Aptitude Daily

Question 1An express A starts at 2.30 pm from Karnal station and travel towards Sirsa station at a speed of 80 km per hour. Another express B starts at 4.30 pm from Karnal to Sirsa at the speed of 100 km per hour. How far away from station Karnal will the two trains meet?a) 400 km b) 100 km c) 800 km d) 450 kmAnswer :c) 800 km.Solution :Train A starts at 2.30 pm from KARNAL to SIRSA.Train B starts at 4.30 pm from KARNAL to SIRSA.Let they meet H hours after 2.30 pm.We have to find the distance traveled by A in H hours or by B in (H-2) hours.Given that, speed of A = 80 km/hrDistance traveled by A in H hours at 80km/hr = (distance = speed x time) = 80H km. (1)Speed of B = 100 km/hr.Note that, it starts at 4.30 pm, that is 2 hours later than A.Therefore, we would find the distance in (H-2) hours.Distance traveled by B in (H-2) hours at 100km/hr = 100(H-2) km (2)We know that trains meet after H hours and this means distance covered by train A in H hours equals distance covered by train B in H-2 hours.Therefore, from (1) and (2), we get 80H = (H-2)10020H = 200H = 10 hours.That is, they meet after 10 hours from2.30 pm.Hence, the required distance = 80H = 80 x 10 = 800 km.Question 2A train P starts from Delhi at 8 pm, reaches its destination UP at 1 am. Another train Q starts from UP at 8 pm and reaches Delhi at 2 am. The two trains P and Q will cross each other at :a) 10.45 pm b) 12.45 pm c) 11.45 pm d) none of these.Answer :a) 10.45 pm.Solution :Let the distance between UP and Delhi be X km.Time taken by P to reach UP (to cover X km) = 8 pm to 1 am = 5 hours.Speed of P = distance / time = X/5 km/hr.Time taken by Q to reach Delhi (to cover X km) = 8 pm to 2 am = 6 hours.Speed of Q = X/6 km/hr.Let them meet Y hours after 8 pm.Distance covered by P at Y hours at X/5 km/hr = XY/5 kmDistance covered by Q at Y hours at X/6 km/hr = XY/6 km.Since they travel in opposite directions, during the time when they cross each other, the sum of the distance covered by the first and second trains will be equal to the distance between Delhi and destination at UP.i.e., XY/5 + XY/6 = XXY (1/5 + 1/6 ) = XY (1/5 + 1/6 ) = 1Y (11/30) = 111Y = 30.Y = 30/11 = 2.72 hoursTherefore, they meet after 2.72 hours from 8 pm.That is, 8 pm + 2 hours + 45 minutes (nearly).Hence, the answer is 10.45 pm.Question 3Chennai express starts at 8 am from A towards B and Mumbai express starts at 8 am from B towards A. After their meet, Chennai express takes 18 hours to reach B and Mumbai express takes 8 hours to reach A. Then which of the following is their speeds ratio?a) 1 : 2 b) 2 : 3 c) 2 : 5 d) 1 : 3Answer :b) 2 : 3.Solution :Note: If two trains start at the same time from A and B towards each other and after they take a and b hours in reaching B and A respectively, thenAs speed : Bs speed = sqrt(b) : sqrt (a).Here they start at same time ( 8 am) and travels in opposite direction.(After their meet), Chennai express takes 18 hours to reach B and Mumbai express takes 8 hours to reach A.b = 8 and a = 18.By using the above note, required ratio = sqrt(8) : sqrt(18) = 2 sqrt(2) : 3 sqrt(2) = 2 : 3.Hence, the answer is 2 : 3.Question 1Find X's age which equals the number of grand children of a man who has 4 sons and 4 daughters. Each daughter of the man's wife have 3 sons and 4 daughters and each son of the man's wife have 4 sons and 3 daughters.a) 40 b) 56 c) 64 d) none of theseAnswer :b) 56Solution :We have to find the number of grand children of the man.Given that, he had 4 sons and 4 daughters.Each son has 4 sons and 3 daughters and each daughter has 3 sons and 4 daughters.Therefore total number of grandsons = 4x4 + 4x3 = 16 + 12 = 28And total number of grand daughters = 4x3 + 4x4 = 28Total number of grandchildren is 28+28 = 56.Hence the required age is 56.Question 2A man have many daughters, each daughter have as many sons as her sisters. The product of the number of daughters and grandsons of the man lies between 40 and 50. Find the number of daughters of the man.a) 6 b) 4 c) 3 d) 8Answer : b) 4Solution:Let N be the number of daughters of the man.Then each daughter has N-1 sisters.Given that, each daughter has as many sons as her sisters.That is, each of them has N-1 sons.Number of grandsons of the man = N(N-1)And the required product = N(N-1) x N = N x N(N-1) which lies between 40 and 50.If N = 1 then N x N(N-1) = 0.f N = 2 then N x N(N-1) = 2 x 2(1) = 4If N = 3 then N x N(N-1) = 3 x 3 x 2 = 18If N = 4 then N x N(N-1) = 4 x 4 x 3 = 48If N = 5 then N x N(N-1) = 5 x 5 x 4 = 100Therefore the possible value of N is 4.Hence the man has 4 daughters.Question 3Two men A and B have equal number of daughters and A have 1 more son than B. Each son and daughter of A have 2 sons and 2 daughters and each son and daughter of B have 3 sons and 3 daughters. If A and B have equal number of grandchildren then find the number of sons of B.a) 3 b) 4 c) 5 d) none of these.Answer :d) none of theseSolution :Let D be the number of daughters of A and B.Let S be the number of sons of A.Then S-1 be the number of sons of B.Now, number of grandsons of A = 2S + 2D.Number of granddaughters of A = 2S + 2DNumber of grand children of A = 2S + 2D + 2S + 2D = 4S + 4D.Number of grandsons of B = 3(S-1) + 3D = 3S + 3D - 3Number of granddaughters of B = 3(S-1) + 3D = 3S + 3D - 3Number of grand children of A = 3S + 3D - 3 + 3S + 3D - 3 = 6S + 6D - 6And, 4S + 4D = 6S + 6D - 62S + 2D = 6S + D = 3Then the possibilities of (S,D) are (1,2), (2,1), (0,3), (3,0)We have to find the value of S-1.S-1 cannot be a negative number, so (0,3) is not possible.Therefore, possible S-1 values are 0,1,2.Hence the answer is option d.Question 1If Mr.X, Mr.Y and Mr.Z are three siblings and following three statements are true1) Mr.X has one older brother and three younger sisters.2) Mr.Y has two older brothers and two younger sisters3) Mr.Z has three older brothers and one younger sister.Then the least possible number of siblings in their family is:a) 5 b) 6 d) 7 d) 8Answer :b)6Solution :First, we need to think about X, Y and Z as who is the oldest among the three and who is the youngest among the three.X has one older brother while Y has two. Therefore, X is older than Y. Similarly Y is older than Z. Therefore, among X,Y and ZX is the oldest, Y is next and Z is the youngest.X has an older brother and Z has a younger sister.Try making a chart laying out all the three conditions given in question, and mark the gender of each known sibling.Then, it is clear that X is male and Y is female.MMFMFF1st2nd3rd4th5th6thXYZBased on the above chart, there should be at least 6 siblings.Question 2If A, B and C are triblet. A,C are males and B is female,1) A & C say " We have 2 more brothers than the number of sisters"2) B says " I have 2 more brothers than twice the sisters "If the above two statements are true then the number of siblings in their family is:a) 6 b) 12 c) 8 d) 9Answer :b) 12Solution :Let g be the number of girls in the family.Let b be the number of boys in the family.Consider statement 1, A and C would have b-2 brothers (excluding them). This b-2 is equal to two more than the number of sisters they have.Therefore, b-2 = g+2Or, b-g = 4 -----(1)Now, from statement 2, B would have g-1 sisters (excluding herself) and b brothers. She has 2 more brothers than twice the number of sisters.Therefore, Twice the number of sisters = 2(g-1) which is equal to 2 + number of brothers.Then we have, 2(g-1) + 2 = b2g - 2 + 2 = bb-2g = 0 -------(2)Solving (1) and (2), we have g = 4 and b = 8.Hence the number of siblings = 4+8 = 12.Question 3Aravind and Bavya are friends.1)Aravind's sister has thrice as many sisters of Bavya as brothers.2)Aravind has 1 more brother than than that of Bavya3)Twice Bavya's brother's sisters equals the number of boys in Aravind's family.If the above all 3 statements are true then the number of brothers what Aravind has more than Bavya is:a) 1 b) 2 c) 3 d) none of theseAnswer :a) 1Solution :Let the number of boys in Aravind's family be aLet the number of girls in Aravind's family be bLet the number of boys in Bavya's family be cLet the number of girls in Bavya's family be d.consider statement 1, "Aravind's sister has thrice as many sisters of Bavya as brothers" :Number of Bavya's sisters = d-1 (excluding her)Thrice the Bavya's sisters = 3(d-1)Aravind's sister has 'a' number of brothers.Therefore, 3(d-1) = ad-1 = a/3....(1)Consider statement2,"Aravind has 1 more brother than that of Bavya" :Number of Aravind's brothers = a-1Number of Bavya's brother = cSince, Aravind has 1 more brother than Bavya then a-1 = c+1a-c = 2...(2)Consider statement3,"Twice Bavya's brother's sisters equals the number of boys in Aravind's family"The number of sisters of Bavya's brother = dNumber of boys in Aravind's family = aTherefore 2d = aOr d = a/2Then, d-1 = a/2 - 1 ...(3)Now, let us solve equations we obtained in previous steps :From (1) and (3),a/3 = a/2 - 1a = 6Put a value in (2), we have c = 4Put a value in (3), we have d = 3Therefore Aravind has a-1 = 6-1 = 5 brothersAnd Bavya has c = 4 brothers.Hence, Aravind has 1 more brother than Bavya.

