The figure above represents the floor plan of an art gallery that has

Question

https://gmatclub.com/forum/download/file.php?id=44840 The figure above represents the floor plan of an art gallery that has a lobby and 18 rooms. If Lisa goes from the lobby into room A at the same time that Paul goes from the lobby into room R, and each goes through all of the rooms in succession, entering by one door and exiting by the other, which room will they be in at the same time ? (1) Lisa spends 2x minutes in each room and Paul spends 3x minutes in each room. (2) Lisa spends 10 minutes less time in each room than Paul does. floor.jpg

DangerPenguin · Accepted Answer

statement 1: sufficient
I drew a sketch on a piece of paper and then kept adding 2 minutes for each room lisa was and did the same for Paul and then saw that they would meet in room K because Lisa would be arriving there in 20 minutes and Paul would be there in 21 minutes

statement 2: insufficient
If Lisa spends 2 minutes in each room and Paul spends 12 minutes in each room, they meet in room P. If Lisa spends 14 minutes in each room and Paul spends 24 minutes in each room, they meet in room L.

Answer A

fireagablast · Answer

I think bottom line is that this is a ratio problem.

(1) ratio is 2x:3x, no matter what the ratio stays 2:3 for any value of x
(2) ratio is x-10:x, and the ratio changes significantly per step
11-10:10 --> 1:10
12-10:12 -->2:12=1:6
3:13
4:14 -->2:7

etc

This means that A should be the answer, even though I went with B specifically because it seemed to make sense generally but when you're dealing with a fixed paths and rooms it changes the result.

beattheheat · Answer

Statement 1: Lisa spends 2x minutes in each room and Paul spends 3x minutes in each room.
We can deduct that by the time Lisa passes from 3 rooms, Paul can pass through only 2 rooms.
With this information we can identify the room where they will meet - Sufficient

Statement 2: Lisa spends 10 minutes less time in each room than Paul does.
We don't have enough information regarding the amount of time Lisa or Paul spends in each room - Insufficient

Answer A

GMATBusters · Answer

Statement 1: Lisa spends 2x minutes in each room and Paul spends 3x minutes in each room.
We can deduct that by the time Lisa passes from 3 rooms, Paul can pass through only 2 rooms.
This way we get a  movement pattern , hence we can find the room in which they meet
Sufficient

Statement 2: Lisa spends 10 minutes less time in each room than Paul does.
lets paul takes 200 min, Lisa takes 190 min
or, paul takes 20 min and Lisa takes 10 min.
hence the movement pattern in both cases would be different.
NOT Sufficient

avigutman · Answer

Apply  Bunuel  's solution from here to see why statement 1 is sufficient on its own:
 https://gmatclub.com/forum/two-trains-x-and-y-started-simultaneously-from-opposite-ends-of-a-168061.html#p1337527

Bambi2021 · Answer

The most important condition to be sure about here is whether there is only one way to walk through all the rooms. And it seems to be the case that not one of the rooms is connected to more than one other room. Therefore we can safely conclude that this art gallery would not have passed the fire precautions.

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Posted from my mobile device

avigutman · Answer

Video solution from Quant Reasoning starts at 7:49 Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1 https://www.youtube.com/watch?v=KTVdGztdDrw

GianKR · Answer

Lisa and Paul are walking towards each other, shrinking together a total distance of 18 rooms.

The moment they meet they would have spent exactly the same time walking towards each other at their respective speeds.

Since the time is the same, we know that the speed and the distance will vary directly. Therefore: SL/SP= DL/DP (S is the speed and D is the distance; in this case the number of rooms).

Since we know the total distance that is 18 rooms,  if we know the ratio of lisa's distance to Paul's distance (DL/DP) or the ratio of Lisa's speed to Paul's speed (SL/SP), we will be able to answer this question.

Statement 1  tells us that 2xDL = 3xDP (time spent on each room multiplied by the number of rooms for Lisa and Paul are equal because they spent the same time moving towards each other when they met).
Hence, DL/DP=3x/2x=3/2.
We can stop here and state that this statement is sufficient without having to make any calculations.
But for the sake of clarity, let's find out where they exactly meet:

DL : DP : Total
3 : 2 : 5 
Since the actual total is 18, using the unknown multiplier, we get:
DL : DP : Total
10,8 : 7,2 : 18

They will therefore meet in room K .  Statement 1 is sufficient.

An alternative method to solve this is to use algebra by solving the following system of two equations, to find out the values of DL and DP: 
2xDL=3xDp ---> 2DL=3DP (dividing by x)
DL+DP=18
After solving, you will get DP=7,2 and DL= 10,8.

Statement 2  tells us that Lisa spends 10 minutes less time in each room than Paul does. That tells us that Lisa is faster than Paul, so we know that they are going to meet somewhere between J and R but we don't know where exactly. 
For different values of the time spent in each room by Lisa and Paul, we can get different values for the ratio of their distances, and therefore different answers to where they meet. Hence  statement 2 is not sufficient.

Correct answer is A

AK1995 · Answer

There is only one route from room A to room R and vice versa so IMO, the best way to approach this question is to think of it as two buses (or cars), one leaving from city A to city B and the other leaving from city B to city A and find out where they meet in the middle.

We know the total distance between the two. We know that the time taken by both of them is going to be the same.

Hence, to understand how much distance each of them would cover before meeting, we need the ratio of the ditances they covered, which means we need the ratio of their speeds (since they both travel for the same time).

