solution:

LCM (48,72,108) = 432 seconds

= 7 minutes 12 seconds

The next change will be at 10:27:12 hours

Problems on Numbers

Raja had 85 currency notes in all, some of which

were of Rs. 100 denomination and the remaining

of Rs. 50 denomination. The total amount of all

these currency notes was Rs. 5000. How much

amount did she have in the denomination of Rs.

50?

Problems on Numbers

Solution:

Let x be the number of Rs. 50 notes

50x + 100 (85-x) = 5000

Solving the above equation,

Amount = Rs. 3500

Problems on Numbers

Mr. Raja is on tour and he has Rs. 360 for his

expenses. If he exceeds his tour by 4 days, he

must cut down his daily expenses by Rs. 3. For

how many days is Mr. Raja on tour?

Problems on Numbers

Solution: Let x be the number of days

360 / x – 360 / (x+4) = 3


The number of days of the tour x = 20 days

Problems on Numbers

If the numerator of a fraction is increased by

25% and the denominator is decreased by 20%

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

Problems on Numbers

Solution:

NR = x + 25x/100 DR = y -20y/100

(5x/4)/(4y/5) = 5/4

Solving the equation,

The fraction is 4/5

Problem on Numbers

There is a circular pizza with negligible

thickness that is cut into X pieces by 4 straight

line cuts. What is the maximum and minimum

value of x respectively?

Problem on Numbers

Answer:

Maximum = 11.

Minimum = 5.

Problem on Numbers

A man was engaged on a job for 30 days on

condition that he would get a wage of Rs. 10 for

the day he works but he has to pay a fine of Rs.

2 for each day of his absence. If he gets Rs. 216

at the end. Find the number of days he was

absent?

Problem on Numbers

Solution:

10x – 2(30 –x) = 216

x = 23

Absent for = (30 – 23) = 7 days

Problems on Numbers

I bought a car with a peculiar 5-digit number

license plate which on reversing could still be

read. On reversing the value is increased by

78633. What is the original number if all the

digits were different from 0 - 9?

Problems on Numbers

Solution:

Only 0,1,6,8,9 can be reversed. Hence the

number is 89601. It’s reverse is 10968 and

difference is 78633

Problems on Numbers

Naval collected 8, spiders and beetles in a little

box. When he counted the legs he found that

they were altogether 54. How many beetles and

how many spiders did he collect?

Problems on Numbers

Solution:

Spiders have 8 legs and beetles have 6 legs.

S + B = 8, 8S + 6B = 54.

Solving the above equations

The number of spiders = S = 3

The number of beetles = B = 5

Father’s age is three times the sum of the

ages of his two children, but twenty years

hence his age will be equal to sum of their

ages ?

Problems on Ages

Solution:

Father age = 3(x+y)F+20 = x+y+403x+3y+20 = x+y+403x-x+3y – y =202x+2y = 20 x +y = 10F = 3*10 =30 The father’s age is 30.

Problems on Ages

Problems on Ages

A man’s age is 125% of what it was 10 years

ago, but 83 1/3% of what it will be after ten

years. What is his present age?

Problems on Ages

Solution:Let the present age be x years,Then, 125% of (x-10) = x83 1/3% of (x+10) = x125% of (x-10) = 83 1/3 % of (x+10)5/4 (x-10) = 5/6 (x+10)5x/12 = 250 / 12 x = 50 years

Problem on Ages

My age three years hence multiplied by 3 and

from that subtracted three times my age three

years ago will give you my exact age. Find my

age?

Problem on Ages

Solution:

(x+3)3 – 3(x-3) = x

x =18

Problem on Ages

A boy asks his father “ What is the age of grand

father? Father replied “ He is x years old in x2

years” and also said “ We are talking about 20 th

century” What is the year of birth of grand

father?

Problem on Ages

Solution:

20th century means 1900 – 2000 year

Perfect square 44 * 44 = 1936 (present)

Year of birth = 1936 - 44 = 1892

Problem on Ages

A wizard named Nepo says “ I am only three

times my son's age. My father is 40 years more

than twice my age. Together the three of us are a

mere 1240 years old. How old is Nepo?

