Aptitude Numerical Reasoning Blue Lotus Aptitude Numerical Reasoning
Numerical Reasoning Problems on Numbers Problems on Ages Ratio and Proportion Alligation or Mixture Chain Rule Partnership Venn Diagram
Numerical Reasoning Area and Volume Probability Time and Work (Pipes) SI and CI Average Permutation and Combination Percentage
Numerical Reasoning Boats and Streams Time and Distance (Trains) Data Sufficiency Profit and Loss Calendar Clocks Data Interpretation Cubes
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.
Problems on Numbers Sn = a(rn – 1)/(r-1); Arithmetic Progression: The nth term of A.P. is given by Tn = a + (n – 1)d; Sum of n terms of A.P Sn = n/2 *(a + L) or n/2 *[2a+(n-1)d)] a = 1st term, n = number of term, d= difference, Tn = nth term Geometrical Progression: Tn = arn – 1. Sn = a(rn – 1)/(r-1); Where a = 1st term , r = 1st term / 2nd term
Basic Formulae 1. ( a+b)2 = a2 + b2 + 2ab 2. (a-b)2 = a2 +b2 -2ab 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)
Problems on Numbers Three numbers are in the ratio 3:4:5. the sum of the largest and the smallest equal to the sum of the third and 52. Find the smallest number ?
Problems on Numbers Solution: Let the numbers be 3x, 4x and 5x Then 5x+3x = 4x +52 8x – 4x = 52 4x = 52 x = 52/4 x = 13 The smallest number = 3x = 3*13 = 39.
Problems on Numbers What is one half of two third of three fourths of four fifths of five sixth of six sevenths of seven eights of eight ninth of nine tenths of thirty?
Problems on Numbers Solution: = ½ * 2/3 *3/4 * 4/5 * 5/6*6/7*7/8*8/9*9/10 *30 = 3
Problem on Numbers If the operation ^ is defined by the equator x ^ y = 2x + y what is the value of a in 2 ^ a = a ^ 3? (Sathyam)
Solution: 2(2) = a ^ 3 4 + a = 2a + 3 a = 1 Problem on Numbers Solution: 2(2) = a ^ 3 4 + a = 2a + 3 a = 1
Problem on Numbers There are 150 weight some are 1 kg weight and some are 2 kg weights. The sum of the weights is 260. what is the number of 1 kg weight. (TCS)
Problem on Numbers Solution: X + 2Y = 260 X + Y = 150 On Solving Two Equations Y = 110 X = 150 – 110 = 40 Kg
Problem on Numbers The cost of 1 pencil, 2 pens and 4 erasers is Rs. 22, while the cost of five pencils, four pens and two eraser is 32. how much will 3 pencils, 3 pens and 3 eraser? (TCS)
Problem on Numbers Solution: Let Pencil be x, Pens be y, Erasers be z x + 2y + 4z = 22 5x + 4y + 2z = 32 Adding we get 6x+6y+6z = 54 3x + 3y + 3z = 27 3 Pencil, 3 Pens and 3 Eraser is Rs. 27.
Problem on Numbers If the numerator of a fraction is increased by 25% and denominator decrease by 20%, the new value is 5/4. what is the original value? (TCS)
Problem on Numbers Solution: ( x + 25x/100) / (y – 20y/100) = 5/4
Problem on Numbers The difference between two numbers is 1/7 of the sum of these two numbers. What is the ratio of the two numbers? (Wipro)
Problem on Numbers (x- y ) = 1/7 (x+y) 7( x- y) =( x + y)
Problem on Numbers A fraction has a denominator greater than its numerator by 4. but if you add 10 to the denominator, the value of the fraction would then become 1/8, what is the fraction? (Caritor)
Problem on Numbers Solution: x/(x+4+10) = 1/8 x/(x+14) = 1/8 7x = 14; hence x = 2 x/(x+4 )= 2/(2+4 )= 2/6
Problems on Ages The ages of two persons differ by 10 years. If 5 years ago, the elder one be 2 times as old as the younger one, find their present ages.
Problems on Ages Solution: x - y = 10; x = 10 + y x - 5 = 2(y-5) y + 10 -5 = 2y -10 y+5 = 2y -10 2y- y = 15 y=15 and x = 25 Their present ages are 15 years and 25 years.
Problems on Ages The present ages of three persons are in the proportion of 4:7:9. Eight years ago, the sum of their ages was 56. Find their present ages ?
Problems on Ages Solution: Three person’s ratio = 4:7:9 Total = 4+7+9 = 20 Sum of their age = 56, after 8 years their sum = 56 +24 = 80 A’s age = 4/20 *80 = 16 B’s age = 7/20 *80 = 28 C’s age = 9/20 *80 = 36 Their present ages are 16, 28 and 36.
Problems on Ages Father’s age is 5 times his son's age.4 years back the father was 9 times older than his son. Find the father's present age? (TCS)
Problems on Ages Solution: F = 5S F – 4 = 9(S-4) F – 5s = 0 Father age = 40 years
Problems on Ages One year ago Pandit was three times his sister’s age. Next year he will be only twice her age. How old will Pandit be after five years? (TCS)
Problems on Ages Solution: (P-1) = 3(S -1) P + 1 = 2( s+1) After 5 years = 12
Problems on Ages A father is 30 years older than his son, however he will be only thrice as old as his son after 5 years what is father’s present age?
Problems on Ages Solution: F = S + 30 F + 5 = 3(S+5)
Problems on Ages A father is three times as old as his son after 15 years the father will be twice as old as his son’s age at that time. What is the father’s present age ? (TCS)
Solution: F = 3S F + 15 = 2(S +15) Father’s age = 45, Son’s age = 15 Problems on Ages Solution: F = 3S F + 15 = 2(S +15) Father’s age = 45, Son’s age = 15
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 Proportion If Rs. 1260 is divided among A, B, C in the ratio 2 : 3 : 4 what is C’s share?
