UPSC ONLINE ACADEMY

Raashid Shah

PIPES SET4

1. A tank is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the tank in the same time during which the tank is filled by the third pipe alone. The second pipe fills the tank 5 hours faster than the first pipe and 4 hours slower than the third pipe. Find the time required by the first pipe to fill the tank ?A) 10 hoursB) 15 hoursC) 17 hoursD) 18 hoursAns. BSuppose, first pipe alone takes x hours to fill the tank . Then, second and third pipes will take (x -5) and (x – 9) hours respectively to fill the tank. As per question, we get 1x+1x-5=1x-9=>x-5+xx(x-5)=1x-9=>(2x-5)(x-9)=x(x-5)=>x2-18x+45=01x+1x-5=1x-9=>x-5+xx(x-5)=1x-9=>(2x-5)(x-9)=x(x-5)=>x2-18x+45=01x+1x-5=1x-9=>x-5+xx(x-5)=1x-9=>(2x-5)(x-9)=x(x-5)=>x2-18x+45=0After solving this equation, we get (x-15)(x+3) = 0,As value can not be negative, so x = 15 2. One pipe can fill a tank three times as fast as another pipe. If together the two pipes can fill the tank in 36 minutes, then the slower pipe alone will be able to fill the tank inA) 144 minsB) 140 minsC) 136 minsD) 132 minwAns. ALet the slower pipe alone fill the tank in x minutesthen faster will fill in x/3 minutes. Part filled by slower pipe in 1 minute = 1/xPart filled by faster pipe in 1 minute = 3/x Part filled by both in 1 minute =1x+3x=136=>4x=136x=36*4=144mins1x+3x=136=>4x=136x=36*4=144mins1x+3x=136=>4x=136x=36*4=144mins 3. 12 buckets of water fill a tank when the capacity of each tank is 13.5 litres. How many buckets will be needed to fill the same tank, if the capacity of each bucket is A) 9 litres?B) 15 buketsC) 17 buketsD) 18 buketsAns. CCapacity of the tank = (12*13.5) litres= 162 litresCapacity of each bucket = 9 litres.So we can get answer by dividing total capacity of tank by total capacity of bucket.Number of buckets needed = (162/9) = 18 buckets 4. A tank can be filled by two pipes A and B in 60 minutes and 40 minutes respectively. How many minutes will it take to fill the tank from empty state if B is used for the first half time and then A and B fill it together for the other half.A) 15 minsB) 20 minsC) 25 minsD) 30 minsAns. DLet the total time be x mins.Part filled in first half means in x/2 = 1/40 Part filled in second half means in x/2 =160+140=124 Total = x2*140+x2*124=1=>x2(140+124)=1=>x2*115=1=>x=30mins160+140=124 Total = x2*140+x2*124=1=>x2(140+124)=1=>x2*115=1=>x=30mins160+140=124 Total = x2*140+x2*124=1=>x2(140+124)=1=>x2*115=1=>x=30mins 5. Two pipes A and B can fill a tank in 6 hours and 4 hours respectively. If they are opened on alternate hours and if pipe A is opened first, in how many hours, the tank shall be full ?A) 3 hoursB) 5 hoursC) 7 hoursD) 10 hoursAns. B(A+B)’s 2 hour’s work when opened = 16+14=512(A+B)’s 4 hour’s work=512*2=56Remaining work = 1-56=16Now, its A turn in 5 th hour16 work will be done by A in 1 hourTotal time = 4+1=5hours16+14=512(A+B)’s 4 hour’s work=512*2=56Remaining work = 1-56=16Now, its A turn in 5 th hour16 work will be done by A in 1 hourTotal time = 4+1=5hours16+14=512(A+B)’s 4 hour’s work=512*2=56Remaining work = 1-56=16Now, its A turn in 5 th hour16 work will be done by A in 1 hourTotal time = 4+1=5hours  

