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Basic Time & Work problems

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Basic Time & Work problems [#permalink] New post 23 Oct 2010, 08:42
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

80% (02:03) correct 20% (03:12) wrong based on 4 sessions
Hi,

I was hoping to get the correct approach for the following types of basic problems in Time & Work.

Type I - Extra number of men required/employed

1) An engineer undertakes a project to build a road 15 km long in 300 days with 45 men employed. After 100 days, only 2.5 km of the road has been completed. Find the (approx.) number of extra men needed to complete the work in time.

a) 43
b) 45
c) 55
d) 68
e) 60

OA -
[Reveal] Spoiler:
D


Type II - when amount paid is included

1) A can do a piece of work in 12 days. A and B complete the work together and were paid Rs. 54 and Rs. 81 respectively. How many days must they have taken to complete the work together?

a) 4 days
b) 4.5 days
c) 4.8 days
d) 5 days
e) 4.2 days

OA -
[Reveal] Spoiler:
C


Type 3 - men leave halfway through work

1) In a fort, there was sufficient food for 200 soldiers for 31 days. After 27 days 120 soldiers left the fort. For how many extra days will the rest of the food last for the remaining soldiers?

a) 12 days
b) 10 days
c) 8 days
d) 6 days

OA -
[Reveal] Spoiler:
D


Any pointers would be helpful.
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Re: Basic Time & Work problems [#permalink] New post 23 Oct 2010, 16:44
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Type 3 - men leave halfway through work

1) In a fort, there was sufficient food for 200 soldiers for 31 days. After 27 days 120 soldiers left the fort. For how many extra days will the rest of the food last for the remaining soldiers?

a) 12 days
b) 10 days
c) 8 days
d) 6 days






Food Availability
MenDaysQuantity
Given200316200
Till 27 days200275400
From 28th day80x800
[/quote]

Hence the number of days the remaining quantity will last = 10 days. However the original quantity was to last for only 4 more days if none of the soldiers had left.

Hence extra days is 10-4 = 6 days (D).
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Last edited by ezhilkumarank on 23 Oct 2010, 16:57, edited 1 time in total.
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Re: Basic Time & Work problems [#permalink] New post 23 Oct 2010, 16:56
Type II - when amount paid is included

1) A can do a piece of work in 12 days. A and B complete the work together and were paid Rs. 54 and Rs. 81 respectively. How many days must they have taken to complete the work together?

a) 4 days
b) 4.5 days
c) 4.8 days
d) 5 days
e) 4.2 days

Given A can do the work in 12 days. Hence A's rate is 1/12.

Also amount payment made to A and B is Rs 54 and Rs 81 respectively -- which is the ratio of 2:3.

Viewing only A's data: Since payment is proportional to the amount of work done.

Fraction of the total work done by A * Days in which A would have completed the work alone.

Hence (2/5) * 12 = 4.8 days. Hence A worked for 4.8 days @ a rate of (1/12) along with B to complete the work.

Answer C (4.8).
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Re: Basic Time & Work problems [#permalink] New post 23 Oct 2010, 17:13
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Type I:

To complete 2.5 km in 100 days, we need 45 men
To complete 12.5 km (leftover road) in 100 days, we need 45 x 12.5/2.5 men (It will be 12.5/2.5 and not 2.5/12.5 because more work will need more men. So we have to make 45 greater. Direct variation.) = 225 men

To complete 12.5 km in 100 days we need 225 men.
To complete 12.5 km in 200 days (this is the number of leftover days), we need 225 x 100/200 (Now if we have more days for same work, we need less men so 100/200 to make 225 smaller. Inverse Variation) = 112.5 men i.e. 113 men (112 men cannot finish the work. Even if this was 112.3, I would have made it 113)

We already have 45 men, we need 68 more.

Note: I could have done the above in a single step by saying 45 x 12.5/2.5 x 100/200 = 112.5
I used two steps to clarify.

Type II:
If we are given how much they were paid, we know the ratio of their rate of work. It will be the ratio in which they were paid. This is so because if you are twice as fast as me, you will be paid twice as much as me.
Ratio of rate of work = 54:81 = 2:3
Then, ratio of time taken will be 3:2 (Ratio of rate of work is inverse of ratio of time taken. If this is unclear, just check out work theory)
Then, B can do the work in 8 days.
Together, they will take 8 x 12/ (8 + 12) = 4.8 days

Type III:
Let me say that the amount of food one soldier consumes in one day is 1 soldier-day food.

Total amount of food was 200 x 31 soldier-days = 6200 soldier-days
In 27 days, 200 soldiers consumed the food which is equal to 27 x 200 = 5400 soldier-days.
Left over food = 800 soldier-days
Now 800 soldier-days food will last 80 soldiers 10 days. Since 27 days are already over, it would have lasted only 4 more days but now it is going to last 10 days so it will last an extra 6 days.
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Re: Basic Time & Work problems [#permalink] New post 23 Oct 2010, 17:26
Type I - Extra number of men required/employed

1) An engineer undertakes a project to build a road 15 km long in 300 days with 45 men employed. After 100 days, only 2.5 km of the road has been completed. Find the (approx.) number of extra men needed to complete the work in time.

a) 43
b) 45
c) 55
d) 68
e) 60

Answer:



Work and Rate
RateMenDaysWork completed in meters
First 100 daysx451002.5*1000



We can find out the actual rate (x) at which the men are working.
=> x*45*100 = 2500
=> x = 5/9.

Now we will use the actual rate at which the men are working to find out the extra men required to complete the work in the remaining 200 days. [We should only use the actual rate at which the men are working not the rate which they were expected to work.]




Work and Rate
RateMenDaysWork completed in meters
First 100 days5/9451002.5*1000
Next 100 days5/9x20012.5*1000


(5/9)*x*200 = 12500
=> x = 112.5 ~ 113. So the number of extra men required to complete the work in the remaining 200 days is 113 - 45 = 68.

Answer
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Re: Basic Time & Work problems [#permalink] New post 23 Oct 2010, 21:05
Great! Thank you for the help ezhilkumarank and VeritasPrepKarishma. For Time & Work, i tend to directly solve problems using the one days work approach. And for the above problems that approach did not get me the right answer.
Thank you for the help!
Re: Basic Time & Work problems   [#permalink] 23 Oct 2010, 21:05
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