Ratio and Proportion Solution: C’s Share = 4/9*1260 C’s share = Rs. 560
Ratio and Proportion To 15 liters of water containing 20% alcohol, we add 5 liters of pure water. What is the % of alcohol?
Ratio and Proportion Solution: 15 lit 20 % 20 lit (15+5) x by solving we get = 15% 15% alcohol
Ratio and Proportion What number should be added or subtracted from each term of the ratio 17 : 24 so that it becomes equal to 1 : 2
Ratio and Proportion Solution: Let the number be x. 17 + x/24 + x = 1/2 Solving the above equation, The number to be subtracted is 10.
Ratio and Proportion The ratio of white balls and black balls is 1:2. If 9 gray balls are added it becomes 2:4:3. Then what is the number of black balls ?
Ratio and Proportion Solution: Ratio of all the three balls = 2:4:3 Ratio of two balls before adding gray = 1:2 9 gray ratio =3 3 parts = 9 balls 1 part = 9/3 4 parts =? = 9*4/3 =12 Number of black balls is 12
Ratio and Proportion Rs. 770 was divided among A, B and C such that A receives 2/ 9th of what B and C together receive. Find A’s share?
Ratio and Proportion Solution: A = 2/9 (B+C) B+C =9A/2 A+B+C = 770
Alligation or Mixture (Quantity of cheaper / Quantity of costlier) (C.P. of costlier) – (Mean price) = -------------------------------------- (Mean price) – (C.P. of cheaper)
Alligation or Mixture Cost of Cheaper Cost of costlier c d Cost of Mixture m d-m m-c (Cheaper quantity) : (Costlier quantity) = (d – m) : (m – c)
Alligation or Mixture 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 Mixture Solution: 7 17 10 (17-10) (10-7) 7 : 3 7 17 10 (17-10) (10-7) 7 : 3 The ratio is 7:3 The quantity of 2nd kind = 3/10 of 100kg = 30kg
Alligation or Mixture A 3-gallon mixture contains one part of S and two parts of R. In order to change it to mixture containing 25% S how much R should be added?
Alligation or Mixture Answer: R : S 2 : 1 75% : 25% 3 : 1 2 : 1 75% : 25% 3 : 1 1 gallon of R should be added.
Alligation or Mixture In an examination out of 480 students 85% of the girls and 70% of the boys passed. How many boys appeared in the examination if total pass percentage was 75%
Alligation or Mixture Solution: 85 75 10 5 10 5 Number of Boys = 480 * 10/15 Number of Boys = 320
Alligation or Mixture In two varieties of tea, one costing Rs. 25/kg. and the other costing RS. 30/kg are blended to produce blended variety of tea in ratio 2:3. find the cost price of the mixture ?
Alligation or Mixture Solution: 25 30 x (30- 28) (28-25) 2 : 3 25 30 x (30- 28) (28-25) 2 : 3 Let mixed price be x If you subtract 28 from 30 you will get 2 and if you subtract 25 from 28 you will get 3.
Alligation or Mixture A person has Rs. 5000. He invests a part of it at 3% per annum and the remainder at 8% per annum simple interest. His total income in 3 years is Rs. 750. Find the sum invested at different rates of interest.
Alligation or Mixture Solution: Average rate of interest = 5% per annum 3% 8% 5% 3% 2% Investment at 3% per annum = 3x5000/5= 3000 Investment at 8% per annum = 2x5000/5=2000
Chain Rule Direct Proportion : A B Indirect Proportion: A B
Chain Rule A Garrison of 500 men had provision for 27 days. After 3 days, a reinforcement of 300 men arrived. The remaining food will now last for how many days?
Chain Rule Solution: Men days 500 24 800 x 800X = 500x24 500 24 800 x 800X = 500x24 X =(500x24)/800 =15 days
Chain Rule If 20 men take 15 days to complete a job. In how many days will 25 men finish the work? CTS Question
Chain Rule Solution: Men Days 20 15 25 x x/15=20/25 x = (20*15)/25 =12 20 15 25 x x/15=20/25 x = (20*15)/25 =12 They will take 12 days
Chain Rule If 11.25m of a uniform iron rod weighs 42.75 kg, what will be weight of 6m of the same rod?
Chain Rule Solution: length ( m ) weight ( kg ) 11.25 42.75 6 x 11.25 42.75 6 x Since it is a direct proportion, x 6 6 x 42.75 --------- = -------- x = --------------- 42.75 11.25 11.25 The weight of rods x = 22.8 kg
Chain Rule A stationary engine has enough fuel to run 12 hours when its tank is 4/5 full. How long will it run when the tank is 1/3 full? TCS Question
Chain Rule Answer: Tank hours 4/5 12 1/3 x 4/5 x = 12 * 1/3 4/5 12 1/3 x 4/5 x = 12 * 1/3 It will run for 5 hours
Chain Rule 20 men complete one - third of a piece of work in 20 days. How many more men should be employed to finish the rest of the work in 25 more days?
Chain Rule Solution: Men days work 20 20 1/3 work done = 1/3 x 25 2/3 remaining = 1-1/3=2/3 More work, more men (direct proportion) More days, less men (indirect Proportion) 1/3 *X = (2/3 )*20*(20 /25) X = 800/25 = 32 More men to be employed = (32-20) 12 More people needed to finish the job
Chain Rule 15 men take 21 days of 8 hrs each to do a piece of work. How many days of 6 hrs each would 21 women take, if 3 women do as much work as two men?
