Presentation on theme: "Aptitude Numerical Reasoning"— Presentation transcript:
1Aptitude Numerical Reasoning Blue LotusAptitudeNumerical Reasoning
2Numerical Reasoning Problems on Numbers Problems on Ages Ratio and ProportionAlligation or MixtureChain RulePartnershipVenn Diagram
3Numerical Reasoning Area and Volume Probability Time and Work (Pipes) SI and CIAveragePermutation and CombinationPercentageCubes
4Numerical Reasoning Boats and Streams Time and Distance (Trains) Data SufficiencyProfit and LossCalendarClocksData Interpretation
5Problems 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.
6Problems on Numbers Sn = a(rn – 1)/(r-1); Arithmetic Progression: The nth term of A.P. is given byTn = a + (n – 1)d;Sum of n terms of A.PSn = n/2 *(a + L) or n/2 *[2a+(n-1)d)]Geometrical Progression:Tn = arn – 1.Sn = a(rn – 1)/(r-1);
10Problem 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.
11Problem on Numbers Solution: Side of the square is x + 2 Side of the triangle 2xP = 4(x + 2)= 4x +8Perimeter = 3*2x= 6x4x + 8 = 6xx = 4
12Problem 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)
13Problem on Numbers Solution: Total no of children = 30 * 16 = 480 No. of column of 24 children each = 480 / 24= 20
14Problem 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?
15Problem on Numbers Solution: Let population be = x Males = 5 x/9 Married 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%
16Problems on NumbersHow many terms of the A.P. 1, 4, 7…. are needed to give the sum 715 ?
17Problems on Numbers 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 =22Number of terms needed is 22
18Problems on NumbersThe 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?
19Problems on Numbers Solution: x+ y = 8 --------------------1 10y+x = 10x+y+5410y + x - 10x – y = 54-x + y =Solve equation 1&2x= 1, y = 7Required number = 10*1 +7 =17
20Problems 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.
21Problems on Ages x-y = 10; x = 10 + y x- 5 = 2(y-5) y + 10 -5 = 2y -10 Their present ages are 15 years and 25 years.
22Problems 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 ?
23Problems on Ages Solution: Three person’s age ratio = 4:7:9 Sum of their age = 56, after 8 yearstheir sum of their ages = 80A’s age = 4/20 *80 = 16B’s age = 7/20 *80 = 28C’s age = 9/20 *80 = 36Their present ages are 16, 28 and 36.
24Problems on AgesFather’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.
25Problems on Ages Solution: Father age = 3(x+y) F+20 = x+20+y+20 3x+3y+20 = x+y+403x +3y = x+y3x-x+3y – y =202x+2y = 20x+y = 10F = 3*10 =30The father’s age is 30.
26Problems 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?
27Problems on Ages Solution: J = 2Q 1 (J – 4) – (Q +3) = 12 2 Solve 1 &2 Sum of their ages = = 57
28Problems on AgesTen 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?
29Problems on Ages Solution: Let Chandrawathi’s age 10 years ago x Her mother’s age 10 years ago 4x(4x ) = 2(x+10+10)x = 10Chandrawathi is x +10 = 20 years
30Ratio and ProportionRatio: The Relationship between two variables is ratio.Proportion: The relationship between two ratios is proportion.
31Ratio and Proportion The two ratios are a : b and the sum nos. is x ax bxanda + b a + bSimilarly for 3 numbers a : b : c
32Ratio 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)
34Ratio 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 ?
35Ratio and Proportion Solution: Ajay : Aman = 8 : 15 In next ratio Aman is 5 to make that 15 multiply by 3 that is ratio ofAman : 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 = 8:15:24:30Suman ratio = 24/77
36Ratio and Proportion 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 : 2and alloy B has tin and copper in the ratio 1 : 4,then what would be the amount of tin in the newalloy?
37Ratio and Proportion Solution : The amount of tin in both alloys = (60*2/5) + (100*1/5)= 44 kg
38Ratio and Proportion In a class composed of x girls and y boys what part of the class is composed of girls?(Satyam Question)
39Ratio and ProportionAnswer:Ratio = x : (x+y) = x / x+y
40Ratio and ProportionIn 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.
