Patricia has several coins, each of which is worth either 5 cents or

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) y>2 and x<7
Can be infinite combinations.
Insufficient

Combined..
Various combinations still exist
x=6, 5y=41-6=35...y=7
x=1, 5y=41-1=40...y=8
Possible values: (x,y) = (6,7)=(1,8)
Insufficient

E
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let 5 cents be x and 25 cents be y

From 1 : it says that 5x+25y= 205
so x can be 1 and y 8 or x 21 and y 4 in suffcient

From 2: y>2 and x<7
so for many values of y @6 and x@6 and x 1 and y @ 8
insufficient..

IMO clear E ..

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.

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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