Chain Rule Solution: 3 women = 2 men 21 women = 14 men Men Days Hrs 21 8 14 x 6 x/21=(15/14)/(8/6) x= 30 days
Partnership Types: A invested Rs.x and B invested Rs.y then A:B = x : y A invested Rs. x and after 3 months B invested Rs. y then the share is A:B = x * 12 : y * 9
Partnership Sanjiv started a business by investing Rs. 36000. After 3 months Rajiv joined him by investing Rs. 36000. Out of an annual profit of Rs. 37100 find the share of each. Satyam Question
Partnership Solution : 36000 * 12 : 36000 * 9 4 : 3 36000 * 12 : 36000 * 9 4 : 3 Sanjiv’s share of profit = (4*37100)/7 = 21200 profit = 21200 Rajiv’s share of profit = 15900
Partnership A sum of money is divided among A, B, C such that for each rupee A gets, B gets 65 paise and c gets 35 paise if c’s share is Rs. 560. what is the sum?
Partnership Solution A : B : C 100 : 65 : 35 20 : 13 : 7 Total = 20+13+7 = 40 C’ share = 560 7/40 *X =560 X= 3200
Partnership A starts business with Rs.3500 and 5 months after B joins A as his partner. After a year the profits are divided in the ratio of 2:3. How much did B contribute ?
Partnership Solution: A :B =3500*12 : 7X 42000 : 7X = 2: 3 B’s contribution is Rs.9000
Partnership A and B invest in a business in the ratio 3 : 2. If 5% of the total profit goes to charity and A’s share is Rs. 855 what is the total profit?
Partnership Solution: A : B = 3:2 Let Profit be X X – 5% of X X- 5X/100 = 95X/100 3/5 * 95X/100 = 285 19X= 28500 X = 26500/19 = 1500 Total profit is Rs.1500
Partnership A and B enter into partnership for a year. A contributes Rs.1500 and B Rs. 2000. After 4 months, they admit C who contributes Rs. 2250. If B withdraws his contribution after 9 months, find their profit share ratio at the end of the year?
Partnership Solution: A: B: C = 1500*12: 2000*9: 2250*8 = 18000: 18000: 18000 = 1: 1: 1 Profit share at the end of the year, 1: 1: 1
Time and Work If A can do a piece of work in n days, then A’s 1 day’s work = 1/n If A is thrice as B, then: Ratio of work done by A and B = 3:1
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 One fast typist types same matter in 2 hours and another slow typist types the same matter in 3 hours. If both do combine in how much time will they finish? TCS Question
Time and Work Solution: Fast typist = 1/2 ; slow typist = 1/3 ; Together: = 1/2 + 1/3 = 5/6 so 6/5 hrs The work will be completed in 6/5 Hrs.
Time and Work A and B can finish a piece of work in 30 days, B and C in 40 days, while C and A in 60 days .In how many days A, B and C together can finish the work ?
Time and Work Solution: A + B = 30 days = 1/30 B + C = 40 days = 1/40 C +A = 60 days = 1/60 All work together A+B+C+B+C+A = 1/30 +1/40 +1/60 2(A+B+C) = 1/30+1/40+1/60 = (4+3+2) /120 = 9/120*2 = 9/240 = 3/80 = 26 2/3 A, B and C can finish the work in 26 2/3 days
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 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 20/3 days = 6 2/3 days The work will be completed in 6 2/3 days.
Time and Work A work done by two people in 24 minutes. One of them can do this work alone in 40 minutes. How much time is required to do the same work by the second person? TCS Question
Time and Work Solution : A and B together = 1/24; A = 1/40; B = ? = 1/24 – 1/40 = 2/120 = 1/60 The second person will complete in 60 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: This problem is based on 2nd method. 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 tap can fill a cistern in 8 hours and another can empty it in 16 hours. If both the taps are opened simultaneously, Find the time ( in hours) to fill the cistern
Time and Work (Pipes) Solution: Tap 1 = 1/8 (fill); Tap 2 = 1/16 (empty) = 1/8 – 1/16 = 1 / 16 Total time taken to fill the cistern = 16 hours
Time and Work (Pipes) A water tank is 2/5th full. Pipe A can fill the tank in 10 minutes and the pipe B can empty it in 6 minutes. If both the pipes are open, how long will it take to empty or fill the tank completely?
Time and Work (Pipes) Answer : A = 1/10; B = 1/6 = 1/10 -1/6 = - 1/15 Empty in 15 minutes To empty 2/5 of the tank 2/5 * 15 = 6 Time taken (empty)= 6 minutes
Area and Volume Cube: Let each edge of the cube be of length a. then, Volume = a3cubic units Surface area= 6a2 sq.units. Diagonal = √3 a units.
Area and Volume Cylinder: Let each of base = r and height ( or length) = h. Volume = πr2h Surface area = 2 πr h sq. units Total Surface Area = 2 πr ( h+ r) units.
Area and Volume Cone: Let radius of base = r and height=h, then Slant height, l = √h2 +r2 units Volume = 1/3 πr2h cubic units Curved surface area = πrl sq.units Total surface area = πr (l +r)
Area and Volume Sphere: Let the radius of the sphere be r. then, Volume = 4/3 πr3 Surface area = 4 π r2sq.units
Area and Volume Circle: A= π r 2 Circumference = 2 π r Square: A= a 2 Perimeter = 4a Rectangle: A= l x b Perimeter= 2( l + b)
Area and Volume Triangle: A = ½*base*height Equilateral = √3/4*(side)2 Area of the Scalene Triangle S = (a+b+c)/ 2 A = √ s*(s-a) * (s-b)* (s-c)
Area and Volume What is the cost of planting the field in the form of the triangle whose base is 2.8 m and height 3.2 m at the rate of Rs.100 / m2
Area and Volume Solution: Area of triangular field = 1/2 * 3.2 * 2.8 m2 = 4.48 m2 Cost = Rs.100 * 4.48 = Rs.448.
Area and Volume Area of a rhombus is 850 cm2. If one of its diagonal is 34 cm. Find the length of the other diagonal?
Area and Volume Solution: 850 = ½ * d1 * d2 = ½ * 34 * d2 = 17 d2 = 50 cm Second diagonal = 50cm
Area and Volume A grocer is storing small cereal boxes in large cartons that measure 25 inches by 42 inches by 60 inches. If the measurement of each small cereal box is 7 inches by 6 inches by 5 inches then what is maximum number of small cereal boxes that can be placed in each large carton ?