Problem on Ages

Solution:

N = 3S

F = 40 + 2N

F + N + S = 1240

Solve 3 equations

Age of Nepo = 360 years

Problems on Ages

Joe’s father will be twice his age 6 years from

now. His mother was twice his age 2 years

before. If Joe will be 24 two years from now,

what is the difference between his father’s &

mothers age? (TCS)

Problems on AgesSolution:F + 6 = 2(J +6)M-2 = 2(J-2)Joe is now 22, F = 2(22) +12 – 6 = 50M = 2(22) – 4+2 = 42Difference = 50 -42 =8

Problems on Ages

One year ago the ratio of Baskaran’s and

Saraswati’s age was 6 : 7 respectively. Four

years hence, this ratio would become 7 : 8. How

old is Saraswathi?

Problems on Ages

Solution:

One year ago,

Baskaran’s age = 6x Saraswathi’s age = 7x

Four years hence

[(6x + 1) + 4] / [(7x+1) + 4] = 7 / 8

Simplifying x =5

Saraswathi’s present age is 7x +1 = 36 years

Puzzle

Can you make 120 by using 5 zeros?

Puzzle

Solution:

Fact (0! + 0! + 0! + 0! + 0!)

= Fact (5)

= 5!

= 120

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


A sporting goods store ordered an equal

number of white and yellow balls. The tennis

ball company delivered 45 extra white balls

making the ratio of white balls to yellow balls

1/5 : 1/6. How many white tennis balls did the

store originally order for? (TCS Question)


Solution: Let the number of yellow balls be x

(x + 45) : x = 1/5 : 1/6


The number of white balls originally ordered

would be = 225 balls


The ratio of the rate of flow of water in pipes

varies inversely as the square of the radius of

the pipes. What is the ratio of the rates of flow

in two pipes of diameters 2 cm and 4 cm

respectively?


Solution:R1 = 1st PipeR2 = 2nd PipeR1 α 1/r12

R2 α 1/r22

R1: R2 = 1/r12: 1/r22

= 1/12: 1/22

= 1/1: 1 / 4 = 4:1 Ratio of rates of flow is 4:1


I participated in a race. 1/5th of the

participants were before me and 5/6th of

them behind me. Find the total number of

participants.

(Infosys Question)


Solution:

Let the total number of participants be x.

(x – 1)/5 + 5 (x-1)/6 = x


The total number of participants would be = 31


If A : B = 2 : 3, B : C = 4 : 5 and C : D = 6 : 7,

find A : D?


Solution :

(A /B *B/C*C/D) =2/3*4/5*6/7

A : D = 16 : 35


A bag contains 50 paise, 25 paise and 10 paise

coins in the ratio 5 : 9 : 4, amounting to Rs. 206.

Find the number of coins of each type.


Solution : The actual ratio would be5 * 50/100 : 9 * 25/100 : 4 * 10/100 = 50 : 45 : 8Value of 50 paise coins=206*50/103=10025paise coins =206*45/103=9010paise coins =206*8/103 =16Deriving from this ratio,Number of 50 paise coins = 100*2=200Number of 25 paise coins = 90*4 =360Number of 10 paise coins = 16*10 =160


There are 20 poles with a constant distance

between each pole. A car takes 24 seconds to

reach the 12th pole. How much time will it take

to reach the last pole?


Solution:

11 : 19 = 24 : x

By Solving

x = 41.45 seconds


Annual incomes of A and B is in the ratio 5 : 4

and their annual expenses bear a ratio of 4 : 3. If

each of them saves Rs. 500 at the end of the year

then find the annual income.


Solution:

IncomeA : B = 5x : 4x

Expenses A : B = 4y : 3y

5x – 4y = 500 and 4x – 3y = 500

Solving the above equations,

A’s annual income = Rs. 2500

B’s annual income = Rs. 3500

Puzzle

Can you draw

three concentric circles with

a line passing through their center

without lifting hand?

•(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 quantity of 2nd kind = 3/10 of 100kg = 30kg


In what ratio two varieties of tea one costing Rs. 27 per kg and the other costing Rs. 32 per kg should be blended to produce a blended variety of tea worth Rs. 30 per kg. How much should be the quantity of second variety of tea, if the first variety is 60 kg?


Solution:

27 32

30

2 3

Quantity of cheaper tea = 2

Quantity of superior tea 3

Quantity of cheaper tea =2*x/5 = 60 , x=150

Quantity of superior tea = 3 * 150/5 = 90 kg



A man buys two cows for Rs. 1350 and sells one

so as to lose 6% and the other so as to gain 7.5%

and on the whole be neither gains nor loss. What

does each cow cost?


