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well, my problem is with (2)Each house on Kermit Lane that has a front porch does not have a front yard. not knowing that we have only front porch and only front yard, and knowing only (1) we can assume that there are some houses with both front porches and front yards.

that is why I thought that the answer is C _________________

Happy are those who dream dreams and are ready to pay the price to make them come true

1 can be rephrased as- Every house has a back yard. So there are 40 houses in total.

20 = front porches 20 = front yards 40 = back yards

According to 2, if a house has a front porch, then it does not have a front yard. So the first two categories, above, don't overlap. There can be at least 40 houses (20 with front porches and 20 with front yards - and these could overlap completely with the 40 that have backyards to give 40 houses total) There could be 80 houses: 20 with only front porches, 20 more with only front yards, and 40 more with only back yards.

Re: Of all the houses on Kermit Lane, 20 have front porches, 20 [#permalink]
08 Oct 2013, 05:19

I chose D

St (1) No house is without a backyard ----> so we know that every house must have a backyard. so the number of houses must equal 40. Sufficient

St (2) Each house that has a front porch CAN NOT have a front yard. This means that the 20 houses with front porches are separate and distinct from the other 20 houses that have front yards. there is no overlap here. so total houses= 20 front porches + 20 front yards = 40. Sufficient

Note that there are no houses that have neither. SO it is one big set of 40 houses that encompass 2 distinct sets of 20 front porches and 20 front yards

Both statements are sufficient D _________________

Re: Of all the houses on Kermit Lane, 20 have front porches, 20 [#permalink]
08 Oct 2013, 07:03

2

This post received KUDOS

SaraLotfy wrote:

I chose D

St (1) No house is without a backyard ----> so we know that every house must have a backyard. so the number of houses must equal 40. Sufficient

St (2) Each house that has a front porch CAN NOT have a front yard. This means that the 20 houses with front porches are separate and distinct from the other 20 houses that have front yards. there is no overlap here. so total houses= 20 front porches + 20 front yards = 40. Sufficient

Note that there are no houses that have neither. SO it is one big set of 40 houses that encompass 2 distinct sets of 20 front porches and 20 front yards

Both statements are sufficient D

I don't think it is D. Since 2nd statement gives you information that there are atleast 40 houses on the lane. But doesn't give you any information about the houses that have backyards.

The situation might be that 20 have front porches, 20 have front yards and all of these 40 houses have backyards. But it can also be that 20 have front porches, 20 have front yards and none of these 40 houses have backyards. Therefore in that case there will be 80 houses in total..

Hence you get 2 different situations with the statement and hence answer will only be A.

Re: Of all the houses on Kermit Lane, 20 have front porches, 20 [#permalink]
17 May 2014, 04:33

Expert's post

jlgdr wrote:

I don't understand any of the explanations above. How are we supposed to know which categories overlap and which do not?

Thanks Cheers! J

I'm betting on you B1

Of all the houses on Kermit Lane, 20 have front porches, 20 have front yards, and 40 have back yards. How many houses are on Kermit Lane?

(1) No house on Kermit Lane is without a back yard. This implies that ALL house on Kermit Lane are with a back yard. Since we know that there are 40 houses with a back yard, then there must be 40 houses on the street. Sufficient.

(2) Each house on Kermit Lane that has a front porch does not have a front yard. This implies that houses with a front porch (20) and the houses with a front yard (20) does not overlap. Thus there must be at least 20 + 20 = 40 houses (20 with a front porch and a back yard + 20 with a front yard and a back yard) and at most 20 + 20 + 40 = 80 houses (in case the houses with a back yard does not overlap with other categories). Not sufficient.

Re: Of all the houses on Kermit Lane, 20 have front porches, 20 [#permalink]
24 Jul 2014, 03:05

Bunuel wrote:

jlgdr wrote:

I don't understand any of the explanations above. How are we supposed to know which categories overlap and which do not?

Thanks Cheers! J

I'm betting on you B1

Of all the houses on Kermit Lane, 20 have front porches, 20 have front yards, and 40 have back yards. How many houses are on Kermit Lane?

(1) No house on Kermit Lane is without a back yard. This implies that ALL house on Kermit Lane are with a back yard. Since we know that there are 40 houses with a back yard, then there must be 40 houses on the street. Sufficient.

(2) Each house on Kermit Lane that has a front porch does not have a front yard. This implies that houses with a front porch (20) and the houses with a front yard (20) does not overlap. Thus there must be at least 20 + 20 = 40 houses (20 with a front porch and a back yard + 20 with a front yard and a back yard) and at most 20 + 20 + 40 = 80 houses (in case the houses with a back yard does not overlap with other categories). Not sufficient.

Answer: A.

Hope it's clear.

I don't understand A Granted the total no is 40, but the no of houses can still overlap Say there are 16 houses with front porches and 19 with front yards, implying that there are 5 that have both in common. Thus the total no of houses could be 35 or some other variation. What am i missing here?

Of all the houses on Kermit Lane, 20 have front porches, 20 [#permalink]
24 Jul 2014, 03:41

1

This post received KUDOS

Expert's post

mahendru1992 wrote:

Bunuel wrote:

jlgdr wrote:

I don't understand any of the explanations above. How are we supposed to know which categories overlap and which do not?

Thanks Cheers! J

I'm betting on you B1

Of all the houses on Kermit Lane, 20 have front porches, 20 have front yards, and 40 have back yards. How many houses are on Kermit Lane?

(1) No house on Kermit Lane is without a back yard. This implies that ALL house on Kermit Lane are with a back yard. Since we know that there are 40 houses with a back yard, then there must be 40 houses on the street. Sufficient.

(2) Each house on Kermit Lane that has a front porch does not have a front yard. This implies that houses with a front porch (20) and the houses with a front yard (20) does not overlap. Thus there must be at least 20 + 20 = 40 houses (20 with a front porch and a back yard + 20 with a front yard and a back yard) and at most 20 + 20 + 40 = 80 houses (in case the houses with a back yard does not overlap with other categories). Not sufficient.

Answer: A.

Hope it's clear.

I don't understand A Granted the total no is 40, but the no of houses can still overlap Say there are 16 houses with front porches and 19 with front yards, implying that there are 5 that have both in common. Thus the total no of houses could be 35 or some other variation. What am i missing here?

The total number of houses cannot be 35 or any other number but 40. How can it be 35 if we know that there are 40 houses with a back yard?

We are given that there are 40 houses with a back yards. From (1) we can deduce that ALL house are with a back yard (there are no house without it). If ALL houses are with a back yard and there are total of 40 houses with a back yard, there must be total of 40 houses.