quantitative ability - level b

NARENSINSTITUTE FOR COMPETITIVE EXAMINATIONS

QUANTITATIVE ABILITY LEVEL - B

Numerical ReasoningProblems on NumbersProblems on AgesRatio and ProportionAlligation or MixtureChain RulePartnershipVenn Diagram

Numerical ReasoningArea and VolumeProbabilityTime and Work (Pipes)SI and CIAveragePermutation and CombinationPercentageCubes

Numerical ReasoningBoats and StreamsTime and Distance (Trains)Data SufficiencyProfit and LossCalendarClocksData Interpretation

Problems on NumbersDivision 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.

Problems on NumbersArithmetic 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);

Basic Formulae1. ( 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 NumbersA 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 NumbersSolution: = 17 + 1= 18 hours

Problem on NumbersThe 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 2xP = 4(x + 2) = 4x +8Perimeter = 3*2x = 6x4x + 8 = 6xx = 4

Problem on NumbersOn 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 NumbersSolution:Total no of children = 30 * 16 = 480No. of column of 24 children each = 480 / 24 = 20

Problem on Numbers5/9 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 NumbersSolution: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 NumbersHow many terms of the A.P. 1, 4, 7. are needed to give the sum 715 ?

Problems on NumbersSolution:Sum of n terms of A.PSn = 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 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 NumbersSolution: x+ y = 8 --------------------110y+x = 10x+y+5410y + x - 10x y = 54-x + y = 6 ---------------------2Solve equation 1&2x= 1, y = 7 Required number = 10*1 +7 =17

Problems on AgesThe 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 + yx- 5 = 2(y-5)y + 10 -5 = 2y -10y+5 = 2y -102y- y = 15y=15: x = 25Their present ages are 15 years and 25 years.

Problems on AgesThe 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 AgesSolution:Three persons age ratio = 4:7:9Sum of their age = 56, after 8 years their sum of their ages = 80As age = 4/20 *80 = 16Bs age = 7/20 *80 = 28Cs age = 9/20 *80 = 36Their present ages are 16, 28 and 36.

Problems on AgesFathers 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 AgesSolution: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 fathers age is 30.

Problems on AgesJalia 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 AgesSolution:J = 2Q 1(J 4) (Q +3) = 12 2Solve 1 &2 Sum of their ages = 19 + 38 = 57

Problems on AgesTen years ago, Chandrawathis 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 AgesSolution:Let Chandrawathis age 10 years ago xHer mothers age 10 years ago 4x(4x + 10 + 10) = 2(x+10+10)x = 10Chandrawathi is x +10 = 20 years

Ratio and Proportion Ratio: The Relationship between two variables is ratio.

Proportion: The relationship between two ratios is proportion.

Ratio and 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

Ratio and ProportionConcentration 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)

Ratio and ProportionAnswer :10 : 20 : 302 : 3 : xMultiplying 20 : 60 : 30x20 + 60 + 30x = 230x = 5

Ratio and ProportionAjay, Aman, Suman and Geetha rented a house and agreed to share the rent as followAjay : Aman = 8:15Aman : Suman = 5:8Suman: geetha = 4:5The part of rent paid by Suman will be ?

Ratio and ProportionSolution:

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

Ratio and Proportion60 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?

Ratio and Proportion Solution :

The amount of tin in both alloys= (60*2/5) + (100*1/5)= 44 kg

Ratio and Proportion

In a class composed of x girls and y boys what

part of the class is composed of girls?

(Satyam Question)

Ratio and ProportionAnswer:

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

Ratio and Proportion 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.

Ratio and ProportionSolution:Let the number of males is 5x and females is 3x and Total is 8x 5x 3x = 40 x = 20Total number of workers in the factory = 8x =8*20 = 160

Ratio and Proportion 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 ?

Ratio and ProportionSolution:

Let quantity of milk be 5x, and water be x then, 5x = 5x + 5 2Hence x=5Quantity of milk = 5x5 = 25 liters.

Alligation or Mixture

(Quantity of cheaper / Quantity of costlier)

(C.P. of costlier) (Mean price)= -------------------------------------- (Mean price) (C.P. of cheaper)


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?

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

Alligation or MixtureA 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

Alligation or MixtureSolution:

BA49 72 3Brand 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?

Alligation or MixtureSolution:

-6%7.5% 07.5 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?

