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555-605 (Medium)|   Work and Rate Problems|                     
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Working together, Rafael and Salvador can tabulate a certain set of da 04 Dec 2022, 15:08
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avigutman
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Hello guys ,
lets say time needed for Salvador=x and time needed for Rafael=x-3 so
1/x+1/x-3=1/2 solving this I get x^2-7x+6=0 so x could be 1 or 6 and we reject 1 because 1-3=negative number.
If I had chosen to set Rafael's time x and Salvador's x+3 i would end up with a quadratic that would give me one answer choice (with the first approach I almost lost the question because i thought that I had 2 valid options for A) . What I want to ask is the following, is there a way to choose beforehand the option that will require the least additional steps at least in this type of questions?
ThatDudeKnows avigutman I have the same doubt. MartyTargetTestPrep
gmexamtaker1 and Elite097 no, I can't think of a way to know ahead of time whether it's better to go with x & (x+3) or (x-3) & x.
Having said that, in my humble opinion we shouldn't put pen to paper at all for this question. In a 2-minute-per-question kind of test, this question should be solved conceptually. We know how long the job takes them together, and with statement (1) on its own, if we were to draw a number line, we could plot the amounts of time each would take on his own as two tick marks with 3 hours in between. The question is, essentially, can those two tick marks be moved up or down the number line without changing the amount of time they take to do the job together (since that's already given)? The answer is no: if we push the tick marks to the right, the job is going to take them longer if they were to work together, and vice versa if we push the tick marks to the left. Therefore, knowing the delta in their individual times is sufficient to answer the question.
I would evaluate statement (2) conceptually as well. If you find yourself building a quadratic equation on the GMAT, it's likely that you're doing something inappropriate.
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avigutman

Hi my friend! I hope that you had a great weekend.

I see why your approach makes the most sense. However, if I am not able to think like this in the moment on an official test, I would be so appreciative for your thoughts on my algebra question below:

First step:
I did (1/R + 1/S)*2=1 --> then did 1/R + 1/S = 1/2

Second step:
I got lost when I introduced the "T" for time when solving individually...
So, for Rafael: (1/R)*T=1
And for Salvador: (1/S)*(T+3)=1

I was then really struggling to see how I could solve with three variables. I see that another expert simplified the number of variables, but I was not able to follow that approach. Counting on the quant reasoning whiz to help me here :)
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Working together, Rafael and Salvador can tabulate a certain set of da 05 Dec 2022, 05:10
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woohoo921
avigutman
I see why your approach makes the most sense. However, if I am not able to think like this in the moment on an official test, I would be so appreciative for your thoughts on my algebra question below:

First step:
I did (1/R + 1/S)*2=1 --> then did 1/R + 1/S = 1/2

Second step:
I got lost when I introduced the "T" for time when solving individually...
So, for Rafael: (1/R)*T=1
And for Salvador: (1/S)*(T+3)=1

I was then really struggling to see how I could solve with three variables. I see that another expert simplified the number of variables, but I was not able to follow that approach. Counting on the quant reasoning whiz to help me here :)
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Sorry, woohoo921. I want to help you, and I feel strongly that helping you in this case means not helping you. If you're not able to reason your way through this in the moment (which is totally understandable by the way), the way to maximize your GMAT score is NOT to fall back on algebra. It's to guess and move on.
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avigutman
We know how long the job takes them together, and with statement (1) on its own, if we were to draw a number line, we could plot the amounts of time each would take on his own as two tick marks with 3 hours in between. The question is, essentially, can those two tick marks be moved up or down the number line without changing the amount of time they take to do the job together (since that's already given)? The answer is no: if we push the tick marks to the right, the job is going to take them longer if they were to work together, and vice versa if we push the tick marks to the left. Therefore, knowing the delta in their individual times is sufficient to answer the question.
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Hi avigutman – 2 theories BEFORE I ask my question

Quote:

If A – B = 1 and B x A = 12
(A, B) is either (4,3) or (-3,-4) ONLY
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ii)

Quote:

If A + B = 7 and B x A = 12
Hence (A, B) is either (3,4) or (4,3) ONLY
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So thats the theory.

That being said - I don’t think you can do the yellow highlight however BECAUSE the combined time is the

Product of the individual times = 2
-------------------------------------------------------------------------
Sum of the individual times = 1


Two things are in play: Product and Sum

The biggest issue is that these two things are being Divided

So the ratio could be (6/3) or (5/2.5) or (100/50) or (40/20)

Hence - there are multiple variations.
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Working together, Rafael and Salvador can tabulate a certain set of da 05 Dec 2022, 18:59
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jabhatta2
avigutman
We know how long the job takes them together, and with statement (1) on its own, if we were to draw a number line, we could plot the amounts of time each would take on his own as two tick marks with 3 hours in between. The question is, essentially, can those two tick marks be moved up or down the number line without changing the amount of time they take to do the job together (since that's already given)? The answer is no: if we push the tick marks to the right, the job is going to take them longer if they were to work together, and vice versa if we push the tick marks to the left. Therefore, knowing the delta in their individual times is sufficient to answer the question.

