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At the bakery Lew spent a total of 6$ for one kind of

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At the bakery Lew spent a total of 6$ for one kind of [#permalink] New post 15 Apr 2012, 21:29
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Question Stats:

50% (02:23) correct 50% (01:45) wrong based on 42 sessions
At the bakery Lew spent a total of 6$ for one kind of cupcake and one kind of doughnut. How many donuts did he buy?

(1) Price of 2 doughty was $.10 less than 3 cupcakes
(2) Average price of 1 doughnut and 1 cupcake was $.035
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Re: Word problem: doughnut [#permalink] New post 15 Apr 2012, 21:56
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Smita04 wrote:
At the bakery lew spent a total of 6$ for one kind of cupcake and one kind of doughnut. How many donuts did he buy?
1) price of 2 doughty was $.10 less than 3 cupcakes
2) average price of 1 doughnut and 1 cupcake was $.035


let
cupcakes purchased= x
doughnut purchased= y

price of one cupcake and doughnut be c and d respectively,
then

cx+dy = 6

we need to find y.

statement 1) 2d= 3c - 0.1
INsufficient

statement 2) (c+d)/2 = 0.035
c+d = 0.035*2

Insufficient

1) and 2)

we can have c and d but we have no info about x and y.

Insufficient.

hence E

Hope this helps..!!
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Re: Word problem: doughnut [#permalink] New post 05 Feb 2014, 13:41
kraizada84 wrote:
Smita04 wrote:
At the bakery lew spent a total of 6$ for one kind of cupcake and one kind of doughnut. How many donuts did he buy?
1) price of 2 doughty was $.10 less than 3 cupcakes
2) average price of 1 doughnut and 1 cupcake was $.035


let
cupcakes purchased= x
doughnut purchased= y

price of one cupcake and doughnut be c and d respectively,
then

cx+dy = 6

we need to find y.

statement 1) 2d= 3c - 0.1
INsufficient

statement 2) (c+d)/2 = 0.035
c+d = 0.035*2

Insufficient

1) and 2)

we can have c and d but we have no info about x and y.

Insufficient.

hence E

Hope this helps..!!


But I guess still one has to prove whether 'x' and 'y' could be found using the equation Dx + Cy, remember that x,y standing for the number of Donuts and Cupcakes have to be integers and must satisfy Dx + Cy = 6. Therefore, the next question is, how can we do this in the most efficient way?

Experts, any thoughts?
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Re: At the bakery Lew spent a total of 6$ for one kind of [#permalink] New post 05 Feb 2014, 21:54
Expert's post
Smita04 wrote:
At the bakery Lew spent a total of 6$ for one kind of cupcake and one kind of doughnut. How many donuts did he buy?

(1) Price of 2 doughty was $.10 less than 3 cupcakes
(2) Average price of 1 doughnut and 1 cupcake was $.035


There is an error in this question.
"Average price of 1 doughnut and 1 cupcake was $0.35"
It doesn't make sense if the average price is 3.5 cents. The numbers don't work. It must be 35 cents. Convert everything to cents.

Using both statements together, you get
3C - 2D = 10 (C is the price of cupcakes and D is the price of doughnuts)
C + D = 35*2

Solving them simultaneously, you get C = 30, D = 40.

If Nc is number of cupcakes and Nd is number of doughnuts,
30*Nc + 40*Nd = 600
3*Nc + 4*Nd = 60
The first solution I get is Nd = 3, Nc = 16
We will get many more solutions such as Nd = 6, Nc = 12. Also, Nd = 9, Nc = 8 etc
Hence both statements together are not sufficient.
Answer (E)

Note that in this equation: 3*Nc + 4*Nd = 60, if instead of 60, the total price were say 18, there would have been only one solution.
3*Nc + 4*Nd = 18
Nc = 2, Nd = 3
Other possible solutions will be Nc = 6, Nd = 0; Or Nc = -2, Nd = 6 etc
Since we cannot buy negative or 0 number of items (she spends money on doughnuts AND cupcakes so she must have bought at least one of each), no other solution works.

Refer to this post for further details on this method: http://www.veritasprep.com/blog/2011/06 ... -of-thumb/
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Re: At the bakery Lew spent a total of 6$ for one kind of   [#permalink] 05 Feb 2014, 21:54
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