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

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

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)]

Geometrical Progression: Tn = arn – 1.

Sn = a(rn – 1)/(r-1);

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

Problem on Numbers

A monkey starts climbing up a tree 20 feet tall.

Each hour it claims 3 feet and slips back 2 feet.

How much time would it take the monkey to

reach the top? (Satyam)

Problem on Numbers

Solution:

= 17 + 1= 18 hours

Problem on Numbers

The length of the side of a square is represented

by x + 2. the length of the side of an equilateral

triangle is 2x, if the square and the equilateral

triangle have equal perimeter the find the value

of x.

Problem on NumbersSolution:

Side of the square is x + 2

Side of the triangle 2x

P = 4(x + 2)

= 4x +8

Perimeter = 3*2x

= 6x

4x + 8 = 6x

x = 4

Problem on Numbers

On sports day if 30 children were made to

stand in a column, 16 column could be formed

if 24 children were made to stand in a column.

How many columns could be formed?

(Satyam)

Problem on Numbers

Solution:

Total no of children = 30 * 16 = 480

No. of column of 24 children each = 480 / 24

= 20

Problem on Numbers

5/9 part of the population in a village are males

if 30% of the males are married what is the

percentage of unmarried females in the total

population?

Problem on Numbers

Solution:Let population be = xMales = 5 x/9Married Men =30%(5x/9)= x/6 (Married females) Total females = x – 5x/9 = 4x/9Unmarried females = 4x/9 – x/6 = 5x/18Percentage = (5x / 18)*100% x

= 250 / 9%

Problems on Numbers

How many terms of the A.P. 1, 4, 7…. are needed to give the sum 715 ?

Solution:Sum of n terms of A.P

Sn = n/2 *[2a+(n-1)d)]

Sn = 715, n =? a = 1 d = 3715 = n/2 *[2*1+(n-1)3]715 =n/2 *[2+(n-1)31430 = n[2+3n -3]1430 = n[3n – 1]1430 = 3n2 -n3n2 – n – 1430 = 0 Solving the Quadratic Equation, n =22 Number of terms needed is 22

Problems on Numbers

Problems on Numbers

The sum of the digits of a two-digits number

is 8. if the digits are reversed the number is

increased by 54. Find the number?

Problems on Numbers

Solution:

x+ y = 8 --------------------1

10y+x = 10x+y+54

10y + x - 10x – y = 54

-x + y = 6 ---------------------2

Solve equation 1&2

x= 1, y = 7

Required number = 10*1 +7 =17

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

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: 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. 8 years ago, the sum of

their ages was 56. Find their present ages ?

Problems on Ages

Solution:

Three person’s age ratio = 4:7:9

Sum of their age = 56, after 8 years

their sum of their ages = 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

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, find the age of Father.

Problems on Ages

Solution:Father age = 3(x+y)F+20 = x+20+y+203x+3y+20 = x+y+403x +3y = x+y+40 -203x-x+3y – y =202x+2y = 20x+y = 10F = 3*10 =30 The father’s age is 30.

Problems on Ages

Problems on Ages

Jalia is twice older than Qurban. When Jalia was

4 years younger, Qurban was 3 years older the

difference between their ages is 12 what is the

sum of their ages?

Problems on Ages

Solution:

J = 2Q 1

(J – 4) – (Q +3) = 12 2

Solve 1 &2

Sum of their ages = 19 + 38 = 57

Problems on Ages

Ten years ago, Chandrawathi’s mother was 4

times older than her daughter. After 10 years the

mother will be twice older than daughter. What

is the present age of Chandrawathi?

Problems on Ages

Solution:

Let Chandrawathi’s age 10 years ago x

Her mother’s age 10 years ago 4x

(4x + 10 + 10) = 2(x+10+10)

x = 10

Chandrawathi is x +10 = 20 years

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


Concentration of three wines A, B, and C are

10, 20, and 30 percent respectively. They are

mixed in the ratio 2 : 3 : x resulting in a 23%

concentration solution. Find x.

(Caritor Question)


Answer :

10 : 20 : 30

2 : 3 : x

Multiplying 20 : 60 : 30x

20 + 60 + 30x = 230

x = 5


Ajay, Aman, Suman and Geetha rented a

house and agreed to share the rent as follow

Ajay : Aman = 8:15

Aman : Suman = 5:8

Suman: geetha = 4:5

The part of rent paid by Suman will be ?