Solution:

-6 7.5

0

7.5 6

Ratio = 5 : 4

Cost of first cow = 1350*5/9=Rs. 750

Cost of second cow = Rs. 600

Three types of tea A,B,C costs Rs. 95/kg, Rs.

100/kg. and Rs 70/kg respectively. How many

kg of each should be blended to produce 100

kg of mixture worth Rs.90/kg given that the

quantities of B and C are equal?

Sathyam Question


Answer:B+C/2 A85 95

905 5

Ratio is 1:1 so A = 50 , B + C = 50

The quantity would be 50 : 25 : 25


In what proportion water must be added to

spirit to gain 20% by selling it at the cost price?


Solution:Profit%=20%Let C.P =S.P= Rs.10 Then CP=100/(100+P%)SP =25/3

0 10 25/3

5/3 25/3

The ratio is 1: 5


Chain Rule

Direct Proportion :

A B

A B

Indirect Proportion:

A B

A B

Chain Rule

If 36 men can do a piece of work in 25 days then

in how many days 15 men will do it?

Chain Rule

Solution:

Men Days

36 25

15 x (Indirect)

x = 36 * 25/15

x = 60 days

If 12 carpenters working 6 hours a day can make

460 chairs in 24 days, how many chairs will 18

carpenters make in 36 days, each working 8 hours

a day?

Chain Rule

Solution:

Men Days Hours Chairs

12 24 6 460

18 36 8 x

x/460=18*8*36/12*6*24

x=1380

They will make 1380 chairs.

Chain Rule

Chain Rule

If 1 man can load 1 box in a truck in 9 minutes

and a truck can hold up to 8 boxes. How many

trucks completely are loaded when 18 men

work for 1.5 hours? ( CTS )

Chain Rule

Solution:

Men Box Min Trucks

1 1 9 1

18 8 90 x

x = 18 *1* 90/1*8 * 9 = 22.5

The number of trucks completely loaded = 22

Chain Rule

A garrison of 3300 men has provisions for 32 days when given at a rate of 850 grams per head. At the end of 7 days reinforcement arrives and it was found that now the provisions will last 8 days less when given at the rate of 825 grams per head. How many more men can it feed? (Sathyam question)

Chain Rule

Solution:Men Days Grams3300 17 850x 25 825

x = 3300 * 25 * 850/17 * 825 = 50005000 – 3300 = 1700

It can feed 1700 more men.

Chain RuleIf 3 men or 6 boys can do a piece of work in

10 days, working 7 hours a day. How many

days will it take to complete a piece of work

twice as large with 6 men and 2 boys working

together for 8 hours a day?

Chain Rule

Solution:One man’s work = 2 boys’ work

So, 6 men and 2 boys = 6 (2) + 2 = 14 boysBoys Hours Days Work6 7 10 114 8 x 2

x = 6 * 7 * 10 * 2/14 * 8They will finish the work in 7 ½ 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

Partnership

A and B are two partners in a business. A

contributes Rs. 1200 for 5 months and B Rs. 750

for 4 months. If total profit is Rs. 450, find the

respective shares.

Partnership

Solution:

Ratio is1200 * 5 : 750 * 4 = 2 : 1

Profit share of A = 2/3 * 450 = Rs. 300

Profit share of B = Rs. 150

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:

Let the total profit be Rs. 100

After paying charity A’s share = 3/5 *95 = 57

If A’s share is Rs. 57, the total profit is 100

If A’s share is Rs. 855, the total profit is

100 * 855/57 = Rs. 1500

The total profit = Rs. 1500

Partnership

A,B,C entered into a partnership by making an

investment in the ratio of 3 : 5 : 7. After a year

C invested another Rs. 337600 while A

withdrew Rs. 45600. The ratio of investments

then changed into 24: 59 : 167. How much did A

invest initially?

Partnership

Solution:

Let the investments of A, B, and C be 3x, 5x, 7x

(3x – 45600) : 5x : (7x + 337600) = 24 : 59 : 167

(3x – 45600)/5x = 24/59

x = 47200

Initial investment of A = 47200 * 3 = Rs. 141600

A,B,C started a business each investing Rs.

20000. After 5 months A withdrew Rs. 5000, B

withdrew Rs. 4000 and C invested Rs. 6000

more. At the end of the year a total profit of Rs.

69900 was recorded. Find the share of each.