41Ratio and Proportion Solution: Let the number of males is 5x and females is 3x and Total is 8x5x – 3x = 40x = 20Total number of workers in the factory = 8x=8*20 = 160
42Ratio and ProportionA 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 ?
43Ratio and Proportion Solution: Let quantity of milk be 5x, and water be x then,5x = 5xHence x=5Quantity of milk = 5x5 = 25 liters.
44Alligation or Mixture (Quantity of cheaper / Quantity of costlier) (C.P. of costlier) – (Mean price)=(Mean price) – (C.P. of cheaper)
45Alligation or Mixture Cost of Cheaper Cost of costlier c d Cost of Mixturemd-m m-c(Cheaper quantity) : (Costlier quantity) = (d – m) : (m – c)
46Alligation or MixtureA 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?
47Alligation or Mixture Solution: 7 17 10 (17-10) (10-7) 7 : 3 10(17-10) (10-7):The ratio is 7:3The quantity of 2nd kind = 3/10 of 100kg= 30kg
48Alligation 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
49Alligation or Mixture Solution: B A 4 9 7 2 3 4 972 3Brand A required = 3*40 / 5 = 24kg
50Alligation or MixtureA man buys two cows for Rs 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?
51Alligation or Mixture Solution: -6% 7.5% 7.5 6 Ratio is 5 : 4 -6% 7.5%Ratio is 5 : 4The cost of first cow = 5*1350/9 =Rs. 750The cost of second cow = Rs. 600
52Alligation or MixtureHow 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 per liter?
53Alligation or Mixture Solution: The cost of water is 0. W M 0 540 420 420Ratio is 4 : 14The amount of water added for 14 liters is 4 liters.
54Alligation or Mixture There are 65 students in a class, 39 rupees are distributed among them so that each boy gets80 paise and girl gets 30 paise. Find the numberof boys and girls in that class.
55Alligation or Mixture Solution: “Money per boy or girl” is considered. Per student = 3900/65 = 60 paise.Girls BoysGirls : Boys = 2 : 3Number of boys = 39Number of girls = 26
56Chain RuleDirect Proportion :A BIndirect Proportion:A B
57Chain Rule A man completes 5/8 of a job in 10 days. At this ratio how many more days will it take forhim to finish the job?
58Chain Rule Answer: Jobs Days 5/8 10 3/8 x 5/8 103/8 x5x / 8 = (10*3) / 8 (Direct )The number of days to complete = 6
59Chain 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 workto be finished. How many men were there originally?(Satyam Question)
60Chain Rule Solution: Men Days x 100 (x-10) 110 (indirect ) The number of men present originally = 110
61Chain 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)
62Chain Rule Solution: 3 women = 2 men Men Days Hours 15 21 8 14 x 6 14 x 6x = 15*8*6Sx = 30The number of days required is 30.
63Chain Rule 6 cats kill 6 rats in 6 minutes. How many cats will be needed to kill 100 rats in 50 minutes?(Satyam question)
64Chain Rule Solution: Cats Rats Minutes 6 6 6 x 100 50 xx/6 = (100*6)/(6*50)The number of cats needed would be 12.
65Chain Rule 39 persons can repair a road in 12 days, working 5 hours a day. In how many days will 30persons working 6 hours a day, complete thesame work?
66Chain Rule Solution : - Men days hours 39 12 5 30 x 6 xMore hours less days ( inverse proportion )Less men more days ( inverse proportion )x * 39 * 12 = * x =* 30 x = 13 days.
67Partnership 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
68PartnershipA 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)
69Partnership Solution : A : B : C 100 : 65 : 35 100 : 65 : 35Total = = 200C’s share = 560(35*x)/200 = 560X = 3200The sum invested is Rs. 3200
70PartnershipA 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 ?
