Three machines can finish a cerain work in 36 hours at the same rate. If a 4th machine with the same rate is added in, what is the time to finish the work?Ans:(1/108+1/108+1/108+1/108)x=1 4x/108=1 x=108/4 x=27.

There are three tailors - A,B and C who can stich a shirt in 1hr , 1.5hrs and 2 hrs respectively. If there is and order of 150 shirts, in how many hours can they fufil the order working together? Any shirt is worked on only by one of the tailors , ie two or more tailors do not work on the same shirt.

a)66 9/11 hrs (this is a mixed fraction)b) 69 hrsc) 69 3/13 hrs ("").d) 70 hrsAns:

Take LCM of 1, 1.5 & 2 so, LCM will be 3 Double it we get 6

A stitches 6 shirts in 6 hrsB stitches 4 shirs in 6 hrsC stitches 3 shirts in 6 hrs so,

6 hrs----------13 shirts

In 66 hrs -----143 shirts will be made.

Now 7 shirts more to be made,

Tailor A will make 3 shirts in 3 hrsTailor B will make 2 shirts in 3 hrsTailor c will make 1.5 shirts in 3 hrs so Tailor C will take 1 hr more to complete 1/2 shirt

So, Total time taken by all three to complete 7 shirts = 4 hr

Total time taken to complete 150 shirts= 66+4=70 Ans(D)John can dig constantly at 15 inches per minute and Linda can dig constantly at 45 inches per minute. A certain hole can be dug by John alone in 12 hours. If hole is dug by John for half the time and both together for rest the time, how many minutes does it take to dig the hole?Ans:say it takes x minutesso, (x/2)*15 + (x/2)*15 +(x/2)*45=12*60*15=> x*15 + x*15 + x*45 =2*12*60*15=> 30x + 45x =2*12*60*15=> 75x=2*12*60*15=> x=288

Two cyclists are moving towards each other at 10 miles/hour. They are now 50 miles apart. At this instance a fly starts from one

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cyclist and move towards other and moves to and fro till the two cyclist meet each other. If the fly is moving at 15 miles/hour, what is the total distance covered by the fly?

Ans:Relative speed=10+10=20mile/hrTime to travel 50 miles = 50/20 = 5/2 hrDistance travelled by fly in 2.5 hrs = 5/2 * 15 = 37.5 miles

Mike and James start at the same time and paint their apartment in 3 hrs. James always rests for 5 minutes after painting for 20min, and Mike rest 5 min after painting for 30 min:

Col A

One half the number of min they work together

Col BThe number of minutes they don't work together.Ans:let Mike and James start working at 12:00.

MIke:12:00-12:20 , 12:20- 12:25, 12:25-12:45: 12:45-12:50,12:50-1:10,1:10-1:15,...... 3:00( bold times are brakes)

James:12:00-12:30,12:30-12:35,12:35-1:05, 1:05-1:10,1:10-1:40,1:40-1:45,.......3:00.

It can be observed that, Mike's brake starts every 25 minutes , and James's brake starts every 35 minutes.

Number of Mike's brakes in 3 hours is 180/25=7 Numbers of James's brake in 3 hours is 180/35=5Number of brakes they had together is 180/(LCM of (25 and 35))=180/175=1therfore, total number of minutes they are not working together is5*(7+5-1)=55.

Total number of minutes they are working together is 180-55=125

Col A: 125/2=62.5 > Col B:55

answer is A.

A group of people can do a work in certain number of days. After they worked for 12 days 1/4 of them have left. To complete the remaining work the remaining men have taken as many days as the initial no. of men would have taken to complete the entire work. In how days did they finally complete the work.

Ans:say group contributes S amount of work per dayand completes the task in D daysso, total work amount is S*Dthen,12*S+D*(3/4)*S=D*S=>D=48

so they spend total (48+12)=60 days to complete the task

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a freight track 1.2 miles long is travelling 40 miles per hour on a track parallel to a highway. if a car is travelling 60 mph in the same dirn as the train, how many miles will the car travel while passing the train?

Ans:Let after Time t car passes the freight.