Below are three different problems on mixtures.Question 1How much water must be added to reduce the quantity of milk to 20% in the 16 litres mixture of 30% milk?a) 10 litres b) 8 litres c) 12 litres d) 6 litresAnswer :b) 8 litres.Solution :Given that, 16 litres of mixture contains 30% milk.That means we have 30 percent milk and 70 percent water.Now,30 percent of 16 litres = 30 x 16/100 = 4.8 litres.So we have 4.8 litres of milk and 11.2 litres of water (16 - 4.8 = 11.2).Now we want to add water until the quantity of milk is reduced to 20 percent.Note that the amount of milk(4.8 litres) will remain constant. We are only changing the amount of water.That is, we want 4.8 litres of milk to be the 20 percent of the entire quantity.Set up a proportion to find out how much is 100 percent of the quantity.(20 percent)/(4.8 litres) =(100 percent)/(X litres)X = 24 litres.That is, 20 percent of 24 litres of mixture is 4.8 litres of milk.Given quantity of mixture is 16 litres.Then the amount of water to be add = 24-16 litres = 8 litres.Question 2There are three bottles A, B and C of same size and they contain the mixture of gasoline and alcohol in the ratio 4:3, 8:5 and 3:1. The contents of A, B and C are poured into a new bottle then the ratio of the new mixture is:a) 361:344 b) 705:364 c) 364:387 d) 235:129Answer :d) 235:129Solution :The ratio of gasoline and alcohol in A = 4:3Then the proportion of gasoline in A = 4/7And the proportion of alcohol in A = 3/7Then, the proportion of gasoline in three bottles = 4/7, 8/13, 3/4And the proportion of alcohol in three bottles = 3/7, 5/13, 1/4Therefore, total proportion of gasoline in new bottle = 4/7 + 8/13 + 3/4 = (4x13x4)/364 + (8x4x7)/364 + (3x7x13)/364= (208+224+273)/364 = 705/364And the total proportion of alcohol in new bottle = 3/7 + 5/13 + 1/4 = (156+140+91)/364 = 387/364Now, the required ratio = 705/364 : 387/364 = 705:387 = 235 : 129Question 350% and 25% of good quality of wheat of X and Y are mixed to obtain 42 kg of 32 1/7 % of good quality wheat. Then the amount of Y to be mixed to form such mixture is :a) 30 kg b) 12 kg c) 40 kg d) 9 kgAnswer :a) 30 kgSolution :Given that, the amount of mixture = 42 kgLet the amount of 50% of good quality wheat of X be A kg and 25% of Y be B kg.Then, A + B = 42 kg ....(1)And, 50% of A + 25% of B = 32 1/7 % of 4250A/100 + 25B/100 = 32 1/7 % of 42 = 225/7 x 1/100 x 4250A + 25B = 42 x 225/72A + B = 42 x 9/7 = 378/714A + 7B = 378....(2)Solving (1) and (2), A = 12 and B = 30.Hence the required amount of Y is 30 kg.

Below are three problems on Time and distance, dealing with different speed's time calculations.Question 1A train travelled a distance of 610 km such that 'd' km by 40km/hr and remaining by 90km/hr and takes a total of 9 hours then the distance travelled at 90km/hr is:a)542.5km b)450km c)543.5km d)544.5kmAnswer :b)450kmSolution :Total distance travelled by the train = 610 km And d km at 40km/hrTime taken to cross d km is d/40 hours ...(1)Remaining distance 610 - d km is travelled at 90km/hrTime taken to cross 610 - d km = 610 - d / 90 hours ...(2)Total time taken = 9 hoursi.e., d/40 + (610 - d)/90 = 99 = [9d + (610 - d)4]/3609 = 9d + 2440 - 4d / 3603240 = 5d + 24403240 - 2440 = 5d800 = 5dd = 800/5d = 800/5 = 160kmDistance travelled at 90km/hr = 610 - d = 610 - 160 = 450kmQuestion 2The distance between state A ans B is 3500 kms and a train left from a station in state B at 10 am to a station in state A. The train travelled at the speed of 160 km/hr for 20 hours and the speed was reduced to 120 km/hr. Find at what time the train will reach state A.a)10.30 am b)12.05 pm c)3.55 am d)10.15 pmAnswer :a)10.30amSolution :Distance between state A and B = 3500 kmsTrain started at 10 am from state BDistance covered in 20 hours at 160km/hr = 160 x 20 km = 3200 kmsThen the remaining distance at 120km/hr is = 3500 - 3200 = 300kmTime taken to cross 300km is = 300 / 120 = 5/2 hours = 2 hours and 30 minutes.So the train will reach state A after 20 hours + 2 hours + 30 minutes from 10 ami.e., 10 am + 20 hours + 2 hours + 30 minutes = 10.30 amHence the required time is 10.30 am.Question 3A train travels at S kmph and 2S kmph for first 1/4 hour and 3/4 hour respectively. And then again travels at its actual speed. Now what will be the value of S if the train covers totally 360 kms in 3/2 hours?a) 150 kmph b) 160 kmph c) 120 kmph d) 125 kmphAnswer :b) 160 kmphSolution:Given that the train travels at S km/hr for 1/4 hours.Then the distance covered in 1/4 hour = S x 1/4 km = S/4kmThen travels at 2S kmph for 3/4 hour.Then the distance covered in 3/4 hour = 2S x 3/4 = 6S/4 kmSince the actual speed of the train = S km/hrand given that the total time taken = 3/2 hours.then the remaining time at S km/hr = 3/2 - (1/4 +3/4) = 1/2 hour.Then the last covering distance = S x 1/2 km = S/2 kmTotal distance = S/4 + 6S/4 + S/2 kmThen, S/4 + 6S/4 + S/2 km = 360 km (given)9S/4 = 360kmS = 160Hence the required speed is 160 kmph