71Partnership Solution: A and B = 3 : 2 Let Profit be x x – 5% x- 5x/100 = 95x/100A’s share is = 3/5 * 95x/100 = 85519x/100 =28519x = 28500x = 26500/19 = 1500Total profit is Rs.1500
72PartnershipA,B,C subscribe Rs. 50,000 for a business. A subscribes Rs more than B and B subscribes Rs more than C. Out of a total profit of Rs. 35,000 how much does A receive?(TCS Question)
73Partnership Answer : Ratio A B C 9000 + x : 5000 + x : x On solving we get x = 12000Ratio A B C21 : 17 :12 Total = 50A receives = (21* 35000)/50 = Rs
74PartnershipA began a business with Rs. 450 and B jointed with Rs When did B join if the profit at the end of the year was divided between them in the ratio 2 : 1?(Caritor Question)
75Partnership Answer : Ratio A : B 450x12 : 300 (12 – x ) On solving x = 3B invested money after 3 months
76PartnershipA and B enter into partnership for a year. A contributes Rs.1500 and B Rs After 4 months, they admit C who contributes Rs If B withdraws his contribution after 9 months, find their profit share at the end of the year? (In the ratio)
77Partnership Solution: A: B: C = 1500*12 : 2000*9 : 2250*8 = : : 18000= 1: 1 : 1Profit share at the end of the year is 1: 1: 1
78Time and WorkIf A can do a piece of work in n days, then A’s 1 day’s work = 1 / nIf A is thrice as B, then:Ratio of work done by A and B = 3 : 1Ratio of times taken by A and B= 1 : 3
79Pipes 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
80Pipes and CisternsP1 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/yP1 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
81Time and Work10 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 ?
82Time and Work Solution: 10 men = 15 days means 1day work = 1/15 10 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.
83Time 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 ?
84Time and Work Solution: A + B = 30 days = 1/30 B + C = 40 days = 1/40 C+ A = 60 days = 1/60All work togetherA+B+C+B+C+A = 1/30 +1/40 +1/602(A+B+C) = 1/30+1/40+1/60= /120 = 9/120= 9/240 = 3/80= 26 2/3A, B and C can finish the work in 26 2/3 days
85Time 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?
86Time and Work Solution: A = 1/30, B = 1/50, C = 1/40 A+C = 1/30+1/40 = 7/120Work done by A & B and A & C = ( 8/ /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 /7518th day A&C are working = 120 / 7 * 4 / 75 = 32/35They will finish the work in 17 32/35 days
87Time and Work A can do a piece of work in 12 days. B is 60% more efficient than A. Find the number of daysB takes to do the same piece of work.
88Time and Work Answer : Work Days 100 12 160 x xOn solving , x =( 100*12) / 160 = 7 ½ DaysB will take 7½ days.
89Time and Work Seven men can complete a work in 12 days. They started the work and after 5 days, twomen left. In how many days will the work becomplete the remaining work.
90Time and Work Solution: (7*12) men can complete the work in 1 day 1 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
91Time 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)
92Time and Work (Pipes)Answer:The time taken is 17 1/7 minutes.
93Time 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 ?
94Time and Work (Pipes) Solution: All the tap work together = 1/12 + 1/15 - 1/20= 5/60 + 4/60 – 3/60= 6/60= 1/10The waste pipe can empty the cistern in 10 minutes.
95Time 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 ?
96Time 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.
97Time and Work (Pipes) Two pipes A and B can fill a tank in 20 min. and 40 min. respectively. If both the pipes are openedsimultaneously, after how much time A should beclosed so that the tank is full in 10 minutes?
98Time and Work (Pipes) Let B be closed after x minutes. Then, 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 =x = 20 min.A must be closed after 20 minutes.
99Time and Work (Pipes)Two taps A and B can fill a tank in 6 hours and 4hours respectively. If they are opened on alternatehours and if pipe A is opened first, in how manyhours, the tank shall be full?
100Time and Work (Pipes) 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 filledby A in 1 hour.Total time = 4+1 =5
101Area and Volume Cube: Let each edge of the cube be of length” a”then, Volume = a3cubic unitsSurface area= 6a2 sq.units.Diagonal = √3 a units.
102Area and Volume Cylinder: Let each of base = r and height ( or length) = h.Volume = πr2hSurface area = 2 πr h sq. unitsTotal Surface Area = 2 πr ( h+ r) units.
