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At the beginning of the day, a store had grills and smokers in a ratio

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At the beginning of the day, a store had grills and smokers in a ratio  [#permalink]

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New post 14 Mar 2018, 21:22
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At the beginning of the day, a store had grills and smokers in a ratio of 3 to 4. If no new items were added to the store and the only items that left were those that were sold, what was the ratio of grills to smokers at the end of the day?

(1) There were 12 grills and 16 smokers sold

(2) One third of the grills were sold

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Re: At the beginning of the day, a store had grills and smokers in a ratio  [#permalink]

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New post 14 Mar 2018, 21:52
G:S=3:4=3X:4X
1. g:s=3X-12:4X-16 ------------->NS
2. g:s=3X-X:4X =2X:4X =1:2 ------------->S

Thus I would like to go for B
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Re: At the beginning of the day, a store had grills and smokers in a ratio  [#permalink]

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New post 14 Mar 2018, 22:28
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Bunuel wrote:
At the beginning of the day, a store had grills and smokers in a ratio of 3 to 4. If no new items were added to the store and the only items that left were those that were sold, what was the ratio of grills to smokers at the end of the day?

(1) There were 12 grills and 16 smokers sold

(2) One third of the grills were sold


Let initial number of grills = 3x and initial number of smokers = 4x.

Statement 1:
new ratio of grills : smokers = (3x-12) : (4x-16)
Whats interesting here is that the grills and smokers sold are also in the ratio 3:4 only (12:16 = 3:4). Since initially they were in the ratio 3:4 and they are sold in the ratio 3:4 so the new ratio will also be 3:4.

Algebraically, we can see this way = (3x-12)/(4x-16)
Taking 3 common from numerator and 4 common from denominator we have = 3*(x-4) / 4*(x-4)
x-4 cancels from both numerator/denominator leaving 3/4 only, or the new ratio also is 3:4. Sufficient.


Statement 2:
1/3 grills sold, so grills left = 2/3 * 3x = 2x. But there is no information about number of smokers sold. So not sufficient.


Hence A answer
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Re: At the beginning of the day, a store had grills and smokers in a ratio  [#permalink]

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New post 14 Mar 2018, 22:35
amanvermagmat wrote:
Bunuel wrote:
At the beginning of the day, a store had grills and smokers in a ratio of 3 to 4. If no new items were added to the store and the only items that left were those that were sold, what was the ratio of grills to smokers at the end of the day?

(1) There were 12 grills and 16 smokers sold

(2) One third of the grills were sold


Let initial number of grills = 3x and initial number of smokers = 4x.

Statement 1:
new ratio of grills : smokers = (3x-12) : (4x-16)
Whats interesting here is that the grills and smokers sold are also in the ratio 3:4 only (12:16 = 3:4). Since initially they were in the ratio 3:4 and they are sold in the ratio 3:4 so the new ratio will also be 3:4.

Algebraically, we can see this way = (3x-12)/(4x-16)
Taking 3 common from numerator and 4 common from denominator we have = 3*(x-4) / 4*(x-4)
x-4 cancels from both numerator/denominator leaving 3/4 only, or the new ratio also is 3:4. Sufficient.


Statement 2:
1/3 grills sold, so grills left = 2/3 * 3x = 2x. But there is no information about number of smokers sold. So not sufficient.


Hence A answer



I completely get my mistake of overlooking 12:16=3:4 for statement
But for statement 2, why we have to assume that there must be some numbers of Smokers sold.
Can't we assume 'nothing mentioned =0 sold' ?
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Re: At the beginning of the day, a store had grills and smokers in a ratio  [#permalink]

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New post 14 Mar 2018, 22:43
u1983 wrote:
amanvermagmat wrote:
Bunuel wrote:
At the beginning of the day, a store had grills and smokers in a ratio of 3 to 4. If no new items were added to the store and the only items that left were those that were sold, what was the ratio of grills to smokers at the end of the day?

(1) There were 12 grills and 16 smokers sold

(2) One third of the grills were sold


Let initial number of grills = 3x and initial number of smokers = 4x.

Statement 1:
new ratio of grills : smokers = (3x-12) : (4x-16)
Whats interesting here is that the grills and smokers sold are also in the ratio 3:4 only (12:16 = 3:4). Since initially they were in the ratio 3:4 and they are sold in the ratio 3:4 so the new ratio will also be 3:4.

Algebraically, we can see this way = (3x-12)/(4x-16)
Taking 3 common from numerator and 4 common from denominator we have = 3*(x-4) / 4*(x-4)
x-4 cancels from both numerator/denominator leaving 3/4 only, or the new ratio also is 3:4. Sufficient.