Alligation or MixtureSolution: The cost of water is 0.WM0540 420

120420Ratio is 4 : 14The amount of water added for 14 liters is 4 liters.


There are 65 students in a class, 39 rupees aredistributed among them so that each boy gets 80 paise and girl gets 30 paise. Find the number of boys and girls in that class.

Alligation or MixtureSolution: Money per boy or girl is considered.Per student = 3900/65 = 60 paise.GirlsBoys3080 602030 Girls : Boys = 2 : 3Number of boys = 39Number of girls = 26

Chain RuleDirect Proportion : A B A BIndirect 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 RuleAnswer: JobsDays5/8103/8 x 5x / 8 = (10*3) / 8 (Direct )

The number of days to complete = 6

Chain RuleA 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 RuleSolution:MenDaysx100(x-10)110 (indirect ) 100x = 110(x 10 )The number of men present originally = 110

Chain Rule15 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 RuleSolution: 3 women = 2 men Men DaysHours 15 21 8 14 x 6x = 15*8 21 14*6Sx = 30The number of days required is 30.

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 RuleSolution:

CatsRatsMinutes666x10050

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 RuleSolution : -Mendayshours 39 12 5 30 x 6More 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.

Partnership

Types: A invested Rs.X and B invested Rs.Y then

A : B = X : YA invested Rs.X and after 3 months B invested Rs.Y then the share isA : 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 Cs share is Rs. 560, what is the sum?(TCS Question)

Partnership

Solution :A :B:C100:65:35

Total = 100 + 65 +35 = 200Cs share = 560 (35*x)/200 = 560X = 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 As share is Rs.855,What is the total profit ?

Partnership

Solution:A and B = 3 : 2Let Profit be xx 5%x- 5x/100 = 95x/100As share is = 3/5 * 95x/100 = 85519x/100 =28519x = 28500x = 26500/19 = 1500Total 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 C9000 + x : 5000 + x : x 14000+3x = 50000On solving we get x = 12000Ratio AB C 21 : 17 :12 Total = 50A 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 / 1On solving x = 3B invested money after 3 months

PartnershipA 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)

PartnershipSolution:A: B: C = 1500*12 : 2000*9 : 2250*8 = 18000 : 18000 : 18000 = 1: 1 : 1Profit share at the end of the year is 1: 1: 1

Time and WorkIf A can do a piece of work in n days, then As 1 days work = 1 / nIf 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

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

Pipes and Cisterns 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

Time and Work

10 men can complete a piece of work in 15 days and 15 women can complete the same work in 12 days. If all the 10 men and 15 women work together, in how many days will the work get completed ?

Time and WorkSolution: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 WorkA 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 WorkSolution: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 WorkA 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 WorkSolution:A = 1/30, B = 1/50, C = 1/40A+B = 1/30+1/50 = 8/150A+C = 1/30+1/40 = 7/120 Work done by A & B and A & C = ( 8/150 + 7/120) = 67/600For 16 days , work done = 67x8 / 600 = 536/600 =67/75Work left = 1-67/75 = 8/7517th 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

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 WorkAnswer : Work Days 100 12 160 xOn solving , x =( 100*12) / 160 = 7 DaysB will take 7 days.

Time and WorkSeven 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 WorkSolution:(7*12) men can complete the work in 1 day1 mans 1 days work = 1/847 mens 5 days work = 1/12 *5 = 5/12Remaining work = (1-5/12) = 7/125 mens 1 days 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

Time and Work (Pipes)

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.

Time and Work (Pipes)

A cistern has two taps which fill it in 12 minutes and 15 minutes respectively. There is also a waste pipe in the cistern. When all the pipes are opened, the empty cistern is filled in 20 minutes. How long will a waste pipe take to empty a full cistern ?

Time and Work (Pipes)Solution:

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.

Time and Work (Pipes)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 ?

Time and Work (Pipes)Solution:In one day (A +B) can do = 1/20 + 1/25 = 9/100In 5 days (A +B ) can do = 5x9/100 = 9/20Work left = 1 9 /20 =11/20A alone can do the remaining = ( 11/20) x 20 = 11The cistern will be filled in 11 minutes.

Time and Work (Pipes)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.

Time and Work (Pipes)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?

Time and Work (Pipes)Solution:As work in hour = 1/6, Bs 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 As turn and 1/6 part is filled by A in 1 hour.Total time = 4+1 =5

Area and VolumeCube:Let each edge of the cube be of length athen,Volume = a3cubic unitsSurface area= 6a2 sq.units.Diagonal = 3 a units.

