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A certain building has only 2-room and 3-room apartments to rent. If

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A certain building has only 2-room and 3-room apartments to rent. If  [#permalink]

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A certain building has only 2-room and 3-room apartments to rent. If the range of the rental fees of all the apartments in the building is $1,000 and the range of the rental fees for 3-room apartments is $600, what is the range of the rental fees for 2-room apartments?

(1) The lowest rental fee for a 3-room apartment in the building is $200 more than the highest rental fee for a 2-room apartment in the building.
(2) The lowest rental fee for a 2-room apartment is $800 in the building.

 

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Re: A certain building has only 2-room and 3-room apartments to rent. If  [#permalink]

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New post 23 Jul 2019, 08:18
A certain building has only 2-room and 3-room apartments to rent. If the range of the rental fees of all the apartments in the building is $1,000 and the range of the rental fees for 3-room apartments is $600, what is the range of the rental fees for 2-room apartments?

(1) The lowest rental fee for a 3-room apartment in the building is $200 more than the highest rental fee for a 2-room apartment in the building.
(2) The lowest rental fee for a 2-room apartment is $800 in the building.


Each alone is sufficient. As range will be 200 always.
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Re: A certain building has only 2-room and 3-room apartments to rent. If  [#permalink]

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New post 23 Jul 2019, 08:21
IMO : C

A certain building has only 2-room and 3-room apartments to rent. If the range of the rental fees of all the apartments in the building is $1,000 and the range of the rental fees for 3-room apartments is $600, what is the range of the rental fees for 2-room apartments?

(1) The lowest rental fee for a 3-room apartment in the building is $200 more than the highest rental fee for a 2-room apartment in the building.
(2) The lowest rental fee for a 2-room apartment is $800 in the building.



Sol:

we need the lowest of the lowest fee to see the difference and for the 2 we will need the info from the option 2.
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Re: A certain building has only 2-room and 3-room apartments to rent. If  [#permalink]

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New post 23 Jul 2019, 08:31
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A certain building has only 2-room and 3-room apartments to rent. If the range of the rental fees of all the apartments in the building is $1,000 and the range of the rental fees for 3-room apartments is $600, what is the range of the rental fees for 2-room apartments?

This is an easy question.
Let's analyze the stem first.
Lets LABEL
2 room Lowest X
2 room Highest Y
3 room Lowest Z
3 room Highest T

Its is given to use that the range combined is 1000
and T-Z = 600 ....1

We are looking for Y - X

Let's analyze the statements now.

(1) The lowest rental fee for a 3-room apartment in the building is $200 more than the highest rental fee for a 2-room apartment in the building.
This gives us Z = Y +200 ....2
This tells us that the highest of the 2 room is still lower than the lowest of the 3 room. Which helps us establish that T - X = 1000 ....3
Y - X = (Puting 2 & 3 here ) Z-200 - T+ 1000 = Z-T +800 . Now put 1 here =T- 600- T + 800 = 200 is the answer.

Hence this statement is correct.


(2) The lowest rental fee for a 2-room apartment is $800 in the building.
This statement is insufficient.

Hence the answer is A.
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A certain building has only 2-room and 3-room apartments to rent. If  [#permalink]

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New post Updated on: 23 Jul 2019, 09:29
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Quote:
A certain building has only 2-room and 3-room apartments to rent. If the range of the rental fees of all the apartments in the building is $1,000 and the range of the rental fees for 3-room apartments is $600, what is the range of the rental fees for 2-room apartments?

(1) The lowest rental fee for a 3-room apartment in the building is $200 more than the highest rental fee for a 2-room apartment in the building.
(2) The lowest rental fee for a 2-room apartment is $800 in the building.


Given: Range of all = $1000, and Range of 3r = 600;
Rule: Range = Greatest - Lowest value
total range = 1000
3r range = x - y = 600
2r range = a - b = ?

(1) y=a+200; 600=x-a+200… x=400+a,
now the lowest y = highest a, so b = lowest of total and x = highest of total,
total range 1000 = x-b, or… 1000=(400+a)-b,… 600=a-b, sufficient.
(2) lowest 2r b = 800: range 2r = a-800, insufficient.

Answer (A).