Area and Volume Solution: No. of small boxes = (25*42*60 ) / ( 7*6*5 ) = 300 300 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 the area in percentage?
Area and Volume Solution: = D1 + D2 – D1*D2 /100 = 10 + 10 – 10*10/100 = 20 -1 = 19% New area changed = 19%.
Area and Volume A circular wire of radius 42 cm is bent in the form of a rectangle whose sides are in the ratio of 6:5. Find the smaller side of the rectangle?
Area and Volume Solution: length of wire = 2 πr = (22/7*14*14)cm Perimeter of Rectangle = 2(6x+5x) cm = 22xcm 22x =264 x = 12 cm Smaller side = (5*12) cm = 60 cm
Area and Volume A man is running around a rectangle. It takes time, 2 times in traveling length than traveling width and the total perimeter is 300 m. Find the Area?.
Area and Volume Solution: Breadth = X Length = 2X Area = 2X*X = 2X2 Perimeter 6X = 300m X = 50m Area = 2x2 =2*50*50 Area = 5000 sq.m.
Probability Probability: P(E) = n(E) / n(S) Addition theorem on probability: n(AUB) = n(A) + n(B) - n(AB) Mutually Exclusive: P(AUB) = P(A) + P(B) Independent Events: P(AB) = P(A) * P(B)
Probability There are 19 red balls and One black ball. Ten balls are placed in one jar and remaining in one jar. What is probability of getting black ball in right jar ? (Infosys -2008)
Probability Answer: Probability is 1/2.
Probability There are 5 red shoes 4 green shoes. If one draws randomly a shoe what is the probability of getting a red shoe? CTS Question
Probability Answer: The probability is 5/9
Probability A bag contains 2 red, 3 green and 2 blue balls are to be drawn randomly. Two balls are drawn at random. What is the probability that the balls drawn contain only blue balls ?
Probability Answer : The probability is 1/21
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 ?
Probability Answer: The Probability of Selecting = 1/7*7 = 1/49 The Probability of not Selecting = 1-1/49 = 48/49
Probability The letters of the word SOCIETY are placed in a row. What is the probability that the three vowels come together?
Probability Answer: Required Probability = (5!*3! )/7! = 1/7
Simple / Compound Interest Simple Interest = PNR / 100 Amount A = P + PNR / 100 When Interest is Compound annually: Amount = P (1 + R / 100)n Compound Interest = A - P
Simple / Compound Interest Half-yearly C.I.: Amount = P (1+(R/2)/100)2n Quarterly C.I. : Amount = P (1+(R/4)/100)4n
Simple/compound interest Difference between C.I and S.I for 2 years = P*(R/100)2. Difference between C.I and S.I for 3 years = P{(R/100)3+3 (R/100)2 }
Simple / Compound Interest What is the S.I. on Rs. 3000 at 18% per annum for the period from 4th Feb 1995 to 18th April 1995 Sathyam Question
Simple / Compound Interest Answer: Time = 24+ 31+17 = 73 days = 73/365 = 1/5 P = 3000; R = 18%; = PNR/100 = 3000*1*18/100*5 The simple interest is Rs. 108
Simple / Compound Interest A sum of money doubles itself at C.I. in 15 years. In how many years will it become eight times? Satyam Question
Simple / Compound Interest Solution: A = P(1+R/100)15 2P =P(1+R/100) 15 ; 2 = (1+R/100)15 If A = 8P 8P = P(1+R/100)n 23 = ( 1+R/100) [(1+R/100) 15]3 = (1+R/100)n n = 3*15 It will take a period of 45 years.
Simple / Compound Interest Raja borrowed a certain money at a certain rate of S.I. After 5 years, he had to pay back twice the amount that he had borrowed. What was the rate of interest? TCS Question
Simple / Compound Interest Solution: A = 2P A = P + PNR/100 2P = P(1+NR/100) 2 = (1+5*R/100) 1 = R/20 The rate of interest is 20%
Simple/compound interest In simple interest what sum amounts to Rs. 1120 in 4 years and Rs. 1200 in 5 years? CTS Question
Simple/Compound interest Answer : Interest for 1 year = 1200 – 1120 = 80 Interest for 4 year = 80*4 = 320 A = 1120 P = A – P = 1120 – 320 The Principal is Rs. 800
Simple/Compound interest A simple interest amount for Rs. 5000 for 6 months is Rs. 200. What is the annual rate of interest? CTS Question
Simple/Compound interest Solution: P = 5000; N = 6/12 = ½ I = 200 R = I *100 / P*N =200*100*2/5000*1 = 40/5 = 8% The annual rate of interest is 8%
Simple/Compound interest A man earns Rs. 450 as an interest in 2 years on a certain sum invested with a company at the rate of 12% per annum. Find the sum invested.
Simple/Compound interest Solution: P = I*100/R*N = 450*100/12*2 Principal = Rs. 1875
Simple/Compound interest If Rs. 85 amounts to Rs. 95 in 3 years, what Rs. 102 will amount in 5 years at the same rate percent?
Simple/Compound interest Solution: Let P = Rs. 85; A = Rs. 95; I = 10/3 in 1 year Rate = I*100/P*N =4% ( app) Amount = P+PNR/100 = 102+20 =122 Hence the amount in 5 years = Rs. 122
Simple/Compound interest 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 interest Solution: C.I – S.I = P (R/100)2 = 4500(5/100)2 = 11.25 Difference = Rs. 11.25
Simple/Compound interest What will be the C.I on Rs. 15625 for 2½ years at 4% per annum?
Simple/Compound interest Solution: A = P(1+R/100)n A = 15625 [(1+4/100)2 ( 1+(4*1/2) / 100)] = 17238 C.I = A – P C.I = 17238 - 15625 Compound interest = Rs. 1613
Average Average is a simple way of representing an entire group in a single value. “Average” of a group is defined as: X = (Sum of items) / (No of items)
Average Total temperature for the month of September is 840C. If the average temperature of that month is 28C. Find out the days in the month of September?
