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A certain building has only 2room and 3room apartments to rent. If
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23 Jul 2019, 08:00
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A certain building has only 2room and 3room 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 3room apartments is $600, what is the range of the rental fees for 2room apartments? (1) The lowest rental fee for a 3room apartment in the building is $200 more than the highest rental fee for a 2room apartment in the building. (2) The lowest rental fee for a 2room apartment is $800 in the building.
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Re: A certain building has only 2room and 3room apartments to rent. If
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23 Jul 2019, 08:18
A certain building has only 2room and 3room 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 3room apartments is $600, what is the range of the rental fees for 2room apartments? (1) The lowest rental fee for a 3room apartment in the building is $200 more than the highest rental fee for a 2room apartment in the building. (2) The lowest rental fee for a 2room 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 2room and 3room apartments to rent. If
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23 Jul 2019, 08:21
IMO : C
A certain building has only 2room and 3room 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 3room apartments is $600, what is the range of the rental fees for 2room apartments?
(1) The lowest rental fee for a 3room apartment in the building is $200 more than the highest rental fee for a 2room apartment in the building. (2) The lowest rental fee for a 2room 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 2room and 3room apartments to rent. If
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23 Jul 2019, 08:31
A certain building has only 2room and 3room 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 3room apartments is $600, what is the range of the rental fees for 2room 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 TZ = 600 ....1
We are looking for Y  X
Let's analyze the statements now.
(1) The lowest rental fee for a 3room apartment in the building is $200 more than the highest rental fee for a 2room 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 ) Z200  T+ 1000 = ZT +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 2room apartment is $800 in the building. This statement is insufficient.
Hence the answer is A.



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A certain building has only 2room and 3room apartments to rent. If
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Updated on: 23 Jul 2019, 09:29
Quote: A certain building has only 2room and 3room 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 3room apartments is $600, what is the range of the rental fees for 2room apartments?
(1) The lowest rental fee for a 3room apartment in the building is $200 more than the highest rental fee for a 2room apartment in the building. (2) The lowest rental fee for a 2room 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=xa+200… x=400+a, now the lowest y = highest a, so b = lowest of total and x = highest of total, total range 1000 = xb, or… 1000=(400+a)b,… 600=ab, sufficient. (2) lowest 2r b = 800: range 2r = a800, 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 2room and 3room apartments to rent. If
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23 Jul 2019, 08:32
2) The lowest rental fee for a 2room 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 yx. yhighest.xlowest. 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 2room and 3room apartments to rent. If
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23 Jul 2019, 08:37
Quote: A certain building has only 2room and 3room 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 3room apartments is $600, what is the range of the rental fees for 2room apartments?
(1) The lowest rental fee for a 3room apartment in the building is $200 more than the highest rental fee for a 2room apartment in the building. (2) The lowest rental fee for a 2room apartment is $800 in the building. 2BHK> Min X , Max Y 3BHK>Min A , Max B Given BX=1000 1 BA=6002 1. A=200+Y3 Solving 2 & 3 B200Y=600 BY=400 B=400+Y 4 Solving 1 and 4 400+YX=1000 we get YX=600 2. X=800 But no info abt Y Insufficient Hence A
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Re: A certain building has only 2room and 3room apartments to rent. If
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23 Jul 2019, 08:46
IMO answer is C:
from the root we know that H3L3 = 600$, we need to find H2L2 range of all apartments is 1000$, so this can be H3L2 or H2L2 or H2L3, many possibilitiesare possible.
from 1: we know H3L2 = 1000, clearly we do not know H2 from this. from 2: we know L2 = 800, therefore, H3 = 1800 we do not know H2
1+2 > L3 = 200+H2 > H2 = L3200 H3 = 1800, L3 = 1200 H2 = 1000 H2L2 = 200$ suff



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A certain building has only 2room and 3room apartments to rent. If
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Updated on: 23 Jul 2019, 19:03
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 3room apartment in the building is $200 more than the highest rental fee for a 2room 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, Min3Max2 = 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 2room 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 2room and 3room apartments to rent. If
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23 Jul 2019, 08:53
Answer is A.
(1) The lowest rental fee for a 3room apartment in the building is $200 more than the highest rental fee for a 2room apartment in the building. (2) The lowest rental fee for a 2room apartment is $800 in the building.
Statement 1: This says that the range of the 2room apartment is definitely below the range of the 3room apartment, and we can even deduce the actual range based on the information given ($1000$600$200) = $200 is the range of the 2room apartment. Sufficient.
Statement 2: Insufficient, because we have the lower limit of the 2room 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 2room and 3room apartments to rent. If
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Updated on: 23 Jul 2019, 23:48
Let, range of the rent of 2room apartments be: P – Q Let, range of the rent of 3room 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 3room apartment in the building is $200 more than the highest rental fee for a 2room apartment in the building. This verifies that the Lowest Rent of 3room apartment > Highest Rent of 2room 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 2room 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 2room and 3room apartments to rent. If
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23 Jul 2019, 08:59
let 2 room highest ; Ha and lowest La 3 room highest ; Hb and lowest be Lb given HbLa=1000 HbLb=600 need to determine HaLa #1 The lowest rental fee for a 3room apartment in the building is $200 more than the highest rental fee for a 2room apartment in the building. Lb=200+Ha we know La=Hb1000 and Ha=Lb200 so HaLa = Lb200Hb1000 and HbLb = 600 so HaLa = 600200+1000 ; 200 sufficient #2 The lowest rental fee for a 2room 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 2room and 3room 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 3room apartments is $600, what is the range of the rental fees for 2room apartments?
(1) The lowest rental fee for a 3room apartment in the building is $200 more than the highest rental fee for a 2room apartment in the building. (2) The lowest rental fee for a 2room apartment is $800 in the building.