103Area and Volume Cone: Let radius of base = r and height=h, then Slant height, l = √h2 +r2 unitsVolume = 1/3 πr2h cubic unitsCurved surface area = πr l sq.unitsTotal surface area = πr (l +r)
104Area and Volume Sphere: Let the radius of the sphere be r. then, Volume = 4/3 πr3Surface area = 4 π r2sq.units
105Area and Volume Circle: A= π r 2 Circumference = 2 π r Square: A= a 2 Perimeter = 4aRectangle: A= l x bPerimeter= 2( l + b)
106Area and Volume Triangle: A = ½ * base * height Equilateral = √3/4*(side)2Area of the Scalene TriangleS = (a + b + c) / 2A = √ s*(s-a) * (s - b)* (s - c)
107Area 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
108Area and Volume Solution: Area of triangular field = ½ * 3.2 * 2.8 m2 Cost = Rs.100 * 4.48= Rs.448..
109Area and VolumeArea of a rhombus is 850 cm2. If one of its diagonal is 34 cm. Find the length of the other diagonal.
110Area and Volume Solution: 850 = ½ * d1 * d2 = ½ * 34 * d2 = 17 d2 = 50 cmSecond diagonal = 50cm
111Area and VolumeA 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 ?
112Area and Volume Solution: No. of Boxes = (25*42*60) /( 7*6*5) = 300 300 boxes of cereal box can be placed.
113Area and Volume If the radius of a circle is diminished by 10%, what is the change in area in percentage?
114Area and Volume Solution: x = - 10 , y = - 10 = x + y +xy/100 % = – (10*10/100) %= - 19%Area changed = 19%.
115Area and Volume A circular wire of radius 42 cm is bent in the form of a rectangle whose sides are in the ratioof 6:5. Find the smaller side of the rectangle?
116Area and Volume Solution: length of wire = 2 πr = (22/7*14*14)cm Perimeter of Rectangle = 2(6x+5x) cm= 22xcm22x =264x = 12 cmSmaller side = (5*12) cm = 60 cm
117Area and Volume If the length of a rectangle is increased by 30% decreased by 20%, then what percent of its areawould increase?(TCS Question)
118Area and Volume Solution: x = 30 y = - 20 (Decreased ) Using formula = x + y + (xy/100)%The increase would be 4%
119Area 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)
120Area and Volume Answer: The equation is 1 / 2*x(x+10) = 5500
121Area and Volume The length and breadth of a rectangular plot are in the ratio 7 : 5. If the length is reduced by 5 mand breadth is increased by 2 m then the area isreduced by 65 m2. Find the length and breadth ofthe rectangular plot.(Caritor Question)
122Area and Volume Answer: Difference in areas = 65 (7x *5x) – (7x – 5) (5x + 2) = 65x = 5The length of the rectangle = 7x = 7*5 = 35 mThe breadth of the rectangle = 5x = 5*5= 25 m
126ProbabilitySam 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 ?
127Probability 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- 1/49The probability = 48/49
128ProbabilityThere 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
162Average The Average of 50 numbers is 38. If two numbers namely 45 and 55 are discarded, Whatis the average of the remaining numbers?
163Average Solution: Total of 50 numbers = 50*38=1900
164Permutations and Combinations Factorial Notation:n! = n(n-1)(n-2)….3.2.1Number of Permutations:n!/(n-r)!Combinations:n!/r!(n –r)!
165Permutations and Combinations The 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
166Permutations and Combinations 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 …
167Permutations and Combinations Solution:n = 10r = 3n P r = n!/(n-r)!= 10! / (10-3)!= 10! / 7!= 8*9*10= 720Number of ways is 720.
168Permutations and Combinations 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 ?
169Permutations and Combinations Solution:It is selection so use combination formulaProgrammers and managers = 6C3 * 4C2= 20 * 6 = 120Total number of ways = 120 ways.
170Permutations and Combinations A man has 7 friends. In how many ways canhe invite one or more of them to a party?
171Permutations and Combinations 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= 127Number of ways is 127
172Permutations 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?
173Permutations and Combinations Solution:5! = 120 ways
174Permutations and Combinations In a chess board there are 9 vertical and 9 horizontal lines. Find the number of rectangles formed in the chess board.
175Permutations and Combinations Solution:9C2 * 9C2 = 1296
176Permutations 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?
177Permutations and Combinations 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
178Percentage By a certain Percent, we mean that many hundredths. Thus, x Percent means x hundredths, written as x%
179PercentageIf Length is increased by X% and Breadth is decreased by Y% What is the PercentageFinding out of Hundred.Increase or Decrease in Area of the rectangle?Formula: X+Y+ XY/100 %Decrease 20% means -20
180PercentageTwo numbers are respectively 30% & 40% less than a third number. What is second number as a percentage of first?