Average Solution Number of days= 840/28=30 days
Average The painter is paid x rupees for painting every 10m of a wall and y rupees for painting every extra meter. During one week, he painted 10m on Monday, 13m on Tuesday, 12m on Wednesday, 11m on Thursday and 12m on Friday. What is his average daily earning in rupees for the five day week?
Average Solution: Day 1 = x, Day 2 = x+3y, Day 3 = x+2y, Average = (x+ x+3y+ x+2y+ x+y+ x+2y) / 5 = 5x+8y / 5 = 5x/5 + 8y / 5 Average for 5 days is x+ (8y/5)
Average The average of 11 observations is 60. If the average of 1st five observations is 58 and that of last five is 56, find sixth observation?
Average Solution: 5 observations average = 58 Sum = 58*5 = 290 Last 5 observation average = 56 Sum = 56*5 = 280 Total sum of 10 numbers = 570 (290 + 280) Total sum of 11 numbers = 660 (11*60) 6th number = 90 (660 -570)
Average The average of age of 30 students is 9 years. If the age of their teacher included, it becomes 10 years. Find the age of the teacher?
Average Solution: Total age of 30 students = 30*9 = 270 Including the teacher’s age = 31*10 = 310 Difference is = 310-270 = 40 years
Permutations and Combinations Factorial Notation: n! = n(n-1)(n-2)….3.2.1 Number of Permutations: n!/(n-r)! Combinations: n!/r!(n –r)!
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 …
Permutations and Combinations Solution: n = 10 r = 3 n P r = n!/(n-r)! = 10! / (10-3)! = 10! / 7! = 8*9*10 = 720 Number of ways is 720.
Permutations and Combinations To fill a number of vacancies, an Employer must hire 3 programmers from 6 applicants, and two managers from 4 applicants. What is total number of ways in which she can make her selection ?
Permutations and Combinations Solution: It is selection so use combination formula Programmers and managers = 6C3 * 4C2 = 20 * 6 = 120 Total number of ways = 120 ways.
Permutations and Combinations In how many ways can the letters of the word BALLOON be arranged so that two Ls do not come together?
Permutations and Combinations Solution: Total arrangement = 7! / 2!*2! (L and O occurred twice) =1260 Ls come together (BAOON) (LL) = 6! / 2! = 3* 4* 5*6* = 360 Ls not come together 1260 – 360 = 900 Number of ways = 900.
Permutations and Combinations A man has 7 friends. In how many ways can he invite one or more of them to a party?
Permutations and Combinations Solution: In this problem, the person is going to select his friends for party, he can select one or more person, so = 7C1 + 7C2+7C3 +7C4 +7C5 +7C6 +7C7 = 127 Number of ways is 127
Percentage By a certain Percent, we mean that many hundredths. Thus, x Percent means x hundredths, written as x%
Percentage Two numbers are respectively 30% and 40% less than a third number. What is the second number as a percentage of first?
Percentage Solution: Let 3rd number be x. 1st number = x – 30% of x = x – 30x/100 = 70x/ 100 = 7x/10 2nd number = x – 40% of x = x – 40x/100 = 60x/ 100 = 6x/10 Suppose 2nd number = y% of 1st number 6x / 10 = (y/100 )* 7x /10 y = 600 / 7 y = 85 5/7 Hence it is 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. What is the total amount of money spent on machinery and the raw materials?
Percentage Solution: Let total salary =100% Spending: Machinery + Raw material + staff = 35%+40%+10% = 85% Remaining percentage = 100 %– 85% = 15% 15 % of X = 60000 X = 4, 00,000 In this 4, 00,000 75% for machinery and raw material = 4, 00,000* 75/100 = 3, 00,000
Percentage If the number is 20% more than the other, how much percent is the second number less than the first?
Percentage Solution: Let X =20 = X / (100+X) *100% = 20 /120 *100% =16 2/3% The percentage is 16 2/3%
Percentage An empty fuel tank of a car was filled with A type petrol. When the tank was half empty, it was filled with B type petrol. Again when the tank was half empty, it was filled with A type petrol. When the tank was half – empty again, it was filled with B type petrol. What is the percentage of A type petrol at present in the tank?
Percentage Solution: Let capacity of the tank be 100 liters. Then, Initially: A type petrol = 100 liters After 1st operation: A = 100/2 = 50 liters, B = 50 liters After 2nd operation: A = 50 / 2+50 = 75 liters, B = 50/2 = 25 liters After 3rd operation: A = 75/2= 37.5 liters, B = 25/2 +50 = 62.5 liters Required Percentage of Type A is = 37.5%
Percentage Find the percentage increase in the area of a Rectangle whose length is increased by 20% and breadth is increased by 10%
Percentage Answer: Percentage of Area Change=( X +Y+ XY/100)% =20+10+20*10/100 =32%
Percentage If A’s income is 40% less than B’s income, then how much percent is B’s income more than A’s income?
Percentage Answer: Percentage = R*100%/(100-R) = (40*100)/ (100-40) =66 2/3%
Percentage One side of a square is increased by 30%. To maintain the same area by how much percentage the other side will have to be decreased?
Percentage Answer: Percentage = r*100%/(100+r) = (30*100) / 130 = 23 1/3%
Boats and streams Up stream – against the stream Down stream – along the stream u = speed of the boat in still water v = speed of stream Down stream speed (a)= u+v km / hr Up stream speed (b)=u-v km / hr u = ½(a+b) km/hr V = ½(a-b) km / hr
Boats and streams A man can row Upstream at 12 kmph and Downstream at 16 kmph. Find the man’s rate in still water and rate of the current?
Boats and streams Solution: Rate in still water = 1/2 (16 + 12) = 14 Kmph Rate of Current = 1/2 (16 – 12 ) = 2 Kmph
Boats and streams A Boat is rowed down a river 40 km in 5 hr and up a river 21 km in 7 hr. Find the speed of the boat and the river?
