There was a vague discussion about this problem some months ago but no real resolution was reached.
[spoiler]I think this is a real nasty one, simply because under time pressure it leads your thinking in the wrong direction and has you solving pointless equations. Doh![/spoiler]
At the bakery, Lew spent a total of $6 for one kind of cupcake and one kind of doughnut.
How many doughnuts did he buy?
1) The price of 2 doughnuts was $0.10 less than the price of 3 cupcakes
2) The avergage price of 1 doughnut and 1 cupcake was $0.35
Answer: E
[spoiler] From statement 1:
Let d be the cost of a single doughnut
Let c be the cost of a single cupcake
An expression for statement 1 is:
2d = 3c - $0.10 ; rearranged becomes
3c - 2d = $0.10
2 variables and 1 equation, therefore NOT SUFFICIENT.
From statement 2:
The average price of 1 doughnut and 1 cupcake is $0.35
In my view an algebraic expression for this is:
(c + d)/2 = $0.35 ; rearranged becomes
c + d = $0.70
2 variables and 1 equation, therefore NOT SUFFICIENT.
Statement 1 and 2 Together:
We have 2 distinct equations so it looks as if we should be able solve this problem.
3c - 2d = $0.10 - eq 1
c + d = $0.70 - eq 2
multiply eq 2 by 2, add the two equations together to isolate c.
3c - 2d = $0.10 +
2c + 2d = $1.40
5c = $1.50
c = $0.30.
From equation 2 we can determine the cost of a doughnut as $0.40.
Although we have determined the cost of a cupcake and a doughnut we still haven't answered the question of how many doughnuts Lew has purchased. An equation describing this relationship would be:
Where x = number of doughnuts
y = number of cupcakes
0.4x + 0.3y = $6.00
Unfortunately we have two variables and only 1 equatons. NOT SUFFICIENT. Therefore E is the answer.
[/spoiler]
[spoiler]I think this is a real nasty one, simply because under time pressure it leads your thinking in the wrong direction and has you solving pointless equations. Doh![/spoiler]
At the bakery, Lew spent a total of $6 for one kind of cupcake and one kind of doughnut.
How many doughnuts did he buy?
1) The price of 2 doughnuts was $0.10 less than the price of 3 cupcakes
2) The avergage price of 1 doughnut and 1 cupcake was $0.35
Answer: E
[spoiler] From statement 1:
Let d be the cost of a single doughnut
Let c be the cost of a single cupcake
An expression for statement 1 is:
2d = 3c - $0.10 ; rearranged becomes
3c - 2d = $0.10
2 variables and 1 equation, therefore NOT SUFFICIENT.
From statement 2:
The average price of 1 doughnut and 1 cupcake is $0.35
In my view an algebraic expression for this is:
(c + d)/2 = $0.35 ; rearranged becomes
c + d = $0.70
2 variables and 1 equation, therefore NOT SUFFICIENT.
Statement 1 and 2 Together:
We have 2 distinct equations so it looks as if we should be able solve this problem.
3c - 2d = $0.10 - eq 1
c + d = $0.70 - eq 2
multiply eq 2 by 2, add the two equations together to isolate c.
3c - 2d = $0.10 +
2c + 2d = $1.40
5c = $1.50
c = $0.30.
From equation 2 we can determine the cost of a doughnut as $0.40.
Although we have determined the cost of a cupcake and a doughnut we still haven't answered the question of how many doughnuts Lew has purchased. An equation describing this relationship would be:
Where x = number of doughnuts
y = number of cupcakes
0.4x + 0.3y = $6.00
Unfortunately we have two variables and only 1 equatons. NOT SUFFICIENT. Therefore E is the answer.
[/spoiler]