Originally posted by exc4libur on 23 Jul 2019, 08:31.
Last edited by exc4libur on 23 Jul 2019, 09:29, edited 1 time in total.
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Re: A certain building has only 2-room and 3-room apartments to rent. If  [#permalink]

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New post 23 Jul 2019, 08:32
2) The lowest rental fee for a 2-room apartment is $800 in the building.

Stmt 2 :
lowest of 2 room is 800 therefore
for 3 room highest is 1800 and lowest is 1200 .
but does not tells the highest of 2room . Not sufficient.

stmt1:
let range of 2 room be y-x. y-highest.x-lowest.
lowest of 3 =200+y
highest of 3 = 1000+x
range given =600
200+y-(1000+x)=600
sufficient
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A certain building has only 2-room and 3-room apartments to rent. If  [#permalink]

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New post 23 Jul 2019, 08:37
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Quote:
A certain building has only 2-room and 3-room apartments to rent. If the range of the rental fees of all the apartments in the building is $1,000 and the range of the rental fees for 3-room apartments is $600, what is the range of the rental fees for 2-room apartments?

(1) The lowest rental fee for a 3-room apartment in the building is $200 more than the highest rental fee for a 2-room apartment in the building.
(2) The lowest rental fee for a 2-room apartment is $800 in the building.



2BHK-> Min X , Max Y
3BHK->Min A , Max B

Given B-X=1000 ----1
B-A=600------2

1. A=200+Y----3
Solving 2 & 3
B-200-Y=600
B-Y=400
B=400+Y ----4

Solving 1 and 4
400+Y-X=1000
we get Y-X=600

2. X=800
But no info abt Y
Insufficient

Hence A
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Re: A certain building has only 2-room and 3-room apartments to rent. If  [#permalink]

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New post 23 Jul 2019, 08:46
IMO answer is C:

from the root we know that H3-L3 = 600$, we need to find H2-L2
range of all apartments is 1000$, so this can be H3-L2 or H2-L2 or H2-L3, many possibilitiesare possible.

from 1: we know H3-L2 = 1000, clearly we do not know H2 from this.
from 2: we know L2 = 800, therefore, H3 = 1800
we do not know H2

1+2 --> L3 = 200+H2 --> H2 = L3-200
H3 = 1800, L3 = 1200
H2 = 1000
H2-L2 = 200$ suff
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A certain building has only 2-room and 3-room apartments to rent. If  [#permalink]

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New post Updated on: 23 Jul 2019, 19:03
2
From the statement we have the following:

Range is Max - Min = 1000

Range for 3 - room = Max3 - Min3 =600


From (1) The lowest rental fee for a 3-room apartment in the building is $200 more than the highest rental fee for a 2-room apartment in the building, we have that:

Min3 - Max2 = 200, then the range will have to be Max3 - Min2 =1000

but since we have Max3 - Min3 =600, Min3-Max2 = 200, the difference between Max2 - Min 2 must be 200 in order to complete the range.

So, sufficient.

From (2) The lowest rental fee for a 2-room apartment is $800 in the building, we have that:


We have the lowest rental fee for a 2 room but we do not anything about the max value of 2, so clearly insufficient.

(A) is our answer

Originally posted by Mizar18 on 23 Jul 2019, 08:47.
Last edited by Mizar18 on 23 Jul 2019, 19:03, edited 1 time in total.
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Re: A certain building has only 2-room and 3-room apartments to rent. If  [#permalink]

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New post 23 Jul 2019, 08:53
1
Answer is A.


(1) The lowest rental fee for a 3-room apartment in the building is $200 more than the highest rental fee for a 2-room apartment in the building.
(2) The lowest rental fee for a 2-room apartment is $800 in the building.

Statement 1:
This says that the range of the 2-room apartment is definitely below the range of the 3-room apartment, and we can even deduce the actual range based on the information given ($1000-$600-$200) = $200 is the range of the 2-room apartment. Sufficient.

Statement 2:
Insufficient, because we have the lower limit of the 2-room apartment range, but we don't have any information about the upper limit. Also, we are given the actual rent amount but this is irrelevant because there is nothing to compare it to.