Below are four problems to find the next number you have to use some arithmetic calculation with digits of the given expression.Question 1If 16 + 32 = 95, 24 + 48 = 64, 31 + 69 = 64 then what will be 38 + 76 = ?a)29 b)19 c)30 d)35Answer :b)19Solution :Let ab + cd be the pattern of the given expression.The logic used in deriving the answer is, the first digit of the answer is last digit of the (a + b)2and second digit of the answer is last digit of (c + d)2i.e., 16 + 32 => (1 + 6)2(3 + 2)2= 7252= last digit of 49 and last digit of 25i.e., 16 + 32 = 95And 24 + 48 => (2 + 4)2(4+8)2= 62122= last digit of 36 and last digit of 14424 + 48 = 6431 + 69 => (3 + 1)2(6 + 9)2= 42152= last digit of 16 and last digit of 22531 + 62 = 65Then 38 + 76 => (3 + 8)2(7 + 6)2= 112132= last digit of 121 and last digit of 16938 + 76 = 19Hence the answer is 19.Question 2If 21 + 34 = 112, 16 + 48 = 461, 70 + 38 = 507 then what will be 28 + 49 = ?a)289 b)928 c)682 d)582Answer :d)582Solution :Let ab + cd be the given pattern.The logic used is, the 1st digit of the answer is the difference between c and d and the 2nd, 3rd of the answer is reverse of ab.(i.e., ba)i.e., 21 + 34 = (4 - 3)12 = 11216 + 48 = (8 - 4)61 = 46170 + 38 = (8 - 3)07 = 50728 + 49 = (9 - 4)82 = 582Hence the answer is 582.Question 3If 12 + 21 = 14, 28 + 82 = 464, 41 + 14 = 161 then 25 + 52 = ?a)425 b)465 c)475 d)485Answer :a)425Solution :Let ab + cd = XY be the given pattern where X = ad and Y = bci.e.,12 + 21 = (1 x 1)(2 x 2) = 1428 + 82 = (2 x 2)(8 x 8) = 46441 + 14 = (4 x 4)(1 x 1) = 16125 + 52 = (2 x 2)(5 x 5) = 425Hence the required answer is 425.Question 4If 53 + 42 = 14, 94 + 33 = 18 and 73 + 96 = 12 then 84 + 72 = ?a)5 b)10 c)15 d)20Answer :b)10Solution :Let ab + cd = sum of the digits in (ab) and (cd)i.e., 53 + 42 = sum of the digits in (5 x 3) and (4 x 2) = sum of the digits in 15 and 8 = 1 + 5 + 8 = 1494 + 33 = sum of the digits in (9 x 4) and (3 x 3) = sum of the digits in 36 and 9 = 3 + 6 + 9 = 1873 + 96 = sum of the digits in (7 x 3) and (9 x 6) = sum of the digits in 21 and 54 = 2 + 1 + 5 + 4 = 1284 + 72 = sum of the digits in (8 x 4) and (7 x 2) = sum of the digits in 32 and 14 = 3 + 2 + 1 + 4 = 10

Below are three problems dealing with percentage calculations.Question 1If 100/3 % of A's work is same as much as 75% of B's work and if B makes turn out of Rs.15000 per month then A's work in terms of making turn out is:a)Rs.50,000 b)Rs.33,750 c)Rs.42,000 d)Rs.55,000Answer :b)Rs.33,750Solution :Let A's work be XThen 100/3% of X = 75% of 15000100/3 x 1/100 x X = 75/100 x 15000X = 75 x 150 x 3 x 100/100X = 75 x 150 x 3X = 33,750.Hence A's turn out is Rs.33,750 per month.Question 2A business man saves 30% of his profit. After 6 months when his profit is increased by 50% he could save the same amount as 30% of his old profit.Now, he wants to calculate the % of his increased expenditure, Can you help?a)83 3/7 % b)75 5/7% c)71 3/7 % d)93 3/7 %Answer :c)71 3/7%Solution :Let original profit = Rs.100Then saving (30%) = Rs.30And expenditure (70%) = Rs.70New profit (after 6 months) = Rs.150New saving = Rs.30New expenditure = Rs.(150 - 30) = Rs.120Increase in expenditure = New expenditure - old expenditure = Rs.(120 - 70) = Rs.50Increase percentage = Increase in expenditure / old expenditure x 100 %= (50/70) x 100% = 500/7% = 71 3/7 %.Question 3Mr.A reduces his consumption by 40% when the price of rice is increased by 40%. Then find A's percentage decrease in expenditure.a) 16% b) 55% c) 35% d) 67%Answer :a) 16%Solution :Let the original consumption be 100 unitsAnd let original price (rice) be Rs.100 per unitoriginal expenditure for 100 units = price per unit x 100 units = (100 x 100)= Rs.10,000After decreasing the consumption to 40%, his new consumption = 60 units.After increasing the price of rice to 40%, the new price per unit = Rs.140 per unit.Then new expenditure = new price per unit x 60 units = 140 x 60 = Rs.8400Decrease in expenditure = Rs.(10000 - 8400) = Rs.1600Percentage decrease in expenditure = Decrease in expenditure / Original Expenditure x 100% = 1600/10000 x 100 = 16%Hence the required % = 16%

Below are four problems on average, dealing with age calculations for placement tests.Question 1There is a group of 5 members. If the average age of last 3 is 20% of the average age of the 1st person and the eldest of that three. The total age of the 2nd one and the youngest of that three is 39 years. If the age of first person is 26 years, what is the age of middle one of that three?a)12 years b)10years c)none of these d)data inadequateAnswer :d)data inadequateSolution :Let the age of the five persons be A, B, C, D and E.Without loss of generality assume that, C is eldest and E is youngest and D is middle one among C, D and E, we have to find the d.Given that, C + D + E / 3 = 20% 0f (A + C)20(A + C)/100 = (C + D + E)/35(C + D + E) = 3(A + C)Given that A = 26, then above eqn becomes 5(C + D + E) = 3(26 + C)5C + 5(D + E) = 78 + 3C2C + 5D + 5E = 78 ....(1)The total age of the 2nd one and the youngest of that three is 39 yearsi.e.,(B + E) = 39 => E = 39 - B ....(2)Substitute eqn (2) in (1)2C + 5D + 5(39 - B) = 78.However, the age of C and B or the total or age of the group is not given.Hence the given data is inadequate.Question 2The average age of 5 members of a team 3 years ago is 17 years. The average age of the team remains same, when a new person added to the team. Find the present age of the new person.a)3 years b)5/2 years c)2 years d)3/2 yearsAnswer :c)2 yearsSolution :The average age of 5 members(3 years ago) = 17 years.Then the total age of 5 members(3 years ago) = 17 x 5 = 85 years.And the total age of 5 members now = 85 + (5 x 3) = 85 + 15 = 100 years ...(1)After adding the new person, the average age of 6 members = 17 yearsThen the total age of 6 members (now) = 17 x 6 = 102 years ...(2)Then the age of new person = 102 - 100 = 2 years.Hence the required answer is 2 years.Question 3In an entrance exam of an Engineering college, the average score of result of candidates is 40 in the Round I.The average score of result of 120 candidates of Round II is 32.If we add the average score of two rounds, the result is decreased by 4 when compared to Round I. Find the total number of candidates participated in the exam.a)240 b)150 c)120 d)300Answer :a)240Solution :Let the number of candidates in Round I be X and their average score is 40.Total score of candidates in Round I = 40XNumber of candidates in Round II = 120 and their average score is 32Total score of candidates in Round II = 32 x 120 = 3840Total number of candidates ( Round I + Round II) = X + 120Total score of candidates (Round I + Round II) = 40X + 3840New average score (Round I + Round II) is decreased by 4 when compared to Round I, i.e., 40 - 4 = 36Total score / Total number of candidates = New Average40X + 3840 / (X + 120) = 3640X + 3820 = 36(X + 120)40X + 3840 = 36X + 432040X - 36X = 4320 - 38404X = 480X = 120Number of candidates in Round I = 120 candidates.Total number of candidates ( Round I + Round II) = 120 + 120 = 240Question 4The average age of the children and adults in a family is 29/2 years and 29.8 years. If there are twice as many children as adults then find the average age(in months) of the whole family.a)254 b)235 c)234 d)298Answer :b)235Solution :Let the number of children in a family be 2XThe average age of children in the family is 29/2 yearsThen the total age of children = (29/2) x 2X years = 29XThe number of adults in the family is XThe average age of adults in the family is 29.8 yearsThen the total age of adults = (29.8)X years.Average age of the family = 29X + 29.8X / 2X + X58.8X / 3X = 58.8 / 3 = 19.6 years.Now, 19.6 years = 12 x 19.6 months = 235.2 monthsHence the required answer is 235 months(approximately)