181Percentage 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 /10y = 600 / 7y = 85 5/785 5/7%
182Percentage After having spent 35% of the money on machinery, 40% on raw material, 10% onstaff, a person is left with Rs.60,000. Thetotal amount of money spent on machineryand the raw material is?
183Percentage Solution: Let total salary =100% Salary = 100 Spending: Machinery + Raw material + staff = = 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, 00,000Rs. 3, 00,000
184Percentage If the number is 20% more than the other, how much percent is the second number lessthan the first?
185Percentage Solution: Let x =20 = x / (100+x) *100% = 20 / 120 *100% = 16 2/3%The percentage is 16 2/3 %
186Percentage An empty fuel tank of a car was filled with A type petrol. When the tank was half empty, itwas filled with B type petrol. Again when thetank was half empty, it was filled with A typepetrol. When the tank was half – empty again, it was filled with B type petrol. What is thepercentage of A type petrol at present in the tank?
187Percentage Solution: Let capacity of the tank be 100 liters. Then, Initially: A type petrol = 100 litersAfter 1st operation:A = 100/2 = 50 liters, B = 50 litersAfter 2nd operation:A = 50 / 2+50 = 75 liters, B = 50/2 = 25 litersAfter 3rd operation:A = 75 / 2 = 37.5 liters, B = 25/2 +50 = 62.5 litersRequired Percentage = 37.5%
188Percentage Find the percentage increase in the area of a rectangle whose length is increased by 20% andbreadth is increased by 10%
189Percentage Answer: Percentage of Area Change=( X +Y+ XY/100)% = *10/100=32%Increase 32%
190Percentage If A’s income is 40% less than B’s income, then how much percent is B’s income more than A’sincome?
192PercentageFresh 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?
193Percentage Solution: Fresh grapes contain 10% pulp 20 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.
194PercentageIf A is 20% of C and B is 25% of C then what percentage is A of B?
195Percentage 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
196Boats and streams Up stream – against the stream Down stream – along the streamu = speed of the boat in still waterv = speed of streamDown stream speed (a)= u+v km / hrUp stream speed (b)=u-v km / hru = ½(a+b) km/hrV = ½(a-b) km / hr
197Boats and StreamsA 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.
198Boats and Streams Solution: Downstream speed (a) = 40/5 = 8 km/h Upstream speed (b)= 21/7 = 3 km/hSpeed of the boat =1/2(a+b)= 5.5 km/hSpeed of the river =1/2(a-b) = 2.5 km/h
199Boats and Streams A boat goes 40 km upstream in 8 hours and 36 km downstream in 6 hours. Find the speed of theboat in still water in km/hr?
200Boats and Streams Solution: Speed of the boat in upstream = 40/8 = 5 km/hrSpeed of the boat in downstream= 36/6 =6km/hrSpeed of the boat in still water = (5+6) / 2= 5.5 km/hr
201Boats and streamsA 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 ?
202Boats 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 = Timex/8 + x/2 = 7 ½x/8 +4x/8 = 15/25x / 8 = 15/25x = 15/2 * 85x = 60x =12
203Boats and StreamsA 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?
204Boats 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
205Boats and StreamsA 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?
206Boats 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) km5(x+1) = 7(x-1)x = 6Speed of the swimmer in still water = 6 kmph.
207Time and Distance Speed:- Distance covered per unit time is called Speed.Speed = Distance / TimeDistance = Speed * TimeTime = Distance / Speed
208Time and Distance Distance covered α Time (direct variation). Distance covered α speed (direct variation).Time α 1/speed (inverse variation).
209Time 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 traveledTotal time taken
210Time and DistanceIf 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
211Time and Distance Solution: x/5-x/6 = 12/60 ( Difference 7+5=12) Solving the equation,x = 6The distance covered is 6 km.
212Time and DistanceThe 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?
213Time 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
214Time and DistanceIf 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
215Time and Distance Solution: Average Speed = 2xy / (x + y) The average speed is 48 kmph.
