At the bakery, Lew spent a total of $6.00

[sharmin.karim]

At the bakery, Lew spent a total of $6.00 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 average (arithmetic mean) price of 1 doughnut and 1 cupcake was $0.35.

I initially answered A b/c I used the equation in 1 to substitute into the equation from the questions. But now looking back I realize b/c we are looking for the quantity and not the dollar value spent, the answer would have to be E. Can someone please confirm?

[RonPurewal]

At the bakery, Lew spent a total of $6.00 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 average (arithmetic mean) price of 1 doughnut and 1 cupcake was $0.35.

I initially answered A b/c I used the equation in 1 to substitute into the equation from the questions. But now looking back I realize b/c we are looking for the quantity and not the dollar value spent, the answer would have to be E. Can someone please confirm?

yeah, this is (e).

why did you pick (a)? you can't find any unique values of anything from just that equation.

if you have the two equations together, you can find out the price of each item (specifically, 40 cents for a doughnut and 30 cents for a cupcake). however, this still doesn't tell you how many of each item lew bought, since there are many, many ways to get a total of $6.00 from units of 40 cents and 30 cents.

[vinversa]

http://www.postimage.org/image.php?v=gxT0oD0

there are many, many ways to get a total of $6.00 from units of 40 cents and 30 cents.

4#d + 3#c = $60

http://www.postimage.org/image.php?v=gxT0oD0

[RonPurewal]

http://www.postimage.org/image.php?v=gxT0oD0

there are many, many ways to get a total of $6.00 from units of 40 cents and 30 cents.

4#d + 3#c = $60

http://www.postimage.org/image.php?v=gxT0oD0

yeah.

although that's a hugely unnecessary amount of work -- notice that as soon as you get TWO different answers, you're DONE and the answer is "insufficient".
if you keep grinding out more possibilities after you've got two of the answers highlighted in yellow, you are just wasting time -- you're already proved "insufficient".

[divineacclivity]

At the bakery, Lew spent a total of $6.00 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 average (arithmetic mean) price of 1 doughnut and 1 cupcake was $0.35.

I initially answered A b/c I used the equation in 1 to substitute into the equation from the questions. But now looking back I realize b/c we are looking for the quantity and not the dollar value spent, the answer would have to be E. Can someone please confirm?

yeah, this is (e).

why did you pick (a)? you can't find any unique values of anything from just that equation.

if you have the two equations together, you can find out the price of each item (specifically, 40 cents for a doughnut and 30 cents for a cupcake). however, this still doesn't tell you how many of each item lew bought, since there are many, many ways to get a total of $6.00 from units of 40 cents and 30 cents.

Ron,
How can I ,looking at the equation, tell that an equation is going to have more than one set of solutions without really having to find more than one solution. Is that possible?
For example, I arrived at this one following the information given in the question:
2D = 2C-0.10
D+C = 2 x 0.35
=> C = 0.16, D = 0.54 ......... (1)

dD+cC = 6 ; where d = #of D & c #of C, and D & C are respective costs ...............(2)

from (1) & (2),
27 d + 8 c = 300

possible solutions (d,c) are: (4,24), (6,16) and there could be more...

[RonPurewal]

How can I ,looking at the equation, tell that an equation is going to have more than one set of solutions without really having to find more than one solution. Is that possible?

[/quote]

In some cases it will be possible; in others it won't.

You'll probably do a lot better by just investigating the situation. It won't take much time at all, and you can actually trust the results that you'll get. (By contrast, trying to make up "rules" for when multiple solutions will exist is really, really hard and full of pitfalls -- as we can see by the fact that you're asking here about how to do it!)

[RonPurewal]

For example, I arrived at this one following the information given in the question:
2D = 2C-0.10
D+C = 2 x 0.35

Watch it with the first of these. The given statement is
2d = 3c - 0.10
(not 2c)

=> C = 0.16, D = 0.54 ......... (1)

What is this meant to be a solution for?

It solves the second thing (average = 0.35). However, it's not a solution to your first equation (2d = 2c - 0.10) ... nor is it a solution to the correct first equation (2d = 3c - 0.10). I'm confused as to exactly what you're doing here.

dD+cC = 6 ; where d = #of D & c #of C, and D & C are respective costs ...............(2)

This is too many letters for my brain to handle.

I'd just write (whole #)D + (other whole #)C = 6, and start thinking about testing cases.

from (1) & (2),
27 d + 8 c = 300

How'd you get this?

[divineacclivity]

For example, I arrived at this one following the information given in the question:
2D = 2C-0.10
D+C = 2 x 0.35

Watch it with the first of these. The given statement is
2d = 3c - 0.10
(not 2c)

=> C = 0.16, D = 0.54 ......... (1)

What is this meant to be a solution for?

It solves the second thing (average = 0.35). However, it's not a solution to your first equation (2d = 2c - 0.10) ... nor is it a solution to the correct first equation (2d = 3c - 0.10). I'm confused as to exactly what you're doing here.

dD+cC = 6 ; where d = #of D & c #of C, and D & C are respective costs ...............(2)

This is too many letters for my brain to handle.

I'd just write (whole #)D + (other whole #)C = 6, and start thinking about testing cases.

from (1) & (2),
27 d + 8 c = 300

How'd you get this?

Correcting the equations:
2D = 3C - 0.1 (statement 1)
D+C = 1.4 (statement 2)
--------------------
Solving the above two equations gives D =0.4, C = 0.3

Now, 0.3 x (#of cakes) + 0.4 (# of donuts) = 6
0.3 c + 0.4 d = 6
or 3c + 4d = 60

I don't recall the question exactly but it was sometime discussed on this forum:
For an eqn ax + by = C,
if C > 2ab, then there'd be more than 1 solutions to the eqn

[RonPurewal]

I don't recall the question exactly but it was sometime discussed on this forum:
For an eqn ax + by = C,
if C > 2ab, then there'd be more than 1 solutions to the eqn

Assuming you're talking about whole numbers, of course. If you don't have to have whole numbers, then there are always infinity number of solutions to this.