Boats and streams Solution: Speed of the Boat Downstream = 40/7 = 8 (a) Speed of the Boat Upstream = 21/7 = 3 (b) Speed of the Boat = 1/2 ( a + b ) = 1/2 ( 8+3 ) = 5.5 Kmph Speed of the River = 1/2 ( a – b ) = 1/2 (8 – 3) = 2.5 kmph
Boats and streams A boat’s crew rowed down a stream from A to B and up again in 7 ½ hours. If the stream flows at 3km/hr and speed of boat in still water is 5 km/hr. find the distance from A to B ?
Boats and streams Solution: Down Stream = Sp. of the boat + Sp. of the stream = 5 +3 =8 Up Stream = Sp. of the boat – Sp. of the stream = 5-3 = 2 Let distance be X Distance/Speed = Time X/8 + X/2 = 7 ½ X/8 +4X/8 = 15/2 5X / 8 = 15/2 5X = 15/2 * 8 5X = 60 X =12
Boats and Streams A boat goes 40 km upstream in 8 hours and 36 km downstream in 6 hours. Find the speed of the boat in still water in km/hr?
Boats and Streams Solution: Speed of the boat in upstream = 40/8 = 5 km/hr Speed of the boat in downstream= 36/6 =6 km/hr Speed of the boat in still water = (5+6 ) / 2 = 5.5 km/hr
Boats and Streams A man rows to place 48 km distant and back in 14 hours. He finds that he can row 4 kmph with the stream in the same time as 3 Kmph against the stream. Find the rate of the stream?
Boats and Streams Solution: Down stream 4 km in x hours. Then, Speed Downstream = 4/x km/hr, Speed Upstream = 3/x km/hr 48/ (4/x) + 48/(3/x) = 14 x = 1/2 Speed of Downstream = 8, Speed of upstream = 6 Rate of the stream =1/2 (8-6) km/hr = 1 km/hr
Time and Distance Speed:- Distance covered per unit time is called speed. Speed = distance/time (or) Distance = speed*time (or) Time = distance/speed
Time and Distance Distance covered α Time (direct variation). Distance covered α speed (direct variation). Time α 1/speed (inverse variation).
Time and Distance Speed from km/hr to m/sec - (Multi by 5/18). Speed from m/sec to km/h, - (Multi by 18/5). Average Speed:- Average speed = Total distance traveled Total time taken
Time and Distance From height of 8 m a ball fell down and each time it bounces half the distance back. What will be the distance traveled? Sathyam Question
Time and Distance Solution: = 8 + 4 + 4+2+2+1+1+0.5+0.5 +….etc. The total distance traveled is 24 m
Time and Distance Two cars are 15 km apart. One is running at a speed of 50 kmph and the other at 40 kmph. How much time will it take for the two cars to meet? Sathyam Question
Time and Distance Solution: Time taken =Distance / (S1 – S2) = 15 / (50 – 40) = 15 / 10 = 1.5 It will take 1½ hours.
Time and Distance The center of a storm shifts 22.5 miles in 1 hour. At the same rate what time will it take to move 60 miles? TCS Question
Time and Distance Answer: For 22.5 miles it takes 1 hour It means for 60 miles T = D / S Time taken = 60 / 22.5 It will take 2 2/3 hours.
Time and Distance By walking at ¾ of his usual speed, a man reaches office 20 minutes later than usual. Find his usual time?
Time and Distance Solution: Usual time = Numerator * late time = 3*20 = 60
Time and Distance A man on motorcycle rides 110 miles in 330 minutes. What is his average speed in miles per hour? TCS Question
Time and Distance Answer: Speed = D / T =110*60 /330 The average speed = 20 miles/hour
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 train S1 / S2 = √T2/T1 160/x = √16/9 160/x = 4/3 x = 120 The speed of second train is 120km/hr.
Time and Distance (Trains) Two hours after a freight train leaves Delhi a passenger train leaves the same station traveling in the same direction at an average speed of 16 km/hr. After traveling 4 hours the passenger train overtakes the freight train. What was the average speed of the freight train? Wipro Question
Time and Distance (Trains) Solution : Speed of Passenger train = 16 kmph Distance = 16*4 = 64 Speed of freight train = Distance / ( S1 + S2 ) = 64 / (4+2) = 64/6 = 10.6 km/hr The average speed = 10.6 km/hr
Time and Distance (Trains) There are 20 poles with a constant distance between each pole. A train takes 24 sec to reach the 12 pole. How much time will it take to reach the last pole ?
Time and Distance (Trains) Solution: To cross 11 poles it is taking 24 sec To cross 19 poles it will take x time Poles time 11 24 19 x 11x = 19 * 24 x = 19* 24 /11 x = 41.45 sec It reaches the last pole in 41.45sec
Time and Distance (Trains) 120 m long train crosses the pole after 2½ sec. Find how much time it takes to cross 140 m long platform? Caritor Question
Time and Distance (Trains) Solution: To cross 120 m it is taking 2 ½ sec. (5/2sec) To cross (120 +140)=260 m it will take x sec 120x = 260*5/2 (apply chain rule) = 5 5/12 It takes 5 5/12 seconds.
Time and Distance (Trains) A train X speeding with 120 kmph crossed another train Y, running in the same direction, in 2 minutes. If the lengths of the trains X and Y be 100 m and 200m respectively, what is the speed of train Y?
Time and Distance (Trains) Solution: Let the speed of train Y be x km/hr Relative Speed of X to Y = (120 –x) km/hr = [(120 –x)*5/18] m/sec =( 600 – 5x) / 18 m/sec T = D / Rel. Speed 300 / (600 – 5x /18) = 120 ( 2 Minutes ) 5400 = 120 (600 -5x) x = 111 m/sec.