216Time and Distance By walking at ¾ of his usual aped, a man reaches office 20 minutes later than usual.Find his usual time?
217Time and Distance Solution: Usual time = Numerator * late time = 3*20 = 60
218Time and DistanceIf a person walks at 14 km/hr instead of 10 km/hr, he would have walked 20 km more. Find his actual distance traveled?
219Time and Distance Solution: Let the distance be x km. Then, x/10 = (x+20) / 1414x = 10x +2004x = 200x = 50 km.
220Time 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 ?
221Time and Distance (Trains) Solution:Let x be the speed of the second trainS1 / S2 = √T2/T1160/x = √16/9160/x = 4/3x = 120The speed of second train is 120km/hr.
222Time 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
223Time 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
224Time 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 ?
225Time and Distance (Trains) Solution:To reach 11 poles it takes 24 secFor 19 poles it will take x timePoles timex11x = 19 * 24x = 19* 24 /11x = secIt reaches the last pole in sec
226Profit and Loss 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.
227Profit and Loss A man buys an article for Rs. 27.50 and sells it for Rs Find his gain percentage.Sathyam Question
229Profit and LossMr. Ravi buys a cooler for Rs For how much should he sell so that there is a gain of 8%
230Profit and Loss Answer: C.P = 4500 Profit = 8/100*4500 = 360 S.P = C.P + Profit = = 4860He should sell it for Rs
231Profit 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 ?
232Profit and Loss Solution: C.P of two CDs =Rs.380 S.P of 1st CD = x – 22 x /100S.P of 2nd CD = y + 12y /100SP1 = SP2x – 22x/100 = y – 12y/100x = 56/39 yx + y = 38056/39y + y = 380y = 156x = 224Cost of the two CDs are Rs. 224 and Rs.156
233Profit 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?
234Profit and Loss Solution: Let us take C.P of goods = 100 The Marked price = 130=1/2* /4*0.85* /4 *0.7*130=15.375%
235Profit and LossRajesh 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?
236Profit and Loss Solution: S.P = Rs.660 Profit = 10 % C.P = 110 / 100*660 = Rs.600Let x be M.P75% of x = 600x = 800Marked Price = 800
237Profit and Loss A shopkeeper gains the cost of 8 meters of thread by selling 40 meters of thread. Findhis gain percentage?
252Calendar Month code: Ordinary year J = 0 F = 3 M = 3 A = 6 M = 1 J = 4 J = A = 2S = O = 0N = D = 5Month code for leap year after Feb. add 1.
253Calendar Ordinary year =( A + B + C + D) -2 take remainder7Leap year = (A + B + C + D) – 3take remainder
254CalendarWhat is the day of the week on 30/09/2007?
255Calendar Solution: A = 2007 / 7 = 5 B = 2007 / 4 = 501 / 7 = 4 ( A + B + C + D) -2=7= ( ) -2= 14/7 = 0 = Sunday
256CalendarWhat was the day of the week on 13th May, 1984?
257Calendar Solution: A = 1984 / 7 = 3 B = 1984 / 4 = 496 / 7 = 6 (A + B + C + D) -3=7( ) -3= = 14/7= 0, Sunday.
258CalendarOn what dates of April 2005 did Sunday fall?
259CalendarSolution: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( ) -2= = 12 / 7 = 5 = Friday.1st is Friday means Sunday falls on 3,10,17,24
261Calendar A = 1986 / 7 = 5 B = 1986 / 4 = 496/7 = 6 C = 5 / 7 = 5 D = 0 Solution:A = 1986 / 7 = 5B = 1986 / 4 = 496/7 = 6C = 5 / 7 = 5D = 0A + B + C + D -2=7= 14 / 7 = Sunday
262ClocksClock: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
263ClocksFind the angle between the minute hand and hour hand of a clock when the time is 7:20.
265ClocksHow many times in a day, the hands of a clock are straight?
266ClocksSolution:In 12 hours, the hands coincide or are in opposite direction 22 times a day.In 24 hours, the hands coincide or are in opposite direction 44 times a day.
267ClocksHow many times do the hands of a clock coincide in a day?
268ClocksSolution:In 12 hours, the hands coincide or are in opposite direction 11 times a day.In 24 hours, the hands coincide or are in opposite direction 22 times a day.