Hi avigutman – 2 theories BEFORE I ask my question

Quote:

If A – B = 1 and B x A = 12
(A, B) is either (4,3) or (-3,-4) ONLY

ii)

Quote:

If A + B = 7 and B x A = 12
Hence (A, B) is either (3,4) or (4,3) ONLY

So thats the theory.
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So far, so good, jabhatta2.
jabhatta2
That being said - I don’t think you can do the yellow highlight however BECAUSE
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You're about to explain your objection to the yellow highlight, but you never actually engage with my argument in the yellow highlight. You don't explain how two tick marks can be moved up or down the number line without changing the amount of time they take to do the job together.
jabhatta2
combined time is the

Product of the individual times = 2
-------------------------------------------------------------------------
Sum of the individual times = 1


Two things are in play: Product and Sum

The biggest issue is that these two things are being Divided

So the ratio could be (6/3) or (5/2.5) or (100/50) or (40/20)

Hence - there are multiple variations.
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You're ignoring one important piece of information that we have: the difference between them is 3.
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Working together, Rafael and Salvador can tabulate a certain set of da 05 Dec 2022, 20:29
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avigutman
You're about to explain your objection to the yellow highlight, but you never actually engage with my argument in the yellow highlight. You don't explain how two tick marks can be moved up or down the number line without changing the amount of time they take to do the job together.

You're ignoring one important piece of information that we have: the difference between them is 3.
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Hi avigutman – I think my primary issue with the yellow highlight above is that there is division going on behind the scenes (Product of Individual Times / Sum of Individual Times) when combined times are being calculated.

Now - In order to keep a division constant (in this case - 2 /1 ) - Numerators and Denominators have to change by the same (% change)

Example#1
(2/1) vs (4/2) are the same
Reason - Numerator increased by 100 % (From 2 to 4) and the denominator increased by 100 % (from 1 to 2)

Example#2
(6/3) vs (7/3.5) are the same
Reason - Numerator increased by 116.67 % (from 6 to 7) and the denominator increased by 116.67 % (from 3 to 3.5)

----------------------------------------

Now if there are 2 tick marks on the number line (3 hours and 6 hour) and we move each tick mark by 4 units each

So the new tick marks are 7 hours and 10 hours

- the New Sum will change
- the New Product will change

Whats to say the Numerator and the Denominator wont increase by the same (% change) ?

Perhaps the % change is the same (for the Numerator vs for the Denominator)

How can you say that the % change is going to different (in the Numerator vs in the Denominator) -- everytime you shift the two tick marks (3 hours and 6 hours) by 4 units each or 5 units each or 6 units each.

At-least thats what is nagging me

Thank you
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Working together, Rafael and Salvador can tabulate a certain set of da 06 Dec 2022, 18:13
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jabhatta2
In order to keep a division constant (in this case - 2 /1 ) - Numerators and Denominators have to change by the same (% change)

Example#1
(2/1) vs (4/2) are the same
Reason - Numerator increased by 100 % (From 2 to 4) and the denominator increased by 100 % (from 1 to 2)

Example#2
(6/3) vs (7/3.5) are the same
Reason - Numerator increased by 116.67 % (from 6 to 7) and the denominator increased by 116.67 % (from 3 to 3.5)
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Careful, jabhatta2, the percent increase in example#2 is 16.67%, not 116.67%.
jabhatta2
Whats to say the Numerator and the Denominator wont increase by the same (% change) ?
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Well, the yellow highlight proves it, but I'm sensing that you're not satisfied with that (you should reflect on why that is - it's a perfectly valid proof). I've put together an algebraic proof for you as well, in the attached image. You'll see that I'm comparing change factors (interchangeable with percent change of course). On the left side I have the change factor for the numerator (the product of the times ab) when moving those times up the number line by k units, and on the right side I have the change factor for the denominator (the sum of the times a+b when moving those times up the number line by k units). My objective was to prove that the left hand side is greater than the right hand side for all positive values of a, b, and k. From line 3 to line 4 I cross multiply (I'm allowed to since we're dealing with positive numbers). In line 5 I'm canceling out (subtracting from both sides) the green circles. In the last line we have a positive left hand side vs. zero on the right hand side.
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Thank you so much Avi - that really helped !

Based on the above - % change in the Numerator vs % change in the Denominator are always going to be different

Hence the division of 2/1 will change

That being said, I am sure you didnt think about this specific algebraic solution when solving (S1)

When doing (s1) - was your logic more HIGH LEVEL? I think this might be more high level logic

Quote:
(A) If JD takes 6 hours to finish a project whereas Avi takes 3 hours to finsh a project -- both of them working together will take 2 hours.

Now
(A-1) If JD takes 7 hours to finish a project whereas Avi takes takes 3 hours to finish a project --- both of them working together should take MORE THAN 2 hours.
(A-2) If JD takes 6 hours and Avi takes 4 hours ---- both of them working together should take MORE THAN 2 hours.
(A-3) If JD takes 7 hours and Avi takes 4 hours ---- both of them working together should take MORE THAN 2 hours.
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Working together, Rafael and Salvador can tabulate a certain set of da 07 Dec 2022, 12:00
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jabhatta2
Thank you so much Avi - that really helped !