Solution:

Ajay : Aman = 8 : 15 In next ratio Aman is 5 to make that 15 multiply by 3 that is ratio of Aman : Suman = 15:24Suman and Geetha contribution is 4:5 but Suman previous contribution is 24 to make Suman contribution in 24 multiply by 6Suman: Geetha = 4*6:5*6 = 24:30Ratio of Ajay:Aman:Suman:Geetha = 8:15:24:30 Suman ratio = 24/77


60 kg of an alloy A is mixed with 100 kg of alloy

B. If alloy A has lead and tin in the ratio 3 : 2

and alloy B has tin and copper in the ratio 1 : 4,

then what would be the amount of tin in the new

alloy?


Solution :

The amount of tin in both alloys= (60*2/5) + (100*1/5)

= 44 kg


In a class composed of x girls and y boys what

part of the class is composed of girls?

(Satyam Question)


Answer:

Ratio = x : (x+y) = x / x+y


In a factory ,the ratio of male workers to female workers was 5:3. If the number of female workers was less by 40.What was the total number of workers in the factory.


Solution:

Let the number of males is 5x and

females is 3x and Total is 8x

5x – 3x = 40

x = 20

Total number of workers in the factory = 8x

=8*20 = 160


A Mixture contains milk and water in the ratio 5:1.On adding 5 liters of water, the ratio of milk to water becomes 5:2. What is the quantity of milk in the original mixture ?


Solution:

Let quantity of milk be 5x, and water be x then,

5x = 5

x + 5 2

Hence x=5

Quantity of milk = 5x5 = 25 liters.

•(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 certain type of mixture is prepared by mixing

brand A at Rs. 9/kg with brand B at Rs. 4/kg. If

the mixture is worth Rs. 7/kg how many kg of

brand A are needed to make 40 kgs of the

mixture? Satyam Question


Solution:

B A4 9

72 3

Brand A required = 3*40 / 5 = 24kg


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 he neither gains nor

loses. How much does each cow cost?


Solution:

-6% 7.5% 0

7.5 6

Ratio is 5 : 4

The cost of first cow = 5*1350/9 =Rs. 750The cost of second cow = Rs. 600


How much water be added to 14 liters of milk

worth Rs. 5.4 per liter so that the value of the

mixture may be Rs. 4.20 per liter?


Solution: The cost of water is 0.W M0 540 420

120 420Ratio is 4 : 14The amount of water added for 14 liters is 4 liters.



There are 65 students in a class, 39 rupees are

distributed among them so that each boy gets

80 paise and girl gets 30 paise. Find the number

of boys and girls in that class.


Solution: “Money per boy or girl” is considered.Per student = 3900/65 = 60 paise.

Girls Boys30 80 6020 30 Girls : Boys = 2 : 3

Number of boys = 39Number of girls = 26

Chain Rule

Direct Proportion :

A B

A B

Indirect Proportion:

A B

A B

Chain Rule

A man completes 5/8 of a job in 10 days. At

this ratio how many more days will it take for

him to finish the job?

Chain Rule

Answer: Jobs Days

5/8 103/8 x

5x / 8 = (10*3) / 8 (Direct )

The number of days to complete = 6

Chain Rule

A certain number of men can finish a piece of work in 100 days. If however there were 10men less it will take 10 days more for the work to be finished. How many men were there originally?

(Satyam Question)

Chain Rule

Solution:

Men Days

x 100

(x-10) 110 (indirect )

100x = 110(x – 10 )

The number of men present originally = 110

15 men take 21 days of 8 hours each to do a piece

of work. How many days of 6 hours each would it

take for 21 women if 3 women do as much work

as 2 men?

(Satyam Question)

Chain Rule

Solution:

3 women = 2 men

Men Days Hours

15 21 8

14 x 6

x = 15*8 21 14*6S

x = 30

The number of days required is 30.

Chain Rule

Chain Rule

6 cats kill 6 rats in 6 minutes. How many cats

will be needed to kill 100 rats in 50 minutes?

(Satyam question)

Chain Rule

Solution:

Cats Rats Minutes6 6 6x 100 50

x/6 = (100*6)/(6*50)The number of cats needed would be 12.

Chain Rule

39 persons can repair a road in 12 days, working

5 hours a day. In how many days will 30

persons working 6 hours a day, complete the

same work?

Chain Rule

Solution : -Men days hours 39 12 5 30 x 6

More hours less days ( inverse proportion )Less men more days ( inverse proportion )

x 5 39 5 * 39 * 12 ------- = ---- * ---- x = --------------- 12 6 30 6 * 30

x = 13 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

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?

