Aptitude Problems Solved Examples 1) Pranav went to the bank at the speed of 60 kmph while returning for his home he covered the half of the distance at the speed of 10 kmph, but suddenly he realized that he was getting late so he increased the speed and reached the home by covering rest half of the distance at the speed of 30 kmph.The average speed of the Pranav in the whole length of journey is? Answer: 24 kmph Explanation: Distance between home and Bank – x km Total distance = x + x = 2x Total time taken = x/60 + (x/2)/10 + (x/2)/30 = x/12 Average speed = 2x/(x/12) = 24 kmph 2) The average expenditure of Sharma for the January to June is Rs. 4200 and he spent Rs. 1200 in January and Rs.1500 in July. The average expenditure for the months of February to July is: Answer: 4250 Explanation: Total Expenditure(Jan – June) = 4200 * 6 = 25200 Total Expenditure(Feb – June) = 25200 – 1200 = 24000 Total Expenditure(Feb – July) = 24000 + 1500 = 25500/6 = 4250 3) The average weight of all the 11 players of CSK is 50 kg. If the average of first six lightest weight players of CSK is 49 kg and that of the six heaviest players of CSK is 52 kg. The average weight of the player which lies in the sixth position in the list of players when all the 11 players of CSK are arranged in the order of increasing or decreasing weights. Answer: 56 kg Explanation: Average of First six players = 49 * 6 = 294 Average of Last six players = 52 * 6 = 312 Average of all players = 50 * 11 = 550 Average weight of sixth player = 294 + 312 – 550 = 56 4) The average presence of students of a class in a College on Monday, Tuesday and Wednesday is 32 and on the Wednesday, Thursday, Friday and Saturday is 30. if the average number of students on all the six days is 26 then the number of students who attended the class on Wednesday is? Answer: 60 Explanation: 32 * 3 + 30 * 4 – 26 * 6 = 96 + 120 – 156 = 60
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5) Suresh started his journey from P to Q by his bike at the speed of 40 kmph and then, the same distance he travelled on his foot at the speed of 10 kmph from Q to R. Then he returned from R to P via Q at the speed of 24 kmph. The average speed of the whole trip is: Answer:19.2 kmph Explanation: Average speed from P to R = 2 * 40 * 10 / (40 + 10) = 16 kmph Average Speed = 2 * 16 * 24 / (16 + 24) = 19.2 kmph 6) Ramesh walked 6 km to reach the station from his house, then he boarded a train whose average speed was 60 kmph and thus he reached his destination. In this way he took a total time of 3 hours. If the average speed of the entire journey was 32 kmph then the average speed of walking is: Answer:4 kmph Explanation: Total Distance = 32 * 3 = 6 + 60 * x x = 1.5 hour ; Walking Speed = 6/1.5 = 4 kmph 7) Bala travels first one-third of the total distance at the speed of 10 kmph and the next one-third distance at the speed of 20 kmph and the last one – third distance at the speed of 60 kmph. What is the average speed of Bala? Answer:18 kmph Explanation: = 3 * 10 * 20 * 60 / (200 + 1200 + 600) = 18 kmph 8) The average income of Arun, Bala and Chitra is Rs. 12,000 per month and average income of Bala, Chitra and David is Rs. 15,000 per month. If the average salary of David be twice that of Arun, then the average salary of Bala and Chitra is in Rs? Answer:13500 Explanation: Arun + Bala + Chitra = 12000*3 Bala + Chitra + David = 15000*3 David – Arun = 3000*3 = 9000 David = 2Arun David = 18000 and Arun = 9000 Average salary of Bala and Chitra, = (45000-18000)/2 = 13,500 9) The distance of the School and house of Suresh is 80km. One day he was late by 1 hour than the normal time to leave for the college, so he increased his speed by 4km/h and thus he reached to college at the normal time. What is the changed speed of Suresh? Answer: 20 kmph Explanation: 80/x – 80/(x+4) = 1 x(x+20) – 16(x+20) = 0 x = 16kmph Increased speed = 20 kmph 10) Anita goes to College at 20 km/h and reaches college 4 minutes late. Next time she goes at 25 km/h and reaches the college 2 minutes earlier than the scheduled time. What is the distance of her school? Answer: 10 km
