gmat prep -- donuts Mmmmmmm....
This topic has expert replies
-
- Senior | Next Rank: 100 Posts
- Posts: 30
- Joined: Sun Apr 27, 2008 4:22 pm
Timer
00:00
Your Answer
A
B
C
D
E
Global Stats
Using either equaion we can get to know the price of each Cake and Doughnut.
But we cant find the quantity purchased. Tricky question.. seemd to be (D) is the firt instance.
Original equation can be wrtten as [(PriceC * QtyofC)+(PriceOfD*Qty of D) =6]
But we cant find the quantity purchased. Tricky question.. seemd to be (D) is the firt instance.
Original equation can be wrtten as [(PriceC * QtyofC)+(PriceOfD*Qty of D) =6]
- Stuart@KaplanGMAT
- GMAT Instructor
- Posts: 3225
- Joined: Tue Jan 08, 2008 2:40 pm
- Location: Toronto
- Thanked: 1710 times
- Followed by:614 members
- GMAT Score:800
Timer
00:00
Your Answer
A
B
C
D
E
Global Stats
Let's rely on our old DS buddy, "# of equations vs # of unknowns".mav800rick wrote: Original equation can be wrtten as [(PriceC * QtyofC)+(PriceOfD*Qty of D) =6]
From the original, we have 1 equation and 4 unknowns. To solve the entire system, we need 3 more equations.
We're only asked to solve for D, so it's possible we can get away with fewer equations if they eliminate multiple variables in one shot. For example, "the revenue from cupcakes is $4" would eliminate two variables and we'd be left with PriceD * D = 2.
(1) one equation involving 2 of our variables, but not in a way that allows us to eliminate them both. Insufficient.
(2) one equation involving 2 of our variables, but not in a way that allows us to eliminate them both. Insufficient.
Together: 3 distinct linear equations for 4 unknowns, without anything special to allow us to sneakily solve for D: insufficient.
Stuart Kovinsky | Kaplan GMAT Faculty | Toronto
Kaplan Exclusive: The Official Test Day Experience | Ready to Take a Free Practice Test? | Kaplan/Beat the GMAT Member Discount
BTG100 for $100 off a full course
GMAT/MBA Expert
- Ian Stewart
- GMAT Instructor
- Posts: 2621
- Joined: Mon Jun 02, 2008 3:17 am
- Location: Montreal
- Thanked: 1090 times
- Followed by:355 members
- GMAT Score:780
Timer
00:00
Your Answer
A
B
C
D
E
Global Stats
They aren't all linear equations; in the first you multiply variables together (price of doughnuts times number of doughnuts). While it doesn't matter much with this question, in general one must be more cautious about applying 'n equations, n unknowns' rules with non-linear equations. The standard GMAT-like example:Stuart Kovinsky wrote: Together: 3 distinct linear equations for 4 unknowns
If ab = 1, what is the value of a?
1) bc = 1
2) ac = 1
Using both statements, you have three distinct equations, three unknowns. They are not linear equations, however. You still don't have enough information to find any of the unknowns; a, b and c can all be equal to 1, or they can all be equal to -1.
- Stuart@KaplanGMAT
- GMAT Instructor
- Posts: 3225
- Joined: Tue Jan 08, 2008 2:40 pm
- Location: Toronto
- Thanked: 1710 times
- Followed by:614 members
- GMAT Score:800
Timer
00:00
Your Answer
A
B
C
D
E
Global Stats
You're correct.
In this particular question, we can ignore positive/negative solutions (since real world things like items and prices will never be negative), but we do need to be careful.
In this particular question, we can ignore positive/negative solutions (since real world things like items and prices will never be negative), but we do need to be careful.
Stuart Kovinsky | Kaplan GMAT Faculty | Toronto
Kaplan Exclusive: The Official Test Day Experience | Ready to Take a Free Practice Test? | Kaplan/Beat the GMAT Member Discount
BTG100 for $100 off a full course
Timer
00:00
Your Answer
A
B
C
D
E
Global Stats
Hello,
Edit: here's my work. Could anyone tell me what I'm missing?:
Stem:
C=cost of cupcake
D=cost of doughnut
1C + 1D = 6. What is qty of D?
Statement 1:
2D = 3C - 0.10
multiply both sides by 10 to get rid of the fraction:
20D = 30C - 1
Divide both sides by 20:
D=3/2C - 1/20
Multiply 3/2 by 10 to get a common denominator:
D=30/20C - 1/20
D=29/20C
This give me the cost of D in terms of C. I have no info about the quantity. Not Sufficient. Scratch out A and D.
Statement 2:
(C + D) / 2 = 0.35
I'm not sure what to do here. It just doesn't seem sufficient. Scratch out B.
Together:
It seems that if I plug the results from statement 1 into the equation for statement 2, I still don't get a quantity for D. Could someone please explain a more efficient approach for this question?
Thanks so much
Edit: here's my work. Could anyone tell me what I'm missing?:
Stem:
C=cost of cupcake
D=cost of doughnut
1C + 1D = 6. What is qty of D?
Statement 1:
2D = 3C - 0.10
multiply both sides by 10 to get rid of the fraction:
20D = 30C - 1
Divide both sides by 20:
D=3/2C - 1/20
Multiply 3/2 by 10 to get a common denominator:
D=30/20C - 1/20
D=29/20C
This give me the cost of D in terms of C. I have no info about the quantity. Not Sufficient. Scratch out A and D.
Statement 2:
(C + D) / 2 = 0.35
I'm not sure what to do here. It just doesn't seem sufficient. Scratch out B.
Together:
It seems that if I plug the results from statement 1 into the equation for statement 2, I still don't get a quantity for D. Could someone please explain a more efficient approach for this question?
Thanks so much
-
- GMAT Instructor
- Posts: 2630
- Joined: Wed Sep 12, 2012 3:32 pm
- Location: East Bay all the way
- Thanked: 625 times
- Followed by:119 members
- GMAT Score:780
Timer
00:00
Your Answer
A
B
C
D
E
Global Stats
That equation is problematic. If you make C the cost of each cupcake, you'd need to multiply that by the number of cupcakes you have to find the total cost of the cupcakes.Poisson wrote:Hello,
Edit: here's my work. Could anyone tell me what I'm missing?:
Stem:
C=cost of cupcake
D=cost of doughnut
1C + 1D = 6. What is qty of D?
For instance, if I buy 3 cupcakes and 2 donuts, my cost is 3c + 2d. But if I buy 10 cupcakes and 12 donuts, my cost in 10c + 12d.
Since we don't know how many cupcakes or donuts Lew bought, we'd have to assign variables to each (say x and y), then make the total cost of a function of them, something like xc + yd = $6.
From there, we need to puzzle out four variables (x, y, c, and d).