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# a certain pet shop, 1/3 of pets are dogs, 1/5 of pets are

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a certain pet shop, 1/3 of pets are dogs, 1/5 of pets are [#permalink]  06 Jan 2008, 06:58
a certain pet shop, 1/3 of pets are dogs, 1/5 of pets are birds. how many of pets are dogs

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

OA is D, how come that second statement can be sufficient to solve that problem? am I missing some thing?

I thought that only first st is sufficient... how do you think?
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Re: Pet shop problem [#permalink]  06 Jan 2008, 07:31
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certain pet shop, 1/3 of pets are dogs, 1/5 of pets are birds. how many of pets are dogs

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

You need to find out the total number of pets to solve this question. If you let x=total pets, D= 1/3 * x, B= 1/5 * x, then you get:

1) 1/5*x=30, so you know that x=150, SUFFICIENT

2) d=b+20
(1/3 *x)=(1/5 *x) +20
x/3= x/5 +20
5x/15 - 3x/15=20
2x/15 = 20
2x = 300
x= 150 SUFFICIENT
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Re: Pet shop problem [#permalink]  06 Jan 2008, 08:22
bsjames2 wrote:
Quote:
certain pet shop, 1/3 of pets are dogs, 1/5 of pets are birds. how many of pets are dogs

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

You need to find out the total number of pets to solve this question. If you let x=total pets, D= 1/3 * x, B= 1/5 * x, then you get:

1) 1/5*x=30, so you know that x=150, SUFFICIENT

2) d=b+20
(1/3 *x)=(1/5 *x) +20
x/3= x/5 +20
5x/15 - 3x/15=20
2x/15 = 20
2x = 300
x= 150 SUFFICIENT

can't understand the logics behind, why 20 is written separate, shouldn't it be in the numerator?
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Re: Pet shop problem [#permalink]  06 Jan 2008, 13:48
kazakhb wrote:
a certain pet shop, 1/3 of pets are dogs, 1/5 of pets are birds. how many of pets are dogs

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

OA is D, how come that second statement can be sufficient to solve that problem? am I missing some thing?

I thought that only first st is sufficient... how do you think?

From the stem we know that Dogs=Total/3 and Birds=Total/5
From S1 we know Birds=30. you calculate the total number and then the number of dogs.
From S2 we know that Dogs=Birds+20. Without resolving the problem you hace three variables (Dogs,Total and Birds) and three equations (one from S2 and two from the question stem). Therefore S2 is SUFF

B
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Re: Pet shop problem [#permalink]  06 Jan 2008, 15:34
kazakhb wrote:
a certain pet shop, 1/3 of pets are dogs, 1/5 of pets are birds. how many of pets are dogs

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

OA is D, how come that second statement can be sufficient to solve that problem? am I missing some thing?

I thought that only first st is sufficient... how do you think?

1: obvs suff.
2: x=birds y=total.

so x=1/5y x+20=1/3y two equations two unknowns --> 1/5y+20=1/3y --> 1/3-1/5 --> 2/15=20

y=150 so 1/3 of 150 is 50.

D
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Re: Pet shop problem [#permalink]  06 Jan 2008, 18:36
kazakhb wrote:
a certain pet shop, 1/3 of pets are dogs, 1/5 of pets are birds. how many of pets are dogs

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

OA is D, how come that second statement can be sufficient to solve that problem? am I missing some thing?

I thought that only first st is sufficient... how do you think?

for [2] , 3 equations 3 unknowns.
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Re: Pet shop problem [#permalink]  06 Jan 2008, 23:46
automan wrote:
kazakhb wrote:
a certain pet shop, 1/3 of pets are dogs, 1/5 of pets are birds. how many of pets are dogs

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

OA is D, how come that second statement can be sufficient to solve that problem? am I missing some thing?

I thought that only first st is sufficient... how do you think?

From the stem we know that Dogs=Total/3 and Birds=Total/5
From S1 we know Birds=30. you calculate the total number and then the number of dogs.
From S2 we know that Dogs=Birds+20. Without resolving the problem you hace three variables (Dogs,Total and Birds) and three equations (one from S2 and two from the question stem). Therefore S2 is SUFF

B

nice explanation, thanks, in such way it is more easier to see the equations
thanx to all who tried to explain
Re: Pet shop problem   [#permalink] 06 Jan 2008, 23:46
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