Profit and Loss Cost Price. - CP Selling Price. - SP Profit or Gain. - P = SP – CP Loss. - L = CP - SP
Profit and Loss . 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 Anu bought a necklace for Rs. 750 and sold it for Rs. 675. Find her Loss percentage?.
Profit and Loss Solution: CP = 750, SP = 675 L = 750 – 675 = 75 Loss% = loss /CP *100 = 75*100/750 = 10% Loss 10%
Profit and Loss A shopkeeper bought a watch for Rs. 400 and sold it for Rs. 500. What is his profit percentage? TCS Question
Profit and Loss Solution: CP = 400; SP = 500 P = 500 – 400 = 100 Profit % = Profit /CP *100 = 100*100/400 = 25% Profit 25%
Profit and Loss By Selling 15 Mangoes , a Fruit vendor recovers the cost price of 20 Mangoes. Find the profit percentage?
Profit and Loss Solution: The expenditure and the revenue are equated, Percentage of profit =Goods left*100 Goods sold = ( 5*100 ) /15 = 33.3%
Profit and Loss A shopkeeper loses 7% by selling a cricket ball for Rs. 31. for how much should he sell the ball so as to gain 5%
Profit and Loss Solution: First case S.P = Rs. 31 and Loss% = 7% C.P = [100/(100 – loss%)]*S.P = (100*31) / (100-7) = 100 / 3 Second case, C.P = Rs 100 / 3 and gain% = 5% S.P = [(100+gain 5%) / 100]*C.P = [ (100+5 ) / 100] * 100/3 = Rs. 35
Profit and Loss What is the selling price of a Toy car if the cost of the car is Rs. 60 and a profit of 10% over selling price is earned? CTS Question
Profit and Loss Solution: Profit = 60*10/100 = 6 Selling Price = C.P + Profit = 60 + 6 Selling Price = Rs. 66
Profit and Loss Find the single discount to a series discount
Profit and Loss Answer: SP = [( 80*90*95 )/ 100*100*100 ]* CP Discount = (1 – 0.684) * 100% = 0.316 *100 % Discount = 31.6%
Calendar Odd days: 0 = Sunday 1 = Monday 2 = Tuesday 3 = Wednesday 4 = Thursday 5 = Friday 6 = Saturday
Calendar Month code: Ordinary year J = 0 F = 3 M = 3 A = 6 M = 1 J = 4 J = 6 A = 2 S = 5 O = 0 N = 3 D = 5 Month code for leap year after Feb. add 1.
Calendar Ordinary year = (A + B + C + D )-2 -----------------------take remainder 7 Leap year = (A + B + C + D) – 3 ------------------------- take remainder
Calendar What is the day of the week on 30/09/2007?
Calendar Solution: A = 2007 / 7 = 5 B = 2007 / 4 = 501 / 7 = 4 ( A + B + C + D )-2 = ----------------------- 7 = ( 5 + 4 + 2 + 5) -2 ----------------------- = 14/7 = 0 = Sunday
Calendar What was the day of the week on 13th May, 1984?
Calendar Solution: A = 1984 / 7 = 3 B = 1984 / 4 = 496 / 7 = 6 ( A + B + C + D) -3 = ----------------------- 7 = 14/7= 0, Sunday.
Calendar On what dates of April 2005 did Sunday fall?
Calendar Solution: You should find for 1st April 2005 and then you find the Sundays date. A = 2005 / 7 = 3 B = 2005 / 4 = 501 / 7 = 4 C = 1 / 7 = 1 D = 6 (A + B + C + D) -2 = ----------------------- 7 3 + 4 + 1 + 6 -2 ----------------------- = 12 / 7 = 5 = Friday. 1st is Friday means Sunday falls on 3, 10, 17, 24
Calendar What was the day on 5th January 1986?
Calendar A = 1986 / 7 = 5 B = 1986 / 4 = 496/7 = 6 C = 5 / 7 = 5 D = 0 Solution: A = 1986 / 7 = 5 B = 1986 / 4 = 496/7 = 6 C = 5 / 7 = 5 D = 0 (A + B + C + D) -2 = ----------------------- 7 5 + 6 + 5 + 0-2 = ----------------------- = 14 / 7 = Sunday
Clocks Clock: In every Hour, the minute hand gains 55 minutes on the hour hand In every hour both the hands coincide once. = (11m/2) – 30h (hour hand to min hand) = 30h – (11m/2) (min hand to hour hand) If you get answer in minus, you have to subtract your answer with 360 o
Clocks 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 = 100 Angle between 7: 20 is 100o
Clocks How many times in a day, the hands of a clock are straight?
Clocks Solution: In 12 hours, the hands coincide or are in opposite direction 22 times a day. In 24 hours, the hands coincide or are in opposite direction 44 times a day.
Clocks How many times do the hands of a clock coincide in a day?
Clocks Solution: 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.
Clocks At what time between 7 and 8 o’clock will the hands of a clock be in the same straight line but, not together?
Clocks Solution: h = 7 = 30h – 11m/2 180 = 30 * 7 – 11 m/2 On simplifying we get , 5 5/11 min past 7
Clocks At what time between 5 and 6 o’clock will the hands of a clock be at right angles?
Clocks Solution: h = 5 90 = 30 * 5 – 11m/2 Solving 10 10/11 minutes past 5
Clocks Find the angle between the two hands of a clock at 15 minutes past 4 o’clock
Clocks Solution: Angle = 30h – 11m/2 = 30*4 – 11*15 / 2 The angle is 37.5o
Clocks At what time between 5 and 6 o’clock are the hands of a clock together?
Clocks Solution: h = 5 O = 30 * 5 – 11m/2 m = 27 3/11 Solving 27 3/11 minutes past 5
Data Interpretation In interpretation of data, a chart or a graph is given. Some questions are given below this chart or graph with some probable answers. The candidate has to choose the correct answer from the given probable answers.
