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# Solving Work Problems

Solving Work Problems

Algebraic equations have sought wide applications in solving various word problems . One of its common applications are done in solving work problems. In this article, I have selected four work problems presenting with their solutions.

Problem Number One :

A farmer can plow a field in 5 days using a tractor. A hired man can plow the same field in 7 days using a smaller tractor. How many days will it take for both of them to plow the same field together?

Solution :

Let X = number of days it will take for both the farmer and the hired man to plow the field together.

1/5 = the part plowed in one day by the farmer

1/7 = the part plowed in one day by the hired man

Working equartion: (1/5 + 1/7) X = 1 1 stands for one complete job

. (X/5 + X/7 = 1 ) 35

7X + 5X = 35

(12X = 35)1/12

X = 35/12 or 2 and 11/12 days

It will take 2 and 11/12 days for both of them to plow the field together.

Problem Number Two : If in the previous problem , the hired man worked one day with the smaller machine and joined by the farmer with the larger one, how many more days it will take both of them to finish plowing?

Solution : Since the hired man plowed one-seventh of the field in a day, six-sevenths (6/7) remained unplowed.

Let X = number of days it will take the two to finish the job

X/5 = the part plowed by the farmer

X/7 = the part plowed by the hired man

We now have the working equation, X/5 + X/7 = 6/7

( X/5 + X/7 = 6/7 ) 35

7X + 5X = 30

(12X = 30) 1/12 X = 2and ½

The farmer and the hired man have still to work for 2 and ½ days to finish plowing the field.

Problem Number Three : Arthur can dig a tunnel 5 days longer than Michael can dig one. If they both work together, they can dig a tunnel in 8 days. How fast can Michael dig a tunnel ?

Let X = number of days it will take Mihcael to dig one tunnel

X + 5 = number of days it will take Arthur to dig a tunnel

Working equation: ( 1/X + 1/(X+5) ) 8 = 1

(8/X + 8/(X+5) = 1 ) (X) (X + 5)

8(X + 5) + 8X = X^2 + 5X

8X + 40 + 8X = X ^ 2 + 5X

16X + 40 = X^2 + 5X

X^2 = 11X - 40 = 0 Using quadratic formula to solve for X

X = ( 11 + SQRT(121 + 4(40) )) div 2 X = (11 + 16.76)/2 = 13.88 days It will take Michael 13.88 days to dig a tunnel.

Problem Number Four : A swimming pool can be filled in 6 hours and requires 9 hours to drain. If the drain was accidentally left open for 6 hours while the pool was being filled, how long did filling the pool require ?

1/6 = part of the pool filled after one hour

1/9 = part of the pool drained after an hour

Working equation : (1/6 - 1/9) 6 + X/6 = 1

After six hours 1/3 of the pool was filled : (1/6 - 1/9) 6 = 1/3

2/3 of the pool is still to be filled after 6 hours

(X/6 = 2/3)6 = X = 4 hours 6 + 4 = 10 hours It takes 10 hours for the pool to be filled.

## Comments

tnk u ,,now i know:)

BOBO KA

salamat ani kahibalo nako ani

arigatou thank you nnahimo ko homework ko

salamat sa iu nagawa ko assignment ko

thanks poh

thnx po... nagawa ko ang aking project...

thnx poh...marmi akong natutunnan sa iyo...

tama si owais ....

thank you .po

thank you po!

..it is very useful for students like me...it doesn't take hard to understand,,and each problem is different from one another..thank you. :)

thank you now i know about it

i think owais you are wrong he did it right

i think problem number 4 has some problem here...y is this x/6 added in the working equation....if 1/3 of the pool is filled in 6 hours it doesn't mean it stopped draining....for them remaining 2/3 part you also have to include the draining timing....i suppose it can be solved easily just with dis equation (1/6-1/9)X=1 for the value of x....you get 18...and that sounds realistic....

It is doing and so fun! It took only one hour for me to understand.I got it!

THANK YOU

It is doing and so fun! It took only one hour for me to understand.I got it!

THANK YOU

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