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At a restaurant, five friends each purchased a sandwich. Did the sum

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At a restaurant, five friends each purchased a sandwich. Did the sum  [#permalink]

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New post 06 Oct 2018, 06:10
2
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A
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E

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Question Stats:

57% (01:01) correct 43% (00:43) wrong based on 117 sessions

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At a restaurant, five friends each purchased a sandwich. Did the sum of the five sandwiches exceed $70?

(1) The price of the least expensive sandwich exceeded $13.

(2) The price of the most expensive sandwich exceeded $18.

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Re: At a restaurant, five friends each purchased a sandwich. Did the sum  [#permalink]

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New post 06 Oct 2018, 06:44
GMATYoda wrote:
At a restaurant, five friends each purchased a sandwich. Did the sum of the five sandwiches exceed $70?

(1) The price of the least expensive sandwich exceeded $13.

(2) The price of the most expensive sandwich exceeded $18.



Lets analyze each of the statements:

Statement 1:

Suppose each friend at paid the minimum amount of $13 to buy each sandwich then total price =13*5= $65
Since each sandwich price will be a liitle more than $13 so minimum price of total sandwich will be > $65

This can be 66,67,70,80,90 anythings

So clearly this statement is not sufficient.

Now lets see statement 2:

Clearly this doesnt give any information about minimum value so not sufficient.

4 sandwich can be price $1 and 5th can price $18 .(Will imply price less than 70)
4 sandwich 15 and 5h sanwich price price is $18 .(Will imply price greater than 70)

Now lets see both the statements together

Minmum price is $13 and Max price is $18
So lets calculate the minimum total price of 5 sandwich

4*13+ 18*1=70 (since prices are little more than 13 and little more than 18) so net minimum price will be little more
than 70 .

Hence this together answers the question whether price will be greater than $70.

So answer is C.


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Re: At a restaurant, five friends each purchased a sandwich. Did the sum  [#permalink]

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New post 07 Oct 2018, 04:20
If we consider the minimum price, 13*4 +18*1 , it would turn up $70. Thus it will not exceed $70, but equals to $70. How we are concluding IMO as C. I think I'm missing somewhere.
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Re: At a restaurant, five friends each purchased a sandwich. Did the sum  [#permalink]

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New post 07 Oct 2018, 05:16
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sghoshgt wrote:
If we consider the minimum price, 13*4 +18*1 , it would turn up $70. Thus it will not exceed $70, but equals to $70. How we are concluding IMO as C. I think I'm missing somewhere.


Please notice the statement 1 carefully .(It is very smartly crafted)

1.The price of the least expensive sandwich exceeded $13.


So the price is not exactly $13 but just a little more than $13.So based on this the minmum price 13(little greater than 13)*4 +18*1 will be little greater than $70



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Re: At a restaurant, five friends each purchased a sandwich. Did the sum  [#permalink]

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New post 03 Dec 2018, 03:49
Hi there!
Can someone explain to me why, in order to calculate the minimum total price, we do 13*4+18*1 instead of 13*5?

Thanks a lot!
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Re: At a restaurant, five friends each purchased a sandwich. Did the sum  [#permalink]

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New post 03 Dec 2018, 04:14
oblivion441 wrote:
Hi there!
Can someone explain to me why, in order to calculate the minimum total price, we do 13*4+18*1 instead of 13*5?

Thanks a lot!



The purpose of the question is to ask the minimum value is greather or less than equal to 70.

Now to your question.

If you assume all the sandwiches are priced $13 then you have the value which says prices are greather than $13*5 ...now this can be $66 or $78..
Say there can be various sets.

13,13,13,13,13---13 is least
13,14,15,16,17--13 is least
13,50,100,200,1000--13 is least..
But total value in al above scenario for 5 elemets will be diffrenet.So we cannot just assume $13 as least for all.


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Re: At a restaurant, five friends each purchased a sandwich. Did the sum  [#permalink]

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New post 03 Dec 2018, 04:30
Ok thanks,

sorry so, by reading the question, do I have to assume that at least one person bought the sandwich that exceeded 18?

Because I intended that the prices that "exceeded" were the ones on the menu, not the ones bought by the "5 friends".

Thank you very very much!
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At a restaurant, five friends each purchased a sandwich. Did the sum  [#permalink]

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New post 03 Dec 2018, 04:47
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sorry so, by reading the question, do I have to assume that at least one person bought the sandwich that exceeded 18?

Yes

This is explicitly stated by statement one :)
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Re: At a restaurant, five friends each purchased a sandwich. Did the sum  [#permalink]

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New post 03 Dec 2018, 08:41
GMATYoda wrote:
At a restaurant, five friends each purchased a sandwich. Did the sum of the five sandwiches exceed $70?

(1) The price of the least expensive sandwich exceeded $13.

(2) The price of the most expensive sandwich exceeded $18.



chetan2u

Statement 1 & Statement 2 are clearly not enough by themselves.

(1)+(2) tells us that cheapest sandwich is $13 and most expensive sandwich is $18.

Now, we don't if there are any other sandwiches that cost between $13 & $18, not do we know how many $13 sandwiches are bought or how many $18 sandwiches are bought. All we know there are total 5 sandwiches.

Now, lets say 4 sandwiches that cost $18 were bought and 1 sandwich costing $13 is bought. Here the total is $85 for 5 sandwiches, clearly exceeding $70.

The question does not tell us nor does give us the details of each sandwich bought. The question only gives the range.

Hence, IMO since we dont know the breakup of each sandwich, we cannot say the total cost of 5 sandwiches will be under $70.

Hence, answer should be (E) instead of (C).
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Re: At a restaurant, five friends each purchased a sandwich. Did the sum  [#permalink]

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New post 03 Dec 2018, 18:01
At a restaurant, five friends each purchased a sandwich. Did the sum of the five sandwiches exceed $70?

(1) The price of the least expensive sandwich exceeded $13.
Now the least possible of each will be just above 13, so 13.01
If all are priced equally, 13.01*5=65.05
But the others could you easily cost 30 each, then >70
Insufficient

(2) The price of the most expensive sandwich exceeded $18.
So least price of most expensive can be 18.01
If all coated 18, the total would exceed 70..
If others are say just 5 each, then no
Insufficient

Combined..
Let us find the least..
So most expensive =18.01
And let all four be just greater than 13..
So total >(13*4)+18....total>52+18.....total >70
Sufficient

C
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