Area and VolumeCylinder:Let each of base = r and height ( or length) = h. Volume = r2hSurface area = 2 r h sq. unitsTotal Surface Area = 2 r ( h+ r) units.

Area and VolumeCone:Let radius of base = r and height=h, thenSlant height, l = h2 +r2 unitsVolume = 1/3 r2h cubic unitsCurved surface area = r l sq.unitsTotal surface area = r (l +r)

Area and VolumeSphere:Let the radius of the sphere be r. then,Volume = 4/3 r3Surface area = 4 r2sq.units

Area and VolumeCircle: A= r 2

Circumference = 2 r

Square: A= a 2

Perimeter = 4a

Rectangle: A= l x b

Perimeter= 2( l + b)

Area and VolumeTriangle:A = * base * heightEquilateral = 3/4*(side)2 Area of the Scalene Triangle S = (a + b + c) / 2 A = s*(s-a) * (s - b)* (s - c)

Area and VolumeWhat 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 VolumeSolution:Area of triangular field = * 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 VolumeSolution: 850 = * d1 * d2 = * 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 VolumeSolution: No. of Boxes = (25*42*60) /( 7*6*5) = 300300 boxes of cereal box can be placed.

Area and Volume

If the radius of a circle is diminished by 10%, what is the change in area in percentage?

Area and VolumeSolution: 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 VolumeSolution:length of wire = 2 r = (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 VolumeSolution: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 is5500 m2. Find the equation to determine its area.

(Caritor Question)

Area and VolumeAnswer:

The equation is 1 / 2*x(x+10) = 5500x2 + 10x 11000 = 0

Area and VolumeThe 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 VolumeAnswer: Difference in areas = 65(7x *5x) (7x 5) (5x + 2) = 65x = 5The length of the rectangle = 7x = 7*5 = 35 m

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

ProbabilityProbability:

P(E) = n( E) / n( S)Addition theorem on probability:

n(AUB) = n(A) + n(B) - n(AB)Mutually Exclusive:

P(AUB) = P(A) + P(B)Independent Events:

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

Probability

There are 19 red balls and 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)

ProbabilityAnswer:

Probability is 1/2.

Probability

Sam and Jessica are invited to a dance party. If there are 7 men and 7 women in total at the dance and 1 woman and 1 man are chosen to lead the dance, what is the probability that Sam and Jessica will not chosen to lead the dance ?

ProbabilitySolution:Probability of selecting Sam and Jessica = 7C1 * 7C1 = 49n(s) = 49n () = 1 ( selecting Sam and Jessica)P () = n () / n(s) = 1/49Not selecting Sam and Jessica = 1- 1/49The probability = 48/49

ProbabilityThere 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

ProbabilityAnswer:

12 times

Probability

In a simultaneous throw of two coins, what is the probability of getting at least one head?

TCS Question

ProbabilityAnswer :

Probability = 3/4

Probability

Three unbiased coins are tossed. What is the probability of getting at least 2 heads?

ProbabilityAnswer :

The probability is 4/8 = 1/2

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?

ProbabilityAnswer:A does not solve the problem = 1- 90/100 = 1/10B does not solve the problem = 1- 70/100 = 3/10Either A or B to solve = 1 (1/10)*(3/10) = 1- 3/100 = 97/100

ProbabilityTwo unbiased Dice are thrown. Find the probability that neither a doublet nor a total of 10 will appear ?

ProbabilitySolution: 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) =1/36Probability = 1 P( Au B) = 1 ( P(A) + P(B) - P(AnB) ) = 7 / 9

Simple / Compound Interest Simple Interest = PNR / 100Total amount A = P + PNR / 100When Interest is Compound annually:Amount = P (1 + R / 100)n

Simple / Compound InterestHalf-yearly C.I.: Amount = P (1+(R/2)/100)2n

Quarterly C.I. :

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

Simple/compound interestDifference 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 }

Simple / Compound InterestThe 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

Simple / Compound InterestSolution:R = N and I = 16/25 P R = I * 100 / P*N R*N = 64 where R = N R = 8The rate of interest is 8%

Simple / Compound InterestA sum of S.I. at 13% per annum amounts to Rs. 2502.5 after 4 years. Find the sum.Sathyam Question

Simple / Compound InterestAnswer: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 interestAnswer : C.I S.I = P(R/100)2 80 = P*(8*8/100*100) The sum is Rs.12,500

Simple/Compound interest

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

Simple/Compound interestSolution:

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?