(TCS Question)

Partnership

Solution :

A : B : C

100 : 65 : 35

Total = 100 + 65 +35 = 200

C’s share = 560

(35*x)/200 = 560

X = 3200

The sum invested is Rs. 3200

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 and B = 3 : 2Let Profit be xx – 5%x- 5x/100 = 95x/100A’s share is = 3/5 * 95x/100 = 85519x/100 =28519x = 28500x = 26500/19 = 1500

Total profit is Rs.1500

Partnership

A,B,C subscribe Rs. 50,000 for a business. A

subscribes Rs. 4000 more than B and B

subscribes Rs. 5000 more than C. Out of a total

profit of Rs. 35,000 how much does A receive?

(TCS Question)

Partnership

Answer :

Ratio A B C

9000 + x : 5000 + x : x

14000+3x = 50000

On solving we get x = 12000

Ratio A B C

21 : 17 :12 Total = 50

A receives = (21* 35000)/50 = Rs. 14700

Partnership

A began a business with Rs. 450 and B jointed

with Rs. 300. When did B join if the profit at the

end of the year was divided between them in the

ratio 2 : 1?

(Caritor Question)

Partnership

Answer :

Ratio A : B

450x12 : 300 (12 – x )

5400 / ( 300( 12 – x )) = 2 / 1

On solving x = 3

B invested money after 3 months

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 is 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

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

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 do 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 = 9/240 = 3/80 = 26 2/3 A, B and C can finish the work in 26 2/3 days

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. In how many

days alternatively work will be completed?

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

Work done by A & B and A & C = ( 8/150 + 7/120) = 67/600

For 16 days , work done = 67x8 / 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 = 120 / 7 * 4 / 75 = 32/35

They will finish the work in 17 32/35 days

Time and Work

Time and Work

A can do a piece of work in 12 days. B is 60%

more efficient than A. Find the number of days

B takes to do the same piece of work.

Time and Work

Answer :

Work Days

100 12

160 x

On solving , x =( 100*12) / 160 = 7 ½ Days

B will take 7½ days.

Time and Work

Seven men can complete a work in 12 days.

They started the work and after 5 days, two

men left. In how many days will the work be

complete the remaining work.

Time and Work

Solution:(7*12) men can complete the work in 1 day1 man’s 1 day’s work = 1/847 men’s 5 day’s work = 1/12 *5 = 5/12Remaining work = (1-5/12) = 7/125 men’s 1 day’s work =5*( 1/84 )= 5/845/84 work is done by them in 1 day7/12 work is done by them in 84/5 *7/12 = 49 / 5 = 9 4/5 days

Pipe A can fill a cistern in 20 minutes and pipe B

in 30 minutes and pipe C can empty the same in

40 minutes. If all of them work together, find the

time taken to fill the tank.

(Satyam Question)

Time and Work (Pipes)

Answer:

The time taken is 17 1/7 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.


A cistern is provided by two taps A and B. A

can fill it in 20 minutes and B in 25 minutes.

Both taps are kept open for 5 minutes and then

the second is turned off. How many minutes

will be taken to fill the tank ?


Solution:

In one day (A +B) can do = 1/20 + 1/25 = 9/100

In 5 days (A +B ) can do = 5x9/100 = 9/20

Work left = 1 – 9 /20 =11/20

A alone can do the remaining = ( 11/20) x 20

= 11

The cistern will be filled in 11 minutes.



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?

Time and Work (Pipes) Solution:Let B be closed after x minutes. Then,Part filled by (A+B) in x min. + part 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 = 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 = ½ * 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 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 = - 10 , y = - 10

= x + y +xy/100 %

= -10 - 10 – (10*10/100) %

= - 19%

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

If the length of a rectangle is increased by 30%

decreased by 20%, then what percent of its area

would increase?

(TCS Question)

Area and Volume

Solution:

x = 30 y = - 20 (Decreased )

Using formula = x + y + (xy/100)%

The increase would be 4%

Area and Volume

The sides of the right triangular field containing

the right angle are x and x + 10 and its area is

5500 m2. Find the equation to determine its area.

(Caritor Question)

Area and Volume

Answer:

The equation is 1 / 2*x(x+10) = 5500

x2 + 10x – 11000 = 0

Area and Volume

The length and breadth of a rectangular plot are

in the ratio 7 : 5. If the length is reduced by 5 m

and breadth is increased by 2 m then the area is

reduced by 65 m2. Find the length and breadth of

the rectangular plot.