Hence the answer is A, 1 alone is sufficient.
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A certain building has only 2-room and 3-room apartments to rent. If  [#permalink]

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New post Updated on: 23 Jul 2019, 23:48
1
Let, range of the rent of 2-room apartments be: P – Q
Let, range of the rent of 3-room apartments be: A – B

Given that-
Highest Rent among all rooms – Lowest Rent among all rooms = 1000 -> (a)
A – B = 600 -> (b)

We need to find the value of P – Q

(1) The lowest rental fee for a 3-room apartment in the building is $200 more than the highest rental fee for a 2-room apartment in the building.
This verifies that the Lowest Rent of 3-room apartment > Highest Rent of 2-room apartment
Therefore, (a) converts to A – Q = 1000 -> (c)
Given, B – P = 200 -> (d)

From (c)(d) we get,
A – Q – (B - P) = 1000 - 200
A – Q – B + P = 800
(A – B) + (P – Q) = 800
600 + (P – Q) = 800 [From (a)]
P - Q = 200

Sufficient

(2) The lowest rental fee for a 2-room apartment is $800 in the building.
Given, Q = 800
There is no way of determining the value of P with the available information.

Not Sufficient

Answer A

Originally posted by Sayon on 23 Jul 2019, 08:54.
Last edited by Sayon on 23 Jul 2019, 23:48, edited 1 time in total.
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Re: A certain building has only 2-room and 3-room apartments to rent. If  [#permalink]

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New post 23 Jul 2019, 08:59
1
let
2 room highest ; Ha and lowest La
3 room highest ; Hb and lowest be Lb
given
Hb-La=1000
Hb-Lb=600
need to determine
Ha-La
#1
The lowest rental fee for a 3-room apartment in the building is $200 more than the highest rental fee for a 2-room apartment in the building.
Lb=200+Ha
we know
La=Hb-1000
and Ha=Lb-200
so Ha-La = Lb-200-Hb-1000
and Hb-Lb = 600
so Ha-La = -600-200+1000 ; 200
sufficient
#2
The lowest rental fee for a 2-room apartment is $800 in the building.
La= 800
no info about Ha
we can find Hb and Lb which would be 1800 and 1200
insufficient
IMO A

A certain building has only 2-room and 3-room apartments to rent. If the range of the rental fees of all the apartments in the building is $1,000 and the range of the rental fees for 3-room apartments is $600, what is the range of the rental fees for 2-room apartments?

(1) The lowest rental fee for a 3-room apartment in the building is $200 more than the highest rental fee for a 2-room apartment in the building.
(2) The lowest rental fee for a 2-room apartment is $800 in the building.
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Re: A certain building has only 2-room and 3-room apartments to rent. If  [#permalink]

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New post 23 Jul 2019, 09:03
1
1
A certain building has only 2-room and 3-room apartments to rent. If the range of the rental fees of all the apartments in the building is $1,000 and the range of the rental fees for 3-room apartments is $600, what is the range of the rental fees for 2-room apartments?

(1) The lowest rental fee for a 3-room apartment in the building is $200 more than the highest rental fee for a 2-room apartment in the building.
(2) The lowest rental fee for a 2-room apartment is $800 in the building.

Please see the solution below in image.

IMO A
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2019-07-23_213200.jpg
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Re: A certain building has only 2-room and 3-room apartments to rent. If  [#permalink]

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New post 23 Jul 2019, 09:04
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A certain building has only 2-room and 3-room apartments to rent. If the range of the rental fees of all the apartments in the building is $1,000 and the range of the rental fees for 3-room apartments is $600, what is the range of the rental fees for 2-room apartments?

(1) The lowest rental fee for a 3-room apartment in the building is $200 more than the highest rental fee for a 2-room apartment in the building.

Sufficient.

Say, the cheapest room costs 100 and the most expensive one costs 1100.
We are given relation between the highest rental fee for a 2-room apartment and the lowest rental fee for a 3-room apartment.
x - the highest rental fee for a 2-room apartment
x+200 - the lowest rental fee for a 3-room apartment
x+200+600 - the highest rental fee for a 3-room apartment
x+800=1100
x=300
the cheapest 2-room apartment is 100
Thus the range is 300-100=200

(2) The lowest rental fee for a 2-room apartment is $800 in the building.
Unfortunately, this statement does not help us since we don't know relation with the cheapest/the highest 3-room apartment.
Thus, not sufficient.

