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Re: Patricia has several coins, each of which is worth either 5 cents or [#permalink]
Let the number of 5 cents coins be x and 25 cents coins be y
(I) total value is 205 cents
5x + 25y = 205
x + 5y=41

Notice that x and y MUST BE integers, since they represent the number of coins, which cannot be decimal.
Although you have 2 unknowns and 1 equation you have to test some numbers before you conclude this statement insufficient because you have the restriction of x and y to be integers.

let's start with x=1 --> y=8 OK ,acceptable since both x and y are integers
x=2 --> y=7,8 NOT OK, this cannot be the answer because y is not an int
x=3 --> y =7,6 NOT OK, y is not int (notice that for each +1 of x we get -0,2 of y, so you could think to try x=6 after that)
x=6 --> y= 7 OK, acceptable since both x and y are integers

Because we have 2 acceptable values of x and y we say that statement I is insufficient.

(II) x<7 and y>2
insufficient.
x could be 1 or 2 or 3...

Combined (I)+(II)
we proved at (I) that if x=1 (<7) then y=8 (>2) and if x=6 (<7) then y=7 (>2). 2 possible answers thus insufficient.
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Re: Patricia has several coins, each of which is worth either 5 cents or [#permalink]
chetan2u wrote:
dcummins wrote:
Patricia has several coins, each of which is worth either 5 cents or 25 cents. How many coins does she have?

(1) The total value of Patricia's coins is 205 cents.

(2) Patricia has more than 2 coins worth 25 cents, and fewer than 7 coins worth 5 cents.


Let 5 cents coins be X and 25 cents coins be y..
(I) total value is 205 cents.
So 5x + 25y = 205
x + 5y=41..
Various combinations of X,y possible ....1,8 or 6,7 or 11,6 and so on.
Insufficient

(II) X>2 and y<7
Can be infinite combinations.
Insufficient

Combined..
Various combinations still exist
11,6 or 16,5 or 21,4 or 26,3 or 31,2 or 36,1 or 41,0
Thus insufficient

E


Hi chetan2u

(2) Patricia has more than 2 coins worth 25 cents, and fewer than 7 coins worth 5 cents

in your post you have taken 5 cents as x and 25 cents as y. So shouldn't x < 2 and y > 7?
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Re: Patricia has several coins, each of which is worth either 5 cents or [#permalink]
Expert Reply
Hoozan wrote:
chetan2u wrote:
dcummins wrote:
Patricia has several coins, each of which is worth either 5 cents or 25 cents. How many coins does she have?

(1) The total value of Patricia's coins is 205 cents.

(2) Patricia has more than 2 coins worth 25 cents, and fewer than 7 coins worth 5 cents.


Let 5 cents coins be X and 25 cents coins be y..
(I) total value is 205 cents.
So 5x + 25y = 205
x + 5y=41..
Various combinations of X,y possible ....1,8 or 6,7 or 11,6 and so on.
Insufficient

(II) X>2 and y<7
Can be infinite combinations.
Insufficient

Combined..
Various combinations still exist
11,6 or 16,5 or 21,4 or 26,3 or 31,2 or 36,1 or 41,0
Thus insufficient

E


Hi chetan2u

(2) Patricia has more than 2 coins worth 25 cents, and fewer than 7 coins worth 5 cents

in your post you have taken 5 cents as x and 25 cents as y. So shouldn't x < 2 and y > 7?


Yes, there is an error. It will be y>2 and x<7
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Re: Patricia has several coins, each of which is worth either 5 cents or [#permalink]
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Re: Patricia has several coins, each of which is worth either 5 cents or [#permalink]
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