(Caritor Question)

Area and Volume

Answer:

Difference in areas = 65

(7x *5x) – (7x – 5) (5x + 2) = 65

x = 5

The length of the rectangle = 7x = 7*5 = 35 m

The breadth of the rectangle = 5x = 5*5= 25 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

Probability

There are 19 red balls and 1 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.

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

Solution:Probability of selecting Sam and Jessica = 7C1 * 7C1 = 49n(s) = 49n (є) = 1 ( selecting Sam and Jessica)P (є) = n (є) / n(s) = 1/49Not selecting Sam and Jessica = 1- 1/49

The probability = 48/49

Probability

There are 5 distinct pairs of white socks and 5 pairs

of black socks in a cupboard. In the dark, how many

socks do I have to pull out to ensure that I have at

least one correct pair of white socks?

TCS Question

Probability

Answer:

12 times

Probability

In a simultaneous throw of two coins, what is

the probability of getting at least one head?

TCS Question

Probability

Answer :

Probability = 3/4

Probability

Three unbiased coins are tossed. What is the

probability of getting at least 2 heads?

Probability

Answer :

The probability is 4/8 = 1/2

Probability

Probability

A can solve 90% of the problems given in a book and B can solve 70%. What is the probability that at least one of them will solve a problem selected at random from the book?

Probability

Answer:

A does not solve the problem = 1- 90/100 = 1/10

B does not solve the problem = 1- 70/100 = 3/10

Either A or B to solve = 1 – (1/10)*(3/10)

= 1- 3/100

= 97/100

Probability

Two unbiased Dice are thrown. Find the

probability that neither a doublet nor a total of

10 will appear ?

Probability

Solution:

A= Doublet (1,1),(2,2),(3,3),(4,4),(5,5),(6,6)

B =(6,4),(4,6),(5,5)

P(A)= 6 / 36 , P(B) = 3 / 36 , P(A n B) =1/36

Probability = 1 – P( Au B)

= 1 – ( P(A) + P(B) - P(AnB) )

= 7 / 9

Simple Interest = PNR / 100

Total amount A = P + PNR / 100

When Interest is Compound annually:

Amount = P (1 + R / 100)n

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 }

The simple interest on a certain sum is 16 / 25 of

the sum. Find the rate percent and time if both are

numerically equal.

Sathyam Question


Solution:

R = N and I = 16/25 P

R = I * 100 / P*N

R*N = 64 where R = N

R = 8

The rate of interest is 8%


A sum of S.I. at 13½% per annum amounts to

Rs. 2502.5 after 4 years. Find the sum.

Sathyam Question


Answer:

A = P + PNR / 100

A = P (1 +NR/100)

The principal is Rs. 1625


Simple/compound interest

The difference between the compound and simple interest on a certain sum for 2 years at the

rate of 8% per annum is Rs.80,What is the sum?

Wipro Question

Simple/Compound interest

Answer :

C.I – S.I = P(R/100)2

80 = P*(8*8/100*100)

The sum is Rs.12,500


If a sum of money compounded annually amounts of thrice itself in 3 years, in how many years will it become 9 times itself?

CTS Question


Solution:

A = 3P and find n when A = 9P

The number of years required = 6 years


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


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

CI = A –P = 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)

The average weight of 5 persons sitting in

boat is 38 Kg .If the average weight of the

boat and the persons sitting in the boat is

52Kg,What is the weight of the boat?

Average

Soluton:

Average weight of 5 persons = 38 Kg

Total weight of the 5 persons =38*5=190

Total weight including the boat weight is

=52*6=312

Weight of the Boat = 312-190= 122 Kg

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 = 58Sum = 58*5 = 290Last 5 observation average = 56Sum = 56*5 = 280Total 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

Diff= 310-270=40 years

Average

What is the average of even numbers from 1

to 81?

Average

Solution:

Average = (last even + 2) / 2

= (80 + 2 ) / 2

= 41

Average

The Average of 50 numbers is 38. If two

numbers namely 45 and 55 are discarded, What

is the average of the remaining numbers?

Average

• Solution:

Total of 50 numbers = 50*38=1900

Total of 48 numbers = 1900-(45+55)=1800

Average =1800/ 48 = 37.5

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 also the person 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


There are 5 gentlemen and four ladies to dine at a round table. In how many ways can they seat themselves so that no two ladies are together?