1. The following table gives the distribution of students according to professional courses: __________________________________________________________________ Courses Faculty ___________________________________ Commerce Science Total Boys girls Boys girls ___________________________________________________________ Part time management 30 10 50 10 100 C. A. only 150 8 16 6 180 Costing only 90 10 37 3 140 C. A. and Costing 70 2 7 1 80 On the basis of the above table, answer the following questions:
Data Interpretation The percentage of all science students over Commerce students in all courses is approximately: (a) 20.5 (b) 49.4 (c) 61.3 (d) 35.1
Data Interpretation Answer: Percentage of science students over commerce students in all courses = 35.1%
Data Interpretation What is the average number of girls in all courses ? (a) 15 (b) 12.5 (c) 16 (d) 11
Data Interpretation Answer: Average number of girls in all courses = 50 / 4 = 12.5
Data Interpretation What is the percentage of boys in all courses over the total students? (a) 90 (b) 80 (c) 70 (d) 76
Data Interpretation Answer: Percentage of boys over all students = (450 x 100) / 500 = 90%
Data Sufficiency Find given data is sufficient to solve the problem or not. If statement I alone is sufficient but statement II alone is not sufficient If statement II alone is sufficient but statement I alone is not sufficient If both statements together are sufficient but neither of statement alone is sufficient. If both together are not sufficient
Data Sufficiency What is John’s age? In 15 years will be twice as old as Dias would be Dias was born 5 years ago. (Wipro)
Data Sufficiency Answer: c) If both statements together are sufficient but neither of statement alone is sufficient.
Data Sufficiency What is the distance from city A to city C in kms? City A is 90 kms from city B. City B is 30 kms from city C
Answer: d) If both together are not sufficient Data Sufficiency Answer: d) If both together are not sufficient
If A, B, C are real numbers, Is A = C? A – B = B – C A – 2C = C – 2B Data Sufficiency If A, B, C are real numbers, Is A = C? A – B = B – C A – 2C = C – 2B
Answer: D . If both together are not sufficient Data Sufficiency Answer: D . If both together are not sufficient
Data Sufficiency What is the 30th term of a given sequence? The first two term of the sequence are 1, ½ The common difference is -1/2
Data Sufficiency Answer: If statement I alone is sufficient but statement II alone is not sufficient
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.
Answer: D. If both together are not sufficient Data Sufficiency Answer: D. If both together are not sufficient
What is the value of A if A is an integer? A4 = 1 A3 + 1 = 0 Data Sufficiency What is the value of A if A is an integer? A4 = 1 A3 + 1 = 0
Data Sufficiency Answer: B. If statement II alone is sufficient but statement I alone is not sufficient
Cubes A cube object 3”*3”*3” is painted with green in all the outer surfaces. If the cube is cut into cubes of 1”*1”*1”, how many 1” cubes will have at least one surface painted?
Cubes Answer: 3*3*3 = 27 All the outer surface are painted with colour. 26 One inch cubes are painted at least one surface.
Cubes A cube of 12 mm is painted on all its sides. If it is made up of small cubes of size 3mm, and if the big cube is split into those small cubes, the number of cubes that remain unpainted is
Cubes Answer: = 8
Cubes A cube of side 5 cm is divided into 125 cubes of equal size. It is painted on all 6 sides. 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?
Cubes Answer: 54 36 8 27
Cubes A cube of 4 cm is divided into 64 cubes of equal size. One side and its opposite side is coloured painted with orange. A side adjacent to this and opposite side is coloured red. A side adjacent to this and opposite side is coloured green? Cont..
Cubes 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?
Cubes Answer: 8
Cubes A 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 ?
Cubes Answer: 27 18 4 12
Venn Diagram If X and Y are two sets such that X u Y has 18 elements, X has 8 elements, and Y has 15 elements, how many element does X n Y have?
Venn Diagram Solution: We are given n (X uY) = 18, n (X) = 8, n (Y) =15. using the formula. n( X n Y) = n (X) + n (Y) - n ( X u Y) n( X n Y) = 8 + 15 – 18 n( X n Y) = 5
Venn Diagram If S and T are two sets such that S has 21elemnets, T has 32 elements, and S n T has 11 elements, how many element elements does S u T have?
Venn Diagram Answer: n (s) = 21, n (T) = 32, n ( S n T) = 11, n (S u T) = ? n (S u T) = n (S) + n( T) – n (S n T) = 21 + 32 – 11 = 42
Venn Diagram If A and B are two sets such that A has 40 elements, A u B has 60 elements and A n B has 10 elements, how many element elements does B have?
Venn Diagram Answer: n ( A) = 40, n ( n B) = 60 and n ( A n B) = 10, n ( A u B) = n ( A) + n (B) – n ( A n B) 60 = 40 + n (B) – 10 n (B) = 30
Venn Diagram In a group of 1000 people, there are 750 people who can speak Hindi and 400 who can speak English. How many can Speak Hindi only?
Answer: n( H u E) = 1000, n (H) = 750, n (E) = 400, n( H u E) = n (H) + n (E) – n( H n E) 1000 = 750 +400 – n ( H n E) n ( H n E) = 1150 – 100 = 150 No. of people can speak Hindi only _ = n ( H n E) = n ( H) – n( H n E) = 750 – 150 = 600
Venn Diagram In a class of 100 students, the number of students passed in English only is 46, in maths only is 46, in commerce only is 58. the number who passes in English and Maths is 16, Maths and commerce is 24 and English and commerce is 26, and the number who passed in all the subject is 7. find the number of the students who failed in all the subjects.
Venn Diagram Solution: No. of students who passed in one or more subjects = 11+ 9 + 13 + 17 + 15 + 19 + 7 = 91 No of students who failed in all the subjects = 100 -91 = 9
Venn Diagram In a group of 15, 7 have studied Latin, 8 have studied Greek, and 3 have not studied either. How many of these studied both Latin and Greek?
Answer: 3 of them studied both Latin and Greek. Venn Diagram Answer: 3 of them studied both Latin and Greek.