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Explanation: 20*25/(25-20)*6/60 =10. 11) Two places R and S are 800 km apart from each other. Two persons start from R towards S at an interval of 2 hours. Whereas A leaves R for S before B. The speeds of A and B are 40 kmph and 60 kmph respectively. B overtakes A at M, which is on the way from R to S. What is the extra time taken by A to reach at S? Answer: 6hrs 40 minutes Explanation: Time taken by A to reach at Q = 800/40 = 20 hours Time taken by B to reach at Q = 800/60 = 13 hours and 20 min A takes 6hr 40 minutes extra time to reach at Q. 12) The driver of an ambulance sees a college bus 40 m ahead of him after 20 seconds, the college bus is 60 meter behind. If the speed of the ambulance is 30 km/h, what is the speed of the college bus? Answer: 6hrs 40 minutes Explanation: Relative Speed = (Total distance)/total time = (60+40) /20 = 5 m/s = (5*18)/5 = 18 kmph Relative Speed = (speed of ambulance – speed of College bus) Speed of College bus = speed of ambulance – relative speed. = 30-18 = 12 kmph. 13) Two rabbits start running towards each other, one from A to B and another from B to A. They cross each other after one hour and the first rabbit reaches B, 5/6 hour before the second rabbit reaches A. If the distance between A and B is 50 km. what is the speed of the slower rabbit? Answer: 20 kmph Explanation: Let second rabbit takes x hr with speed s2 First rabbit takes x-5/6 hr with speed s1 Total distance = 50km S1 = 50/(x-(5/6)) S2= 50/x As they cross each other in 1hr. Total speed = s1 + s2 Now, T = D / S 50/(s1+s2) = 1 x = 5/2, 1/3 Put x= 5/2 in s2 –> 20km/hr 14) Pranav walked at 5 kmph for certain part of the journey and then he took an auto for the remaining part of the journey travelling at 25 kmph. If he took 10 hours for the entire journey, what part of journey did he travelled by auto if the average speed of the entire journey be 17 kmph Answer: 150 km Explanation: Total distance = 17*10 =170 Let Journey travelled by auto in x hr 25 * x + (10-x ) 5 = 170 25 x + 50 – 5x = 170 x=6 Required Distance = 6 * 25 = 150 km
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15) Aravind started for the station half a km from his home walking at 1 km/h to catch the train in time. After 3 minutes he realised that he had forgotten a document at home and returned with increased, but constant speed to get it succeded in catching the train. Find his latter speed in kmph? Answer: 11/9 Explanation: Distance covered in 3 minutes = 3*(1000/60) = 50 Now he has to cover (500+50)m in (30-3) minutes Required speed = (550/1000)/(27/60) = 11/9 km/h 16) A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both hearts. Find the Probability of the lost card being a heart? Answer: 11/50 Explanation: Total cards = 52 Drawn cards(Heart) = 2 Present total cards = total cards-drawn cards =52-2=50 Remaining Card 13-2 = 11 Probability = 11/50 17) There are three boxes each containing 3 Pink and 5 Yellow balls and also there are 2 boxes each containing 4 Pink and 2 Yellow balls. A