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# Last year, 36 houses in a certain development had roof repairs and 48

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Re: Last year, 36 houses in a certain development had roof repairs and 48 [#permalink]
Bunuel wrote:
Last year, 36 houses in a certain development had roof repairs and 48 houses were repainted. If 20 houses in the development had roof repairs but were not repainted last year, how many houses were repainted but did not have roof repairs?

(A) 12
(B) 16
(C) 20
(D) 28
(E) 32

Number of houses that had roof repairs = 36
Some of these could have been repainted as well (while the others only had roof repairs)

Number of houses that had roof repairs but were not repainted (i.e. only had roof repairs) = 20

Thus, number of houses that had roof repairs and were also repainted
= (Number of houses that had roof repairs) - (Number of houses that had roof repairs but were not repainted)
= 36 - 20 = 16

Number of houses that were repainted = 48
Some of these could have had roof repairs as well (while the others had only been repainted)

Thus, number of houses that were repainted but did not have roof repairs
= (Number of houses that were repainted) - (Number of houses that had roof repairs and were also repainted)
= 48 - 16
= 32

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Re: Last year, 36 houses in a certain development had roof repairs and 48 [#permalink]
Bunuel wrote:
Last year, 36 houses in a certain development had roof repairs and 48 houses were repainted. If 20 houses in the development had roof repairs but were not repainted last year, how many houses were repainted but did not have roof repairs?

(A) 12
(B) 16
(C) 20
(D) 28
(E) 32

We can create the equation:

36 - both = 20

both = 16

Thus, the number of houses that were repainted only was 48 - 16 = 32.

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Re: Last year, 36 houses in a certain development had roof repairs and 48 [#permalink]

I solved this using a double matrix method.

Might help anyone who's trying to solve it by the same approach.

Here's how i did it. Hope this helps
Attachments

IMG_20200812_130659__01.jpg [ 1.47 MiB | Viewed 3964 times ]

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