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Re: Pet shop problem [#permalink]
06 Jan 2008, 12:48

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

Re: Pet shop problem [#permalink]
06 Jan 2008, 22:46

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: a certain pet shop, 1/3 of pets are dogs, 1/5 of pets are [#permalink]
25 Nov 2013, 02:39

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Re: At a certain pet shop, 1/3 of the pets are dogs and 1/5 of t [#permalink]
25 Nov 2013, 02:48

Expert's post

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 shop --> \frac{1}{5} of the pets are birds --> \frac{1}{5}t=30, where t is the # of pets at the pet shop --> t=150 --> \frac{1}{3} of the pets are dogs --> there are \frac{1}{3}*150=50 dogs at the pet shop. Sufficient.

(2) There are 20 more dogs than birds at the pet shop --> d=b+20 --> \frac{1}{3}t=\frac{1}{5}t+20, where t is the # of pets at the pet shop --> t=150 --> \frac{1}{3} of the pets are dogs --> there are \frac{1}{3}*150=50 dogs at the pet shop. Sufficient.