Presentation on theme: "Aptitude Numerical Reasoning"— Presentation transcript:
1 Aptitude Numerical Reasoning Blue LotusAptitudeNumerical Reasoning
2 Numerical Reasoning Problems on Numbers Problems on Ages Ratio and ProportionAlligation or MixtureChain RulePartnershipVenn Diagram
3 Numerical Reasoning Area and Volume Probability Time and Work (Pipes) SI and CIAveragePermutation and CombinationPercentageCubes
4 Numerical Reasoning Boats and Streams Time and Distance (Trains) Data SufficiencyProfit and LossCalendarClocksData Interpretation
5 Problems on Numbers Division Algorithm: Dividend = the number to be divided.Divisor = the number by which it is divided.Dividend / Divisor = Quotient.Quotient * Divisor = Dividend.Quotient * Divisor + Remainder = Dividend.
6 Problems 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);
10 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.
11 Problem 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
12 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)
13 Problem on Numbers Solution: Total no of children = 30 * 16 = 480 No. of column of 24 children each = 480 / 24= 20
14 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?
15 Problem 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%
16 Problems on NumbersHow many terms of the A.P. 1, 4, 7…. are needed to give the sum 715 ?
17 Problems 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
18 Problems 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?
19 Problems 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
20 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.
21 Problems 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.
22 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 ?
23 Problems 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.
24 Problems 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.
25 Problems 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.
26 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?
27 Problems on Ages Solution: J = 2Q 1 (J – 4) – (Q +3) = 12 2 Solve 1 &2 Sum of their ages = = 57
28 Problems 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?
29 Problems 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
30 Ratio and ProportionRatio: The Relationship between two variables is ratio.Proportion: The relationship between two ratios is proportion.
31 Ratio 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
32 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)
33 Ratio and Proportion Answer : 10 : 20 : 30 2 : 3 : x Multiplying 20 : 60 : 30xx = 230x = 5
34 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 ?
35 Ratio 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
36 Ratio 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?
37 Ratio and Proportion Solution : The amount of tin in both alloys = (60*2/5) + (100*1/5)= 44 kg
38 Ratio and Proportion In a class composed of x girls and y boys what part of the class is composed of girls?(Satyam Question)
39 Ratio and ProportionAnswer:Ratio = x : (x+y) = x / x+y
40 Ratio 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.
41 Ratio 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
42 Ratio 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 ?
43 Ratio and Proportion Solution: Let quantity of milk be 5x, and water be x then,5x = 5xHence x=5Quantity of milk = 5x5 = 25 liters.
44 Alligation or Mixture (Quantity of cheaper / Quantity of costlier) (C.P. of costlier) – (Mean price)=(Mean price) – (C.P. of cheaper)
45 Alligation or Mixture Cost of Cheaper Cost of costlier c d Cost of Mixturemd-m m-c(Cheaper quantity) : (Costlier quantity) = (d – m) : (m – c)
46 Alligation 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?
47 Alligation 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
48 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
49 Alligation or Mixture Solution: B A 4 9 7 2 3 4 972 3Brand A required = 3*40 / 5 = 24kg
50 Alligation 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?
51 Alligation 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
52 Alligation 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?
53 Alligation 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.
54 Alligation 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.
55 Alligation 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
56 Chain RuleDirect Proportion :A BIndirect Proportion:A B
57 Chain 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?
58 Chain 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
59 Chain Rule A certain number of men can finish a piece of work in 100 days. If however there were 10men less it will take 10 days more for the workto be finished. How many men were there originally?(Satyam Question)
60 Chain Rule Solution: Men Days x 100 (x-10) 110 (indirect ) The number of men present originally = 110
61 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)
62 Chain 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.
63 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)
64 Chain 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.
65 Chain 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?
66 Chain 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.
67 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
68 PartnershipA 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)
69 Partnership Solution : A : B : C 100 : 65 : 35 100 : 65 : 35Total = = 200C’s share = 560(35*x)/200 = 560X = 3200The sum invested is Rs. 3200
70 PartnershipA 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 ?
71 Partnership 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
72 PartnershipA,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)
73 Partnership 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
74 PartnershipA 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)
75 Partnership Answer : Ratio A : B 450x12 : 300 (12 – x ) On solving x = 3B invested money after 3 months
76 PartnershipA 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)
77 Partnership 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
78 Time 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
79 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
80 Pipes 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
81 Time 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 ?
82 Time 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.
83 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 ?
84 Time 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
85 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?