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PIPES SET3

1. A leak in the bottom of a tank can empty the full tank in 6 hours. An inlet pipe fills water at the rate of 4 litres a minute. When the tank is full, the inlet is opened and due to the leak the tank is empty in 8 hours. The capacity of the tank (in litres) isA) 5780 litresB) 5770 litresC) 5760 litresD) 5750 litresAns. C Work done by the inlet in 1 hour = 16-18=124Work done by inlet in 1 min=124*160=11440=>Volume of 1/1440 part = 4 liters16-18=124Work done by inlet in 1 min=124*160=11440=>Volume of 1/1440 part = 4 liters16-18=124Work done by inlet in 1 min=124*160=11440=>Volume of 1/1440 part = 4 liters Volume of whole = (1440 * 4) litres = 5760 litres. 2. An electric pump can fill a tank in 3 hours. Because of a leak in the tank, it took 3 hours 30 min to fill the tank. In what time the leak can drain out all the water of the tank and will make tank empty ?A) 10 hoursB) 13 hoursC) 17 hoursD) 21 hoursAns. DWork done for 1 hour without leak = 1/3Work done with leak = 312=72Work done with leak in 1 hr= 27Work done by leak in 1 hr =13-27=121312=72Work done with leak in 1 hr= 27Work done by leak in 1 hr =13-27=121312=72Work done with leak in 1 hr= 27Work done by leak in 1 hr =13-27=121So tank will be empty by the leak in 21 hours. 3. A tank can be filled by a tap in 20 minutes and by another tap in 60 minutes. Both the taps are kept open for 10 minutes and then the first tap is shut off. After this, the tank will be completely filled in what time ?A) 10 minsB) 15 minsC) 20 minsD) 25 minsAns. C 4. A cistern can be filled in 9 hours but due to a leak at its bottom it takes 10 hours. If the cistern is full, then the time that the leak will take to make it empty will be ?A) 20 hoursB) 19 hoursC) 90 hoursD) 80 hoursAns. CPart filled without leak in 1 hour = 1/9Part filled with leak in 1 hour = 1/10 Work done by leak in 1 hour=19-110=190=19-110=190=19-110=190 So total time to empty the cistern is 90 hours 5. Taps A and B can fill a bucket in 12 minutes and 15 minutes respectively. If both are opened and A is closed after 3 minutes, how much further time would it take for B to fill the bucket?A) 8 min 15 secB) 7 min 15 secC) 6 min 15 secD) 5 min 15 secAns. APart filled in 3 minutes = 3*(112+115)=3*960=920Remaining part =1-920=1120=>115:1120=1:X=>X=1120*151=>X=8.25mins3*(112+115)=3*960=920Remaining part =1-920=1120=>115:1120=1:X=>X=1120*151=>X=8.25mins3*(112+115)=3*960=920Remaining part =1-920=1120=>115:1120=1:X=>X=1120*151=>X=8.25mins So it will take further 8 mins 15 seconds to fill the bucket.    