Statement 1 gives that while Statement 2 doesn't. Hence option A.

Palpable · Answer

Total number of rooms =18

Time spent per room
Lisa = 2x
Paul = 3x

Rooms covered
Lisa =y 
Paul=18-y

Same time passed when they meet so it can be equated 
2x *y= 3x *(18-y)

X cancelled, single variable equation. Y is solvable

A sufficient

Time spent in each room
Lisa = x
Paul = x-10

Same equation again
Xy= (x-10)*(18-y)

Two variables one equation
Insufficient

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sayan640 · Answer

MartyMurray   Bunuel   GMATNinja   KarishmaB  , Can you please provide a solution to this question ?

KarishmaB · Answer

It is a relative speed problem in a different context, that is all. You don't need to solve it once you can picture it as a relative speed problem. You would just know that you can solve it using statement 1 alone.

If I give you the following, will you be able to solve it?

Lisa and Paul start from 2 opposite ends of a path. The length of the path is 18 metres. 
Lisa is covering 1 metre per 2x mins and Paul is covering 1 metre per 3x mins. After how many metres will they meet?

Can you solve it? Sure. Do you need to, no. But if you were required to, this is how you would:

Time taken to meet =  \frac{18}{Relative Speed} = \frac{18}{(1/2x + 1/3x)} = \frac{18*6x}{5}

Distance covered by Lisa in this time = \frac{18*6x}{5} * \frac{1}{2x} = 10.8 meters  
So Lisa has covered 10 complete metres and is now covering the 11th metre when she meets Paul.

or simply use Ratio of their speeds is 3:2 and carry on from there. You can avoid using x completely then.

In our original question, this is exactly the data we have using statement 1 alone. Each metre is just each room.  
She is in the 11th room which is K and hence statement 1 alone is sufficient. 
Statement 2 alone does not give us the ratio of their speeds or time taken, just the difference in the time taken so we cannot solve using this data.

Answer (A)

It is similar to other such problems that also use relative speed in different contexts. For example,
Two jars - one full of water with a drain and other empty with a faucet. When will both have the same level of water?

Adit_ · Answer

I dont really understand what the problem conveys. What does "which room will they be in at the same time" convey? So given the information should I be able to tell that at a particular time they would be in room x for one and room y for another, or should I assume both of them land in the same room in that time frame or what does it actually mean? Considering that I need to predict which room they will be in lets say they take 2 mins and 3 mins respectively and after 3 mins they will be in some rooms. But if they take 4 mins and 6 mins respectively, after the same 3 mins they will be in their original rooms right? I am confused. Kindly help me out.

Edit: I think the answers all seem to indicate them meeting in the same room at some particular time. But "which room will they be in at the same time" could also mean that I am supposed to predict that at some time x that they are always in (a,b) room respectively. Which is why I thought C is right because you would need both the exact values to know where they exactly are at any exact same time.

Usernamevisible · Answer

Scaling does not change the structure

---

Core idea

If every moving part is multiplied by the same thing:
the pattern stays the same
only the size/time changes

---

Example

Suppose:
Lisa = 2 min/room
Paul = 3 min/room

Ratio = 2:3

Now multiply both by 1000:
Lisa = 2000 min/room
Paul = 3000 min/room

Ratio is still 2:3

So relative movement is unchanged.

---

What changes vs what stays fixed

Changes:
actual clock time
total duration
minutes spent

Stays fixed:
relative speed
meeting location
order of movement

---

Why the meeting room stays same

Every action stretches proportionally.

Like slow-motion playback:
people still collide at same spot
events still happen in same order
only timing changes

---

GMAT translation

Whenever you see:
ax and bx
or quantities sharing same variable factor

Ask:
Does the variable cancel?

If yes:
actual value may not matter
only ratio matters

Example:
2x / 3x = 2/3

x disappears.

---

Case 1: Same multiplier → structure fixed

Examples:
2x and 3x
5k and 7k
x/2 and x/3

Ratios fixed.
Usually sufficient.

---

Case 2: Additive difference → structure changes

Examples:
x+2 and x+100
2x+1 and 2x+7

Ratios change depending on x.

Example:
1:101
vs
1000:1100

Very different behavior.
Usually insufficient.

---

Why statement (2) failed

“Lisa spends 10 fewer minutes than Paul”

L = P - 10

Possible cases:

Lisa 5, Paul 15 → ratio 1:3
Lisa 20, Paul 30 → ratio 2:3
Lisa 90, Paul 100 → ratio 9:10

Completely different movement structures.
So meeting room changes.

---

GMAT heuristic

When you see variables, ask:

Does the variable scale both quantities equally?

If yes:
ax / bx = a / b

Variable disappears.

Strong sign of DS sufficiency.

---

Common GMAT areas where this appears

rates/speeds
work problems
mixture ratios
geometry similarity
probability ratios
depreciation/growth
compounded comparisons

---

Fast DS recognition

Usually sufficient:
3x and 5x
7k and 2k
x/2 and x/3

because ratio fixed

Usually not sufficient:
x+2 and x+9
2x+1 and 2x+7

because ratio changes with x

---

Intuition training

Fixed structure:
2x / 5x = 2/5

x disappears.
Good sign.

Variable structure:
(x+2)/(x+5)

Try values:
x=1 → 3/6
x=100 → 102/105

Huge change.
Bad sign for DS sufficiency.

The figure above represents the floor plan of an art gallery that has

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