Below are three problems dealing with the calculations of the digits of the number.Question 1In how many ways, can a three digit number formed if the sum of the first and last digit is less than or equal to the middle and the middle digit is one-third of 27?a)22 b)25 c)45 d)50Answer :c)45Solution :Let the required three digit number be XYZGiven that the middle digit Y = 1/3 of 27 = 9And X + Z < = Yi.e., X + Z < = 9Then X + Z would belong to {0,1,2,3,4,5,6,7,8,9}Since We have to find the three digit number, the first digit cannot be zero also X + Z cannot be zero.If X + Z = 1 then (X,Z) = (1,0)i.e., the number of ways to get X + Z = 1 is 1If X + Z = 2 then (X,Z) = (1,1) or (2,0)i.e., the number of ways to get X+Z = 2 is 2If X + Z = 3 then (X,Z) = (1,2) or (2,1) or (3,0)i.e., the number of ways to get X + Z = 3 is 3Similarly,The number of ways to get X + Z = 4 is 4.........The number of ways to get X + Z = 9 is 9Therefore the total number of ways to get such a three digit number is (1+2+3+...+9=) 45.Hence the required number of ways is 45.Question 2There is a password of four digit number, if the 2nd digit is 5, the sum of the 1st and 3rd digit is equal to the 2nd digit and the last digit is equal to the fisrt digit then find the number of ways to access a password which has an odd number be the sum of its digits.a)1 b)2 c)3 d)0Answer :c)3Solution :Let the four digit password be abcd.Given that b = 5And a + c = 5The first digit of a password may be zero.Then the possibilities of (a,c) are (0,5), (1,4), (2,3), (3,2), (4,1) and (5,0).Given that the last digit of the password is equal to the first digit and the sum of the digits is an odd number.Then the possibilities are 0550, 1541, 2532, 3523, 4514 and 5505 and their respective sum are 10, 11, 12, 13, 14 and 15.Therefore the possible password are 1541, 3523 and 5505.Hence the required number of ways is 3.Question 3In how many ways can a man access his pin number of four digits if the sum of the 1st and 3rd digit is less than 6 and the sum of the second and fourth digit is less than 8 but greater than 6?a)120 b)168 c)29 d)none of theseAnswer :b)168Solution :Let the four digit pin number be abcd.Given that a + c < 6 and 6 < b + d < 8So a + c belong to {0,1,2,3,4,5}If a + c = 0 then (a,c) = (0,0)i.e., the number of ways to get a + c = 0 is 1If a + c = 1 then (a,c) = (0,1) or (1,0)i.e., the number of ways to get a + c = 1 is 2If a + c = 2 then (a,c) = (0,2) or (1,1) or (2,0)i.e., the number of ways get to a + c = 2 is 3Similarly,The number of ways to get a + c = 3 is 4The number of ways to get a + c = 4 is 5The number of ways to get a + c = 5 is 6Therefore the total number of ways to get a + c is less than 6 is (1 + 2 +...+ 6 =)21.Now, consider 6 < b + d < 8 then obviously b + d would be 7.If b + d = 7 then (b,d) = (0,7),(1,6),(2,5),(3,4),(4,3),(5,2),(6,1) and (7,0)The number of ways to get b + d = 7 is 8Therefore the total number of ways to get such a four digit pin number is 21 x 8 = 168Hence the required number is 168.

Below are three problems on time and speed which you can expect in placement tests.Question 1Madhuri went to a shopping mall and she want to go to the down floor through escalator. The length of the stairway in the escalator is 46 steps. If Madhuri takes X steps, then she would reach down floor in 60 seconds and if she takes Y steps, then she would reach in 36 seconds. Find the values of X and Y.a) 13,17 b) 15,21 c) 19,30 d) 26,34Answer :d)26,34Solution :The stairway have totally 46 steps.She covers 46 - X steps in 60 seconds.Then speed = 46 - X / 60 steps/sec (speed = distance/time)And she covers 46 - Y steps in 36 seconds.Here, speed = 46 - Y / 36We have to find the values of X and Y.Assume that speed in either case is same.Then we would have,46 - X / 60 = 46 - Y / 3646 - X / 5 = 46 - Y / 33(46 - X) = 5(46 - Y)230 - 138 = -3X + 5Y92 = -3X + 5Y .....(1)Now, substitute the given options in eqn(1), to find the answer.If X = 13 and Y = 17 then -3X + 5Y = -3(13) + 5(17) = 46If X = 15 and Y = 21 then -3X + 5Y = -3(15) + 5(21) = -45 + 105 = 60If X = 19 and Y = 30 then -3X + 5Y = -3(19) + 5(30) = -57 + 150 = 93If X = 26 and Y = 34 then -3X + 5Y = -3(26) + 5(34) = -78 + 170 = 92.Hence, the required answer is option d.Question 2In airport, Vikash has to catch the flight so he chooses the escalator to reach the top. The escalator is moving at the constant speed, if he takes 25 steps he would reach the top in 9 seconds. If he takes 7 steps he would reach in 15 seconds. Find the total number of steps in the escalator.a) 52 b) 42 c) 68 d) 62Answer :a) 52Solution :Let the escalator have n stepsNow, Vikash covers n - 25 steps in 9 seconds and n - 7 steps in 15 seconds.Given that the speed in either case is same.Then we would have (n - 25)/9 = (n - 7)/15(n - 25)/3 = (n - 7)/55n - 3n = 125 - 21 = 104.n = 104/2n = 52 stepsHence the required answer is 52.Question 3Avinash went to a shopping mall in Bangalore. He reaches top in 13 seconds taking 16 steps. If he takes 36 steps he would reach in 8 seconds. Find the speed of the escalator in steps per second.a) 3 b) 4 c) 5/7 d) 2/3Answer :b) 4.Solution :Let the number of steps in escalator be n.Given that, he covers n - 16 steps in 13 seconds and n - 36 steps in 8 seconds.Then the required speed is either (n - 16)/13 or (n - 36)/8.Now, we have (n - 16)/13 = (n - 36)/88(n - 16) = 13(n - 36)13n - 8n = 468 - 1285n = 340n = 68 stepsTherefore speed, n - 16/13 steps per second = 68 - 16 / 13 = 4Hence, the answer is 4 steps per second.

Below are three different problems dealing with ratio and proportion calculations.Question 1The price of Machine D equals the sum of the prices of machine A, B and C whose price are in the ratio 2:3:4 respectively. If weights of A, B, C and D varies as square of its individual price and difference of weight of D and A, B and C together is 9880 kg. Then what is the weight of D?a) 15390 kg b) 14790 kg c) 15800 kg d) none of theseAnswer :a) 15390 kgSolution :Note that,"We say that x varies as y, if x = ky for some constant k"Given ratio of prices of A, B and C = 2:3:4Let their price be 2X, 3X and 4X respectively.Then D's price = 2X + 3X + 4X = 9XWeight varies as square of price.Then A's weight = k 4X2B's weight = k 9X2C's weight = k 16X2Sum of their weight = k 29X2And D's weight = k 81X2Therefore, k81X2- k29X2= 9880 kgk52X2= 9880KX2= 190Hence, D's weight = 81kX^2 = 81 x 190 = 15390 kg.Question 2If 6:4:3 is the ratio of volumes of three boxes which are filled by mixtures of two variety rice. The mixture consists type 1 and type 2 rice in the ratio 5:3, 4:3 and 3:1 respectively. The contents of all three boxes are transferred into a new box. The ratio of type 1 and type 2 rice in the new box is:a) 52:19 b) 43:21 c) 51:30 d) 58:33Answer :d) 58:33Solution :Three boxes contains 6X, 4X and 3X kg of mixtures respectively.1st box contains mixture in the ratio 5:3.Type 1 in 1st mixture = 6X x 5/8 = 15X/4 kg.And type 2 in 1st mixture = 6X - 15X/4 = 9X/4 kg2nd box contains mixture in the ratio 4:3.Type 1 in 2nd mixture = 4X x 4/7 = 16X/7 kg.And type 2 in 2nd mixture = 4x - 16X/7 = 12X/7 kg3rd box contains mixture in the ratio 3:1.Type 1 in 3rd mixture = 3X x 3/4 = 9X/4 kg.And type 2 in 3rd mixture = 3X - 9X/4 = 3X/4 kgTotally, type 1 in final mixture = 15X/4 + 16X/7 + 9X/4 = 232X/28 = 58X/7Totally, type 2 in final mixture = 9X/4 + 12X/7 + 3X/4 = 132X/28 = 33X/7Required ratio = 58X/7 : 33X/7 = 58:33Question 3If 4:6 is the ratio of number of girls and boys in a computer coaching class.If 50% of girls and 40% of boys are degree holders then the percentage of the candidates who are non-degree holders is:a) 12% b) 38% c) 56% d) 49%Answer :c) 56%Solution :Given ratio of girls and boys = 4:6 = 2:3Let boys = 3X and girls = 2X.Number of girls who are non-degree holders = (100 - 50)% of 2X = 50% of 2XNumber of boys who are non-degree holders = (100 - 40)% of 2X = 60% of 3X.Total non-degree holders = 50% of 2X + 60% of 3X.= X + 9X/5 = 14X/5Required percentage = [(14X/5) / 5X] x 100 = 14X/5 x 1/5X x 100 = 14 X 4 = 56%.