Simple/Compound interestSolution:

C.I S.I = P (R/100)2

Difference = Rs. 11.25


What will be the C.I on Rs. 15625 for 2 years at 4% per annum?

Simple/Compound interestSolution:A = P(1+R/100)n A = 15625 ((1+4/100)2 (1+4*1/2/100)) = 17238CI = A P = 17238 - 15625Compound interest = Rs. 1613

AverageAverage 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)

AverageThe 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?

AverageSoluton:Average weight of 5 persons = 38 KgTotal weight of the 5 persons =38*5=190Total weight including the boat weight is=52*6=312Weight of the Boat = 312-190= 122 Kg

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?

AverageSolution: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?

AverageSolution:30 students total age = 30*9=270

Including the teachers age = 31*10=310

Diff= 310-270=40 years

Average

What is the average of even numbers from 1 to 81?

AverageSolution:

Average = (last even + 2) / 2 = (80 + 2 ) / 2 = 41

AverageThe Average of 50 numbers is 38. If two numbers namely 45 and 55 are discarded, What is the average of the remaining numbers?

AverageSolution:

Total of 50 numbers = 50*38=1900

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

Average =1800/ 48 = 37.5

Permutations and CombinationsFactorial Notation:

n! = n(n-1)(n-2).3.2.1 Number of Permutations:

n!/(n-r)!

Combinations:

n!/r!(n r)!

Permutations and CombinationsThe number of Combinations of n things taken r at a time in which p particular things will always occur is n-pCr-pThe number of Combinations of n things taken r at a time in which p particular things never occur is n-pCr

Permutations and CombinationsA 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

Permutations and CombinationsSolution:n = 10r = 3n 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 ?Permutations and Combinations

Permutations and CombinationsSolution:It is selection so use combination formulaProgrammers and managers = 6C3 * 4C2 = 20 * 6 = 120Total number of ways = 120 ways.

Permutations and CombinationsA man has 7 friends. In how many ways can he invite one or more of them to a party?

Permutations and CombinationsSolution: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 = 127Number of ways is 127

Permutations and Combinations

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?

Permutations and CombinationsSolution:

5! = 120 ways

Permutations and Combinations

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

Permutations and CombinationsSolution:

9C2 * 9C2 = 1296

Permutations and Combinations 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?

Permutations and CombinationsSolution: 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

PercentageTwo numbers are respectively 30% & 40% less than a third number. What is second number as a percentage of first?

PercentageSolution: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 %

PercentageAn 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%

PercentageAnswer:

Percentage of Area Change=( X +Y+ XY/100)% =20+10+20*10/100 =32%Increase 32%

Percentage

If As income is 40% less than Bs income, then how much percent is Bs income more than As income?

PercentageAnswer: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?

PercentageSolution:Fresh grapes contain 10% pulp20 kg fresh grapes contain 2 kg pulpDry grapes contain 80% pulp.2 kg pulp would contain = 2/0.8 =20/8 = 2.5kgHence 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?

PercentageAnswer:Given A is 20% of C and B is 25% of CA of B is = (20*100%)/25 = 80%

A is 80% of B

Boats and streamsUp stream against the streamDown stream along the streamu = speed of the boat in still water v = speed of stream Down stream speed (a)= u+v km / hrUp stream speed (b)=u-v km / hru = (a+b) km/hrV = (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 StreamsSolution:Downstream speed (a) = 40/5 = 8 km/hUpstream speed (b)= 21/7 = 3 km/h

Speed of the boat =1/2(a+b)= 5.5 km/hSpeed of the river=1/2(a-b) = 2.5 km/h

Boats and StreamsA 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 StreamsSolution:Speed of the boat in upstream = 40/8 = 5 km/hrSpeed of the boat in downstream= 36/6 =6km/hr Speed of the boat in still water = (5+6) / 2 = 5.5 km/hr

Boats and streamsA boats crew rowed down a stream from A to B and up again in 7 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 streamsSolution: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 x/8 +4x/8 = 15/2 5x / 8 = 15/2 5x = 15/2 * 8 5x = 60 x =12

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 StreamsSolution: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 = Speed of Downstream = 8,Speed of upstream = 6Rate of the stream = (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 StreamsAnswer: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 DistanceSpeed:- Distance covered per unit time is called Speed. Speed = Distance / Time Distance = Speed * Time Time = Distance / Speed

Time and Distance Distance covered Time (direct variation).