Let distance travelled by freight = x (this distance is measured from the front end of the freight)Distance travelled by car = 1.2 + x (1.2= length of freight + distance measured from the front end of the freight)

distance = speed * time So dist. by freight = 40t = xdist. travelled by car = 1.2 + x = 60t

Time taken for both, freight n car shud be d same.so x/40 = (1.2 +x)/60solve x= 2.4Therefore, distance travelled by car = 1.2 + 2.4 = 3.6 miles Alternatively,The relative velocity of the car wrt the freight = 60-40 = 20mphThis is bcoz 2 any person on the freight the car will appear to be moving with a velocity of 20mph in the same direction as the freight.So, the time taken 4 the car to cross the freight(as seen by the person on the freight) is = 1.2miles/20mph= 0.06 hrsso, the dist travelled by the car during this time = .06hrs * 60 mph= 3.6 miles

It takes the high-speed train x hours to travel the z miles from Town A to Town B at a constant rate, while it takes the regular train y hours to travel the same distance at a constant rate. If the high-speed train leaves Town A for Town B at the same time that the regular train leaves Town B for Town A, how many more miles will the high-speed train have traveled than the regular train when the two trains pass each other?Ans:Speed of high spd: z/xSpeed of regular train: z/yTime taken for them to meet: z*xy/z(x+y) = xy/x+y (Call this T)

Distance difference (Diff in distance travelled in time T): xy/x+y(z/x - z/y) => z(y-x)/(x+y)

A metel company's old machine makes bolts at a constant rate of 100 bolts per hour. The company's new machine makes bolts at a constant rate of 150 bolts per hour. If both machines start at the same time and continue making bolts simultaneously, how many minutes will it take the two machines to make a total of 300 bolts?Ans:in 1 hr old m/c makes = 100 boltsin 1 hr new m/c makes = 150 boltsin 1 hr old+new m/c make = 100+150=250 bolts

300 bolts can be done in = (300/250)*60 mins = 72 mins

Running at their respective constant rates, machine X takes 2 days longer to produce w widgets than machine Y. At these rates, if the two

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machines together produce 5/4 w widgets in 3 days, how many days would it take machine X alone to produce 2w widgets?Ans:Y : a days ---> W widgetsX : a + 2 days ---> w widgets

so in three days :

3 { w/a + w/a+2 } = 5w/4

a = 4 so machine x takes 4+2 = 6 days to make w widgets

hence machine x will take 12 days to make 2w widgets

Ans:

Machine A rate= 9000 per hourMachine b rate= 7000 per hour

The highest number at which Machine A or B can produce = 8

so lets say machine A worked for 8 hours, therefore it will produce (rate*time) 9000*8=72000

Therefore if Machine A produces 72000 at the maximum time permitted then Machine B will produce the additional 28000 in 4 hours (use fromula work/rate=time, 28000/7000=4hours).

Pumps A and B and C operate at their respective constant rates. Pumps A and B operating simultaneously can fill a certain tank in 6/5 hours; pumps A and C operating simultaneously can fill a certain tank in 3/2 hours; pumps B and C operating simultaneously can fill a certain tank in 2 hours. How many hours does it take pumps A, B and C, operating simultaneously, to fill the tank?Ans:Let a by the number of units/hr of water filled by A, simmillarly goes to b and c,

so a * 6/5 + b * 6/5 = 1 => a + b = 5/6similarly, a + c = 2/3 and b + c = 1/2

Adding all the equations2(a+b+c) = 5/6+2/3+1/2=> 2(a+b+c) = (5+4+3)/6=> 2(a+b+c) = 12 / 6=> (a+b+c) = 1

A distillate flows into an empty 64 gallon drum at spout A and out of the drum at spout B. if the rate of flow through A is 2 gallon per hour, how many gallons per hour must flow out at spout B so that the drum is full in exactly 96 hours?Ans:

Drum fills in 96 hrs

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so,96 hrs----------64 gallonin1 hr ----------64/96 =2/3then,2 - x= 2/3=> x=4/3 gallon/hr(ANS)