(Satyam Question)

Partnership

Solution:

(20000*5+15000*7) : (20000*5+16000*7) :

(20000*5 + 26000*7)

The ratio is 205 : 212: 282

A’s share = 205/699 * 69900 = Rs. 20500

B’s Share = Rs. 21200

C’s share = Rs. 28200

Partnership

A starts business with a capital of Rs. 1200. B

and C join with some investments after 3 and 6

months respectively. If at the end of a year the

profit is divided in the ratio 2 : 3 : 5 respectively

what is B’s investment in the business?

Partnership

Solution:

(1200 * 12) : ( x * 9) : (y * 6) = 2 : 3 : 5

(1200 * 12)/2 = 9x/3

x = 2400

B’s investment = Rs. 2400

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 at the end of the year? (In

the ratio)

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

Ratio of times taken by A and B= 1:3

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

A can do a piece of work in 8 days. A undertook

to do it for Rs. 320. With the help of B, he

finishes the work in 6 days. Find B’s share.

Time and Work

Solution:

(A + B)’s 1 day work = 1/6

A’s 1 day work = 1/8

B’s 1 day work = 1/24

Ratio of their work = 3 : 1

B gets ¼ * 320 = Rs. 80

Time and Work

5 men or 8 women do equal amount of work in a day.

A job requires 3 men and 5 women to finish the job

in 10 days. How many women require to finish the

job in 14 days?

Satyam Question

Time and Work

Solution:

5 men = 8 women, so 1 man = 8/5 women

3m + 5w = 3 (8/5) w + 5 w = 49/5 w

Days Women

10 49/5

14 x

The number of women required = 7

Time and Work

A can do a piece of work in 30 days, and B in 50

days and C in 40 days. If A is assisted by B on

one day and by C on the next day alternatively

work will be completed in ?

Time and Work

Solution:

A = 1/30, B = 1/50, C = 1/40

A+B = 1/30+1/50 = 8/150

A+C = 1/30+1/40 = 7/120

2 days work done by A & B = 8/150 + 7/120) = 67/600

16 days work done = 67 * 8/600 = 536/600 = 67/75

Work left = 1-67/75 = 8/75

17th day A & B are working = 8/75 – 8/150 = 4/75

18th day A&C are working 4/75 = 120/7* 4/75 = 32/35

They will finish the work in 17 + 32/35 = 17 32/35 days

Time and Work

Time and Work

Two men undertake to do a piece of work for

Rs.1400. First man alone can do this work in

7 days while the second man alone can do

this work in 8 days. If they working together

complete this work in 3 days with the help of

a boy, how should money be divided?

Time and Work

Solution:Wages of the first man for 3 days= work done by him in 3 days * Rs. 1400= 3/7 * 1400 = Rs. 600Wage of second man for 3 days = work done by him in 3 days * Rs. 1400 = 3 / 8 *1400 = Rs. 525Wages of the boy for 3 days = Rs. 1400 – Rs. (600 + 525) = Rs. 275 There shares will be 600, 525, 275.

Time and Work

2 men and 3 boys can do a piece of work in 10

days while 3 men and 2 boys can do the same

work in 8 days. In how many days can 2 men

and 1 boy do the work?

Satyam Question

Time and Work

Answer:

1 man’s work = x

1 boy’s work = y

2x + 3y = 1/10 and 3x + 2y = 1/8

Solving the above equations,

The required number of days = 12½ days

Pipes A and B running together can fill a cistern

in 6 minutes. If B takes 5 minutes more than A

to fill the cistern then find the time in which A

and B will fill the cistern separately.

Time and Work (Pipes)

Answer:

1/x + 1/(x+5) = 1/6


Pipe A will fill the cistern in 10 minutes.

Pipe B will fill the cistern in 15 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 ?


Solution:

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.


One fill pipe A is 3 times faster than second

fill pipe B and takes 32 minutes less than the

fill pipe B. when will the cistern be full if both

pipes are opened together ?


Solution:B = xA = 3xB = x (1/3) = x/3x – 32 =x/33x – 96 = x3x – x = 962x = 96x = 96/2 = 48B = 48 -32 = 16A and B together = 1/16 + 1/48 = 1/12

The cistern will be filled in 12 minutes when both the pipes are opened together.



Two pipes A and B can fill a tank in 20 min. and

40 min. respectively. If both the pipes are opened

simultaneously, after how much time A should

be closed so that the tank is full in 10 minutes?