86 Time 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
87 Time 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.
88 Time and Work Answer : Work Days 100 12 160 x xOn solving , x =( 100*12) / 160 = 7 ½ DaysB will take 7½ days.
89 Time 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.
90 Time 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
91 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)
92 Time and Work (Pipes)Answer:The time taken is 17 1/7 minutes.
93 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 ?
94 Time and Work (Pipes) Solution: All the tap work together = 1/12 + 1/15 - 1/20= 5/60 + 4/60 – 3/60= 6/60= 1/10The waste pipe can empty the cistern in 10 minutes.
95 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 ?
96 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.
97 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 openedsimultaneously, after how much time A should beclosed so that the tank is full in 10 minutes?
98 Time 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.
99 Time 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?
100 Time 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
101 Area 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.
102 Area 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.
103 Area 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)
104 Area and Volume Sphere: Let the radius of the sphere be r. then, Volume = 4/3 πr3Surface area = 4 π r2sq.units
105 Area and Volume Circle: A= π r 2 Circumference = 2 π r Square: A= a 2 Perimeter = 4aRectangle: A= l x bPerimeter= 2( l + b)
106 Area 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)
107 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
108 Area and Volume Solution: Area of triangular field = ½ * 3.2 * 2.8 m2 Cost = Rs.100 * 4.48= Rs.448..
109 Area and VolumeArea of a rhombus is 850 cm2. If one of its diagonal is 34 cm. Find the length of the other diagonal.
110 Area and Volume Solution: 850 = ½ * d1 * d2 = ½ * 34 * d2 = 17 d2 = 50 cmSecond diagonal = 50cm
111 Area 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 ?
112 Area and Volume Solution: No. of Boxes = (25*42*60) /( 7*6*5) = 300 300 boxes of cereal box can be placed.
113 Area and Volume If the radius of a circle is diminished by 10%, what is the change in area in percentage?
114 Area and Volume Solution: x = - 10 , y = - 10 = x + y +xy/100 % = – (10*10/100) %= - 19%Area changed = 19%.
115 Area 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?
116 Area 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
117 Area and Volume If the length of a rectangle is increased by 30% decreased by 20%, then what percent of its areawould increase?(TCS Question)
118 Area and Volume Solution: x = 30 y = - 20 (Decreased ) Using formula = x + y + (xy/100)%The increase would be 4%
119 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)
120 Area and Volume Answer: The equation is 1 / 2*x(x+10) = 5500
121 Area 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)
122 Area 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
126 ProbabilitySam 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 ?
127 Probability Solution: Probability of selecting Sam and Jessica = 7C1 * 7C1 = 49n(s) = 49n (є) = 1 ( selecting Sam and Jessica)P (є) = n (є) / n(s)= 1/49Not selecting Sam and Jessica = 1- 1/49The probability = 48/49
128 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
162 Average The Average of 50 numbers is 38. If two numbers namely 45 and 55 are discarded, Whatis the average of the remaining numbers?
163 Average Solution: Total of 50 numbers = 50*38=1900
164 Permutations and Combinations Factorial Notation:n! = n(n-1)(n-2)….3.2.1Number of Permutations:n!/(n-r)!Combinations:n!/r!(n –r)!
165 Permutations 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
166 Permutations 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 …
167 Permutations and Combinations Solution:n = 10r = 3n P r = n!/(n-r)!= 10! / (10-3)!= 10! / 7!= 8*9*10= 720Number of ways is 720.
168 Permutations 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 ?
169 Permutations and Combinations Solution:It is selection so use combination formulaProgrammers and managers = 6C3 * 4C2= 20 * 6 = 120Total number of ways = 120 ways.
170 Permutations and Combinations A man has 7 friends. In how many ways canhe invite one or more of them to a party?
171 Permutations 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
172 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?
173 Permutations and Combinations Solution:5! = 120 ways
174 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.
175 Permutations and Combinations Solution:9C2 * 9C2 = 1296
176 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?
177 Permutations 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
178 Percentage By a certain Percent, we mean that many hundredths. Thus, x Percent means x hundredths, written as x%
179 PercentageIf 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
180 PercentageTwo numbers are respectively 30% & 40% less than a third number. What is second number as a percentage of first?