A cistern has 3 pipes A,B,C .The pipes A & B can fill it in 4 and 5 hours respectively & C can empty it in 2 hours .If pipes r opened at 1 AM , 2 AM, 3 AM respectively , when will the cistern be empty?Ans:assume it will be empty in n hours from 1AM, Then in n hours pipe A will fill n/4 , pipe B will fill (n-1)/5 and pipe C will empty (n-2)/2 , and they all summed up should be 0i.e n/4 + (n-1)/5 - (n-2)/2 = 0 solves to n = 16 hours.So 1 AM + 16 hours = 5PMIt takes 6 technicians a total of 10 hours to build and program a new server from Direct Computer, with each working at the same rate. If six technicians start to build the server at 11:00 AM, and one technician per hour is added beginning at 5:00 PM, at what time will the server be complete?A) 6:40 PMB) 7:00 PMC) 7:20 PMD) 7:40 PME) 8:00 PMAns from 11:00 am to 5:00 pm 6/10 of server was completed so 2/5 was left now 7 technicians can complete the job in 60/7 hrs, 8 in 60/8hrs, 9 in 60/9 hrs so;2/5 - 7/60 = 17/60 left till 6:00 pm 17/60 - 8/60 = 9/60 left till 7 pm so work completed at 8 pm John can complete a given task in 20 days. Jane will take only 12 days to complete the same task. John and Jane set out to complete the task by beginning to work together. However, Jane was indisposed 4 days before the work got over. In how many days did the work get over from the time John and Jane started to work on it together?

A- 6 B- 10 C- 8 D- 7.5 E- 3.5

Ans: Together they do 1/20+1/12=4/30 of the task.

Jane and John started the work together, but only John finished the work because Jane gets sick. So let x be the number of days they worked together.

x*3/40+4*1/20=1

x4/30=4/5 and therefore x =6

So in total they worked 6 days on it together and John worked 4 days on it. So total days spent=10, but if the question is asking how many time did they spend working on the project together, then the answer is 6.

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Jane gave Karen a 5 m head start in a 100 race and Jane was Karen beaten by 0.25m. In how many meters more would Jane have overtaken?

Soln: Jane gave Karen a 5 m head start means Karen was 5 m ahead of Jane. So after the lead, Karen ran 95m and Jane ran 99.75 m when the race ended.

Let speed of Karen and Jane be K and J respectively and lets say after X minutes Jane overtakes Karen.

1st condition: 95/K = 99.75/J2nd condition JX - KX = 0.25

Solving for JX, we get JX=21/4 (substituting for K from 1st condition)Hence Jane needs to run 5.25m more (or total of 105m) to overtake Karen.

A is cycling around a track at a speed of 15 mph. B starts at the same time at a speed of 12 mph. If A and B move at the same direction and the track is 1/2 mile long, how many minutes after they start will A pass B if?

Ans:

A will pass B when A gains 1/2 mile over B.Relative speed = 15 - 12 = 3 mphTo cover 1/2 mile it will take -- (1/2) /3 = 1/6 hrs = 1/6 * 60 = 10 mins.

Lindsay can paint 1/x of a certain room in 20 minutes. What fraction of the same room can Joseph paint in 20 minutes if the two of them can paint the room in an hour, working together at their respective rates?Ans:Assume the fraction is y.

rate(work per minute) for Lindsay = 1/20x

rate for Joseph = y/20

Combined rate = 1/60

Given,

1/60 = 1/20x + y/20 => y = (x 3/3x)

John and Maggie working together complete the job in T days. When John works alone , he takes 9 days more than T. When Maggie does the job alone, she takes 4 days more than T. How many days does it take for the job to be done when John & Maggie work together?Ans:wrk done by John in 1 day ----1/(9+T)wrk done by maggied in 1 day ---1/(4+T)

combined together they finish in T days i.e. in 1 day 1/T

1/(9+T) + 1/(4+T) = 1/T

=> (13+2T)*T = 36+13T + T^2=> 13T + 2T^2 = 36 + 13T + T^2=> T^2 = 36=> T=6 days

It takes three laborers ten hours to install four faucets (each faucet takes the same amount of work to install). How many hours would it take nine laborers to

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install twelve faucets?Ans:3 l take 10 hrs 9 will take 10/3 hrs

10/3 hrs for 4 faucets10 hrs for 12 faucets

OR,

In project management terms, the effort for installing 4 faucets = 30 hours

so effort for installing 12 = 90 hours.