Solution:Let B be closed after x minutes. Then,Part filled by (A+B) in x min + part filled by A in( 10 – x min)=1x (1/20 +1/40) + (10 – x) * 1/20 = 1x (3/40) + 10 – x/20 = 13x/40 + 10 –x/ 20 = 13x + 20 – 2x = 403x – 2x = 40 -20x = 20 min. A must be closed after 20 minutes.


Two taps A and B can fill a tank in 6 hours and

4 hours respectively. If they are opened on alternate

hours and if pipe A is opened first, in how many

hours, the tank shall be full?


Solution:A’s work in hour = 1/6, B’s work in 1 hour = 1/4(A + B)’ s 2 hr work = 1/6+1/4 = 5/12(A + B)’ s 4 hr work = 10/12 = 5/6Remaining Part = 1- 5/6 = 1/6Now, it is A’s turn and 1/6 part is filled by A in

1 hour.Total time = 4+1 =5

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 = πr l 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 = 1/2*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 = ½ * 3.2 * 2.8 m2

= 4.48 m2

Cost = Rs.100 * 4.48

= Rs.448..

Area and Volume

Area and Volume

Find the length of the longest pole that can be

placed in a room 14 m long, 12 m broad, and 8

m high.

Area and Volume

Solution:

Length of the longest pole = Length of the diagonal of the room

= √(142 + 122 + 82)

= √ 404 = 20.09 m

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 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 area in percentage?

Area and Volume

Solution:

= x + y + xy/100

= -10 - 10 + 10*10/100

= -19%

Diminished area = 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

= 264cm

Perimeter of Rectangle = 2(6x+5x) cm

= 22xcm

22x =264 x = 12 cm

Smaller side = (5*12) cm = 60 cm

Area and Volume

A farmer owns a square field with a pole in one

of the corners to which he tied his cow with a

rope whose length is about 10 m. What is the

area available for the cow to graze?

(Caritor Question)

Area and Volume

Answer:

Area = 78.5 m2

Area and Volume

If the length of a rectangle is reduced by 20%

and breadth is increased by 20%. What is the

percentage change in the area?

(Infosys Question)

Area and Volume

Solution:

x + y + (xy/100)%

= - 20 + 20 – 400/100

= -4

The area would decrease by 4%

Area and Volume

A power unit is there by the bank of the river of

750 m width. A cable is made from power unit to

power a plant opposite to that of the river and

1500 m away from the power unit. The cost of

the cable below water is Rs. 15 per meter and

cost of cable on the bank is Rs. 12 per meter.

Find the total cost. (TCS Question)

Area and Volume

Answer:

Cable under water is 750m, on the bank 750m

(750 * 15) + (750 * 12) = 20250

Total cost = Rs. 20250

• 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) Where P(A B)=0

• Independent Events:

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

Probability

There are 6 pairs of gloves of different sizes. In how

many ways can you choose one for the left hand and

one for the right hand such that they are not of the

same pair?

(Caritor Question)

Probability

Solution:

The number of possibilities are 6 * 6 = 36

They should not be from the same pair = 36 - 6

=30

30 ways.

Probability

A speaks the truth 80% of the times, B speaks

the truth 60% of the times. What is the

probability that they tell the truth at the same

time?

(Wipro Question)

Probability

Answer :

P(Both tell truth) = P(A) * P (B)

= 0.8 * 0.6

P = 0.48

Probability

A group consists of equal number of men and

women. Of them 10% of men and 45% of women

are unemployed. If a person is randomly selected

from the group find the probability for the selected

person to be an employee.

(Satyam Question)

Probability

Solution:

Let the number of men is 100 and women be 100

Employed men and women = (100-10)+(100-45)

= 145

Probability = 145 / 200 = 29 / 40

Probability

Probability

The probability of an event A occurring is 0.5 and that of B is 0.3. If A and B are mutually exclusive events. Find the probability that neither A nor B occurs?

Probability

Solution:

It is Mutually exclusive events P(A n B)=0

Probability = 1 – ( P(A) + P (B) – P(A n B) )

= 1 – (0.5 + 0.3 – 0)

= 0.2

Simple Interest = PNR / 100

Amount A = P + PNR / 100

When Interest is Compound annually:

Amount = P (1 + R / 100)n

C.I = 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 }

If the compound interest on a certain sum for

two years is Rs.60.60 and simple interest is

Rs. 60, then the rate of interest per annum is ?