Distance covered speed (direct variation).

Time 1/speed (inverse variation).

Time and DistanceSpeed 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 = 6The 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 km80/x 80/(x+10) = 16/60By 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

By walking at of his usual aped, a man reaches office 20 minutes later than usual. Find his usual time?

Time and DistanceSolution: 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 DistanceSolution: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 ?

Time and Distance (Trains)Solution:Let x be the speed of the second trainS1 / S2 = T2/T1160/x = 16/9160/x = 4/3 x = 120The speed of second train is 120km/hr.

Time and Distance (Trains)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

Time and Distance (Trains)Answer:Relative speed = x y = 45 40 = 5 km/h=5*5/18= 25/18 m/sFor 25/18 m it takes 1 secFor 350 m it takes = 18 * 350 /25 =252Time taken = 252 seconds

Time and Distance (Trains)

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 ?

Time and Distance (Trains)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

Profit and LossGain =S.P-C.PLoss =C.P-S.PLoss 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

A man buys an article for Rs. 27.50 and sells it for Rs. 28.60. Find his gain percentage.

Sathyam Question

Profit and LossAnswer:

Gain 4%

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%


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 LossSundeep 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 LossSolution: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 LossA 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 LossSolution:

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

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 LossSolution:S.P = Rs.660 Profit = 10 %C.P = 110 / 100*660 = Rs.600Let x be M.P75% of x = 600 x = 800Marked Price = 800

Profit and LossA shopkeeper gains the cost of 8 meters of thread by selling 40 meters of thread. Find his gain percentage?

Profit and LossAnswer: 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 LossSolution:Resultant profit = (x + y + xy/100)

The resultant profit is 3.5%

Profit and LossAditya 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 = 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 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 LossSolution: Let C.P be Rs. 100. Hence S.P = Rs. 120Gain = S.P C.P = 360 120 = 260Gain =Rs. 260

Profit and LossA 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 LossSolution:

C.P of 200 kg = (80*13.5 + 120*16) =3000S.P of 200 kg = 116% of C.P = 3480The selling price of 1 kg is Rs. 17.40

Profit and LossA 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


The maximum marks = 100

CalendarOdd days:0 = Sunday1 = Monday2 = Tuesday3 = Wednesday4 = Thursday5 = Friday6 = Saturday

CalendarMonth code: Ordinary yearJ = 0 F = 3M = 3 A = 6M = 1 J = 4J = 6 A = 2S = 5 O = 0N = 3 D = 5 Month code for leap year after Feb. add 1.

CalendarOrdinary year =( A + B + C + D) -2 -----------------------take remainder 7

Leap year = (A + B + C + D) 3 ------------------------- take remainder 7

Calendar

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

Calendar

Solution:A = 2007 / 7 = 5B = 2007 / 4 = 501 / 7 = 4C = 30 / 7 = 2D = 5 ( A + B + C + D) -2 = ----------------------- 7 = ( 5 + 4 + 2 + 5) -2 ----------------------- = 14/7 = 0 = Sunday 7

CalendarWhat was the day of the week on 13th May, 1984?

Calendar

Solution:A = 1984 / 7 = 3B = 1984 / 4 = 496 / 7 = 6C = 13 / 7 = 6D = 2 (A + B + C + D) -3 = ----------------------- 7 (3 + 6 + 6 + 2) -3 = ----------------------- = 14/7= 0, Sunday. 7

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

CalendarWhat 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

Clocks

Clock:In every minute, the minute hand gains 55 minutes on the hour handIn 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

Find the angle between the minute hand and hour hand of a clock when the time is 7:20.

Clocks

Solution: = 30h (11m/2) = 30 (7) 11 20/2 = 210 110 = 100Angle between 7: 20 is 100o

Clocks

How many times in a day, the hands of a clock are straight?

ClocksSolution:

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 oclock will the hands of a clock be in the same straight line but, not together?

ClocksSolution:h = 7 = 30h 11m/2180 = 30 * 7 11 m/2On simplifying we get ,5 5/11 min past 7

Clocks

At what time between 5 and 6 oclock will the hands of a clock be at right angles?