IMO A
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A certain building has only 2-room and 3-room apartments to rent. If  [#permalink]

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New post Updated on: 23 Jul 2019, 21:45
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A certain building has only 2-room and 3-room apartments to rent. If the range of the rental fees of all the apartments in the building is $1,000 and the range of the rental fees for 3-room apartments is $600, what is the range of the rental fees for 2-room apartments?

I found it easy to solve by imagining highest and lowest rental fees of 2 and 3 room apartments to be points on a number line and RANGE as DISTANCE between 2 points.

A_____B______X______Y

A = LOWEST FEES (dont know 2 room or 3 room)
Y = HIGHEST FEES (dont know 2 room or 3 room)
B and X are Lowest or Highest of 2 room or 3 room, we dont know which is which.

Distance AY = 1000
Distance of 3 room highest and lowes is 600.... But we dont know which 2 points they are on the number line

(1) The lowest rental fee for a 3-room apartment in the building is $200 more than the highest rental fee for a 2-room apartment in the building.

Lowest of 3 room is higher than highest of 2 room.... SO ALL 2 ROOMS are LOWER than 3 rooms

A = Lowest 2 room
B = Highest 2 room
X = Lowest 3 room
Y = Highest 3 room

FIND AB

AY = 1000... because total range is 1000
BX = 200...... because lowest 3 room is 200 more than highest 2 room
XY = 600.... because 3 room range is 600

AB = 1000 - 600 - 200 = 200

(1) IS SUFFICIENT

(2) The lowest rental fee for a 2-room apartment is $800 in the building.

Lowest 2 room is 800, BUT, we dont know if 800 is A, or B, or X... it could be either of the 3.

We also dont know which one is Highest 2 room, so what do we find the distance between??

(2) IS NOT SUFFICIENT


ANSWER: A - 1 Alone is sufficient

Originally posted by Vinit1 on 23 Jul 2019, 09:21.
Last edited by Vinit1 on 23 Jul 2019, 21:45, edited 1 time in total.
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Re: A certain building has only 2-room and 3-room apartments to rent. If  [#permalink]

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New post 23 Jul 2019, 09:22
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Given that the Range for all the apaartment is $1000

and the range for 3-room apartment is $600

To find the range for 2-room apartment

Statement 1: lowest for 3-room = $200 + highest for 2-room
since range for all = 1000
range for 3-room = 600
therefore, 1000 - 600 = 400
Range for 2-room = 400 - 200 = 200

Suffcient (AD)

Statement 2: lowest for 2-room = 800
Since no any other information such as the highest for 2-room or the relationship to 3-room

Statement 2 is not sufficient

Hence answer choice A.

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Re: A certain building has only 2-room and 3-room apartments to rent. If  [#permalink]

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New post 23 Jul 2019, 09:29
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A certain building has only 2-room and 3-room apartments to rent. If the range of the rental fees of all the apartments in the building is $1,000 and the range of the rental fees for 3-room apartments is $600, what is the range of the rental fees for 2-room apartments?
Let's assume the ranges are the actual rental fees. The apartments in the building can go up to $1000.
(1) The lowest rental fee for a 3-room apartment in the building is $200 more than the highest rental fee for a 2-room apartment in the building.
We shouldn't assume that a 2 room apartment costs less than any 3 room apartment (maybe the 2 room apartment is really nice!) However, this statement gives us this conclusion. Since we know that the 3 room apartment is always more expensive than a 2 room apartment, we know that the ranges for the 2 and 3 room apartments ranges do not overlap. Since we know the range of the entire apartment complex, we know that the upper limit of the apartments is the 3 bedroom apartment (1000). We also know that the lowest part of the range only corresponds to the 2 bedroom apartment (0). Using our assumption above, the range of the 3 bedroom must be 400 - 1000. The lowest 2 room apartment cost must be 0 and the highest must be 400-200 = 200. The range must be 200
(2) The lowest rental fee for a 2-room apartment is $800 in the building.
This doesn't provide us enough information as we do not know what the upper range of the 2 bedroom apartment is.
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Re: A certain building has only 2-room and 3-room apartments to rent. If  [#permalink]

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New post 23 Jul 2019, 09:29
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A certain building has only 2-room and 3-room apartments to rent. If the range of the rental fees of all the apartments in the building is $1,000 and the range of the rental fees for 3-room apartments is $600, what is the range of the rental fees for 2-room apartments?