Statement 2:
1/3 grills sold, so grills left = 2/3 * 3x = 2x. But there is no information about number of smokers sold. So not sufficient.


Hence A answer



I completely get my mistake of overlooking 12:16=3:4 for statement
But for statement 2, why we have to assume that there must be some numbers of Smokers sold.
Can't we assume 'nothing mentioned =0 sold' ?


Hello

I think we cannot assume that. When nothing is mentioned about smokers, then their sales could be anything from 0 and upwards.
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At the beginning of the day, a store had grills and smokers in a ratio  [#permalink]

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New post Updated on: 20 Mar 2018, 12:15
1
At the Beginning:
Let CM for ratio be x
Grills: 3x
Smokers: 4x

St.1:
Sold- Grills:12, Smokers:16
Left behind- Grills: 3x-12, Smokers: 4x-16
Ratio at the end of the day:
G/S= (3x-12)/(4x-16)= 3(x-4)/4(x-4)= 3/4

Sufficient.

St.2: Grills sold= 3x/3= x
Grills left= 2x. No information on numbers of smokers sold. Insufficient.


Ans A

Originally posted by Shobhit7 on 16 Mar 2018, 04:21.
Last edited by Shobhit7 on 20 Mar 2018, 12:15, edited 2 times in total.
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Re: At the beginning of the day, a store had grills and smokers in a ratio  [#permalink]

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New post 20 Mar 2018, 02:54
St1 3x-12/4x-16=3/4
St2 no info about smokers.

Hence A
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Re: At the beginning of the day, a store had grills and smokers in a ratio  [#permalink]

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New post 23 Mar 2018, 23:12
amanvermagmat wrote:
Bunuel wrote:
At the beginning of the day, a store had grills and smokers in a ratio of 3 to 4. If no new items were added to the store and the only items that left were those that were sold, what was the ratio of grills to smokers at the end of the day?

(1) There were 12 grills and 16 smokers sold

(2) One third of the grills were sold


Let initial number of grills = 3x and initial number of smokers = 4x.

Statement 1:
new ratio of grills : smokers = (3x-12) : (4x-16)
Whats interesting here is that the grills and smokers sold are also in the ratio 3:4 only (12:16 = 3:4). Since initially they were in the ratio 3:4 and they are sold in the ratio 3:4 so the new ratio will also be 3:4.

Algebraically, we can see this way = (3x-12)/(4x-16)
Taking 3 common from numerator and 4 common from denominator we have = 3*(x-4) / 4*(x-4)
x-4 cancels from both numerator/denominator leaving 3/4 only, or the new ratio also is 3:4. Sufficient.


Statement 2:
1/3 grills sold, so grills left = 2/3 * 3x = 2x. But there is no information about number of smokers sold. So not sufficient.


Hence A answer

i made a big mistake thinking that initial inventory would have been 12 and 16 , so post the sale the items left would be zero. but the question stem says, there were items left in the store.
is that the reason we, didnt consider 12 and 16 as initial values.
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At the beginning of the day, a store had grills and smokers in a ratio  [#permalink]

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New post 24 Mar 2018, 06:14
Yeah, but what if there are 12 grills and 16 smokers in the first place?
Then answer A is NOT SUFFICIENT, because, if we sell everything, the ratio is 0:0, or 1:1, certainly not 3:4.
Then the right answer is C, because we now know that in total there were more than 12 grills and 16 smokers.

And yes, 3*(x-4)/4*(x-4) = 3/4, but not if x=4, because we cannot divide by 0

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Re: At the beginning of the day, a store had grills and smokers in a ratio  [#permalink]

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New post 24 Mar 2018, 10:56
urbanoc wrote:
Yeah, but what if there are 12 grills and 16 smokers in the first place?
Then answer A is NOT SUFFICIENT, because, if we sell everything, the ratio is 0:0, or 1:1, certainly not 3:4.
Then the right answer is C, because we now know that in total there were more than 12 grills and 16 smokers.

And yes, 3*(x-4)/4*(x-4) = 3/4, but not if x=4, because we cannot divide by 0

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Hello

the question stem says, "..only items that were left were those that were sold..". So some grills/smokers were left.

And 0:0 doesnt make any sense at all. We cannot take ratios of something that doesnt exist, so I believe we can safely discard that possibility here (or 0 smokers or 0 grills left at the end of day).
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Re: At the beginning of the day, a store had grills and smokers in a ratio  [#permalink]

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New post 25 Mar 2018, 08:22
amanvermagmat wrote:
urbanoc wrote:
Yeah, but what if there are 12 grills and 16 smokers in the first place?
Then answer A is NOT SUFFICIENT, because, if we sell everything, the ratio is 0:0, or 1:1, certainly not 3:4.
Then the right answer is C, because we now know that in total there were more than 12 grills and 16 smokers.