ClocksSolution:h = 5 90 = 30 * 5 11m/2Solving10 10/11 minutes past 5

Clocks

Find the angle between the two hands of a clock at 15 minutes past 4 oclock

ClocksSolution:

Angle = 30h 11m/2 = 30*4 11*15 / 2The angle is 37.5o

Clocks

At what time between 5 and 6 oclock are the hands of a clock together?

ClocksSolution: h = 5

O = 30 * 5 11m/2 m = 27 3/11Solving

27 3/11 minutes past 5

Data InterpretationIn 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.

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 100C. A. only 150 8 16 6 180Costing only 90 10 37 3 140C. A. and Costing 70 2 7 1 80

__________________________________________________________________On the basis of the above table, answer the following questions:

Data InterpretationThe 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 InterpretationAnswer:

Percentage of science students over commerce students in all courses = 35.1%

Data InterpretationWhat is the average number of girls in all courses ? (a) 15 (b) 12.5 (c) 16 (d) 11


Average number of girls in all courses = 50 / 4 = 12.5

Data InterpretationWhat is the percentage of boys in all courses over the total students? (a) 90 (b) 80 (c) 70 (d) 76


Percentage of boys over all students = (450 x 100) / 500 = 90%

Data SufficiencyFind given data is sufficient to solve the problem or not.If statement I alone is sufficient but statement II alone is not sufficientIf statement II alone is sufficient but statement I alone is not sufficientIf both statements together are sufficient but neither of statement alone is sufficient.If both together are not sufficient

Data SufficiencyWhat is Johns age?In 15 years will be twice as old as Dias would beDias was born 5 years ago. (Wipro)

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

Data SufficiencyWhat is the distance from city A to city C in kms?City A is 90 kms from city B.City B is 30 kms from city C

Data SufficiencyAnswer:

d) If both together are not sufficient

Data SufficiencyIf A, B, C are real numbers, Is A = C? A B = B CA 2C = C 2B

Data SufficiencyAnswer:D . If both together are not sufficient

Data SufficiencyWhat is the 30th term of a given sequence?The first two term of the sequence are 1, The common difference is -1/2

Data SufficiencyAnswer:If statement I alone is sufficient but statement II alone is not sufficient

Data SufficiencyWas Avinash early, on time or late for work?He thought his watch was 10 minute fast.Actually his watch was 5 minutes slow.

Data SufficiencyAnswer:D. If both together are not sufficient

Data SufficiencyWhat is the value of A if A is an integer?A4 = 1A3 + 1 = 0

Data SufficiencyAnswer:B. If statement II alone is sufficient but statement I alone is not sufficient

CubesA 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?

CubesAnswer:3*3*3 = 27All the outer surface are painted with colour. 26 One inch cubes are painted at least one surface.

CubesA 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

CubesAnswer:

= 8

CubesA cube of side 5 cm is divided into 125 cubes of equal size. It is painted on all 6 sides.How many cubes are coloured on only one side?How many cubes are coloured on only two side? How many cubes are coloured on only three side?How many cubes are not coloured?

CubesAnswer:5436827

CubesA 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..

CubesHow many cubes are coloured with Red alone?How many cubes are coloured orange and Red alone?How many cubes are coloured with three different colours?How many cubes are not coloured?How many cubes are coloured green and Red alone?

CubesAnswer:88888

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.Find the no. of cubes not coloured?Find the no. of cubes coloured blue alone?Find the no. of cubes coloured blue & pink & yellow?Find the no. of cubes coloured blue & pink ? Find the no. of cubes coloured yellow & pink ?

CubesAnswer:271841212

Venn DiagramIf 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 DiagramSolution: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 18n( X n Y) = 5

Venn DiagramIf 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 DiagramAnswer: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 DiagramIf 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 DiagramAnswer: 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 DiagramIn 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 = 150No. of people can speak Hindi only _ = n ( H n E) = n ( H) n( H n E) = 750 150 = 600

Venn DiagramIn 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 DiagramSolution:No. of students who passed in one or more subjects = 11+ 9 + 13 + 17 + 15 + 19 + 7 = 91No of students who failed in all the subjects = 100 -91 = 9

Venn DiagramIn 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 DiagramAnswer: 3 of them studied both Latin and Greek.

THANK YOUSreenivasa Chowdary Nellore M.B.A, [Director]NARENS Institute for Competitive Examinations,HYDERABAD.E-mail: [email protected]: + 91 99123 31989, + 91 90593 14623

quantitative ability - level b

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