(1) The lowest rental fee for a 3-room apartment in the building is $200 more than the highest rental fee for a 2-room apartment in the building.
(2) The lowest rental fee for a 2-room apartment is $800 in the building.




given


( .lowest rental fee of 2 room (l2)............................... lowest rental fee of 3 room(l3)...........................................highest rental fee of 3 room(h3))

consider above 3 in a numberline ,

given are h3-l2 = 1000 ( total range)
h3-l3 = 600 ....................................1

so l3-l2 = 400 ...................................2

we need h2 (highest rental fee of 2 room (h2)) which can be situated anywhere between h3 and l2 to find h2-l2????? ( range of 2 room apartments) which is asked

now (1) The lowest rental fee for a 3-room apartment in the building is $200 more than the highest rental fee for a 2-room apartment in the building.

so given l3-h2 = 200

so l3 = h2 + 200..
so substituting in equation 2

we get h2+ 200 - l2 = 400

ie h2-l2 = 200 .. = range of 2 room apartments so sufficient


(2) The lowest rental fee for a 2-room apartment is $800 in the building.

l2 = 800 so from this we can find, h3 and l3.. but we dont have any information about h2. which can be between h3 and l2. so clearly insufficient.

so ans is A
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Re: A certain building has only 2-room and 3-room apartments to rent. If  [#permalink]

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New post 23 Jul 2019, 09:37
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A certain building has only 2-room and 3-room apartments to rent. If the range of the rental fees of all the apartments in the building is $1,000 and the range of the rental fees for 3-room apartments is $600, what is the range of the rental fees for 2-room apartments?

Given:
Let say 2-room apartments rents are R21<R22< ......<R2n (n rents are in increasing order)
3-room apartments rents are R31<R32< ......<R3m (m rents are in increasing order)
R3m - R21=1000 --(i)
R3m-R31 = 600---(ii)
(i)-(ii) => R31 -R21 = 400 --(iii)

Question is asking: R2n-R21 =? So if we can get a relationship between R2n & R31, then we can it's sufficient to get the answer

(1) The lowest rental fee for a 3-room apartment in the building is $200 more than the highest rental fee for a 2-room apartment in the building. --> sufficient: R31 = 200 + R2n. So from (iii), 200+R2n-R21=400 => R2n-R21=200
(2) The lowest rental fee for a 2-room apartment is $800 in the building. --> insufficient: R21 = 800, from this we can't get any relation between R2n & R31

So the answer is A
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Re: A certain building has only 2-room and 3-room apartments to rent. If  [#permalink]

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New post 23 Jul 2019, 09:50
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I like to solve these types of problems using equations as it allows me to be very organized and avoid confusions. So what do we know?

We know that highest rent minus lowest rent (or the range) is 1000, in other words \(H-L=1000\) (eq-1)
We are also told that - \(H3-L3=600\) (eq-2)

We have been asked to calculate range of two room apts or \(H2-L2=?\)

(1) The lowest rental fee for a 3-room apartment in the building is $200 more than the highest rental fee for a 2-room apartment in the building.
So all 3 room apts are costlier than the 2 room apts, eq-1 then becomes: \(H3-L2=1000\), also st-1 tells us that: \(L3=200+H2\) (eq-3)

Now if we substitute eq-3 into eq-2, we get, \(H3-(200+H2)=600\) ==>> \(H3=800+H2\) (eq-4). Now substitute eq-4 in to eq-1 we get, \(800+H2-L2=1000\) ==>> \(H2-L2=200\). Sufficient.

(2) The lowest rental fee for a 2-room apartment is $800 in the building.
If \(L2=800\), then we know that \(H3=1800\) and \(L3=1200\) (as ranges are 1000 and 600 respectively) but we have no info the highest rental two room apartment (H2) and no apparent way to figure it out. Insufficient.

Answer is thus A.
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Re: A certain building has only 2-room and 3-room apartments to rent. If   [#permalink] 23 Jul 2019, 09:50

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