Yellow ball is selected at random. Find the probability that Yellow ball is from a box of the first group? Answer: 45/61 Explanation: Probability = (3/5 * 5/8)/([3/5 * 5/8] + [2/5 * 1/3]) = 45/61 18) A committee of five persons is to be chosen from a group of 10 people. The probability that a certain married couple will either serve together or not at all is? Answer: 51/126 Explanation: Five persons is to be chosen from a group of 10 people = 10C 5 = 252 Couple Serve together = 8C3 * 2C2 = 56 Couple does not serve = 8C5 = 56 Probability = 102/252 = 51/126 19) Out of 14 applicants for a job, there are 6 women and 8 men. It is desired to select 2 persons for the job. The probabilty that atleast one of selected persons will be a Woman is? Answer: 77/91 Explanation: Man only = 8C2 = 14 Probability of selecting no woman = 14/91 Probability of selecting atleast one woman = 1 – 14/91 = 77/91 20) Three Bananas and three oranges are kept in a box. If two fruits are chosen at random, Find the probability that one is Banana and another one is orange? Answer: 3/5 Explanation: Total probability = 6C2 = 15 Probability that one is Banana and another one is orange = 3C 1 * 3C1 = 9 probability = 9/15 = 3/5
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21) An elevator starts with 4 passenger and stops at 7 floors of an apartment. Find the probability that all 4 passengers travel to different floor ? Answer: 120/343 Explanation: Total number of outcomes = 7*7*7*7 Favourable outcome = 7*6*5*4 P = (7*6*5*4)/7*7*7*7 = 120/343 22) A five digit number is formed from the numbers 1,3,5,7,8 and 9 without repeating the numbers. Find the probability that the number is divisible by 5 ? Answer: 1/6 Explanation: Total number of outcome = 6*5*4*3*2 Favourable outcome = 5*4*3*2. So, P = 1/6 23) Three coins are tossed simultaneously. Find out the probability that there is at least 2 heads ? Answer: 1/2 Explanation: Total outcomes = {HHH,HTH,HHT,THH,TTT,TTH,THT,HTT} Favourable outcome = HHH, HTH, HHT,THH Probability = 4/8 = 1/2 24) (54679 + 5982 + 32614) – (312 * 69) = ? Answer: 71747 Explanation: 54679 + 5982 + 32614 = 93275 312 * 69 = 21528 =93275 – 21528 = 71747 25) If a pipe A can fill a tank 3 times faster than pipe B and takes 32 minutes less than pipe B to fill the tank. If both the pipes are opened simultaneously, then find the time taken to fill the tank? Answer: 12 minutes Explanation: 3x – x = 32 x = 16 1/16 + 1/48 = 4/48 Time taken to fill the tank = 48/4 = 12 minutes 26) Two pipes P and Q can fill a cistern in 10 hours and 20 hours respectively. If they are opened simultaneously. Sometimes later, tap Q was closed, then it takes total 8 hours to fill up the whole tank. After how many hours Q was closed? Answer: 4 hours Explanation: Pipe P Efficiency = 100/10 = 10% Pipe Q Efficiency = 100/20 = 5% Net Efficiency = 15% 15x + 10(8-x) = 100 x=4
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27) A Cistern has an inlet pipe and outlet pipe. The inlet pipe fills the cistern completely in 1 hour 20 minutes when the outlet pipe is plugged. The outlet pipe empties the tank completely in 4 hours when the inlet pipe is plugged. If both pipes are opened simultaneously at a time when the tank was one-third filled, when will the tank fill thereafter? Answer: 2 hours Explanation: Inlet pipe Efficiency = 100/(8/6) = 75% Outlet pipe Efficiency = 100/(4) = 25% Net Efficiency = 75 – 25 = 50%(1/3)filled 2/3 filled = 100% Required time = 100/50 = 2 hours 28) A Cistern has an inlet pipe and outlet pipe. The inlet pipe fills the cistern completely in 1 hour 20 minutes