Solution:

SI for 2 years

60 = P*2*R/100

PR/ 100 = 30

C.I – S.I = P (R/100)2

P(R/100 * R/100) = 0.60

30 * R/100 =0.6 = 2%

The rate of interest is 2%


Find in what time a given sum of money will

quadruple itself at simple interest at the rate

of one paise per rupee per month ?


Solution:

4P after n years

4P = P+ P*n*12/100

4 = 1+ 12n/100

n = 25 years

The period is 25 years.


If Rs. 1000 is invested at 5% simple interest if

the interest is added to the principal every ten

years, the amount will become Rs. 2000 after

how many years ?


Solution:Amount = Principle + interest for 10 years = 1000 +( 1000*10*5/100 ) = 15001500 after 10 years2000 after n years500 = 1500*5*n/100 n = 6 2/3yearTotal 10 + 6 2/3 = 16 2/3 Total time is 16 2/3 years


Simple/compound interest

A sum of money amounts to Rs. 6690 after 3

years and to Rs. 10035 after 6 years on C.I. Find

the sum.

(Satyam Question)

Simple/Compound interest

Solution:

P(1 + R/100)3 = 6690 -----------1

P(1 + R/100)6 = 10035 -----------2

Dividing (1 + R/100)3 = 10035/6690 = 3/2

Substitute in equation 1 then P(3/2)=6690

P = 6690 * 2/3 = 4460

The sum is Rs.4460


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

Difference = Rs. 11.25

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)

Average

A batsman makes a score of 90 runs in the 17th

innings and then increases his average by 3. Find

his average after 17th innings.

(Infosys Question)

Average

Solution:

Let the average of 17th innings be x

Average after the 16th innings = x-3

16(x-3) + 90 = 17x

x = 42

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:

30 students total age = 30*9=270

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

Difference = 310 – 270 = 40 years

Average

Of the three numbers second is twice the first

and is also thrice the third. If the average of the

three numbers is 44, find the largest number.

Average

Answer:

Let the first number be x

Average = (x + 2x + 2x/3)/3 = 44

Largest number is 72

Average

In a coconut groove, (x+2) trees yield 60 nuts

per year, x trees yield 120 nuts per year and (x-

2) trees yield 180 nuts per year. If the average

yield per year per tree be 100 find x.

Average

Answer:

[(x+2)*60 + (x*120) + (x-2) *180] =100

[x+2+x+x-2]

By simplifying x = 4

Average

In a cricket team of 11 boys one player weighing

42 kg is injured and his place is taken by another

player. If the average weight of the team is

increased by 100 g as a result of this, then what

is the weight of the new player?

Average

Solution:

Average weight of 11 boys is increased by 0.1 kg

The total increase in weight = 0.1 * 11 = 1.1 kg

Weight of the new boy = 42 + 1.1 = 43.1 kg

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)!

• The number of Combinations of ‘n’ things taken

‘r’ at a time in which ‘p’ particular things will

always occur is n-pCr-p

• The number of Combinations of ‘n’ things taken

‘r’ at a time in which ‘p’ particular things never

occur is n-pCr


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 among 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.



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 addition

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

= 127

Number of ways is 127


If nP5 = 20 nP3 then what is the value of n?


Solution:

n!/(n-5)! = 20 * n!/(n-3)!

n = 8


Find the number of different 8 letter words

formed from the letters of the word EQUATION

if each word is to start with a vowel


Solution:

For the words beginning with a vowel, the first

letter can be any one of the 5 vowels, the

remaining 7 places can be filled by

7P7 = 5040

The number of words = 5 * 5040 = 25200


In how many different ways can the letters of the

word TRAINER be arranged so that the vowels

always come together?


Solution:

A,I,E can be arranged in 3! Ways

(5! * 3!) / 2! = 360 ways

Percentage

• By a certain Percent, we mean that many

hundredths.

• Thus, x Percent means x hundredths, written

as x%

•Finding out of Hundred.

If Length is increased by X% and Breadth is decreased by Y% What is the percentage Increase or Decrease in Area of the rectangle?

Formula: X+Y+ XY/100 %

Decrease 20% means -20

Percentage

Percentage

Two numbers are respectively 30% & 40%

less than a third number. What is 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 n umber6x / 10 = y/100 * 7x /10 y = 600 / 7 y = 85 5/7 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. The

total amount of money spent on machinery

and the raw material is?

