### Doubts of Time and work sheet (Pages: 1 2 )

Posted:

**Tue Feb 26, 2013 9:54 am**Please post your doubts of time and work sheet on this thread

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Posted: **Tue Feb 26, 2013 9:54 am**

Please post your doubts of time and work sheet on this thread

Posted: **Thu Feb 28, 2013 1:09 pm**

In Q-11 -- i have solved by estamating the new work being 1.8 , instead of 1...sir, is this the correct approach.?

Q--18 ---- we have to answer the original number of men emploeyed or the new one (+10)?

Q - 27 --- i got upto 20 cycles ..but how come 8:00 pm?

Q--18 ---- we have to answer the original number of men emploeyed or the new one (+10)?

Q - 27 --- i got upto 20 cycles ..but how come 8:00 pm?

Posted: **Thu Feb 28, 2013 1:11 pm**

In Q-11 -- i have solved by estamating the new work being 1.8 , instead of 1...sir, is this the correct approach.?

[quote]

Why did you assume total work as 1.8 instead of 1 - there is nothing wrong in doing so, but there has to be some logic such as - I get the answer fast or I find this number easier to handle etc.

[quote]

Why did you assume total work as 1.8 instead of 1 - there is nothing wrong in doing so, but there has to be some logic such as - I get the answer fast or I find this number easier to handle etc.

Posted: **Thu Feb 28, 2013 1:12 pm**

No, you need to find the original number of men employed in this case

Posted: **Thu Feb 28, 2013 1:12 pm**

27. A cistern has three pipes X, Y and Z. X and Y are filler pipes used in filling the cistern in 4 and 5 hours respectively. Z is an exhaust pipe which empties it in 2 hours. If the pipes are opened in order at 4 a.m., 5 a.m. and 6 a.m. respectively, when will the cistern be empty ?

Solution

In this case let us assume that the number of litres to fill the cistern are

LCM(4,5,2) = 20 L

Rx = 20/ 4 = 5 l/hr Ry = 20/5 = 4 l /hr Rz = 20/-2 = -10 l/hr

From 4 TO 5 only one pipe X is functional = X will fill 5 l in this hour

From 5 to 6 two pipes X and Y are functional - they will fill = 4 + 5 = 9 more l in this hour making the total volume filled so far to 14 l

When the third pipe is operational => the net rate of emptying

= > 4 + 5 -10 = -1 L/HR

At the rate of -1 l/hr the three pipes together will tak 14 hours to empty the 14 litres present in the tank.

14 hours counted from 6 am should make it 8 pm

Answer = 8pm

Solution

In this case let us assume that the number of litres to fill the cistern are

LCM(4,5,2) = 20 L

Rx = 20/ 4 = 5 l/hr Ry = 20/5 = 4 l /hr Rz = 20/-2 = -10 l/hr

From 4 TO 5 only one pipe X is functional = X will fill 5 l in this hour

From 5 to 6 two pipes X and Y are functional - they will fill = 4 + 5 = 9 more l in this hour making the total volume filled so far to 14 l

When the third pipe is operational => the net rate of emptying

= > 4 + 5 -10 = -1 L/HR

At the rate of -1 l/hr the three pipes together will tak 14 hours to empty the 14 litres present in the tank.

14 hours counted from 6 am should make it 8 pm

Answer = 8pm

Posted: **Thu Feb 28, 2013 1:13 pm**

In Q-11 -- i have solved by estamating the new work being 1.8 , instead of 1...sir, is this the correct approach.?

Q--18 ---- we have to answer the original number of men emploeyed or the new one (+10)?

Q - 27 --- i got upto 20 cycles ..but how come 8:00 pm?

Q--18 ---- we have to answer the original number of men emploeyed or the new one (+10)?

Q - 27 --- i got upto 20 cycles ..but how come 8:00 pm?

Posted: **Thu Feb 28, 2013 1:14 pm**

In Q-11 -- i have solved by estamating the new work being 1.8 , instead of 1...sir, is this the correct approach.?

[quote]

Why did you assume total work as 1.8 instead of 1 - there is nothing wrong in doing so, but there has to be some logic such as - I get the answer fast or I find this number easier to handle etc.

[quote]

Why did you assume total work as 1.8 instead of 1 - there is nothing wrong in doing so, but there has to be some logic such as - I get the answer fast or I find this number easier to handle etc.

Posted: **Thu Feb 28, 2013 1:17 pm**

No, you need to find the original number of men employed in this case

Posted: **Thu Feb 28, 2013 1:19 pm**

27. A cistern has three pipes X, Y and Z. X and Y are filler pipes used in filling the cistern in 4 and 5 hours respectively. Z is an exhaust pipe which empties it in 2 hours. If the pipes are opened in order at 4 a.m., 5 a.m. and 6 a.m. respectively, when will the cistern be empty ?

Solution

In this case let us assume that the number of litres to fill the cistern are

LCM(4,5,2) = 20 L

Rx = 20/ 4 = 5 l/hr Ry = 20/5 = 4 l /hr Rz = 20/-2 = -10 l/hr

From 4 TO 5 only one pipe X is functional = X will fill 5 l in this hour

From 5 to 6 two pipes X and Y are functional - they will fill = 4 + 5 = 9 more l in this hour making the total volume filled so far to 14 l

When the third pipe is operational => the net rate of emptying

= > 4 + 5 -10 = -1 L/HR

At the rate of -1 l/hr the three pipes together will tak 14 hours to empty the 14 litres present in the tank.

14 hours counted from 6 am should make it 8 pm

Answer = 8pm

Solution

In this case let us assume that the number of litres to fill the cistern are

LCM(4,5,2) = 20 L

Rx = 20/ 4 = 5 l/hr Ry = 20/5 = 4 l /hr Rz = 20/-2 = -10 l/hr

From 4 TO 5 only one pipe X is functional = X will fill 5 l in this hour

From 5 to 6 two pipes X and Y are functional - they will fill = 4 + 5 = 9 more l in this hour making the total volume filled so far to 14 l

When the third pipe is operational => the net rate of emptying

= > 4 + 5 -10 = -1 L/HR

At the rate of -1 l/hr the three pipes together will tak 14 hours to empty the 14 litres present in the tank.

14 hours counted from 6 am should make it 8 pm

Answer = 8pm

Posted: **Thu Feb 28, 2013 1:20 pm**

sir pls solve this q -11 once