181 Percentage Solution: Let 3rd number be x. 1st number = x – 30% of x = x – 30x/100 = 70x/ 100= 7x/102nd number = x – 40% of x = x – 40x/100 = 60x/ 100= 6x/10Suppose 2nd number = y% of 1st number6x / 10 = y/100 * 7x /10y = 600 / 7y = 85 5/785 5/7%
182 Percentage 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?
183 Percentage 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
184 Percentage If the number is 20% more than the other, how much percent is the second number lessthan the first?
185 Percentage Solution: Let x =20 = x / (100+x) *100% = 20 / 120 *100% = 16 2/3%The percentage is 16 2/3 %
186 Percentage 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?
187 Percentage Solution: Let capacity of the tank be 100 liters. Then, Initially: A type petrol = 100 litersAfter 1st operation:A = 100/2 = 50 liters, B = 50 litersAfter 2nd operation:A = 50 / 2+50 = 75 liters, B = 50/2 = 25 litersAfter 3rd operation:A = 75 / 2 = 37.5 liters, B = 25/2 +50 = 62.5 litersRequired Percentage = 37.5%
188 Percentage Find the percentage increase in the area of a rectangle whose length is increased by 20% andbreadth is increased by 10%
189 Percentage Answer: Percentage of Area Change=( X +Y+ XY/100)% = *10/100=32%Increase 32%
190 Percentage If A’s income is 40% less than B’s income, then how much percent is B’s income more than A’sincome?
192 PercentageFresh 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?
193 Percentage 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.
194 PercentageIf A is 20% of C and B is 25% of C then what percentage is A of B?
195 Percentage Answer: Given A is 20% of C and B is 25% of C A of B is = (20*100%)/25 = 80%A is 80% of B
196 Boats 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
197 Boats 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.
198 Boats 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
199 Boats 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?
200 Boats 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
201 Boats 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 ?
202 Boats and streams Solution: Down Stream = Speed of the boat + Speed of the stream= 5 +3 =8Up Stream = Speed of the boat – Speed of the stream= 5-3 = 2Let distance be xDistance/Speed = Timex/8 + x/2 = 7 ½x/8 +4x/8 = 15/25x / 8 = 15/25x = 15/2 * 85x = 60x =12
203 Boats 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?
204 Boats and Streams Solution: Down stream 4 km in x hours. Then, Speed Downstream = 4 / x km/hr,Speed Upstream = 3 / x km/hr48/ (4 / x) + 48/(3 / x) = 14x = ½Speed of Downstream = 8,Speed of upstream = 6Rate of the stream = ½ (8-6) km/hr = 1 km/hr
205 Boats 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?
206 Boats and Streams Answer: Let the Speed of the Swimmer in still water be xDistance covered in going = 5(x+1) kmDistance covered in returning = 7(x-1) km5(x+1) = 7(x-1)x = 6Speed of the swimmer in still water = 6 kmph.
207 Time and Distance Speed:- Distance covered per unit time is called Speed.Speed = Distance / TimeDistance = Speed * TimeTime = Distance / Speed
208 Time and Distance Distance covered α Time (direct variation). Distance covered α speed (direct variation).Time α 1/speed (inverse variation).
209 Time and Distance Speed from km/hr to m/sec - ( * 5/18). Speed from m/sec to km/h, - ( * 18/5).Average Speed:-Average speed = Total distance traveledTotal time taken
210 Time 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
211 Time and Distance Solution: x/5-x/6 = 12/60 ( Difference 7+5=12) Solving the equation,x = 6The distance covered is 6 km.
212 Time 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?
213 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
214 Time 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
215 Time and Distance Solution: Average Speed = 2xy / (x + y) The average speed is 48 kmph.
216 Time and Distance By walking at ¾ of his usual aped, a man reaches office 20 minutes later than usual.Find his usual time?
217 Time and Distance Solution: Usual time = Numerator * late time = 3*20 = 60
218 Time 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?
219 Time and Distance Solution: Let the distance be x km. Then, x/10 = (x+20) / 1414x = 10x +2004x = 200x = 50 km.
220 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 ?
221 Time 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.
222 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
223 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
224 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 ?
225 Time 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
226 Profit 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.
227 Profit and Loss A man buys an article for Rs. 27.50 and sells it for Rs Find his gain percentage.Sathyam Question
229 Profit and LossMr. Ravi buys a cooler for Rs For how much should he sell so that there is a gain of 8%
230 Profit and Loss Answer: C.P = 4500 Profit = 8/100*4500 = 360 S.P = C.P + Profit = = 4860He should sell it for Rs
231 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 ?
232 Profit 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
233 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?
234 Profit 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%
235 Profit 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?
236 Profit 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
237 Profit and Loss A shopkeeper gains the cost of 8 meters of thread by selling 40 meters of thread. Findhis gain percentage?