Below are three moderate problems on time and work.Question 1In a NSS camp, a work can be completed by a group of students in a certain number of days. After 20 days, 2/5 of the students left from the camp and it was found that the remaining student take time as long as before. What will be the time taken to complete the entire work?a) 50 b) 58 c) 59 d) 52Answer :a) 50Solution :Initially, there were X numbers of students to complete the work in Y days.we have to find the value of Y.After 20 days, X students would complete the work in (Y - 20) days.Then, the work completed by X students in (Y - 20) days = X(Y - 20)If 2/5th of student left, then the remaining students = X - 2X/5Therefore, (X - 2X/5) students work for Y days.Now, the work completed by (X- 2X/5) students in Y days = (X - 2X/5)YThen we would have X(Y - 20) = (X - 2X/5)Y5X(Y - 20) = (5X - 2X)Y5XY - 100X = 3XY.2XY - 100X = 0(Y - 50)= 0Y = 50.Hence the required number of days is 50.Question 2A team of 200 wagers undertakes building work of a bridge. The total time allocated to build the entire bridge is 20 days. After 10 days since start, 200 more wagers join the team and together the team completes the bridge in required time. If the original team do not get those 200 extra wagers, how many days they would be behind schedule to complete building the bridge.a) 10 days b) 20 days c) 15 days d) 1 daysAnswer :a) 10 daysSolution :First, 200 wagers worked for 10 days.Amount of total work completed in 10 days = no of wagers x no of days = 200 x 10= 2000 wager-days ...(1)(here the unit of measurement is wager-days. Unit of work can be expressed in the form of persons x time)After adding 200 more wagers, we have 400 wagers in total for the remaining 10 days.During the second 10 days (i.e from 11 day till the end of 20th day), amount of total work completed will be = 400x10= 4000 wager-days ...(2)For the entire duration of 20 days, total work completed = 2000 + 4000 = 6000 wager-days6000 wager- days amount of work is required to complete the bridgeIf 200 wagers are involved, the number of days they would require to complete the bridge = total wager days / 200 = 6000/200 = 30 daysWe know that the addition of extra 200 wagers helped the team to complete in 20 days.This means, without those 200 additional wagers, the original team would have taken, 30-20 = 10 days more to complete the work.Question 3A team P of 20 engineers can complete a task in 32 days. Another team Q of 16 engineers can complete the same task in 30 days. Then the ratio of working capacity of 1 member of P to that of a member of Q is:a) 3:2 b) 4:3 c) 2:5 d) 3:5Answer :b) 4:3Solution :Given, 20 engineers of P can complete the task in 32 days.This means, to complete the entire task in a single day, number of engineers required would be (20 x 32) engineers = 640 engineers.640 engineers can complete the work in 1 day.Therefore, 1 engineer when working alone will take 640 days to complete the work.In other words, 1 engineer when working alone can finish 1/640 work in 1 day.Similarly, 1 engineer from Q team alone can complete 1/ 480 (1/30 x 16) work in 1 daySo, required ratio = 1/640 : 1/480 = 64:48 = 4:3.

Below are three problems on time calculations.Question 1The time showed by an analog clock at a moment is 11 am then 1234567890 hours later it will show the time as:a)11am b)11pm c)5am d)4pmAnswer :c)5amSolution :We know a day has 24 hours.So 1234567890 hours can be expressed as,1234567890 = 24(51440328) + 18 (i.e., divide the given hour(1234567890) by 24)1234567890 hours = 51440328 days + 18 hoursNow the clock shows 11am, after 1234567890 hours it will show 11 am + 18 hours11am + 12 hours + 6 hours11pm + 6 hours5 amHence the required time is 5 amQuestion 2If a digital clock shows time as 15:30 then after 2096415526 hours it will show:a)13:30 b)15:30 c)12:30 d)19:30Answer :a)13:30Solution :We can express the given 2096415526 as 24(87350646) + 22i.e., 2096415526 hours = (87350646) days + 22hoursNow the time is 15:30, after 2096415526 hours it will show15:30 + 22 hours37:30= 24 + 13:30Hence the required time is 13:30Question 3The time showed by the clock at a moment is 8 am then the respective time of after and before 42896 hours is :a)4 pm & 12 am b)6 am & 4 pm c)5 pm & 7 am d)3 am & 3 pmAnswer :a)4 pm & 12 amSolution :We can express 42896 as42896 = 24(1787) + 8Therefore after 42896 hours the time is 8 am + 8 hours8am + 4 hours + 4 hours12 pm + 4hours4pmAnd before 42896 hours is 8am - 8 hours= 12 amHence the required answer is option a.

Below are three problems on numbers dealing with the calculations on digits.Question 1Square of two more than a two digit number is multiplied and divided by 2 and 5 respectively. If twice of the result is equal to 500 then find the number?a) 45 b) 23 c) 87 d) 47Answer :b) 23Solution :Let the number be XTwo more than X = X + 2Square of the number = (X + 2)2Multiplied and divided by 2 and 5 = 2(X + 2)2/ 5Twice the result is 500 = 2 x 2/5 x (X + 2)24/5 x (X + 2)2= 500(X + 2)2= 500 x 5/4(X + 2)2= 625X + 2 = 25X = 23Hence the required number is 23.Question 22-3 of a two digit number is equal to a number whose ten's place is three less than the ten's place of the two digit number and unit's place is one more than the unit's place of the two digit number.Then find the quotient when the square of the two digit number is divided by 261.a)29 b)12 c)45 d)13Answer :a)29Solution :Let the two digit number be 10x + y2/3 of the number = 2/3 (10x + y)Given that, 2/3 of (10x + y) = (x - 3)10 + (y + 1)(2/3)(10x + y) = (x - 3)10 + (y + 1)2(10x + y) = 3((x - 3)10 + (y + 1))20x + 2y = 3(10x - 30 + y + 1)20x + 2y = 30x - 90 + 3y + 3-10x = y - 87Or, 87 = 10x + yHence the two digit number be 87.Square of the number divided by 261 = 87 x 87 / 261 = 29Hence the answer is 29.Question 3In a two digit number, the digit in the units place is 1 more than 4 times the digit in ten's place. If the difference between the number formed by interchanging the digits of the number and the original number is 34 more than the original number, find the two digit number.a)15 b)29 c)33 d)none of theseAnswer :b)29.Solution :Let the two digit number be 10X + Y ...(1)(ten's digit is X and unit's digit is Y)Unit's digit is 1 more than 4 times the ten's place digitTherefore, i.e Y = 4X + 1 ...(2)Substituting (2) in (1), the two digit number becomes 10X + (4X + 1) ...(3)The number after interchanging the digits is 10(4X + 1) + X ...(4)From question data, we know that the difference between digits interchanged number and actual number is 34 more than the actual numberi.e 10(4X + 1) + X - (10X + (4X + 1)) = 34 + (10X + (4X + 1))40X + 10 + X - 10X - 4X - 1 = 34 + 10X + 4X + 130X - 3X + 9 = 34 + 10X + 4X + 127X + 9 = 34 + 10X + 4X + 127X + 9 = 34 + 10X + 4X + 127X - 14X = 35 - 913X = 26X = 2.Substituting X = 2 in (3), we get the required two digit number 10X + (4X + 1) = 20 + 8 + 1 = 29.Hence the answer is 29.