Solution:

5! = 120 ways


In a chess board there are 9 vertical and 9 horizontal lines. Find the number of rectangles formed in the chess board.


Solution:

9C2 * 9C2 = 1296


In how many ways can a cricket team of 11 players be selected out of 16 players , if one particular player is to be excluded?


Solution:

If one particular player is to be excluded, then selection is to be made of 11 players out of 15.

15C11= 15!/( 11!*4!)=1365 ways

Percentage

• By a certain Percent, we mean that many

hundredths.

• Thus, x Percent means x hundredths, written

as x%

•If Length is increased by X% and Breadth is decreased by Y% What is the Percentage •Finding out of Hundred.

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 number6x / 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?

PercentageSolution:Let total salary =100%Salary = 100Spending:Machinery + Raw material + staff = 35+40+10 = 85Remaining = 100 – 85 = 1515 = 60000100 = ? ( By Chain rule) = 4, 00,000In this 4, 00,000 75% for machinery and raw material = 4, 00,000* 75/100 = 3, 00,000 Rs. 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?

PercentageSolution: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

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%

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:

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

=66 2/3%

Percentage

Fresh grapes contain 90% water by weight while dried grapes contain 20% water by weight. What is the weight of dry grapes available from 20 kg of fresh grapes?

Percentage

Solution:

Fresh grapes contain 10% pulp

20 kg fresh grapes contain 2 kg pulp

Dry grapes contain 80% pulp.

2 kg pulp would contain = 2/0.8 =20/8 = 2.5kg

Hence 2.5 kg Dry grapes.

Percentage

If A is 20% of C and B is 25% of C then what percentage is A of B?

Percentage

Answer:

Given A is 20% of C and B is 25% of C

A of B is = (20*100%)/25 = 80%

A is 80% of B

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

Boats and Streams

A boat is rowed down a river 40 km in 5 hours and up a river 21 km in 7 hours. Find the speed of the boat and the river.

Boats and Streams

Solution:

Downstream speed (a) = 40/5 = 8 km/h

Upstream speed (b)= 21/7 = 3 km/h

Speed of the boat =1/2(a+b)= 5.5 km/h

Speed of the river =1/2(a-b) = 2.5 km/h

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 =6km/hr

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

= 5.5 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 = Speed of the boat + Speed of the stream = 5 +3 =8Up Stream = Speed of the boat – Speed 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 man rows to place 48 km 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 4 km in x hours. Then,Speed Downstream = 4 / x km/hr, Speed Upstream = 3 / x km/hr48/ (4 / x) + 48/(3 / x) = 14x = ½Speed of Downstream = 8,Speed of upstream = 6Rate of the stream = ½ (8-6) km/hr = 1 km/hr

Boats and Streams

A Swimmer can swim a certain distance in the direction of current in 5 hr and return the same distance in 7 hr. If the stream flows at the rate of 1 kmph, find the speed of the swimmer in still water?

Boats and Streams

Answer:Let the Speed of the Swimmer in still water be xDistance covered in going = 5(x+1) kmDistance covered in returning = 7(x-1) km 5(x+1) = 7(x-1)

x = 6Speed of the swimmer in still water = 6 kmph.

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

If a man walks at the rate of 5 kmph he misses a

train by only 7 minutes. However if he walks at

the rate of 6 kmph he reaches the station 5

minutes before the arrival of the train. Find the

distance covered by him to reach the station.

Sathyam Question

Time and Distance

Solution:

x/5-x/6 = 12/60 ( Difference 7+5=12)

Solving the equation,

x = 6

The distance covered is 6 km.

Time and Distance

The J and K express from Delhi to Srinagar was

delayed by snowfall for 16 minutes and made up

for the delay on a section of 80 km travelling with

a speed 10 km per hour higher than its normal

speed. Find the original speed of the express?

Time and Distance

Answer:

The train saves 16 minutes by travelling faster over a section of 80 km

80/x – 80/(x+10) = 16/60

By solving we get x = 50

Time and Distance

If the total distance of a journey is 120 km. If one

goes by 60 kmph and comes back at 40 kmph what is

the average speed during the journey?

Sathyam Question

Time and Distance

Solution:

Average Speed = 2xy / (x + y)

The average speed is 48 kmph.

Time and Distance

Time and Distance

By walking at ¾ of his usual aped, 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

If a person walks at 14 km/hr instead of 10

km/hr, he would have walked 20 km more. Find

his actual distance traveled?

