A paint crew gets a rush order to paint 80 houses in a new

Time for painting houses at mixed rate= (t1)
80-y/1.25x + y/x = 320-4y/5x + y/x or 320-4y/5x + 5y/5x

Simplified = 320 + y/5x

Time for painting houses at constant rate = 80/x (t2)

t1/t2 = 320+y/5x * x/80 ----> 32+y/400

or 0.8 + 0.0025y

original time = 80/x weeks
new time = y/x + (80-y)/1.25x
ratio=y/x + (80-y)/1.25x : 80/x = 1.25y+80 - y:100 = 0.25y+80:100
ratio = 0.8 0.0025y
answer = B

(B) 0.8 + 0.0025y

Pretty straightforward question...

@ x houses/wk , 80 houses
time = 80/x weeks ... (1)

@ x houses/wk, y houses
time = y/x weeks ... (2)

@ 1.25x houses/wk, (80-y) houses,
time = (80-y)/1.25x weeks...(3)

Adding (2) & (3):
time = (y+320)/5x .... (4)

Ratio of (4)/(1)

= ((y+320)/5x) x (x/80)
= 0.08 + 0.0025y

Answer B.

B.

BY TAKING ARBITARARY VALUE FOR X AND Y,

LETS SAY Y = 20,
X = 4,

and then plug in number of x and y, upi get 0.85, which is the same as in B.

I am working on a similar problem with a few differences. Can anyone help?

"A painting crew painted 80 houses. They painted the first y houses at a rate of x houses per week. Then more painters arrived and everyone worked together to paint the remaining houses at a rate of 1.25x houses per week. How many weeks did it take to paint all 80 houses in terms of x and y?

The answer is (y+320)/5x

I decided to find the time it took by adding T1 and T2 (y/x and [80-y]\1.25x respectively) but I'm stuck past there. Does anyone know how I can solve this?

To answer questions such as this one, I do a chart that helps me organizing information

R * T = W
Few Painters x T(1) = y
--------------------------------------
All painters 1.25x T(2) = (80 - y)/1.25x

What is the question? The question ask you to find how many weeks did it take to paint all 80 houses in terms of x and y? So in other terms what is T(1)+T(2)?

Simply rearrange the term T(1) and T(2)
T(1) + T (2) = y/x + (80 - y)/1.25x
T(1) + T (2) = (1.25y + 80 - y)/1.25x ==> I put everything under the same denominator
T(1) + T (2) = (0.25y + 80)/1.25x
T(1) + T (2) = (0.25y*4 +80*4)/1.25x*4 ==> Multiply by 4/4
T(1) + T (2) = (y+320)/5x

Here the "hard" part is to rearrange the term so that it matches the correct answer.

Hope it helps,

Not fair, not fair!

How did we get from here:

(y/x)+(80-y)/1.25x : 80/x
.
.
.
.
.
.
To here: 0.8 + 0.0025y

I'll really appreciate a break down of the solving please, thanks!

Let y=0, implying that NONE of the 80 houses are painted at the original rate of x houses per week, with the result that ALL 80 houses are painted at the greater rate of (5/4)x houses per week.

The total time it takes them to paint all the houses under this scenario is what fraction of the time it would have taken if they had painted all the houses at their original rate?
Time and rate have a RECIPROCAL RELATIONSHIP.
Since the rate for painting all 80 houses is 5/4 the original rate, the time for painting all 80 houses is 4/5 of the original time.

The correct answer must yield 4/5 when y=0.
Only B works:
0.8 + 0.0025y = 0.8 = 4/5.

t=w/r

\(\frac{(80-y)/1.25x+y/x}{80/x}=\frac{320+y}{400}=4/5+y/400=0.8+0.0025y\)

Ans (B)

Hi,

If anyone can explain why number of painters not accounted in the calculation, it will clear my doubt.

Posted from my mobile device

Hi Ahmed9955,

There are two notable 'clues' that the actual number of painters does NOT matter. First, there's no reference/variables that refer to the exact number of painters. Second, the specific question that we're asked to answer does not ask us to calculate anything based on the exact number of painters. We're given the respective rates for how quickly the houses are painted - and the math that we have to do is based on those rates - so no additional factors (re: the number of painters) need to be accounted for.

GMAT assassins aren't born, they're made,
Rich

Answer: Option B

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Here is the breakdown :


The Numerator - (y/x)+(80-y)/1.25x / Denominator - 80/x for the overall fraction.
Solving the numerator part - y/x + (80-y)/1.25x. Take the LCM of x and 1.25 x which is 1.25x.
So the expression in numerator becomes - 1.25y + 80 - y / 1.25x. Further simplyfying - .25y + 80 / 1.25x --- equation (1)

Considering the denominator of the original expression - 80/x ---- equation (2)

So the original expression becomes - Equation 1 / Equation 2,

=> .25y + 80 / 1.25x / 80/x
=> .25y + 80 / 1.25x * x/80

Cancelling the x from the denominator and numerator , the expression becomes
=> .25y + 80 / 1.25 * 80. ----(1.25 * 80 = 100), therefore,
=> .25y + 80 / 100
=> .25y/ 100 + 80/100
=> 0.0025y + .8. - > And B

Hope this helps

Here’s a reasoning-based approach.

