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# DS Simultaneous equations - help!

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DS Simultaneous equations - help! [#permalink]

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21 Apr 2007, 16:15
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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

Regards,
Alex
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Re: DS Simultaneous equations - help! [#permalink]

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21 Apr 2007, 23:34
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

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.

Cheers!
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Re: DS Simultaneous equations - help! [#permalink]

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22 Apr 2007, 12:39
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

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

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]

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22 Apr 2007, 13:49
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

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

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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22 Apr 2007, 18:51
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
22 Apr 2007, 18:51
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