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PIPES SET2

1. Two pipes A and B can fill a tank in 20 and 30 minutes respectively. If both the pipes are used together, then how long it will take to fill the tank ?A) 10 minsB) 12 minsC) 15 minsD) 20 minsAns. BPart filled by A in 1 min = 1/20Part filled by B in 1 min = 1/30 Part filled by (A+B) in 1 min = 1/20 + 1/30= 1/12 So both pipes can fill the tank in 12 mins. 2. A cistern can be filled by a tap in 4 hours while it can be emptied by another tap in 9 hours. If both the taps are opened simultaneously, then after how much time cistern will get filled ?A) 7 hoursB) 7.1 hoursC) 7.2 hoursD) 7.3 hoursAns. C Filled in 1 hour = 1/4Empties in 1 hour = 1/9 Net filled in 1 hour = 1/4 – 1/9= 5/36 So cistern will be filled in 36/5 hours i.e. 7.2 hours Comment on this question 3. A tap can fill a tank in 6 hours. After half the tank is filled then 3 more similar taps are opened. What will be total time taken to fill the tank completely.A) 2 hours 30 minsB) 2 hours 45 minsC) 3 hours 30 minsD) 3 hours 45 minsAns. DHalf tank will be filled in 3 hoursLets calculate remaining half, Part filled by the four taps in 1 hour = 4*(1/6) = 2/3 Remaining part after 1/2 filled = 1-1/2 = 1/2 23:12::1:X=>X=(12*1*32)=>X=34hrs=45 mins23:12::1:X=>X=(12*1*32)=>X=34hrs=45 mins23:12::1:X=>X=(12*1*32)=>X=34hrs=45 mins Total time = 3 hours + 45 mins = 3 hours 45 mins4. A water tank is two-fifth full. Pipe A can fill a tank in 10 minutes and pipe B can empty in 6 minutes. If both the pipes are open, how long will it take to empty or fill the tank completely ?A) 6 min to emptyB) 7 min to fullC) 6 min to fullD) 7 min to emptyAns. AThere are two important points to learn in this type of question,First, if both will open then tank will be empty first. Second most important thing is,If we are calculating filling of tank then we will subtract as (filling-empting) If we are calculating empting of thank then we will subtract as (empting-filling) So lets come on the question now, Part to emptied 2/5 Part emptied in 1 minute = 16-110=115=>115:25::1:x=>25*15=6mins16-110=115=>115:25::1:x=>25*15=6mins16-110=115=>115:25::1:x=>25*15=6mins 5. Pipe A can fill a tank in 5 hours, pipe B in 10 hours and pipe C in 30 hours. If all the pipes are open, in how many hours will the tank be filled ?A) 2.5 hoursB) 2 hoursC) 3.5 hoursD) 3 hoursAns. DPart filled by A in 1 hour = 1/5Part filled by B in 1 hour = 1/10Part filled by C in 1 hour = 1/30 Part filled by (A+B+C) in 1 hour = 15+110+130=1315+110+130=1315+110+130=13 So all pipes will fill the tank in 3 hours.

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PIPES SET1

Q.1 There are three inlets P,Q & R attached with a tank. If P & Q are opened then tank is filled in 120hr & if P & Q are opened then tank is filled in 90 hr. Find the time in which P,Q & R can fill the tank, if these all are opened ?A) 60 hB) 61 hC) 62 hD) 63 h Ans. AShortcut :2xyz / xy + yz + zx2*72*120*90 / 25920 = 60 h Q.2 Because of a hole in a bottom of a tank , a tank becomes empty in 6 min. after opening a tap which fills the tank at the rate of 8 lit/min the tank is emptied in 10 min. Find the capacity of the tank ? A) 117 litre B) 118 litre C) 119 litre D) 120 litre Ans. D Shortcut :abc / c-ai.e. 6*8*10 / 10-6 = 120 litre Q.3 Two pipes can separately fill a tank in 10 hr & 15 hr. Both the pipes are opened to fill the cistern but when the tank is 1/3rd full a leak develops in the tank through which 1/3rd of water supplied but both the pipes leak out . What is the total time taken to fill the tank ? A) 6 hr. B) 10 hr. C) 12 hr. D) 15 hr. Ans. A Time taken by the two pipes to fill the tank:10*15/10+15=6 hr. Q.4 A pipe can fill a bath in 20 minutes and another can fill it in 30 minutes. A person opens both the pipes simultaneously. When the bath should have been full, he finds that the waste pipe was open. He then closes the waste pipe and in 3 more minutes the bath is full. In how much time would the waste pipe empty it?A) 50 minB) 48 minC) 44 minD) 38 min Ans. B First pipe can fill the cistern = 20 min Second pipe fill the cistern = 30 min. But together both pipes can fill it = = 12 minNow according to question 12 = 1 {Where x = Time taken by leak to empty the cistern}1 – = 1x = 48 min Q.5 A supply of water lasts for 15 days if 12 gallons leak off everyday, but only for 100 days if 15 gallons leak off daily . What is the total quantity of water in the supply ?A) 900B) 1125C) 3350D) 1250Ans. Aconsumption be x and supply y:150x + 12*150 = y ….(i)100x + 15*100 = y ….(ii)multiply i by 2 and ii by 3 and equate, 4500 – 3600 = 3y – 2y y = 900    