when the outlet pipe is plugged. The outlet pipe empties the tank completely in 6 hours when the inlet pipe is plugged. If there is a leakage also which is capable of draining out the water from the tank at half of the rate of the outlet pipe, then what is the time taken to fill the empty tank when both the pipes are opened? Answer: 2 hours Explanation: Inlet pipe Efficiency = 100/(8/6) = 75% Outlet pipe Efficiency = 100/(6) = 16.66% Efficiency of leakage = half of the rate of the outlet pipe = 8.33% Net Efficiency = 75 – (16.66 + 8.33) = 50% Required time = 100/50 = 2 hours 29) A pipe can fill a cistern in 8 hours. After half the tank is filled, three more similar taps are opened. What is the total time taken to fill the cistern completely? Answer: 5 hours Explanation: In One hour pipe can fill = 1/8 Time is taken to fill half of the tank = 1/2 * 8 = 4 hours Part filled by four pipes in one hour = (4*1/8) = 1/2 Required Remaining Part = 1/2 Total time = 4 + 1 = 5 30) Three pipes A, B, and C can fill the tank in 10 hours, 20 hours and 40 hours respectively. In the beginning all of them are opened simultaneously. After 2 hours, tap C is closed and A and B are kept running. After the 4th hour, tap B is also closed. The remaining work is done by tap A alone. What is the percentage of the work done by tap A alone? Answer: 35 % Explanation: Pipe A’s work in % = 100/10 = 10% Pipe B’s work in % = 100/20 = 5% Pipe C’s work in % = 100/40 = 2.5%
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All of them are opened for 2 hours + after 2 hours, tap C is closed + After the 4th hour, tap B is also closed = 100 => (10+5+2.5)*2 + (10+5)*2 + X = 100 => 35 + 30 + work by tap A alone = 100 => work by tap A alone = 100-65 = 35% 31) Two pipes A and B can fill a tank in 12 hours and 18 hours respectively. The pipes are opened simultaneously and it is found that due to leakage in the bottom of the tank it took 48 minutes excess time to fill the cistern. When the cistern is full, in what time will the leak empty it? Answer: 72 hours Explanation: Work done by the two pipes in 1 hour = (1/12)+(1/18) = (15/108). Time taken by these pipes to fill the tank = (108/15)hrs = 7 hours 12 min. Due to leakage, time taken = 7 hours 12 min + 48 min = 8 hours Work done by two pipes and leak in 1 hour = 1/8. Work done by the leak in 1 hour =(15/108)-(1/8) =(1/72). Leak will empty the full cistern in 72 hours. 32) A dam has four inlets – A, B, C and D. The dam can be filled in 12 minutes through the first three inlets and it can be filled in 15 minutes through the second, the third and fourth inlet also it can be filled through the first and the fourth inlet in 20 minutes. How much time required to fill up the dam by all the four inlets? Answer: 10 mins Explanation: (1/A + 1/B + 1/C) = 1/12 …(i) (1/B + 1/C + 1/D) = 1/15 …(ii) (1/A + 1/D) = 1/20 …(iii) From eqn (i) and (ii) (1/A – 1/D) = 1/60…(iv) From eqn (iii) and (iv) A=30 D=60. Let the time taken to full the tank = T T(1/A + 1/B +1/C +1/D)= 1 T(1/30 + 1/15) = 1 T = 10 mins 33) Three pipes P, Q and R connected to a Cistern. The first pipe (i.e) P can fill 1/2 part of the tank in one hour, second pipe, Q can fill 1/3 part of the cistern in one hour. R is connected to empty the cistern. After opening all the three pipes 7/12 part of the cistern. Then how much time required to empty the cistern completely? Answer: 4 hours Explanation: In 1 hour, P can fill = 1/2 Part Time taken to fill the Cistern by Pipe P = 2 hours In 1 hour, Q can fill = 1/3 Part Time taken to fill the Cistern by Pipe P = 3 hours [1/2 + 1/3 – 1/R] = 7/12 1/R = 1/4 Time required to empty the Cistern = 4 hours
