X and Y Can Do a Piece of Work in 20 Days and 12 Days Respectively

7/28/2019 X and Y Can Do a Piece of Work in 20 Days and 12 Days Respectively

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X and Y can do a piece of work in 20 days and 12 days respectively. X started the work aloneand then after 4 days Y joined him till the completion of the work. How long did the work last?

X Y

Speed

Time 20 12

distance

LCM (20,12)=60Total distance = 60Update

X Y

Speed

Time 20 12

distance 60 60

Apply STD formula in each column and you get speed of X and Y

X Y

Speed 3 5

Time 20 12

distance 60 60

Given: X started the work alone and then after 4 days Y joined him till the completion of thework


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Means X worked for four days alone.

X Y X started

Speed 3 5 3

Time 20 12 4

distance 60 60 34=12

In 4 days, he covered 12 kms.So remaining work = 60 minus 12 =48.Which was completed by X+Y

X Y X started X+Y

Speed 3 5 3 3+5=8

Time 20 12 4 ??

distance 60 60 34=12 48

Run std on last column

Speed x time = dist8 x time = 48Time =48/ 8 =6 daysFinal table looks like this

X Y X started X+Y finished

Speed 3 5 3 3+5=8

Time 20 12 4 6

distance 60 60 34=12 48


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First question

P & q can do a job in 15 days &10 days respectively . They began the work together but p leavesafter some days and q finishes the remaining job in 5 days. After how many days did p leave?Take LCM of (15, 10)=30 and fill up the table

P Q P+q work together Only q

Speed 2 3 3+2=5 3

xTime 15 10 ?? 5

Distance 30 ?? 15

According to last column, q covered 3x 5= 15 kilmeteres out of total 30 km. Hence the remaining 30 minus 15 = 15 kilometers were covered by P+Q together their combined speed is (3+2)= 5.Run the STD formula on (P+Q) column5 x time = 15Time = 15/5=3Means P+Q worked together for 3 days. And then P left.Final answer: P left after 3 days.

Second Question

Singh & ravi can do a job alone in 10 days and 12 days respectively. Singh starts the work & after 6 days ravi also joins to finish the work together. For how manydays ravi actually worked on the job ?

Same procedure, take LCM of (10,12)=60

After 6 days, Ravi also joins. That means for the first 6 days, Singh worked alone.Fill up the table.

Singh Ravi Singh starts Singh+Ravi

Speed 6 5 6 (6+5)=11


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Time 10 12 6 days ??

Distance 60 36 (60-36)=24

In the first 6 days, Singh finished 36 kilometers. Hence remaining work= 60 minus 36= 24These 24 kilometers were covered by Singh+Ravi together. (last column)Run the STD on last columnSpeed x time = distance11 x Time =24Time =24/11= approx. 2.18Final answer: Ravi worked for 2.18 days.

1. Case#1 : Time equations: set of two people working 2. Case# 2: Time Ratios are given

Case#1 : Time equations: set of two people working

A & B can do a piece of work in 12 days , B and C in 15 , C & A in 20 days. How long wouldeach take separately to do the same work ?

In the STD table, when people come and go, we can do addition and subtraction in the SPEED boxes only and now in the Time boxes. Now fill up the table.

A+B B+C C+A

SPEED

TIME 12 15 20

DISTANCE

TAKE THE LCM (12,15,20) =60. Based on that we can find individual speeds using STDformula.
http://mrunal.org/2012/04/aptitude-std-table-two-complex-cases-of.html#2http://mrunal.org/2012/04/aptitude-std-table-two-complex-cases-of.html#2http://mrunal.org/2012/04/aptitude-std-table-two-complex-cases-of.html#510http://mrunal.org/2012/04/aptitude-std-table-two-complex-cases-of.html#510http://mrunal.org/2012/04/aptitude-std-table-two-complex-cases-of.htmlhttp://mrunal.org/2012/04/aptitude-std-table-two-complex-cases-of.htmlhttp://mrunal.org/2012/04/aptitude-std-table-two-complex-cases-of.htmlhttp://mrunal.org/2012/04/aptitude-std-table-two-complex-cases-of.html#510http://mrunal.org/2012/04/aptitude-std-table-two-complex-cases-of.html#2


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A+B B+C C+A

SPEED 5 4 3

TIME 12 15 20

DISTANCE 60 60 60

Assume that these three entities work together

A+B B+C C+A (A+B)+(B+C)+(C+A)

SPEED 5 4 3 5+4+3=12

TIME 12 15 20 ??

DISTANCE 60 60 60 60

Speed of A+B+C together can be derived using(A+B)+(B+C)+(C+A)

= 2(A+B+C)=5+4+3=12Hence speed of A+B+C together =12/2=6

Now STD formulaSpeed x time = distance6 x time = 60Time = 10Meaning if A+B+C work together, they can finish work in 10 days. [although Time of A+B+C,is not asked in the question, but Ive given it only for explanation]

Now back to the question, weve to f ind individual time. So far our table looks like this-

A+B B+C C+A A+B+C

SPEED 5 4 3 6

TIME 12 15 20 10


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TIME 12 15 20 10 60 20 30

DISTANCE 60 60 60 60 60 60 60

Final answer, time taken by A,B,C if they work individually= 30,20,60

Run a cross verification to see the answer is correct. Plug the answer values.

A B C A+B B+C C+A

Speed 2 3 1 5 4 3

Time 30 20 60 12 15 20

Dist 60 60 60 60 60 60

Yes it satisfies the statements given in the Question.


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Case# 2: Time Ratios are given
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A takes twice as much time as B and thrice as much as C to finish a piece of work. They together finish the work in one day. Find the time taken by each of them to finish the work.

A takes twice as much time as B,meaning if B takes 1 day, A takes 2 daysSo the ratio of time is A:B=2:1Similarly A:C=3:1We can write the same thing as

B:A=1:2A:C=3:1

If we want the three terms ratio, we want mid term (A) to be same in both equation.

In the first eq. A is 2 and in second eq. A is 3. LCM (2,3)=6Multiply something in both equations so that we get 6 in both equation.Meaning, Multiply first eq with (2) and second equation with (3)B:A=3:6A:C=6:2

Now we can make a three terms ratioB:A:C=3:6:2
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Keep in mind, this is ratio and not absolute number So we fill Time boxes as As time = 6x Bs time = 3xCs Time = 2x

Run the STD table, fill the data

A B C A+B+C

Speed 1 2 3 6

Time 6x 3x 2x 1 day

Distance 6x 6x 6x 6x

LCM of (6x,3x,2x)= 6xFrom the STD fomula, we can get the individual speeds of A,B,C,= 1,2,3Together their speed A+B+C=1+2+3=6For the last columnSpeed x time = distance6 multiplied with 1 =6xHence x= 1

Put the value of x=1 in the table and you get individual timeA will take 6 days individually, B will take 3 days and C will take 2 days.

Cross verification (plug the answers we got)

A B C A+B+C

Speed 1 2 3 6

Time 6 3 2 1 day

Distance 6 6 6 6

From the table,A takes twice as much time as B and thrice as much as C to finish a piece of work Yes it satisfies the statement given in the question itself.


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Documents

X and Y Can Do a Piece of Work in 20 Days and 12 Days Respectively