And yes, 3*(x-4)/4*(x-4) = 3/4, but not if x=4, because we cannot divide by 0

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Hello

the question stem says, "..only items that were left were those that were sold..". So some grills/smokers were left.

And 0:0 doesnt make any sense at all. We cannot take ratios of something that doesnt exist, so I believe we can safely discard that possibility here (or 0 smokers or 0 grills left at the end of day).


Hi amanvermagmat, thanks for the reply!

The question stem doesn't say like you wrote " the only items that WERE left, were those that were sold",

the question states: "the only items that left were those that were sold", so it is possible to sell everything, so it possible to have at the end 0 (zero) items in the inventory.
I still don't understand why we should assume that not everything can be sold.
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Re: At the beginning of the day, a store had grills and smokers in a ratio  [#permalink]

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New post 25 Mar 2018, 21:54
urbanoc wrote:
amanvermagmat wrote:
urbanoc wrote:
Yeah, but what if there are 12 grills and 16 smokers in the first place?
Then answer A is NOT SUFFICIENT, because, if we sell everything, the ratio is 0:0, or 1:1, certainly not 3:4.
Then the right answer is C, because we now know that in total there were more than 12 grills and 16 smokers.

And yes, 3*(x-4)/4*(x-4) = 3/4, but not if x=4, because we cannot divide by 0

Experts?


Hello

the question stem says, "..only items that were left were those that were sold..". So some grills/smokers were left.

And 0:0 doesnt make any sense at all. We cannot take ratios of something that doesnt exist, so I believe we can safely discard that possibility here (or 0 smokers or 0 grills left at the end of day).


Hi amanvermagmat, thanks for the reply!

The question stem doesn't say like you wrote " the only items that WERE left, were those that were sold",

the question states: "the only items that left were those that were sold", so it is possible to sell everything, so it possible to have at the end 0 (zero) items in the inventory.
I still don't understand why we should assume that not everything can be sold.


Hello

The only explanation I have to your query is that if everything gets sold, then the number of smokers remaining = number of grills remaining = 0, thus the ratio 0:0 doesnt make any sense. I think such a thing will not be tested in a question, where there is a possibility that we have to take ratio of non-existent quantities.
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Re: At the beginning of the day, a store had grills and smokers in a ratio  [#permalink]

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New post 26 Mar 2018, 02:24
Hello

the question stem says, "..only items that were left were those that were sold..". So some grills/smokers were left.

And 0:0 doesnt make any sense at all. We cannot take ratios of something that doesnt exist, so I believe we can safely discard that possibility here (or 0 smokers or 0 grills left at the end of day).[/quote]

Hi amanvermagmat, thanks for the reply!

The question stem doesn't say like you wrote " the only items that WERE left, were those that were sold",

the question states: "the only items that left were those that were sold", so it is possible to sell everything, so it possible to have at the end 0 (zero) items in the inventory.
I still don't understand why we should assume that not everything can be sold.[/quote]

Hello

The only explanation I have to your query is that if everything gets sold, then the number of smokers remaining = number of grills remaining = 0, thus the ratio 0:0 doesnt make any sense. I think such a thing will not be tested in a question, where there is a possibility that we have to take ratio of non-existent quantities.[/quote]


Hi,

I know what you mean, and if we were in a Problem Solving question I would think the same. But since we are in a DS question, one should analyse all possibilities...
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At the beginning of the day, a store had grills and smokers in a ratio  [#permalink]

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New post 26 Mar 2018, 09:32
Guys!

I have a doubt about the OA. The First statement is correct for all situations except for one. ie. if at the beginning of the day, there were only 12 & 16 , grills & cookers respectively. At the end of the day, in such a scenario, there will be 0 of both. Hence, negating the OA.

Bunuel Please let me know, if there is any flaw in the situation i have mentioned above. I solved it using that logic & chose C as it takes care of the issue.
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At the beginning of the day, a store had grills and smokers in a ratio  [#permalink]

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New post 29 Mar 2018, 12:24
I think selling 12 and 16 is a totally valid situation. Nothing stops it and the fact that it creates a non-sensical 0:0, if anything, sounds like proof that the answer isn't A.

Since (2) helps to guarantee that the inventory isn't going to be totally sold out, the answer should be (C) imo.
At the beginning of the day, a store had grills and smokers in a ratio &nbs [#permalink] 29 Mar 2018, 12:24
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