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Re: Of all the houses on Kermit Lane, 20 have front porches, 20 [#permalink]
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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.

Consider Kudos if the post helped. :-D
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Re: Of all the houses on Kermit Lane, 20 have front porches, 20 [#permalink]
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jlgdr wrote:
LalaB wrote:
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.
(2) Each house on Kermit Lane that has a front porch does not have a front yard.


Do we have a consensus here? I still don't get this question. Let's see.

20FP. 20FY, 40BY

Now Statement 1

No is without BY so Total is >=40. Insuff

Statement 2

FP and BY do not intersect x>=60

Both Statements together

X>=40. Stil not a concrete amount for the total

Hence IMO it is E.
Let us know will ya'?

Cheers!
J :)

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The correct answer to this question is A. Refer to the solutions above.
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Of all the houses on Kermit Lane, 20 have front porches, 20 [#permalink]
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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!
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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 houses 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.

Does this make sense?
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Re: Of all the houses on Kermit Lane, 20 have front porches, 20 [#permalink]
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saikrishna123 wrote:
How come we are ignoring the option of a house not having front porch, front yard or back yard?...I agree its uncommon in US but not uncommon in my home country to not have all 3 :)

What am I missing?


Have you read this: of-all-the-houses-on-kermit-lane-20-have-front-porches-122300.html#p1365333 and this: of-all-the-houses-on-kermit-lane-20-have-front-porches-122300.html#p1386111

(1) says that there are 0 houses on Kermit Lane without a back yard! So, ALL houses are with a back yard.
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Re: Kermit Lane [#permalink]
A

What's your doubt?
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Re: Kermit Lane [#permalink]
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
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Re: Of all the houses on Kermit Lane, 20 have front porches, 20 [#permalink]
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
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Re: Of all the houses on Kermit Lane, 20 have front porches, 20 [#permalink]
LalaB wrote:
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.
(2) Each house on Kermit Lane that has a front porch does not have a front yard.


Do we have a consensus here? I still don't get this question. Let's see.

20FP. 20FY, 40BY

Now Statement 1

No is without BY so Total is >=40. Insuff

Statement 2

FP and BY do not intersect x>=60

Both Statements together

X>=40. Stil not a concrete amount for the total

Hence IMO it is E.
Let us know will ya'?

Cheers!
J :)

Let the Kudos rain begin!!
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Re: Of all the houses on Kermit Lane, 20 have front porches, 20 [#permalink]
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
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Re: Of all the houses on Kermit Lane, 20 have front porches, 20 [#permalink]
How come we are ignoring the option of a house not having front porch, front yard or back yard?...I agree its uncommon in US but not uncommon in my home country to not have all 3 :)

What am I missing?
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Re: Of all the houses on Kermit Lane, 20 have front porches, 20 [#permalink]
blink005 wrote:
1 can be rephrased as-
Every house has a back yard.
So there are 40 houses in total.

.


great and simple solution :-D
I did draw diagram and burn more time than yours :(
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Of all the houses on Kermit Lane, 20 have front porches, 20 [#permalink]
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.


Dear Bunuel, What does the sentence means [ in case the houses with a back yard does not overlap with other categories ] ? Since we know the total number of houses located on Kermit Lane, therefore I thought it is sufficient. But then, why it is insufficient?
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Re: Of all the houses on Kermit Lane, 20 have front porches, 20 [#permalink]
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ziyuenlau 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.


Dear Bunuel, What does the sentence means [ in case the houses with a back yard does not overlap with other categories ] ? Since we know the total number of houses located on Kermit Lane, therefore I thought it is sufficient. But then, why it is insufficient?


I tried to explain it here: of-all-the-houses-on-kermit-lane-20-have-front-porches-122300.html#p1386111

(1) says that there are 0 houses on Kermit Lane without a back yard! So, ALL houses are with a back yard.
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Re: Of all the houses on Kermit Lane, 20 have front porches, 20 [#permalink]
blink005 wrote:
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.




Hi! Could you explain statement 1 a little more. My question is: if no houses on kermit lane is without a backyard, then does that mean the 20 front porches and 20 front yards all have backyards? Why can't we add up all the categories together to be 80 houses?
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Re: Of all the houses on Kermit Lane, 20 have front porches, 20 [#permalink]
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.


What if all of 3 are mutually exclusive, then there would be 80 houses in total?
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Re: Of all the houses on Kermit Lane, 20 have front porches, 20 [#permalink]
lakshya14 wrote:
What if all of 3 are mutually exclusive, then there would be 80 houses in total?

it can't be mutually exclusive in statement 1, as stated "(1) No house on Kermit Lane is without a back yard", so if there are 100 houses with backyard irrespective of front porch or front yard in any number. there are only and only 100 houses in there. which is 40 in case of question.

Your statement holds true for second statement there can be 80 houses (mutually exclusive feature). 40(having all), 60 or any other no.
which is why statement 2 is insufficient
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Re: Of all the houses on Kermit Lane, 20 have front porches, 20 [#permalink]
1) "No house on Kermit Lane is without a back yard." In other words, each house has one backyard, therefore House= Backyard. We have 40 backyards which equals to 40 houses. Sufficient

2) "Each house on Kermit Lane that has a front porch does not have a front yard." In other words, you can either have a house with a front porch or a house with a front yard, but you can't have both. Therefore, we have 20 FP and 20FY, a total of 40 houses, BUT what about backyards?

Maybe each FP house and each FY house has a backyard which keeps the total of 40 houses
OR maybe backyard houses do not have a FP or a FY which brings the total number of houses to 80. (20FP+20FY+40B)

Maybe/ maybe not = insufficient

Answer A)
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