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DS Simultaneous equations - help!

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DS Simultaneous equations - help! [#permalink] New post 21 Apr 2007, 15:15
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Hello!

I recently have gotten stuck on a problem that appears to be a simultaneous equations DS problem:

At a certain pet shop, 1/3 of the pets are dogs, and 1/5 of the pets are birds. How many of the pets are dogs?

(1) There are 30 birds at the pet shop
(2) There are 20 more birds than dogs at the pet shop

Obviously, statement 1 is sufficient. For statement two, I tried setting up these two simultaneous equations:

Let b = # of birds
Let d = # of dogs
Let x = # of birds and dogs

d = b + 20
(1/3)d + (1/5)b = x

I know the answer is D: Both statements alone, but can't figure out why. Please help if you can!

Regards,
Alex
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Re: DS Simultaneous equations - help! [#permalink] New post 21 Apr 2007, 22:34
afiggy1 wrote:
Hello!

I recently have gotten stuck on a problem that appears to be a simultaneous equations DS problem:

At a certain pet shop, 1/3 of the pets are dogs, and 1/5 of the pets are birds. How many of the pets are dogs?

(1) There are 30 birds at the pet shop
(2) There are 20 more birds than dogs at the pet shop

Obviously, statement 1 is sufficient. For statement two, I tried setting up these two simultaneous equations:

Let b = # of birds
Let d = # of dogs
Let x = # of birds and dogs

d = b + 20
(1/3)d + (1/5)b = x

I know the answer is D: Both statements alone, but can't figure out why. Please help if you can!

Regards,
Alex


Well statement 1 is sufficient.
Let us assume that p represents the total no.s of pets in the shop. So form statement 2 we have (p/3)-(p/5)=20. From this you can find total no.s of pets in the shop and hence this statement is also sufficient and hence the answere is D.

Javed.

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Re: DS Simultaneous equations - help! [#permalink] New post 22 Apr 2007, 11:39
afiggy1 wrote:
Hello!

I recently have gotten stuck on a problem that appears to be a simultaneous equations DS problem:

At a certain pet shop, 1/3 of the pets are dogs, and 1/5 of the pets are birds. How many of the pets are dogs?

(1) There are 30 birds at the pet shop
(2) There are 20 more birds than dogs at the pet shop

Obviously, statement 1 is sufficient. For statement two, I tried setting up these two simultaneous equations:

Let b = # of birds
Let d = # of dogs
Let x = # of birds and dogs

d = b + 20
(1/3)d + (1/5)b = x

I know the answer is D: Both statements alone, but can't figure out why. Please help if you can!

Regards,
Alex


Let total number perts be N. So we have N/3 dogs and N/5 birds
from (1) N/5 = 30 so N = 150 and N/3 = 50 SUFFF
from (2) N/3 - N/5 = 20 which again gives answer N = 150 and N/3 = 50 SUFF

Answer D

Btw, I think in (2) the phrase is wrong It has to be 20 birds less. you cant have N/5 - N/3 = 20
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Re: DS Simultaneous equations - help! [#permalink] New post 22 Apr 2007, 12:49
techjanson wrote:
afiggy1 wrote:
Hello!

I recently have gotten stuck on a problem that appears to be a simultaneous equations DS problem:

At a certain pet shop, 1/3 of the pets are dogs, and 1/5 of the pets are birds. How many of the pets are dogs?

(1) There are 30 birds at the pet shop
(2) There are 20 more birds than dogs at the pet shop

Obviously, statement 1 is sufficient. For statement two, I tried setting up these two simultaneous equations:

Let b = # of birds
Let d = # of dogs
Let x = # of birds and dogs

d = b + 20
(1/3)d + (1/5)b = x

I know the answer is D: Both statements alone, but can't figure out why. Please help if you can!

Regards,
Alex


Let total number perts be N. So we have N/3 dogs and N/5 birds
from (1) N/5 = 30 so N = 150 and N/3 = 50 SUFFF
from (2) N/3 - N/5 = 20 which again gives answer N = 150 and N/3 = 50 SUFF

Answer D

Btw, I think in (2) the phrase is wrong It has to be 20 birds less. you cant have N/5 - N/3 = 20


Hey Tech -

You are absolutely right ... i juxtaposed the word dogs and birds in statement 2. Thanks for the catch, and thanks for the help!

Regards,
Alex
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 [#permalink] New post 22 Apr 2007, 17:51
Let number of pets be p.

Then p/3 = dogs, p/5 = birds.

St1:
p/5 =30. Can solve for p/3 then. Sufficient.

St2:
p/5-p/3 = 20 --> can solve for p/3. Sufficient.

Ans D
  [#permalink] 22 Apr 2007, 17:51
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