Based on the above - % change in the Numerator vs % change in the Denominator are always going to be different

Hence the division of 2/1 will change

That being said, I am sure you didnt think about this specific algebraic solution when solving (S1)

When doing (s1) - was your logic more HIGH LEVEL? I think this might be more high level logic

Quote:
(A) If JD takes 6 hours to finish a project whereas Avi takes 3 hours to finsh a project -- both of them working together will take 2 hours.

Now
(A-1) If JD takes 7 hours to finish a project whereas Avi takes takes 3 hours to finish a project --- both of them working together should take MORE THAN 2 hours.
(A-2) If JD takes 6 hours and Avi takes 4 hours ---- both of them working together should take MORE THAN 2 hours.
(A-3) If JD takes 7 hours and Avi takes 4 hours ---- both of them working together should take MORE THAN 2 hours.
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Yes, exactly. That’s what I tried to convey in the yellow highlight, jabhatta2.

Posted from my mobile device
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Working together, Rafael and Salvador can tabulate a certain set of da 18 Aug 2023, 23:58
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Hi, please correct my approach, I solved it as follows :

Taking r and s as the rates of Rafael and Salvador,

W= 2(s+r)

From St1,

W/s - W/r = 3

2(s+r)/s - 2(s+r)/r = 3
2(s+r)(r-s)/rs = 3
2(r^2-s^2)/rs = 3

2r^2 - 2s^2 - 3sr =0

This quadritic equation seems unsolvable. Please help!
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Working together, Rafael and Salvador can tabulate a certain set of da 22 Jul 2024, 07:22
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A way to make the question slightly more practical is to assign a number of units produced. Since we have the numbers 0.5,3 and 6 in the question and the two statements I picked a multiple of them or 12.

The question states that Rafael (R) and Salvador (S) can make 12 units in 2 hours, which means their combined work rate per hour is 6 units.

Statement 1 says that R can produce the 12 units working alone 3h faster than S working alone. From this we can find that R will have a work rate of 4 units per hour and will complete the job in 3h, and that S will have a work rate of 2 units per hour and will complete the job in 6h.

Statement 2 says that R working alone can produce the 12 units in half of the time S can working alone. We would reach the same conclusion as above.
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Working together, Rafael and Salvador can tabulate a certain set of da 19 Mar 2025, 08:56
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Sharing my two cents on this question regarding how to close this question in time limit. A lot of experts already mentioned that we can intuitively know from Statement (1) that it is sufficient without actually forming the equations.

Let's say you do go one step beyond to form an equation, and you see it's a quadratic equation. One of the following are possible:
1. It has two positive real roots
2. It has two unreal roots
3. It has two real roots, one positive and one negative

We need to understand given the problem, how we can eliminate 1 and 2 without the need to solve the entire quadratic equation.

On Data Sufficiency, you don't need the exact numbers. You just need to know whether a unique solution exists.
If:
- You get a quadratic equation,
- And the equation represents the relationship between times to complete a job, you're looking for positive solutions since time cannot be negative.

The quadratic equation comes from realistic work rates, where S must be greater than 3 because R equals S minus 3, and R must be positive. This eliminates negative or zero solutions for S.

If there are two real roots, usually only one of them will make sense in the context we are considering (for instance, S must be greater than 3). If there is only one real root, that is still acceptable because it gives us a unique value for S, which leads to a unique value for R.

The work rate formula combined with a straightforward relationship between R and S always leads to a solvable quadratic equation that produces one valid solution, provided that:
- The relationship is simple and linear,
- The values represent time and are positive.

One quadratic equation + clear linear relationship + positive real-world quantity → Sufficient (Unique Positive Solution)

How Do You Know Roots Will Be Real and Positive in This Context?

- Rates and times are modelled by real-life contexts, using linear or rational equations derived from those contexts.
- The quadratic equations that arise from work-rate or age problems are based on rational relationships that must yield real, positive values to make sense in the real world.

If the quadratic leads to no real solution that would mean:

- Rafael and Salvador couldn't possibly be working at rates that add up to the combined rate already given.

This would contradict the problem setup, which clearly states they together complete the work in 2 hours.

So here, knowing the linear relationship between the two rates and seeing a quadratic equation, we can safely conclude there will be a single solution that's relevant to the problem without the need to actually solve the equation.
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you can assume r as x and take s as a constant, then solve

jayakalra
Hi, please correct my approach, I solved it as follows :

Taking r and s as the rates of Rafael and Salvador,

W= 2(s+r)

From St1,

W/s - W/r = 3

2(s+r)/s - 2(s+r)/r = 3
2(s+r)(r-s)/rs = 3
2(r^2-s^2)/rs = 3

2r^2 - 2s^2 - 3sr =0

This quadritic equation seems unsolvable. Please help!
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