We have to calculate the fraction: [Total time taken under 'this' scenario] / [Time it would have taken if they had painted all the houses at the original rate]

What does ‘this scenario’ mean?
The scenario in which additional painters joined somewhere in the middle.

Numerator: Time taken to paint 80 houses such that a certain number of painters started and some additional painters joined them somewhere in the middle.
Denominator: Time taken to paint 80 houses such that the initial number of painters painted all the houses (no additional painters ever joined).

No matter when they join, as long as additional painters join in, the time taken would be less than the time taken in the situation if additional painters never joined.

i.e., Numerator < Denominator

So, the value of the fraction needs to be less than 1 no matter when the additional painters joined. i.e., for all values of y (between 1 and 79), the fraction will be less than 1.

We can use this understanding to eliminate answer choices.

If I can make the value of an answer choice greater than 1 for 0 < y < 80, the answer would be wrong. (Remember: as long as additional workers join anywhere in the middle, the time taken will be less than if additional workers never join.)

Now,
(A) 0.8(80 – y) --> For a low value of y, e.g. 10, this answer choice will result in a value greater than 1. Rejected.
(B) 0.8 + 0.0025y --> As long as y is less than 80, this fraction will remain less than 1. Looks good. Hold it.
(C) 80/y – 1.25 --> Again, if y = 10, we get a value greater than 1. Rejected.
(D) 80/1.25y --> If y = 10, we get a value greater than 1. Rejected.
(E) 80 – 0.25y --> If y = 10, we get a value greater than 1. Rejected.

Therefore, B is the answer.

1. Why did I choose y = 10? In each answer choice I noticed that in order to make the overall value greater than 1, I needed to keep the value of y small. I could have tried 1 as well. Just that I felt 10 was easier to deal with.
2. Also, I did not calculate the exact values. As soon as I realized the overall value was greater than 1, I had enough to reject the answer.


In fact, we can extend this logic.

What if additional workers came right at the beginning (i.e. at y = 0)?
In that case, the time taken to paint the houses would be 80/1.25x
The fraction would become: (80/1.25x)/(80/x) = x/1.25x = 1/1.25 = 4/5
So, we can plug in y = 0 and reject all answer choices that do not yield the value 4/5.

For y = 0,
(A) 0.8(80 – y) --> Greater than 4/5. Rejected.
(B) 0.8 + 0.0025y --> = 4/5. Hold it.
(C) 80/y – 1.25 --> 80/0?? This one is not defined. We need 4/5. Rejected.
(D) 80/1.25y --> Again not defined. We need 4/5. Rejected.
(E) 80 – 0.25y --> 80. Rejected.


By the same logic, what if additional workers never joined? i.e., all the houses were painted by the initial number of workers in both cases. In this case, y = 80.
In that case, the time taken would be the same in both cases. i.e., the fraction's value would be 1.
So, we can plug in y = 80 and reject all answer choices that do not yield the value 1.

For y = 80,
(A) 0.8(80 – y) --> 0. Rejected.
(B) 0.8 + 0.0025y --> 1. Hold it.
(C) 80/y – 1.25 --> Negative. Rejected.
(D) 80/1.25y --> Less than 1. Rejected.
(E) 80 – 0.25y --> Greater than 1. Rejected.

There are a lot of traditional ways of solution that were discussed in the forum. I had a new approach. 

Let's assume that y=80, Which means by the time they completed the 80 house then they increased the speed. So it won't impact over all timing because there is no house left. 

Hence fraction should be 1. 

So substitute y=80, We are looking for value 1... Which is possible only incase of B.­

Got the answer by just going for it and not thinking about too much. Sometimes works, sometimes doesn't.

(1) Creating fractions to use to answer the question.

They worked on y houses at rate of x houses per week, then increased to 1.25x houses per week

Work = Rate * Time

First: y = x * Time ; Time = y/x
Then: (80-y) = 1.25x * Time ; Time = (80-y) / 1.25x

Combined Time = (y/x) + [(80-y) / 1.25x] = (0.25y +800) / 1.25x

If completed all 80 houses at original rate of x houses per week, then time taken would be 80/x

(2) Setting up the final fraction for the answer.

What we require is ===> Time actually taken / Time taken if they maintained original rate

= [(0.25y +800) / 1.25x] / (80/x)

= (0.25y + 80) / 100

This matches answer B which is 0.8 + 0.0025y­