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34) A Cistern can be filled by an inlet pipe at the rate of 4 litres per minute. A leak in the bottom of a cistern can empty the full tank in 8 hours. When the cistern is full, the inlet is opened and due to the leak, the cistern is empty in 40 hours. How many litres does the cistern hold? Answer: 2400 litre Explanation: Part emptied by the leak in 1 hour = 1/8 part filled by (leak & inlet open) in 1 hour = 1/40 Part filled by the inlet pipe in 1 hour = 1/8 – 1/40 = 1/10 Inlet pipe fills the tank in = 10 hours Inlet pipe fills water at the rate of 4 litres a minute. Capacity of Cistern = 10 * 60 * 4 = 2400 litre 35) In a tank there is a pipe which can be used for filling the tank as well as for emptying it. The capacity of the tank is 1200 m³. The emptying of the tank is 10 m³ per minute higher than its filling capacity and the pump needs 6 minutes lesser to empty the tank than it needs to fill it. What is the filling capacity of the pipe? Answer: 40 m³ / min Explanation: 1200/x – 1200/(x+10) = 6 200/x – 200/(x+10) = 6 x2 + 10x – 2000 = 0 x = 40 36) A retailer sold 12 notes at a profit of 20% and 8 notes at a profit of 10%. If he had sold all the 20 notes at a profit of 15%, then his profit would have been reduced by Rs.36. What is the cost price of each note? Answer: Rs.180 Explanation: Cost Price = x Total Profit = 12x * 20/100 + 8x * 10/100 = 32x/100 = 3.2x --- (i) Total Profit = 20x * 15/100 --- (ii) 3.2x – 3x = 36 0.2x = 36 x = 180 37) The profit Percentage on 3 bikes are 15%, 35% and 10% and the ratio of CP is 5:3:1. Also the ratio of the Bike sold of P, Q and R is 2:3:5. Then the overall approximate Profit Percentage is? Answer: 21% Explanation: 5x * 2y + 3x * 3y + x * 5y = 24xy Total Profit = (5x*2y*15/100)+(3x*3y*35/100)+(x*5y*10/100) = 515xy/100 = 5.15xy Overall Profit Percentage = 5.15xy * 100/24xy = 21.46% 38) A and B both are dealers of Honda Motorcycles. The price of an used Honda Motorcycle is Rs.28,000. A gives a discount of 10% on whole, while B gives a discount of 12% on the first Rs. 20,000 and 8% on the rest Rs. 8000. What is the difference between their selling prices? Answer: Rs.240 Explanation: Discount offer by A = 10 % of 28000 = 2800 Total Discount offer by B = 12% of 20,000 + 8% of 8000
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= 3040 Required difference = 3040 – 2800 = 240 39) The profit percentage of P and Q is same on selling the articles at Rs. 1800 each but A calculates his profit on the selling price while Q calculates it correctly on the cost price which is equal to 20%. What is the difference in their profits? Answer: 60 Explanation: Profit(Calculated on SP) = 20% of 1800 = 360 Profit(calculated on CP) x + x/5 = 1800 x = 1500 Profit = 300 Difference = 360 – 300 = 60 40) A person sold a pen at Rs. 96 in such a way that his percentage profit is same as the cost price of the watch. If he sells it at twice the percentage profit of its previous percentage profit then new selling price will be? Answer: Rs.132 Explanation: CP = x Profit Percentage = x% SP = x(100 + x)/100 x(100 + x)/100 = 96 x = 60 Profit Percentage = 60% New SP = 60 * 220 / 100 = 132 41) A trader mixes 25% of solution A to his Solution B and then he sells the whole mixture at the price of Solution B. If the cost price of Solution A be 50% of the cost price of Solution B, what is the net profit percentage? Answer: 100/9% Explanation: Quantity of Solution B = 100 litre Quantity of Solution A = 25 litre CP of 1 litre Solution B = Rs.10 CP of 1 litre Solution A = Rs.5 CP = 100 * 10 + 25 * 5 = 1125 SP = (100 + 25)*10 = 1250 Profit = 1250 – 1125 = 125% = 125 * 100 / 1125 = 100/9% 42) A scientist mixes 10% water in his solution but he is not content with it so he again mixes 10% more water in the previous mixture. What is the profit percentage of the scientist if he sells it at cost price: Answer: 21% Explanation: Let Initial Quantity of Solution = 100 litre After mixing 10% water, Quantity of the mixture = 110 * 110 / 100 = 121 litre