This much effort can be EXTRACTED from 9 resources in 10 hours

OR,

Rate=4/(3*10)=2/15Now rate for 9 people is = 9*2/15 =18/15

Using the formula Time = Work/RateTime= 12*15/18=10

Two pipes A & B are connected to a tank. Pipe A takes 8 minutes to fill the tank. Pipe B takes 9 minutes to empty the tank. First Pipe A is opened for 1 minute, after which Pipe A is closed and then Pipe B is opened for a minute. This continues until the tank is filled. How long does it take to fill the tank ?a/ 17 minutesb/ 34 minutesc/ 144 minutesd/ 127 minutese/ 120 minutesAns:as in 2 min 1/72 is filled...in 126 min 63/72 = 7/8 is filledin 127 min 7/8 + 1/8 ==> tank is filled.

The moving walkway is a 300-foot long conveyor belt that moves continuously at 3 feet per second. When Bill steps on the walkway, a group of people that are also on the walkway stands 120 feet in front of him. He walks toward the group at a combined rate (including both walkway and foot speed) of 6 feet per second, reaches the group of people, and then remains stationary until the walkway ends. What is Bills average rate of movement for his trip along the moving walkway? Ans:Avg Speed = total distance travelled/total time taken

Now let x be the feets travelled by the ppl on the walkway before Bill catches them=> Bill travels 120+x feet in say t seconds at 6 ft/secnow since t will be same for both(120+x)/6 = x/3=> x = 120 feet

=> t = 240/6 = 40 sec

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let t2 be the time taken to travel the remaining 60 ft of walkway @ 3 ft/sect2 = 60/3 = 20 sec

total time taken = 40+20 = 60 secs

total distance travelled = 300ft

avg speed= 300/60 = 5 ft/sec

Jim can read at the rate of 30 pages/hr and Jack can read at the rate of 40 pages/hr. Jim starts reading at 4:30 pm and Jack starts reading the same book at 5:20 pm. At what time will they be on the same page?Ans:30( t+5/6) = 40 tt= 2 1/2 hrs

So time = 5.20 + 2.30 = 7.50

Elena was working to code protocols for a computer processing company. She did 11/18 of the job and allowed Andy to finish it. They work at the same rate and are paid the same hourly pay. If the difference betweent the amounts that were paid to them was $154, what was the total amount the two were paid for the entire coding job?A) 252B) 269.5C) 369D) 423.5E) 693Ans:Elena completed 11/18 of job so Andy did only 7/18 of it................(1)Lets assume that Elena got $ x foe the job n Andy got $ y, given already that x - y = 154

from eqn (1)... we have (11/18-7/18)p = 154

where p is the rate at which both were paid...on calculations we get p = 693

ABC developers employed 30 men to do a piece of work in 38 days. After 25 days the company employed 5 more men and the work was finished one day earlier.

A) the number f days, by which the work would have delayed of these additional number of men had not been employed,

B) 2Ans:we use the work equation:(1/X) * T1 + (1/Y) * T2 = 1

where X is time it takes to finish one work for a person, and same for y.T is the time spent working

"After 25 days the company employed 5 more men"- dat means 30 workers did their job for 25 days. we still don't know the rate of the workers so we represent it with x.

(30/x) * 25 + ( (30+5)/x ) * 12 = 1

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note: since they finished a day earlier, dey completed their job in 37 days, si it took 37-25 = 12 days to complete the remaining.

solving for x, one gets 30(25)+35(12). T = X/30-> T = ( 30(25) + 35(12) ) /30-> T = 39 days

Therefore, if 30 men worked, it would cost them 1 day of delay.

'A' can finish a piece of work in 10 days. If he works at 80% of his effieciency and does 20% overtime in a day. How many days will he take?Ans:A does (1/10) work in normal dayat 80 % eff he will do (0.8)/10if he works 20% overtime.......(2/25)*1.2hence......25/(2*1.2)

A and B working seperately can do a piece of work in 9 and 12 days respectively. If they work for a day alternatively, A beginning, in how many days the work will be completed?a. 10.5b.10.33c.10.25d.10.66e.11Ans:In a day, A does 1/9 of the work and B 1/12 of the work.

For a single cycle of alternation (that is, 2 days - 1st A and then B), the fraction of work that can be finished is (1/9 + 1/12) = 7/36.