Below are three problems based on Profit and Loss calculations.Question 1Find the actual profit percentage of a milk vendor, if he sells milk using a faulty measure which reads 750 ml as 1000 ml.a) 12% b) 25% c) 33% d) 40%Answer :c) 33%Solution :The vendor sells 750 ml instead of 1000 ml (i.e., 1 litre)Profit made by the vendor is 1000 - 750 = 250 ml.250 ml = 1/3 litreProfit = 1/3Profit % = 1/3 x 100 = 33.33% = 33%(approx)Question 2A seller sold 2 items at Rs.23998.80 each, such that one is sold at a profit of 40% and another one at a loss of 40%. Then what is the net loss?a) Rs.9142.40 b) Rs.8142.40 c) Rs.7142.40 d) Rs.6142.40Answer :a) Rs.9142.40Solution :Selling price of 1st one = Rs.23998.80Profit = 40%Remember that, "if there is a profit of R% then Cost price = [SP /(100 + R)] x 100 and in case of loss of R%, CP = [SP / 100 - R] x 100Here, Cost price of 1st item = 23998.80 x 100 / 140 = Rs.17,142Selling price of 2nd item = Rs.23998.80Here loss = 40%Then cost price of 2nd item = Rs.23998.80/(100-40) x 100 = Rs.23998.80 x 100/60= Rs.39,998Total cost price = Rs.17,142 + Rs.39,998 = Rs.57140Total Selling price = Rs.47997.60Net loss = Rs.57140 - Rs.47997.60 = Rs.9142.40Question 3Two salesmen sales their item for Rs.8000 each. One salesmen calculates his profit % on his Cost price and another calculates his Profit % on selling price. If both claim to have a profit of 30% then what is the difference in their profit amount?a) Rs.564 b) Rs.554 c) Rs.654 d) Rs.614Answer :b) Rs.554Solution:In 1st case,Selling price = Rs.8000Profit = 30% of Cost priceCP = [SP /(100 + R)] x 100 = 8000 / 1.3 = Rs.6153.84Profit = SP - CP = Rs.8000 - Rs.6153.84 = Rs.1846.16In case 2,Selling price = Rs.8000Profit = 30% of selling priceProfit = 30/100 x 8000 = Rs.2400Therefore required difference = Rs.2400 - Rs.1846.16 = Rs.553.84Hence the answer is Rs.554(approx)

Below are three puzzle questions based on time calculations to cross a bridge.Question 1Four friends Mr.A, Mr.B, Mr.C and Mr.D are walking in a park. On the way, they have to cross a narrow bridge in which only two person can cross at a time. They walk at different speed and can cross at 12, 24, 21 and 5 minutes respectively. Find the minimum time taken by them to cross the narrow bridge.a)24 minutes b)33 minutes c)36 minutes d)26 minutesAnswer :b)33 minutes.Solution:Given that only two person can cross the bridge at a time.We have to find the minimum time taken by them.Let C and B be allowed to cross the bridge first.Time taken by C and B to cross the bridge is 21 and 24 minutes.At 21st minute C could reach the other side of the bridge and B has 3 minutes more to cross.Next A is allowed to cross the bridge and the time taken by him to cross is 12 minutes.In the fist 3 minutes of A, B would have reach the other side of the bridge and A has 9 minutes more to cross the bridge.Meanwhile D is allowed to cross and time taken by him is 5 minutes.In the 5th minute D would have reach the end but A has 4 minutes more to reach the end.Total time taken by them to cross = 21 + 3 + 5 + 4 = 33 minutes.Hence the minimum time taken by them to cross the bridge is 33 minutes.Question 2Seven girls namely P, Q, R, S, T, U and V can cross a rope bridge in 22, 34, 14, 4, 13, 29 and 25 minutes respectively. If only three girls can cross the bridge at a time then the minimum time taken by them to cross the bridge is:a)34 minutes b)29 minutes c)47 minutes d)40 minutesAnswer :c)47 minutes.Solution :Given that, P, Q, R, S, T, U and V can cross a rope bridge in 22, 34, 14, 4, 13, 29 and 25 minutes respectively.i.e., S, T, R, P, V, U and Q takes 4, 13, 14, 22, 25, 29 and 34 minutes respectively.Only three girls can cross the bridge at a time.Then allow the first 3 slowest girls to cross the bridge.i.e., allow V, U and Q and they take 25, 29, and 34 minutes respectively.Therefore at 25th minute V will reach the other side of the bridge ....(1)And U, Q has 4 and 9 minutes more to reach the other side.Now, allow the next slowest girl P and she takes 22 minutes.At the 4th minute (after 25 minutes)of P, U would reach the other side of the bridge....(2)Q and P has 5 and 19 minutes more to reach the other side.Now the next slowest girl R is allowed and she takes 14 minutes.In the 5th minute (after 29 minutes) of R, Q would reach the other side of the bridge....(3)R and P respectively has 9 and 13 minutes more to reach the other side.Now, the next slowest girl T is allowed and she takes 13 minutes.In the 9th minute (after 34 minutes)of T, R have reached the other side of the bridge....(4)T and P separately has 4 minutes more to reach the other side.Finally, the last girl S is allowed to cross and she takes 4 minutes to cross the bridge.Now, T, P and S takes 4 minutes to reach the other side.At the 4th minute (after 43 minutes) all seven girls are in the other side of the bridge.Therefore the required time = 25 + 4 + 5 + 9 + 4 = 47 minutes.Hence the minimum time is 47 minutes.Question 3Eight members P, Q, R, S, T, U, V and W can cross a bridge in 3, 13, 21, 17, 5, 9, 1 and 25 minutes respectively. If only two can cross the bridge at a time, then find which of the following group crosses in minimum time?a)(P,Q,V,W) b)(R,S,T,U) c)(Q,S,U,W) d)(P,R,T,V)Answer :d)(P,R,T,V)Solution :First let us find the minimum time taken by group a. (P,Q,V,W)Time taken by P, Q, V and W is 3, 13, 1, 25 minutes respectively.Here, we would allow the two slowest persons Q and W to cross a bridge at a time.Then at 13th minute Q would reach the other side of the bridge.And W has 12 minutes more to cross the bridge.Now allow the third person P to cross the bridge and he takes 3 minutes.In the 3rd minute P would reach the other side of the bridge.And W has 9 more minutes to reach the other side.Now, allow the 4th person V to cross the bridge and he takes 1 minute.Then in the next minute (after the above 3 minutes) V will reach the other side.and W has 8 minutes more and in the next 8th minute W would also reach the other side.Therefore, the minimum time taken by the group a is = 13 + 3 + 1 + 8 = 25 minutes.Similarly,The minimum time taken by the group b (R,S,T,U) is = 17 + 4 + 5 = 26 minutesAnd the minimum time taken by the group c (Q,S,U,W) is = 17 + 8 + 9 = 34 minutesAnd the minimum time taken by the group d (P,R,T,V) is = 5 + 3 + 13 = 21 minutesTherefore, group d takes the minimum time to cross the bridge.Hence the required answer is option d.

