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

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Math Expert
Joined: 02 Sep 2009
Posts: 62380
A certain building has only 2-room and 3-room apartments to rent. If  [#permalink]

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23 Jul 2019, 07:00
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Difficulty:

75% (hard)

Question Stats:

57% (01:50) correct 43% (02:19) wrong based on 232 sessions

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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.

 This question was provided by Math Revolution for the Game of Timers Competition

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

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23 Jul 2019, 07:31
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?

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.

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

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Updated on: 23 Jul 2019, 18:03
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.

Originally posted by Mizar18 on 23 Jul 2019, 07:47.
Last edited by Mizar18 on 23 Jul 2019, 18:03, 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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23 Jul 2019, 08:03
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
Attachments

2019-07-23_213200.jpg [ 781.54 KiB | Viewed 1727 times ]

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

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Updated on: 23 Jul 2019, 20:45
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?

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, 08:21.
Last edited by Vinit1 on 23 Jul 2019, 20:45, edited 1 time in total.
A certain building has only 2-room and 3-room apartments to rent. If   [#permalink] 23 Jul 2019, 08:21
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