252 Calendar 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.
253 Calendar Ordinary year =( A + B + C + D) -2 take remainder7Leap year = (A + B + C + D) – 3take remainder
254 CalendarWhat is the day of the week on 30/09/2007?
255 Calendar Solution: A = 2007 / 7 = 5 B = 2007 / 4 = 501 / 7 = 4 ( A + B + C + D) -2=7= ( ) -2= 14/7 = 0 = Sunday
256 CalendarWhat was the day of the week on 13th May, 1984?
257 Calendar Solution: A = 1984 / 7 = 3 B = 1984 / 4 = 496 / 7 = 6 (A + B + C + D) -3=7( ) -3= = 14/7= 0, Sunday.
258 CalendarOn what dates of April 2005 did Sunday fall?
259 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( ) -2= = 12 / 7 = 5 = Friday.1st is Friday means Sunday falls on 3,10,17,24
261 Calendar 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
262 ClocksClock: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
263 ClocksFind the angle between the minute hand and hour hand of a clock when the time is 7:20.
265 ClocksHow many times in a day, the hands of a clock are straight?
266 ClocksSolution: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.
267 ClocksHow many times do the hands of a clock coincide in a day?
268 ClocksSolution: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.
269 ClocksAt what time between 7 and 8 o’clock will the hands of a clock be in the same straight line but, not together?
270 Clocks Solution: h = 7 = 30h – 11m/2 180 = 30 * 7 – 11 m/2 On simplifying we get ,5 5/11 min past 7
271 Clocks At what time between 5 and 6 o’clock will the hands of a clock be at right angles?
272 Clocks Solution: h = 5 90 = 30 * 5 – 11m/2 Solving 10 10/11 minutes past 5
273 Clocks Find the angle between the two hands of a clock at 15 minutes past 4 o’clock
274 Clocks Solution: Angle = 30h – 11m/2 = 30*4 – 11*15 / 2 The angle is 37.5o
275 Clocks At what time between 5 and 6 o’clock are the hands of a clock together?
276 Clocks Solution: h = 5 O = 30 * 5 – 11m/2 m = 27 3/11 Solving 27 3/11 minutes past 5
277 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.
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:
279 Data Interpretation The percentage of all science students over Commerce students in all courses isapproximately:(a) (b) 49.4(c) (d) 35.1
280 Data Interpretation Answer: Percentage of science students over commerce students in all courses = 35.1%
281 Data InterpretationWhat is the average number of girls in all courses ?(a) (b) 12.5(c) (d) 11
282 Data Interpretation Answer: Average number of girls in all courses = 50 / 4= 12.5
283 Data InterpretationWhat is the percentage of boys in all courses over the total students?(a) (b) 80(c) (d) 76
284 Data Interpretation Answer: Percentage of boys over all students = (450 x 100) / 500= 90%
285 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
286 Data Sufficiency What is John’s age? In 15 years will be twice as old as Dias would beDias was born 5 years ago (Wipro)
287 Data Sufficiency Answer: c) If both statements together are sufficient but neither of statement alone is sufficient.
288 Data 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
289 Answer: d) If both together are not sufficient Data SufficiencyAnswer:d) If both together are not sufficient
290 If 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
291 Answer: D . If both together are not sufficient Data SufficiencyAnswer:D . If both together are not sufficient
292 Data 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
293 Data Sufficiency Answer: If statement I alone is sufficient but statement II alone is not sufficient
294 Data 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.
295 Answer: D. If both together are not sufficient Data SufficiencyAnswer:D. If both together are not sufficient
296 What 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
297 Data Sufficiency Answer: B. If statement II alone is sufficient but statement I alone is not sufficient
298 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?
299 CubesAnswer:3*3*3 = 27All the outer surface are painted with colour.26 One inch cubes are painted at least one surface.
300 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
302 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?
304 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..
305 Cubes 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?
307 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 ?
309 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?
310 Venn Diagram Solution: We are given n (X uY) = 18, n (X) = 8, n (Y) =15. using the formula.n( X n Y) = n (X) + n (Y) - n ( X u Y)n( X n Y) = – 18n( X n Y) = 5
311 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?
312 Venn Diagram Answer: n (s) = 21, n (T) = 32, n ( S n T) = 11, n (S u T) = ?n (S u T) = n (S) + n( T) – n (S n T)= – 11 = 42
313 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?
314 Venn Diagram Answer: n ( A) = 40, n ( n B) = 60 and n ( A n B) = 10, n ( A u B) = n ( A) + n (B) – n ( A n B)60 = 40 + n (B) – 10n (B) = 30
315 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?
316 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 = – 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
317 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.
318 Venn Diagram Solution: No. of students who passed in one or more subjects= = 91No of students who failed in all the subjects= = 9
319 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?
320 Answer: 3 of them studied both Latin and Greek. Venn DiagramAnswer:3 of them studied both Latin and Greek.