269ClocksAt what time between 7 and 8 o’clock will the hands of a clock be in the same straight line but, not together?
270Clocks Solution: h = 7 = 30h – 11m/2 180 = 30 * 7 – 11 m/2 On simplifying we get ,5 5/11 min past 7
271Clocks At what time between 5 and 6 o’clock will the hands of a clock be at right angles?
272Clocks Solution: h = 5 90 = 30 * 5 – 11m/2 Solving 10 10/11 minutes past 5
273Clocks Find the angle between the two hands of a clock at 15 minutes past 4 o’clock
274Clocks Solution: Angle = 30h – 11m/2 = 30*4 – 11*15 / 2 The angle is 37.5o
275Clocks At what time between 5 and 6 o’clock are the hands of a clock together?
276Clocks Solution: h = 5 O = 30 * 5 – 11m/2 m = 27 3/11 Solving 27 3/11 minutes past 5
277Data 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.
278__________________________________________________________________ 1. The following table gives the distribution of students according to professional courses:__________________________________________________________________Courses Faculty___________________________________Commerce Science TotalBoys girls Boys girls___________________________________________________________Part time managementC. A. onlyCosting onlyC. A. and CostingOn the basis of the above table, answer the following questions:
279Data Interpretation The percentage of all science students over Commerce students in all courses isapproximately:(a) (b) 49.4(c) (d) 35.1
280Data Interpretation Answer: Percentage of science students over commerce students in all courses = 35.1%
281Data InterpretationWhat is the average number of girls in all courses ?(a) (b) 12.5(c) (d) 11
282Data Interpretation Answer: Average number of girls in all courses = 50 / 4= 12.5
283Data InterpretationWhat is the percentage of boys in all courses over the total students?(a) (b) 80(c) (d) 76
284Data Interpretation Answer: Percentage of boys over all students = (450 x 100) / 500= 90%
285Data 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
286Data Sufficiency What is John’s age? In 15 years will be twice as old as Dias would beDias was born 5 years ago (Wipro)
287Data Sufficiency Answer: c) If both statements together are sufficient but neither of statement alone is sufficient.
288Data Sufficiency What 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
289Answer: d) If both together are not sufficient Data SufficiencyAnswer:d) If both together are not sufficient
290If A, B, C are real numbers, Is A = C? A – B = B – C A – 2C = C – 2B Data SufficiencyIf A, B, C are real numbers, Is A = C?A – B = B – CA – 2C = C – 2B
291Answer: D . If both together are not sufficient Data SufficiencyAnswer:D . If both together are not sufficient
292Data Sufficiency What is the 30th term of a given sequence? The first two term of the sequence are 1, ½The common difference is -1/2
293Data Sufficiency Answer: If statement I alone is sufficient but statement II alone is not sufficient
294Data Sufficiency Was Avinash early, on time or late for work? He thought his watch was 10 minute fast.Actually his watch was 5 minutes slow.
295Answer: D. If both together are not sufficient Data SufficiencyAnswer:D. If both together are not sufficient
296What is the value of A if A is an integer? A4 = 1 A3 + 1 = 0 Data SufficiencyWhat is the value of A if A is an integer?A4 = 1A3 + 1 = 0
297Data Sufficiency Answer: B. If statement II alone is sufficient but statement I alone is not sufficient
298CubesA 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?
299CubesAnswer:3*3*3 = 27All the outer surface are painted with colour.26 One inch cubes are painted at least one surface.
300CubesA 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
302CubesA 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?
304CubesA 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..
305Cubes How 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?
307CubesA 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 ?
309Venn 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?
310Venn 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) = – 18n( X n Y) = 5
311Venn 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?
312Venn 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)= – 11 = 42
313Venn 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?
314Venn 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) – 10n (B) = 30
315Venn 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?
316Answer:n( H u E) = 1000, n (H) = 750, n (E) = 400,n( H u E) = n (H) + n (E) – n( H n E)1000 = – 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
317Venn 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.
318Venn Diagram Solution: No. of students who passed in one or more subjects= = 91No of students who failed in all the subjects= = 9
319Venn 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?
320Answer: 3 of them studied both Latin and Greek. Venn DiagramAnswer:3 of them studied both Latin and Greek.