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BOATS SET4

1. Man can row 30km upstream & 44km downstream in 10 hrs. Also he run 40km upstream & 55km downstream in 13 hrs. Find the rate of the current & the speed of the man in still water ?A) 3 & 8 km/hrB) 2 & 5 km/hrC) 1 & 3 km/hrD) None of the aboveAns. AUpstream Downstream time30 44 1040 55 13Upstream = 30*55 – 40*44 / 44*13 – 55*10 = 5Downstream = 30*55 – 40*44 / 30*13 – 40*10 = 11Therefore, Rate of current = 11-5/2 = 3 km/hrSpeed of the man = 11+5/2 = 8 km/hr 2. A motor boat takes 2 hrs to travel a distance of a 9km down the current & it takes 6 hrs. to travel the same distance against the current. The speed of the boat(in km/hr) in still water & that of the current are A) 3,2B) 3,1C) 3.5,2.5D) 3,1.5Ans. DLet speed of boat in still water be x km/hr & speed of current be y km/hr9/x+y = 2& 9/ x-y =6X+y=9/2x-y = 9/6 = 3/2therefore, x=3 & y=1/2 km/hr 3. A boat can travel with a speed of 16 km/hr in still water. If the rate of stream is 5 km/hr, then find the time taken by the boat to cover distance of 84 km downstream.A) 4 hoursB) 5 hoursC) 6 hoursD) 7 hoursAns. ASpeed downstream = (16 + 5) = 21 kmph Time = distance/speed = 84/21 = 4 hours 4. A man can row at 5 kmph in still water. If the velocity of the current is 1 kmph and it takes him 1 hour to row to a place and come back. how far is that place.A) 0.4 kmB) 1.4 kmC) 2.4 kmD) 3.4 kmAns. CLet the distance is x kmRate downstream = 5 + 1 = 6 kmphRate upstream = 5 – 1 = 4 kmphthen x/6 + x/4 = 1 [because distance/speed = time]=> 2x + 3x = 12 => x = 12/5 = 2.4 km 5. The speed of a boat in still water is 15 km/hr and the rate of current is 3 km/hr. The distance travelled downstream in 12 minutes isA) 1.6 kmB) 2 kmC) 3.6 kmD) 4 kmAns. CSpeed downstreams =(15 + 3)kmph= 18 kmph.Distance travelled = (18 x 12/60)km= 3.6km       iibm

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BOATS SET3

1. A man can row 913 kmph in still water and finds that it takes him thrice as much time to row up than as to row, down the same distance in the river. The speed of the current is.A) 323kmphB) 423kmphC) 523kmphD) 623kmphAns. BSpeed of current = 1/2(a-b) Let the speed upstream = x kmphThen speed downstream is = 3x kmph speed in still water = 12(a+b)=>12(3x+x)=>2x as per question we know, 2x=913=>2x=283=>x=143 So, Speed upstream = 14/3 km/hr, Speed downstream 14 km/hr. Speed of the current=12[14-143]=143=423kmph 2. A man rows 750 m in 675 seconds against the stream and returns in 7 and half minutes. His rowing speed in still water isA) 4 kmphB) 5 kmphC) 6 kmphD) 7 kmphAns. BRate upstream = (750/675) = 10/9 m/sec Rate downstream (750/450) m/sec = 5/3 m/secRate in still water = (1/2)*[(10/9) + (5/3)] m/sec.= 25/18 m/sec = (25/18)*(18/5) kmph = 5 kmph 3. If a boat goes 7 km upstream in 42 minutes and the speed of the stream is 3 kmph, then the speed of the boat in still water isA) 12 kmphB) 13 kmphC) 14 kmphD) 15 kmphAns. BRate upstream = (7/42)*60 kmh = 10 kmph. Speed of stream = 3 kmph. Let speed in sttil water is x km/hr Then, speed upstream = (x ?3) km/hr. x-3 = 10 or x = 13 kmph 4. A man takes twice as long to row a distance against the stream as to row the same distance in favour of the stream. The ratio of the speed of the boat in still water and stream isA) 3:1B) 1:3C) 2:4D) 4:2Ans. ALet speed downstream = x kmphThen Speed upstream = 2x kmph So ratio will be,(2x+x)/2 : (2x-x)/2 => 3x/2 : x/2 => 3:1 5. A man’s speed with the current is 20 kmph and speed of the current is 3 kmph. The Man’s speed against the current will beA) 11 kmphB) 12 kmphC) 14 kmphD) 17 kmphAns. CSpeed with current is 20,speed of the man + It is speed of the current Speed in still water = 20 – 3 = 17 Now speed against the current will be speed of the man – speed of the current= 17 – 3 = 14 kmph  