Percentage

Solution:Let total salary =100%Salary = 100Spending: Machinery + Raw material + staff = 35+40+10 = 85Remaining percentage = 100 – 85 = 1515 = 60000100 = ? By chain rule, we get 400000In this 400000 , 75% for machinery & raw material = 4, 00,000* 75/100

=Rs 3 00000

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

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 = 37.5%

Percentage

Answer:

Increase % in the area of the new rectangle

= x + y + (xy/100)%

= 20 + 10 + (200/100)

= 32%

Increase 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:

B’s income = 40/(100-40) * 100%

= 66 2/3%

Percentage

Ramesh loses 20% of his pocket money. After spending 25% of the remainder he has Rs. 480 left. What was his pocket money?

Percentage

Answer:

x (1-20/100) (1-25/100) = 480

Solving the equation, x = 800

Pocket money = Rs. 800

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

Boats and Streams

Solution:Down stream 4km in x hours. Then,Speed of Downstream = 4/x km/hr, Speed of Upstream = 3/x km/hr48/ (4/x) + 48/(3/x) = 14 , x = 1/2Speed of Down stream = 8, Speed of upstream = 6Rate of the stream = ½ (8-6) km/hr = 1 km/hr

Boats and Streams

In a stream running at 2 km/h a motor boat goes 10 km upstream and back again to the starting point in 55 minutes. Find the speed of the motor boat in still water.

Boats and Streams

Solution:

Let the speed be x km/h

10/(x+2) + 10/(x-2) = 55/60


x = 22 km/h

Boats and Streams

A swimmer can swim a certain distance in the direction of current in 5 hours and return the same distance in 7 hours. If the stream flows at the rate of 1 km/h, find the speed of the swimmer in still water.

Boats and Streams

Solution:

Let the speed of the swimmer in still water be x

Downstream speed = (x+1) * 5

Upstream speed = (x-1) * 7

5(x+1) = 7(x-1)

Solving the equation, x = 6

Speed of the swimmer = 6 km/h

Time and Distance

•Speed:- Distance covered per unit time is

called speed.

Speed = distance/time

•Distance = speed*time

•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 - ( * 5/18).

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

• Average Speed:-

Average speed = Total distance traveled

Total time taken

Time and Distance

The jogging track in a sports complex is 726m

in circumference. Suresh and his wife start from

the same point and walk in opposite direction at

4.5km/hr and 3.75km/hr respectively. They will

meet for the first time in how many minutes ?

Time and Distance

Solution:

Suresh speed m/hr = 4.5*5/18 *60*60 = 4500

His wife speed m/hr = 3.75*5/18*60*60 = 3750

Meeting time

4500/60 x +3750/60x = 756 m

x = 5.28 m

They will meet after 5.28minutes

Time and Distance

Ram travels from P to Q at 10km/hr and returns at

15km/hr. Shyam travels P to Q and returns at

12.5km/hr. If he takes 12 minutes less than ram

then what is the distance between P and Q ?

Time and Distance

Solution:

Average Speed of Ram = 2(10*15)/25 = 12km/hr

Average Speed of Shyam = 2(12.5 *12.5)/25 = 12.5

Ram’s speed in m/sec = 12*5/18 = 10/3

Shyam’s speed in m/sec = 12.5*5/18 = 625/36

(x/ 10/3) – (x/625/36) = 12*60

3x/10 – 30x/625 = 720

= 30

Distance between P and Q is 30 km

Time and Distance

If I walk 30 miles/hr I reach 1 hour before and if I walk

20 miles/hr I reach 1 hour late. Find the distance between

two points and if the exact time of reaching destination is

11 a.m. then find the speed with which I walk?

Infosys Question

Time and Distance

Solution:

x/30-x/20 = 2

By Solving the equation,

The total distance between two points = 120 miles

TheAverage speed required = 24 miles/hr

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 car travels a certain distance taking 7 hours in forward journey. During the return journey it increased the speed by 12 km/hr and takes the time of 5 hours. What is the distance traveled?

(Satyam Question)

Time and Distance

Solution:

Let speed be x , D = S * T =7x

7x = (x+12) 5


x = 30

D = 7 * 30 = 210

Actual Distance = 210 km

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.


How long will a train 100 m long traveling at

72 km/hr take to overtake another train 200 m

long traveling at 54 km/h?