Below are three problems based on speed and time with some logical calculations.Question 1The straight distance between X and Y is 1000 meters. A taxi starts travelling from X at morning 8 o'clock. There is a taxi for every 30 minutes interval. Each taxi travels at a speed of 25 km/hr. A man travels from Y at a speed of 20 km/hr. If the man starts his journey at 2 pm, find the number of taxi he encounters in his journey.a)17 b)5 c)12 d)8Answer :a)17Solution :Distance between X and Y = 1000 meters = 100kmThe taxi starts from X at 8 am and travels at a speed of 25km/hr.Time required for the taxi to reach Y = 4 hours (100/25 = 4)The man travels from Y at a speed of 20km/hr.Time required to reach X = 5 hours (100/20) and his arrival to X is at 7pm.Between 8 am to 2 pm, 12 taxi's are started from X. (2 x 6 = 12, since every 30 mins a taxi is started (per hour 2 taxi))Now the taxi started at 8 am reaches Y at 12 noon, since the time taken to reach Y is 4hrs.Similarly, taxi starting at 8.30 am would reach Y at 12.30 pm, starting at 9am would reach at 1pm, and so on.At 2 pm, the taxi started at 10 am would have reached Y.By 2 pm, 5 taxi's had arrived to Y.So, remaining(12 - 5) 7 taxi's are on their way to Y and will meet the man along the way.Now, the man starts at 2 pm takes 5 hrs to reach X.Within 5 hours (5 x 2) 10 more taxi's are started, which will also meet the man on his way.so the man would meet (10 + 7) 17 taxis on his way.(i.e., Excluding the taxi's that start at 10 and 7 the answer will be 17.10.30,11,11.30,12,12.30,1,1.30,2,2.30,3,3.30,4,4.30,5,5.30,6,6.30...totally 17.)Question 2A taxi starts travel from P at morning 8 am to Q which is at a distance of 50 km from P. There is a taxi for every 30 minutes from P and each taxi travels at a speed of 25 km/hr. Two men A and B starts at 2 pm from Q at a speed of 12.5 km/hr and 10 km/hr respectively. During their journey,find the number of taxi's B encounter more than A.a)1 b)4 c)2 d)3Answer :c)2Solution :Total distance of journey = 50kmA travels from Q at a speed of 12.5km/hr.Time required by A to reach P = 50/12.5 = 4 hoursTherefore the time of arrival to P is at 6 pmSimilarly, B travels from Q at a speed of 10km/hr.Time required by B to reach P = 50/10 = 5 hoursTherefore the time of arrival to P is at 7pmGiven that there is a taxi for every 30 minutes.And, B need 1 hour more than A. Within this 1 hour period B encounters 2 more taxi's than A.Hence the answer is 2Question 3The distance between two stations A and B is 75 km. A taxi starts from A at morning 7 o'clock. There is a taxi for every 45 minutes interval and each taxi travels at a speed of 25 km/hr. A man travels from B and reaches station A such that 15km at a speed of 15 km/hr and remaining distance by 30 km/hr respectively. If the man starts at 1 pm then find the number of taxi's he encounters during his journey.a)7 b)8 c)6 d)9Answer :a)7Solution:Distance between A and B = 75 kmA taxi starts from A at 7 am and each taxi travels at a speed of 25km/hr.Time required for the taxi to reach Q = 3 hours (75/25 = 3)A man travels from B at a speed of 15 km/hr for 15 km and 30 km/hr for remaining 50 kmTime required by him to reach A = (15/15) hours + (60/30) = 1 + 2 = 3 hoursSo the actual time of arrival is at 4 pm.Since there is a taxi for every 45 minutes intervalBetween 7 am to 1 pm, 8 taxi's are started from A.The taxi started at 7 am reaches B at 10 am, since time taken by taxi to reach B is 3 hrs.Similarly, taxi starting at 7.45am would reach B at 10.45 am, starting at 8.30 am would reach at 11.30 am, and so on.By the time 1 pm, the taxi starting at 10 am would gave reached B.In this process a total of 5 taxi's had already arrived to B by 1 pm.So remaining(8 - 5) 3 taxi's are on their way to B and will meet the man along the way.Now the man starting at 1 pm takes 3 hrs to reach A.Within this 3 hours, 4 more taxi's are released which will also meet the man on his way.so the man would meet (3 + 4) 7 taxi's on his way.(i.e., Excluding the taxi's that start at 10 am and 4 pm the answer will be 710.45,11.30,12.15,1,1.45,2.30 and 3.15totally 7.)

Below are three problems on time and distance with speed and time calculations.Question 1Two trains are travelling at 25 kmph, starting at the same time from city 1 and city 2 respectively which are 75 km apart. There is a bee, which can fly at 50 kmph starts from city1 at the same time the train starts and hits the second train.After hitting the train, takes rest till the trains meet each other. Find the time taken by the fly to rest ?a)15 minutes b)30 minutes c)45 minutes d)65 minutesAnswer :b)30 minutesSolution :Distance between the two cities = 75 kmSpeed of the two trains = 25 km per hour.Speed of the bee = 50 km per hour.The bee rests until the two trains meet each other.Time taken by the bee to rest = Time taken to hit the 2nd train - Time taken by the 1st train to meet the second trainWe know that,"Suppose two trains or two objects bodies are moving in opposite directions at u m/s and v m/s, then their relative speed is = (u + v) m/s.and the time taken by them to meet is = the distance travelled by two objects / their relative speed".Relative speed of the bee and 2nd train = (50 + 25) = 75 km/hrAnd the total distance travelled by them = 75 km (distance between two cities)Then the required time = 75/(50 + 25) = 75/75 hr = 1 hour.Relative speed of two trains = (25 + 25) = 50 km/hrAnd the total distance travelled by them = 75 km (distance between two cities)Then the required time = 75 /(25 + 25) = 75/50 hr = 3/2 hr = 1 and half hour.Therefore the required difference = (3/2 - 1)hr = 1/2 hr = 30 minutes.Hence the bee rests for 30 minutes.Question 2Two buses are travelling at 36 kmph and are 120 km apart. There is a fly in one bus which starts to fly at 40 kmph when the bus is started. It flies between two buses until the buses meet each other. Find the distance travelled by the fly.a)66.6 km b)76.8 km c)56.5 km d)87.5 kmAnswer :a)66.6 kmSolution :From the given question, we can know that the required distance (total distance travelled by the fly at 40kmph) is the distance travelled by the fly in the time period before the two buses meet.Distance between two buses = 120 kmSpeed of buses = 36 km/hrTravelling towards each other, so the relative speed = (36 + 36) = 72 km/hrTime to meet = 120 / 72 hours = 10 / 6 hours = 5 / 3 hours.Given that the speed of the fly = 40 kmphAnd the time period it flies = 5/3 hours.Distance travelled by the fly = speed x time = 40 x 5/3 km = 200/3 km = 66.66 kmHence the required distance is 66.6 km.Question 3Two buses are travelling at 36kmph and are 120 km apart. There is a fly in one bus which starts to fly at 72kmph when the bus is started.It hits the second bus and starts flying between two buses until the buses meet each other.Find the distance travelled by the fly after hitting the second bus is.a)120 km b)60 km c)40 km d)72 kmAnswer :c)40 kmSolution :Here, the distance travelled by the fly after hitting the second bus = the distance travelled by n hours (where n is the difference between the time taken by the fly to hit the 2nd bus and the time taken by the two buses to meet).We have to find the value of nDistance between two buses = 120 kmSpeed of buses = 36 km/hrTravelling towards each other, so the relative speed = (36 + 36) = 72 km/hrTime to meet = 120/72 hours = 10/6 hours = 5/3 hours.Given that the speed of the fly = 72 kmphSpeed of the 2nd bus = 36km/hr and the distance = 120 kmTime taken to hit the 2nd bus = 120/(72 + 36) = 120/108 = 30/27 = 10/9 hoursThe required time difference(n) = 5/3 - 10/9 = 5/9 hour.Now, the required distance = Speed x time = 72 x 5/9 = 40 kmHence the required distance is 40 km

Question 1How can you place 12 coins in 6 lines such that each line should contains 4 coins?Solution :Put them in a star shape which contains 2 different triangles and each side with 4 coins.As shown below:

Question 2How can you make a shape of an isosceles triangle with 9 coins such that if you move 2 coins from the shape then it will become 3x3 grid?Solution :Required shape of isosceles triangle with 9 coins is,

Now move the coin 9 and 5 to make 3x3 grid as follows:

Question 3How can you make a shape of 5 triangles with 5 lines by using 10 coins such that1) Each line should contain 4 coins2) Each triangle should contain 3 coins and the sides of the triangles should lie in three lines which are described in (1)?Solution :Required figure is:

Here, required 5 lines of 4 coins are1) 1-3-6-92) 9-8-7-53) 5-4-3-24) 2-6-8-105)10-7-4-1.And, required 5 triangles are 1-3-4, 5-4-7, 7-10-8, 9-8-6 and 2-6-3 which also satisfies the given condition.For example,Consider the triangle 1-3-4, the sides 1-3, 3-4, 1-4 lies in 3 different lines.Similarly, in the triangle 5-4-7, the sides 5-7, 5-4, 7-4 are lies in 3 different lines.Question 4If you place 9 coins in 4 squares of equal size such that each side of every square should contain 2 coins then the number of coins which are shared by all 4 squares is:a) 5 b) 4 c) 3 d) 6Answer :a) 5Solution:Required figure is:

Here, squares with side of 2 coins are 1-2-5-4, 2-3-6-5, 5-6-9-8, 8-7-4-5 and 2-4-8-6.Required 4 squares of equal size are 1-2-5-4, 2-3-6-5, 5-6-9-8 and 8-7-4-5.The coins that are shared is 2, 5, 8, 6 and 4.Hence the required answer is 5.

Below are three numeric puzzles dealing with some basic arithmetic calculations.Question 1Find X's age which equals the number of grand children of a man who has 4 sons and 4 daughters. Each daughter of the man's wife have 3 sons and 4 daughters and each son of the man's wife have 4 sons and 3 daughters.a) 40 b) 56 c) 64 d) none of theseAnswer :b) 56Solution :We have to find the number of grand children of the man.Given that, he had 4 sons and 4 daughters.Each son has 4 sons and 3 daughters and each daughter has 3 sons and 4 daughters.Therefore total number of grandsons = 4x4 + 4x3 = 16 + 12 = 28And total number of grand daughters = 4x3 + 4x4 = 28Total number of grandchildren is 28+28 = 56.Hence the required age is 56.Question 2A man have many daughters, each daughter have as many sons as her sisters. The product of the number of daughters and grandsons of the man lies between 40 and 50. Find the number of daughters of the man.a) 6 b) 4 c) 3 d) 8Answer : b) 4Solution:Let N be the number of daughters of the man.Then each daughter has N-1 sisters.Given that, each daughter has as many sons as her sisters.That is, each of them has N-1 sons.Number of grandsons of the man = N(N-1)And the required product = N(N-1) x N = N x N(N-1) which lies between 40 and 50.If N = 1 then N x N(N-1) = 0.f N = 2 then N x N(N-1) = 2 x 2(1) = 4If N = 3 then N x N(N-1) = 3 x 3 x 2 = 18If N = 4 then N x N(N-1) = 4 x 4 x 3 = 48If N = 5 then N x N(N-1) = 5 x 5 x 4 = 100Therefore the possible value of N is 4.Hence the man has 4 daughters.Question 3Two men A and B have equal number of daughters and A have 1 more son than B. Each son and daughter of A have 2 sons and 2 daughters and each son and daughter of B have 3 sons and 3 daughters. If A and B have equal number of grandchildren then find the number of sons of B.a) 3 b) 4 c) 5 d) none of these.Answer :d) none of theseSolution :Let D be the number of daughters of A and B.Let S be the number of sons of A.Then S-1 be the number of sons of B.Now, number of grandsons of A = 2S + 2D.Number of granddaughters of A = 2S + 2DNumber of grand children of A = 2S + 2D + 2S + 2D = 4S + 4D.Number of grandsons of B = 3(S-1) + 3D = 3S + 3D - 3Number of granddaughters of B = 3(S-1) + 3D = 3S + 3D - 3Number of grand children of A = 3S + 3D - 3 + 3S + 3D - 3 = 6S + 6D - 6And, 4S + 4D = 6S + 6D - 62S + 2D = 6S + D = 3Then the possibilities of (S,D) are (1,2), (2,1), (0,3), (3,0)We have to find the value of S-1.S-1 cannot be a negative number, so (0,3) is not possible.Therefore, possible S-1 values are 0,1,2.Hence the answer is option d.

Below are three problems based on the concept of number of brothers and sisters, you have to find the number of siblings according to given statements.Question 1If Mr.X, Mr.Y and Mr.Z are three siblings and following three statements are true1) Mr.X has one older brother and three younger sisters.2) Mr.Y has two older brothers and two younger sisters3) Mr.Z has three older brothers and one younger sister.Then the least possible number of siblings in their family is:a) 5 b) 6 d) 7 d) 8Answer :b)6Solution :First, we need to think about X, Y and Z as who is the oldest among the three and who is the youngest among the three.X has one older brother while Y has two. Therefore, X is older than Y. Similarly Y is older than Z. Therefore, among X,Y and ZX is the oldest, Y is next and Z is the youngest.X has an older brother and Z has a younger sister.Try making a chart laying out all the three conditions given in question, and mark the gender of each known sibling.Then, it is clear that X is male and Y is female.MMFMFF1st2nd3rd4th5th6thXYZBased on the above chart, there should be at least 6 siblings.Question 2If A, B and C are triblet. A,C are males and B is female,1) A & C say " We have 2 more brothers than the number of sisters"2) B says " I have 2 more brothers than twice the sisters "If the above two statements are true then the number of siblings in their family is:a) 6 b) 12 c) 8 d) 9Answer :b) 12Solution :Let g be the number of girls in the family.Let b be the number of boys in the family.Consider statement 1, A and C would have b-2 brothers (excluding them). This b-2 is equal to two more than the number of sisters they have.Therefore, b-2 = g+2Or, b-g = 4 -----(1)Now, from statement 2, B would have g-1 sisters (excluding herself) and b brothers. She has 2 more brothers than twice the number of sisters.Therefore, Twice the number of sisters = 2(g-1) which is equal to 2 + number of brothers.Then we have, 2(g-1) + 2 = b2g - 2 + 2 = bb-2g = 0 -------(2)Solving (1) and (2), we have g = 4 and b = 8.Hence the number of siblings = 4+8 = 12.Question 3Aravind and Bavya are friends.1)Aravind's sister has thrice as many sisters of Bavya as brothers.2)Aravind has 1 more brother than than that of Bavya3)Twice Bavya's brother's sisters equals the number of boys in Aravind's family.If the above all 3 statements are true then the number of brothers what Aravind has more than Bavya is:a) 1 b) 2 c) 3 d) none of theseAnswer :a) 1Solution :Let the number of boys in Aravind's family be aLet the number of girls in Aravind's family be bLet the number of boys in Bavya's family be cLet the number of girls in Bavya's family be d.consider statement 1, "Aravind's sister has thrice as many sisters of Bavya as brothers" :Number of Bavya's sisters = d-1 (excluding her)Thrice the Bavya's sisters = 3(d-1)Aravind's sister has 'a' number of brothers.Therefore, 3(d-1) = ad-1 = a/3....(1)Consider statement2,"Aravind has 1 more brother than that of Bavya" :Number of Aravind's brothers = a-1Number of Bavya's brother = cSince, Aravind has 1 more brother than Bavya then a-1 = c+1a-c = 2...(2)Consider statement3,"Twice Bavya's brother's sisters equals the number of boys in Aravind's family"The number of sisters of Bavya's brother = dNumber of boys in Aravind's family = aTherefore 2d = aOr d = a/2Then, d-1 = a/2 - 1 ...(3)Now, let us solve equations we obtained in previous steps :From (1) and (3),a/3 = a/2 - 1a = 6Put a value in (2), we have c = 4Put a value in (3), we have d = 3Therefore Aravind has a-1 = 6-1 = 5 brothersAnd Bavya has c = 4 brothers.Hence, Aravind has 1 more brother than Bavya.