For 5 such cycles (10 days), 35/36 of the work will be finished. The 11th day will be A's turn and he can finish the remaining 1/36 of the job in 1/4 of the day.

Thus the entire job can be finished in 10.25 days.

Two people walked the same distance, one person's speed is between 25 and 45, and if he used 4 hrs, the speed of another people is between 45 and 60, and if he used 2 hrs, how long is distance?

116118124136140Ans:For first person range of speed is given 25 < s1 < 45 and he takes 4 hours so 25 < d/4 < 45 or 100 < d < 180similarly for second person 45 < s2 < 60 or 90 < d < 120now check for the correct option. Both 116 and 118 lie within the range calculated. However, 118 is not divisible by 4 so 116 is the answer

if bob can do a job in 20 days and jane can do the job in 30 days, they work together to do this job and in this period, bob stop work for 2.5

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days and jane stop work for x days, and the job be finished for 14 days, what is x? Ans:Bob: J/20, Jane: J/30. Bob absent: J/30*2.5=J/12Jane absent: J/20*xBoth: (11.5-x) * (J/30+J/20) = (11.5-x) * (J/12) = 11.5/12 J - Jx/12J/12+Jx/20+11.5J/12-Jx/12=J12.5/12-1 = 1/30 x ==> 30*(1/24) = x ==> x=1.25 days.

Three friends depart from the same starting point at 9:00 A.M. Chris travels due west at 60 miles per hour. Brian travels due north. Alan starts driving without paying attention to what direction he is driving. At 10:00 A.M., all three cars stop. Using a GPS navigation system, Chris calculates that he and Brian are exactly 65 miles apart. Alan calculates that he is exactly 75 percent as far from Brian as from the starting point and 60 percent as far from the starting point as Brian is. What is the difference between Brians average speed and Alans average speed?Ans:Chris travels W - 60 mph => 60 miles is the distance at 10 AMBrian travels N and he is 65 miles from Chris..It is a right angled triangle and calc the distance covered by Brian from the Starting point (height of the triangle) = sqrt(65^2 - 60^2) = sqrt((65-60)(65+60)) = 5^2 = 25 milesAlan is 60% far from the starting pt as Brian is => 60% of 25 = 15 (distance covered from the starting point)=> Brian's avg speed - Alan's avg speed = 25-15 = 10

A crew can row a certain course up stream in 84 minutes; they can row the same course down stream in 9 minutes less than they can row in still water. How long would they take to row down with the streamAns:let speed of boat be x & stream be y

d/(x-y) =84 ----(1)

d/(x+y) =d/x -9=> d/x -d/(x+y) =9=> dy/x(x+y) =9substituting value of d from (1)

84y(x-y)/x(x+y) =9=> 3x^2 -25xy + 28y^2=0=> 3x^2 -21xy-4xy+28y^2 =0-> 3x(x-7y) -4y(x-7y)=0=> 3x=4y or x=7y

we have to find, d/(x+y) = 84(x-y)/(x+y) = 84*6y/8y =63min84(x-y)/(x+y) = 84 * y/7y =12 min

Three runners A, B and C run a race, with runner A finishing 12meters ahead of runner B and 18 meters ahead of runner C, while runnerB finishes 8 meters ahead of runner C. Each runner travels the entire distanceat a constant speed. What was the length of the race?Ans:d=st

12/Sb=10/Sc Sb/Sc=12/10

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(d-12)/Sb=(d-18)/Sc Sb/Sc=(d-12)/(d-18) ==> 12/10=(d-12)/(d-18)10*(d-12)=12*(d-18)10d-120= 12d-2162d=96==>d=48

It takes 6 beavers 10 hours to build a certain dam, working at a uniform rate. If six beavers start to build the same dam at noon, and one beaver per hour is added beginning at 6:00 PM, at what time will the dam be complete?Ans:10 hrs 6 man 1 work1 hr 6 man 1/10rowk1 hr 1 man =1/60 workfrom 12-6 total work done =6(hrs)*1/10 work7PM total people =7 work 7*1/60 work8pm =8 worker=8*1/609pm=9 worker 9*1/60total =6/10+7/60+8/60+9/60=1so ans 9pm

