Mr. Grayson wanted to have a boundary wall around his garden. He hired

Question

Mr. Grayson wanted to have a boundary wall around his garden. He hired three laborers of the same efficiency. The three worked together for 30 days, but during their work, they all did not come for a few days as they had to attend festive occasions. One of them did not come for 10 days more than the second laborer, and the third laborer did one-third of the total work. How many more days did the first laborer not come compared to the third laborer?

A. 8
B. 7
C. 6
D. 5
E. 4

A. 8
B. 7
C. 6
D. 5
E. 4

jainendradubey · Answer

so they all have to do 1/3 of the work, 3 people are working on it, together they all worked for 30 days, so if we take 30 as the least and plus 10 as the most then we get 35 as the day the 3rd guy worked, 30 as the 1st guy and 40 as the second guy, making them all not for certain days if 5 days the 3rd guy worked was one where 2nd guy didn't.

culpatenetur · Answer

Reply to above 
 Question states 'but during their work, they  all  did not come for a few days as they had to attend festive occasions' so the 3rd guy could not have worked everyday

jainendradubey · Answer

Let me know if my edited answer makes sense. Honestly I believe the question is of a poor quality as i had to merge Venn diagram concept just to make the answer work.

stne · Answer

Third labourer did  \frac{1}{3}  of the total work, means he came for  10  days because  \frac{10}{30} = \frac{1}{3}

Let second labourer not come for  x  days, means he came for  30 -x  days, so work done by him  =\frac{30-x} {30}

Also it is given that first Labourer did not come for  x+10   days , therefore he came for  30 -(x+10)  days:
Work done by him   = \frac{20-x}{30}

Total work :  \hspace{5mm}  \frac{20-x}{30} + \frac{30-x} {30} + \frac{1}{3} = 1

x= 15

Thus first labourer did not come for  25  days and third labourer did not come for  20  days.

25-20 = 5

Ans D

Hope it helped

jainendradubey · Answer

If you and I are working together for 30 days, you'd better sit right by my side for 30 days, or else I won't count it as working together.

stne · Answer

In that case, you can mark the solution as incorrect. Moderator will be happy to archive or delete it.

gmatshay · Answer

Here's how I did it.

Let's assume they each worked  a ,  b , and  c  days.

From the question, we know that  b = a+10  (since the first laborer worked 10 days more than the second)

Let us assume that the rate is  x

The third laborer did  \frac{1}{3}rd  of the work (let's call it  w ) in  c  days

Therefore,  cx = \frac{w}{3}

The other two must have done the rest of the work, so

ax + bx = \frac{2w}{3}

We only care about the relation between the first and the third labourer, so let's substitute the value of  b  we got earlier

ax + (a+10)x = \frac{2w}{3} = 2cx

we can get rid of  x

2a + 10 = 2c

a + 5 = c

And there we have it! The first laborer worked 5 more days than the third. I didn't really use the 30 days piece of information, it wasn't needed to answer the question.

pal26 · Answer

Here's how I solved it.
Let us name the three laborers as L1, L2 and L3 and assume their rate as m (They work at same efficiency).

Let's L2 was absent for x days, therefore L2 worked for 30-x days.
So L1 was absent for x+10 days (Given), hence L1 worked for {30-(x+10)} =20-x days
Also, let us assume that L3 was absent for y days. So L3 worked for 30-y days.
The question asked us to find how many more days L1 was more absent than L3. So, to find  .....(1)

As given in the question, L3 worked one-third of the total work, so L1 and L2 together completed the 2/3 of the work.

So, 1/3=m*(30-y)......(2)

and 2/3=(20-x)*m+(30-x)*m=(50-2x)*m.....(3)

Dividing (2) by (3),

1/2=(30-y)/(50-2x)

Solving, we get,x-y=-5 and putting this in (1), we get x-y+10=-5+10=5 (Ans- D)

Mr. Grayson wanted to have a boundary wall around his garden. He hired

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Mr. Grayson wanted to have a boundary wall around his garden. He hired

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