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Re: A certain building has only 2room and 3room apartments to rent. If
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23 Jul 2019, 09:03
A certain building has only 2room and 3room 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 3room apartments is $600, what is the range of the rental fees for 2room apartments? (1) The lowest rental fee for a 3room apartment in the building is $200 more than the highest rental fee for a 2room apartment in the building. (2) The lowest rental fee for a 2room apartment is $800 in the building. Please see the solution below in image. IMO A
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Re: A certain building has only 2room and 3room apartments to rent. If
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23 Jul 2019, 09:04
A certain building has only 2room and 3room 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 3room apartments is $600, what is the range of the rental fees for 2room apartments?
(1) The lowest rental fee for a 3room apartment in the building is $200 more than the highest rental fee for a 2room 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 2room apartment and the lowest rental fee for a 3room apartment. x  the highest rental fee for a 2room apartment x+200  the lowest rental fee for a 3room apartment x+200+600  the highest rental fee for a 3room apartment x+800=1100 x=300 the cheapest 2room apartment is 100 Thus the range is 300100=200
(2) The lowest rental fee for a 2room apartment is $800 in the building. Unfortunately, this statement does not help us since we don't know relation with the cheapest/the highest 3room apartment. Thus, not sufficient.
IMO A



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A certain building has only 2room and 3room apartments to rent. If
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Updated on: 23 Jul 2019, 21:45
A certain building has only 2room and 3room 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 3room apartments is $600, what is the range of the rental fees for 2room 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 3room apartment in the building is $200 more than the highest rental fee for a 2room 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 2room 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 2room and 3room apartments to rent. If
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23 Jul 2019, 09:22
Given that the Range for all the apaartment is $1000 and the range for 3room apartment is $600 To find the range for 2room apartment Statement 1: lowest for 3room = $200 + highest for 2room since range for all = 1000 range for 3room = 600 therefore, 1000  600 = 400 Range for 2room = 400  200 = 200 Suffcient (AD) Statement 2: lowest for 2room = 800 Since no any other information such as the highest for 2room or the relationship to 3room Statement 2 is not sufficient Hence answer choice A.
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Re: A certain building has only 2room and 3room apartments to rent. If
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23 Jul 2019, 09:29
A certain building has only 2room and 3room 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 3room apartments is $600, what is the range of the rental fees for 2room 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 3room apartment in the building is $200 more than the highest rental fee for a 2room 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 400200 = 200. The range must be 200 (2) The lowest rental fee for a 2room 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 2room and 3room apartments to rent. If
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23 Jul 2019, 09:29
A certain building has only 2room and 3room 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 3room apartments is $600, what is the range of the rental fees for 2room apartments? (1) The lowest rental fee for a 3room apartment in the building is $200 more than the highest rental fee for a 2room apartment in the building. (2) The lowest rental fee for a 2room 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 h3l2 = 1000 ( total range) h3l3 = 600 ....................................1 so l3l2 = 400 ...................................2 we need h2 (highest rental fee of 2 room (h2)) which can be situated anywhere between h3 and l2 to find h2l2????? ( range of 2 room apartments) which is asked now (1) The lowest rental fee for a 3room apartment in the building is $200 more than the highest rental fee for a 2room apartment in the building.so given l3h2 = 200 so l3 = h2 + 200.. so substituting in equation 2 we get h2+ 200  l2 = 400 ie h2l2 = 200 .. = range of 2 room apartments so sufficient (2) The lowest rental fee for a 2room 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 2room and 3room apartments to rent. If
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23 Jul 2019, 09:37
A certain building has only 2room and 3room 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 3room apartments is $600, what is the range of the rental fees for 2room apartments?
Given: Let say 2room apartments rents are R21<R22< ......<R2n (n rents are in increasing order) 3room apartments rents are R31<R32< ......<R3m (m rents are in increasing order) R3m  R21=1000 (i) R3mR31 = 600(ii) (i)(ii) => R31 R21 = 400 (iii)
Question is asking: R2nR21 =? 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 3room apartment in the building is $200 more than the highest rental fee for a 2room apartment in the building. > sufficient: R31 = 200 + R2n. So from (iii), 200+R2nR21=400 => R2nR21=200 (2) The lowest rental fee for a 2room 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 2room and 3room apartments to rent. If
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23 Jul 2019, 09:50
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 \(HL=1000\) (eq1) We are also told that  \(H3L3=600\) (eq2)
We have been asked to calculate range of two room apts or \(H2L2=?\)
(1) The lowest rental fee for a 3room apartment in the building is $200 more than the highest rental fee for a 2room apartment in the building. So all 3 room apts are costlier than the 2 room apts, eq1 then becomes: \(H3L2=1000\), also st1 tells us that: \(L3=200+H2\) (eq3)
Now if we substitute eq3 into eq2, we get, \(H3(200+H2)=600\) ==>> \(H3=800+H2\) (eq4). Now substitute eq4 in to eq1 we get, \(800+H2L2=1000\) ==>> \(H2L2=200\). Sufficient.
(2) The lowest rental fee for a 2room 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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