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BOATS SET2

1. A man can row upstream 10 kmph and downstream 20 kmph. Find the man rate in still water and rate of the stream.A) 0,5B) 5,5C) 15,5D) 10,5Ans. CIf a is rate downstream and b is rate upstream Rate in still water = 1/2(a+b)Rate of current = 1/2(a-b) => Rate in still water = 1/2(20+10) = 15 kmph= Rate of current = 1/2(20-10) = 5 kmph 2. A man takes 3 hours 45 minutes to row a boat 15 km downstream of a river and 2 hours 30 minutes to cover a distance of 5 km upstream. Find the speed of the current.A) 1 km/hrB) 2 km/hrC) 3 km/hrD) 4 km/hrAns. AFirst of all, we know that speed of current = 1/2(speed downstream – speed upstream) [important] So we need to calculate speed downstream and speed upstream first. Speed = Distance / Time [important] Speed upstream =(15334)km/hr=15×415=4km/hrSpeed Downstream = (5212)km/hr=5×25=2km/hrSo speed of current = 12(4-2)=1km/hr 3. In one hour, a boat goes 11km along the stream and 5 km against it. Find the speed of the boat in still waterA) 6B) 7C) 8D) 9Ans. CWe know we can calculate it by 1/2(a+b) = 1/2(11+5) = 1/2(16) = 8 km/hr 4. A motorboat, whose speed in 15 km/hr in still water goes 30 km downstream and comes back in a total of 4 hours 30 minutes. The speed of the stream (in km/hr) isA) 2 km/hrB) 3 km/hrC) 4 km/hrD) 5 km/hrAns. DLet the speed of the stream be x km/hr. Then,Speed downstream = (15 + x) km/hr,Speed upstream = (15 – x) km/hr So we know from question that it took 4(1/2)hrs to travel back to same point. So, 3015+x-3015-x=412=>900225-x2=92=>9×2=225=>x=5km/hr 5. If Rahul rows 15 km upstream in 3 hours and 21 km downstream in 3 hours, then the speed of the stream isA) 5 km/hrB) 4 km/hrC) 2 km/hrD) 1 km/hrAns. DRate upstream = (15/3) kmph Rate downstream (21/3) kmph = 7 kmph. Speed of stream (1/2)(7 – 5)kmph = 1 kmph    

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BOATS SET1

Q.1 A man takes twice as long to row a distance against the stream as to row the same distance in favour of the stream. The ratio of the speed of the boat (in still water) & stream ?A) 3:1B) 2:1C) 4:1D) 5:1Ans. ALet upstream =2x km/hrDownstream = x km/hrBoat:Stream = (u+v/2) : (u-v/2)(2x+x / 2) : (2x-x / 2) Q.2 Boat covers a certain distance downstream in 1 hr, while it comes back in 1 ½ hrs. If the speed of the stream be 3 km/hr. What is the speed of the boat in still water ?A) 15 km/hrB) 16 km/hrC) 17 km/hrD) 18 km/hrAns. AShortcut:Speed= Z(x+y) / y-xOr 3 (1+ 3/2) / (3/2 – 1)= 15 km/hr Q.3 A man can row 15km/hr in still water. If he takes twice as long to row up as to row down the river. Find the rate of the stream ?A) 5 km/hrB) 6 km/hrC) 7 km/hrD) 8 km/hrAns. AShortcut:15+x = 2 (15 – x)15+x = 30-2xThus, x=5km/hr Q.4 Man can row 8km/hr in still water & the river is running at 2km/hr. If the man takes 1hr to row to a place & back. How far is the place ?A) 3.75 kmB) 4.75 kmC) 3.0 kmD) 4.0 kmAns. A Q.5 Man can row upstream at 10 km/hr & downstream at 16 km/hr. Find the man’s rate in still water & the rate of current ?A) 13 & 3 km/hrB) 12 & 2 km/hrC) 11 & 1 km/hrD) 13 & 2 km/hrAns. ARate in still water = ½ (10+16) = 26/2 = 13 km/hrRate of current = ½ (10-16) = 3 km/hr