Working together, Jose and Jane can complete an assigned task in 20 days. However, if Jose worked alone and complete half the work and then Jane takes over the task and completes the second half of the task, the task will be completed in 45 days. How long will Jose take to complete the task if he worked alone? Assume that Jane is more efficient than Jose.Ans:Assume Jane=A , Jose=B(A*B)/(A+B) = 20 daysA completes 1/2 of the work in say d daysd/A =1/2 => A=2dB completes 1/2 of the work in 45-d days(45-d)/B = 1/2 =>90-2d=B=> 90=A+BSubstituting the value of A+B A*B/90 = 20A*B=1800(90-B)*B=1800=> B^2 - 90B+1800=0solvingB=60 & A=30Ans=60days

Two guns were fired from the same place at an interval of 12min; but a person in the train approaching the place hears the second shot 10min after the first.The speed of the train, if speed of sound is 330m/sAns:Let v is the speed of train10(v+330)=330*12V=330/5 =66m/s(Ans)

A person covers equal distances with three different speeds which are integers in m/min. The sum of his three speeds is 14m/min. His average speed for the whole journey is 210/59 m/min. what was his lowest speed in m/min.(A)2 (B)3 (C)1.5 (D)2.5 (E)7Ans:Given that the 3 different speeds are integers and sum to 14, we can eliminate all answers choices except A and B.

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If the speeds are x, y and z, and distance covered at each speed = d, thenTotal time taken, T = d/x + d/y + d/z and Average speed = 3d/Ti.e., Average speed = 3xyz/(xy+yz+zx) = 210/59=> xyz/(xy+yz+zx) = 70/59At this point I'd be sorely tempted to try xyz = 2*5*7In any case it is unlikely that xyz is a multiple of 3, when 70 is not.

A truck drivers drove for two days. On the second day he drove 3 hours longer and @ an average speed of 15 miles/hour faster than he drove on the first day. If he drove a total of 1020 miles and spent 21 hours dricing during the 2 days, what was his average speed on the first day,in MPH.Ans:If ( a, x ) are the speed and number of hours on the first day then a+15 and x+3 are for the second day.

therefore a * x + (a + 15) * (x +3 ) = 1020 and x + (x+3) = 21;x= 9 and 9a + (a + 15) * 12 = 1020 ; 21 a = 840; a = 40

In a 100m race, A can beat B by 25 m and B can beat C by 4m.In the same race, A can beat C by:a 21 b 26 c 28 d 29Ans:1st set of infoPerson distanceA 100B 75

2nd set of infoPerson distanceB 100C 96

We need to find how much distance is covered by C w.r.t A . i.e. when A covers 100 m. In other words when B covers 75m.

Hence the ratio would be

B C100 9675 xso x = 75 * 96/100 = 72. Or you can say this is what C covers when A covered 100 m.So difference between A and C is 28m i.e. A beats C by 28m

OR,

Let, Speed of A=a Speed of B=b Speed of C=cThen (100/a)Xb=100-25=75...........(i)(100/b)Xc=100-4=96............(ii)(i)X(ii)....(10000/a)Xc=75X96

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So, (100/a)Xc=75X96/100=72=100-28.So, answer=28m .......(C)

If 9 men working 15/2 hrs a day can finish a piece of work in 20 days, then how many days will be taken by 12 men, working 6 hrs a day to finish the work, it being given that 2 men of latter type work as much as 3 men of the former type?1 19/2 b 11 c 25/2 d 13Ans:Work = 9*r*(15/2)*20 = 12*r'*6*d, where r and r' are the respective rates of workr' = (3/2)r

=> 9*15*10 = 6*3*6*d=> d = 9*15*10/(6*3*6) = 25/2 days

the number of cards which are red in colour ( include all hearts) or kings are 26 + 2 ( two kings are blacks as well )= 28 .Ans. 28/52.Two trains are leaving from two stations seperated by 50 miles. With a constant speed of 60 miles per hour, the trains are approaching each other on different tracks. if length of each train is 1/6th of a mile, when they meet how much time do they need to pass each other totally?Ans:relative speed=60+60=120

total distance to be covered= length of train1+length oftrain2 =2x

hence, time taken= distance/speed=2x/120x=1/6, so T=10 seconds

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