Time and Distance

Solution:

Let the distance be x km. Then,

x/10 = (x+20) / 14

14x = 10x +200

4x = 200

x = 50 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.


Two trains 200 m and 150 m long are running on

the parallel rails at the rate of 40 km/hr and 45

km/hr. In how much time will they cross each

other if they are running in the same direction?

Sathyam Question


Answer:

Relative speed = x – y = 45 – 40 = 5 km/h

=5*5/18= 25/18 m/s

For 25/18 m it takes 1 sec

For 350 m it takes = 18 * 350 /25 =252

Time taken = 252 seconds


There are 20 poles with a constant distance

between each pole. A train takes 24sec to

reach the 12 pole. How much time will it take

to reach the last pole ?


Solution: To reach 11 poles it takes 24 secFor 19 poles it will take x time Poles time 11 24 19 x11x = 19 * 24 x = 19* 24 /11 x = 41.45 sec It reaches the last pole in 41.45 sec

• 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

Profit and Loss

A man buys an article for Rs. 27.50 and sells it

for Rs. 28.60. Find his gain percentage.

Sathyam Question

Profit and Loss

Answer:

Gain 4%

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

Answer:

C.P = 4500 Profit = 8/100*4500 = 360

S.P = C.P + Profit = 4500 + 360 = 4860

He should sell it for Rs. 4860.

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:C.P of two CDs =Rs.380S.P of 1st CD = x – 22 x /100S.P of 2nd CD = y + 12y /100 SP1 = SP2x – 22x/100 = y – 12y/100 x = 56/39 y x + y = 380 56/39y + y = 380 y = 156 x = 224 Cost of the two CDs are Rs. 224 and Rs.156

Profit and Loss

A tradesman fixed his selling price of goods at

30% above the cost price. He sells half the stock

at this price, one quarter of his stock at a

discount of 15% on the original selling price and

rest at a discount of 30% on the original selling

price. Find the Gain Percent altogether?

Profit and Loss

Solution:

Let us take C.P of goods = 100

The Marked price = 130

=1/2*130 + 1/4*0.85*130 + 1/4 *0.7*130=15.375%

Profit and Loss

Profit and Loss

Rajesh buys an article with 25% discount on its Marked Price. He makes a profit of 10% by selling it at Rs 660. Find the marked price?

Profit and Loss

Solution:

S.P = Rs.660 Profit = 10 %

C.P = 110 / 100*660 = Rs.600

Let x be M.P

75% of x = 600

x = 800

Marked Price = 800

Profit and Loss

A shopkeeper gains the cost of 8 meters of

thread by selling 40 meters of thread. Find

his gain percentage?

Profit and Loss

Answer:

Gain = 20%

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)

The resultant profit is 3.5%

Profit and Loss

Aditya purchases toffees at Rs. 10 per dozen and sells them at Rs. 12 for every 10 toffees. Find the gain or loss? TCS Question

Profit and Loss

Answer:

Profit = 44

Profit and Loss

Anirudh bought 8 lemons for a rupee but sells only 6 lemons for a rupee. Find his profit percentage.

CTS Question

Profit and Loss

Answer:

Profit 33 1/3%

Profit and Loss

A radio when sold at a certain price gives a gain of 20%. What will be the gain if it sold for thrice the price?

Sathyam Question

Profit and Loss

Solution:

Let C.P be Rs. 100. Hence

S.P = Rs. 120

Gain = S.P – C.P = 360 – 120 = 260

Gain =Rs. 260

Profit and Loss

A grocer purchased 80 kg of rice at Rs. 13.50 per kg and mixed it with 120 kg rice Rs. 16 per kg. At what rate per kg should he sell the mixture to gain 16%?

Sathyam Question

Profit and Loss

Solution:

C.P of 200 kg = (80*13.5 + 120*16) =3000

S.P of 200 kg = 116% of C.P = 3480

The selling price of 1 kg is Rs. 17.40

Profit and Loss

A candidate who gets 20% marks fails by 10 marks but another candidate who gets 42% marks get 12% more than the passing marks. Find the maximum marks.

Sathyam Question

Profit and Loss

Answer:

The maximum marks = 100

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

(3 + 6 + 6 + 2) -3

= ----------------------- = 14/7= 0, Sunday.

7

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?

Venn Diagram

Answer:

3 of them studied both Latin and Greek.

Documents

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