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CP of One litre of Solution = Rs.1 Total CP = Rs.100 Total SP = Rs.121 Profit = 121 – 100 = 21 Profit % = 21 * 100/100 = 21% 43) Arun sells an article at 20% profit to Bala, Bala sells it to Catherine at 10% profit. Catherine sells it to Dinesh at Rs. 16 profit. The difference between the cost price of Dinesh and cost price of Arun was Rs. 500. How much did Bala pay to Arun for the article? Answer: Rs.1815 Explanation: Cost Price of Arun = 100x Cost Price of Bala = 120x Cost Price of Catherine = 132x Cost Price of Dinesh = 132x + 16 132x + 16 – 100x = 500 32x + 16 = 500 32x = 484 x = 484/32 = 15.125 Cost Price of Bala =120x = 120 * 15.125 = 1815 44) Rahul purchased an article for Rs. 8400 and sold it for a loss of 5%. From that money he purchased another article and sold it for a gain of 5%. What is the overall gain or loss? Answer: Loss of Rs.21 Explanation: CP = 8400 SP = 8400 * 95/100 = 7980 CP = 7980 SP = 7980 * 105/100 = 8379 Difference = 8400 – 8379 = 21 45) A Shopkeeper bought 30 kg of rice at the rate of Rs. 40 per kg. He sold 40% of the total quantity at the rate of Rs. 50 per kg. At what price per kg should he sell the remaining quantity to make 25% overall profit? Answer: Rs.50 Explanation: Total CP of Rice = 30 * 40 = 1200 40% of Total Quantity = 40% of 30 = 12 SP = 12*50 = 600 SP = 1200 * 125/100 = 1500 SP of Remaining Quantity = 1500 – 600 = 900 Remaining Quantity = 18kg Rice per Kg = 900/18 = Rs. 50
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46) A shopkeeper sold a smartphone for Rs.15000. Had he offered discount of 10% on the Selling Price, he would have earned a profit of 8%. What is the Cost Price of that Smartphone? Answer: 12500 Explanation: S.P of Smart Phone = Rs.15000 Discount = 10% New SP = 15000 – 1500 = Rs. 13500 Profit = 8% CP = 13500 * 100/108 = 12500 47) Pinkey sold a machine to Shalini at a profit of 30%. Shalini sold this machine to Arun at a loss of 20%. If Pinkey paid Rs.5000 for this machine, then find the cost price of machine for Arun? Answer: 5200 Explanation: R1 = 30% R2 = 20% 5000 * 130/100 * 80/100 = Rs. 5200 48) A TV was purchased for Rs. 54000. Its price was marked up by 40%.It was sold at a discount of 20% on the marked price. What was the profit percent of the cost price? Answer: 12% Explanation: 40 – 20 + [40 * (-20)/100] = 20 – 8 = 12% 49) If the Cost Price of 25 articles is equal to the Selling Price of 20 articles, then what is the gain %? Answer: 25% Explanation: Gain % = (5 / 20) x 100 = 25% 50) Arun went to buy an Android mobile, the shopkeeper told him to pay 20% tax if he asked the bill. Arun manages to get the discount of 5% on the actual sale price of the mobile and he paid the shopkeeper Rs. 8550 without tax. Besides he manages to avoid to pay 20% tax on the already discounted price, what is the amount of discount? Answer: 2250 Explanation: CP = 100, SP (with tax) =120 New SP = 100 – 5 = 95 Discount = 120 – 95 = 25 Discount = 25/95 * 8550 = 2250
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