(Satyam Question)


Solution:

They are traveling in the same direction. Hence the Relative speed is x -y

72 – 54 = 18 km/h

18 * 5/18 = 5 m/sec

To travel 5 m it takes 1 second

To travel 300m it will take300/5 = 60 seconds

It will take 60 seconds or 1 minute.


Rajee starts her journey from Delhi to Bhopal and

simultaneously Sheela starts from Bhopal to Delhi.

After crossing each other they finish their remaining

journey in 5 4/9 hours and 9 hours respectively.

What is Sheela’s speed if Rajee’s speed is 36 km/hr.


Answer:

Rajee’s speed = Square root of (t2/t1)

Sheela’s speed

The speed of Sheela = 28 km/hr


Excluding stoppages, the speed of a train is

54 km/hr and including stoppages it is 45

km/hr. For how many minutes does the train

stop per hour?


Solution:

x

= _____________

x/45 + x/36

Take L.C.M and find the x value

x = 12

It stops 12 minutes/hr


A train 130 m long passes a bridge in 21 seconds

moving with a speed of 90 km/hr. Find the

length of the bridge?


Solution:Let the length of the bridge be xSpeed of the train = (length of the train + length

of the bridge) / Time taken90*5/18 = (130 +x )/21Length of the bridge = 395 m

• Gain =(S.P.)-(C.P.)

• Loss =(C.P.)-(S.P.)

• Loss or gain is always reckoned on C.P.

• 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

Sundeep buys two CDs for Rs.380 and sells one

at a loss of 22% and the other at a gain of 12%.

If both the CDs are sold at the same price, then

the cost price of two CDs is ?

Profit and Loss

Solution:CDs =Rs.3801st CD = x -22x/1002nd CD = y + 12y/100SP1 = SP2x – 22x/100 = y – 12y/100 x = 56y/39 x + y = 380 56y/39 + y = 380 y = 156 x = 224 Cost of the two CDs are Rs. 224 and Rs.156

Profit and Loss

Mr. Ravi buys a cooler for Rs. 4500. For how

much should he sell so that there is a gain of 8%

Profit and Loss

Solution:

S.P = (100+8%/100)/CP

= (108/100) * 4500 = 4860

He should sell it for Rs. 4860.

Profit and Loss

Profit and Loss

If the manufacturer gains 20%, the wholesale dealer 25%, and the retailer 35%, then find the cost of production if the retail price is Rs. 1265.

Profit and Loss

Solution: Let C.P be x

Gain 35% of 25% of 20% of x = 1265

135/100 * 125/100 * 120/100 * x = 1265

x = Rs.624.7

Profit and Loss

Manoj sells a shirt to Yogesh at a profit of 15% and Yogesh sells it to Suresh at a loss of 10%. Find the resultant profit or loss.

Profit and Loss

Solution:

Resultant profit = (x + y + xy/100)

= 15 – 10 – (150/100)

= 3½%

The resultant profit is 3.5%

Profit and Loss

Mr. Verma sold his scooter for Rs. 10500 at a gain of 5%. Find the cost price of the scooter.

Profit and Loss

Solution:

C.P = (100/(100+P%))*S.P

= (100/(100+5) )* 10500

= 10000

Cost price = Rs. 10000

Profit and Loss

A man bought a horse and a cart. If he sold the horse at 10% loss and the cart at 20% gain, he would not lose anything, but if he sold the horse at 5% loss and the cart at 5% gain, he would lose Rs. 10 in the bargain. What is the cost of one horse and one cart?

Satyam Question

Profit and Loss

Solution : C.P of horse = x; C.P of cart = y

10x/100 = 20y/100

5x/100 = 5y/100 + 10

Solving the above two equations,

C.P of horse = x = Rs. 400

C.P of cart = y = Rs. 200

Profit and Loss

A man sells an article at a profit of 25%. If he had bought it at 20% less and sold it for Rs. 10.50 less he would have gained 30%. Find the cost price of the article.

Satyam Question

Profit and Loss

Solution:

First C.P = x, S.P = 125% of x = 5x/4

Second C.P = 80x/100 = 4x/5

S.P = 130/100 * 4x/5 = 26x/5

5x/4 - 26x/25 = 10.50

x = 50

The cost price is Rs. 50

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 minute, 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?

Documents

Blue Lotus Aptitude Numerical Reasoning. Problems on Numbers Problems on Ages Ratio and Proportion Alligation or Mixture Chain Rule Partnership Venn Diagram