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CLOCKS & CALENDAR SET2

Q.1  Find the angle between the two hands of a clock at 4:30 pm ?A) 450B) 460C) 470D) 480Ans. AUse the shortcut & apply the formula:i.e. Angle=30(N-M/5)-M/2=30(4-30/5)-30/2 Q.2 If Monday falls on the first of October. Which day will fall after three days of 20th in that month ? A) Wednesday B) Tuesday C) Monday D) Sunday Ans. AIf Monday falls on 1st of October, then 20th October will be Saturday & after three days of it there will be Wednesday Q.3 Time in a clock is quarter past twelve. If the hour hand points to the east, Which is the direction opposite to the Minute hand ? A) West B) North C) South D) South-West                                                      Ans. B                                                                             East                                12                    North  9                3   South                                       6                                             West   Q.4 If the 1st day of 1999 is a sunday, What is the last day? A)sunday B)monday C) Tuesday D)cannot be determined Ans. A If leap year then Monday else Sunday so its SUNDAY Q.5 A clock is set right at 8 a.m and it gains 10 minutes in 24 hours. What will be the true time when the clock indicates 1 p.m on the following day?  A) 12 o’ clock B) 16 min past 12 C) 48 min past 12 D) 30 min past 12 Ans. C Time from 8 a.m. on a day to 1 p.m. on the following day = 29 hours.          24 hours 10 min. of this clock = 24 hours of the correct clock.          145/6 hrs of this clock = 24 hours of the correct clock.           29 hours of this clock = [24 * (6/145) * 29] hrs of the correct clock= 28 hrs 48 min of the correct clock.          Therefore, the correct time is 28 hrs 48 min. after 8 a.m.          This is 48 min. past 12.      

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CLOCKS & CALENDAR SET1

Q.1 Assume that-1. the hour & minute hands of a clock move without jerking.2. the clock shows a time between 8 O’ clock & 9 O’clock.3. the two hands of the clock are one above the other.After how many minutes (nearest integer) will the two hands be again lying one above the other? A) 60 B) 62 C) 65 D) 67 Ans. C Explanation: Reqd. time=60*60/55=65 5/11= 65 Q.2 A watch showed a time of fourteen min past nine (9 hrs & 14 min.) The position of hour hand of the watch are exactly interchanged . The new time shown by the watch is closest to which of the following ?A) Fifteen minutes to threeB) Fourteen minutes to threeC) Thirteen minutes to threeD) Twelve minutes to threeAns. BMinute pointer when move sixty minutes the hour pointer moves five minutes. When minute pointer moves 14 minutes the hour pointer will move = 14/60 * 5 = 14/12Hence the minute pointer will be act45+14/12 = 277/6 Q. 3 Find the time between 8’o clock & 9’o clock when the two hands of a clock are 4min. apart ?A) 39 3/11 & 48 min.B) 40 3/11 & 47 min.C) 41 3/11 & 46 min.D) 42 3/11 & 45 min.Ans. AShortcut:Use this formula(5x±4) * 12/11(5 * 8±4) * 12/11 (In place of x you have to put 8 as given in question)40 ± 4 *12/1112/11 (36 & 44)432/11 & 48=39 3/11 & 48 min. Q.4 Find the angle between the two hands of a clock at 15min. past 4’o clock ?A) 37.50B) 38.50C) 39.50D) 40.50Ans. AShortcut:Use this formula to find the angle:Angle= 30(N-M/5)+M/2Then, now put values30 (4-15/2)+15/2=30+7.5=37.50 Q.5 Find the time between 8’0 clock & 9’0 clock when the two hands of a clock in the same straight line ?A) 10 10/11 & 43 7/11B) 11 11/11 & 44 7/11C) 12 12/11 & 45 7/11D) None of the aboveAns. AUse the shortcut & apply the formula, i.e(5x-30)*12/11 & 5x*12/11(5*8-30)*12/11 & 5*8*12/11120/11 & 480/1110 10/11 & 43 7/11  

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