If you can remember this fact as a fact, then, well, you're a lot better than me. There's no way I could possibly remember something so obscure, so here's how I would reason it out. (I'll just use the 3 and 4 from this problem, rather than "a" and "b", because the 3 and 4 are more intuitively accessible.)

* Let's say i have 4x + 3y = something more than 24.
* Then either 4x or 3y is more than 12. (If not, then your total won't be more than 24.)
* If 4x > 12, then you can "trade" three x's for four y's. I.e., you can take x down by 3, and put y up by 4, and you'll get the same sum. And x will still be a positive whole number.
* If 3y > 12, then same deal, just the other way around (take 4 away from y, and give 3 extra to x).

E.g., if i have 4x + 3y = 60, then one possible solution is ... um ... x = 3 and y = 16. (I found this by trying x = 1, then x = 2, then x = 3. Nothing too smart.)
So then I can make y into 12, and x into 6, and it will definitely still work.

[RonPurewal]

Much more importantly, make sure that you are not memorizing hundreds and hundreds of obscure number-theory rules for the sake of avoiding lists of numbers. Listing possibilities should still be a primary approach to these problems, even if you know thousands of obscure facts about number theory.

If you have 4x + 3y = 60, for instance, I've got ...
... x = 1 doesn't work
... x = 2 doesn't work
... x = 3, y = 16
... x = 4 doesn't work
... x = 5 doesn't work
... x = 6, y = 12
Done!
Certainly no less efficient than any theory-based approach.

[divineacclivity]

I don't recall the question exactly but it was sometime discussed on this forum:
For an eqn ax + by = C,
if C > 2ab, then there'd be more than 1 solutions to the eqn

Assuming you're talking about whole numbers, of course. If you don't have to have whole numbers, then there are always infinity number of solutions to this.

If you can remember this fact as a fact, then, well, you're a lot better than me. There's no way I could possibly remember something so obscure, so here's how I would reason it out. (I'll just use the 3 and 4 from this problem, rather than "a" and "b", because the 3 and 4 are more intuitively accessible.)

* Let's say i have 4x + 3y = something more than 24.
* Then either 4x or 3y is more than 12. (If not, then your total won't be more than 24.)
* If 4x > 12, then you can "trade" three x's for four y's. I.e., you can take x down by 3, and put y up by 4, and you'll get the same sum. And x will still be a positive whole number.
* If 3y > 12, then same deal, just the other way around (take 4 away from y, and give 3 extra to x).

E.g., if i have 4x + 3y = 60, then one possible solution is ... um ... x = 3 and y = 16. (I found this by trying x = 1, then x = 2, then x = 3. Nothing too smart.)
So then I can make y into 12, and x into 6, and it will definitely still work.

Ron, I kinda lost you after this :( : * Then either 4x or 3y is more than 12. (If not, then your total won't be more than 24.)

[divineacclivity]

Much more importantly, make sure that you are not memorizing hundreds and hundreds of obscure number-theory rules for the sake of avoiding lists of numbers. Listing possibilities should still be a primary approach to these problems, even if you know thousands of obscure facts about number theory.

If you have 4x + 3y = 60, for instance, I've got ...
... x = 1 doesn't work
... x = 2 doesn't work
... x = 3, y = 16
... x = 4 doesn't work
... x = 5 doesn't work
... x = 6, y = 12
Done!
Certainly no less efficient than any theory-based approach.

Ok, Ron, I'll remember this. Thank you very much.

[RonPurewal]

Good. There are plenty of DS problems (especially onse dealing with remainders/primes/divisibility) on which listing cases is the only feasible way to solve the problem. So, if testing cases is not in your arsenal, you just won't be able to solve those.

[divineacclivity]

Good. There are plenty of DS problems (especially onse dealing with remainders/primes/divisibility) on which listing cases is the only feasible way to solve the problem. So, if testing cases is not in your arsenal, you just won't be able to solve those.

:)

[divineacclivity]

I don't recall the question exactly but it was sometime discussed on this forum:
For an eqn ax + by = C,
if C > 2ab, then there'd be more than 1 solutions to the eqn

Assuming you're talking about whole numbers, of course. If you don't have to have whole numbers, then there are always infinity number of solutions to this.

If you can remember this fact as a fact, then, well, you're a lot better than me. There's no way I could possibly remember something so obscure, so here's how I would reason it out. (I'll just use the 3 and 4 from this problem, rather than "a" and "b", because the 3 and 4 are more intuitively accessible.)

* Let's say i have 4x + 3y = something more than 24.
* Then either 4x or 3y is more than 12. (If not, then your total won't be more than 24.)
* If 4x > 12, then you can "trade" three x's for four y's. I.e., you can take x down by 3, and put y up by 4, and you'll get the same sum. And x will still be a positive whole number.
* If 3y > 12, then same deal, just the other way around (take 4 away from y, and give 3 extra to x).

E.g., if i have 4x + 3y = 60, then one possible solution is ... um ... x = 3 and y = 16. (I found this by trying x = 1, then x = 2, then x = 3. Nothing too smart.)
So then I can make y into 12, and x into 6, and it will definitely still work.

Ron, I kinda lost you after this :( : * Then either 4x or 3y is more than 12. (If not, then your total won't be more than 24.)

Ron, could you please answer this one too? Thanks.