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Re: At a sandwich shop, all sandwiches sell for the same price. If the sho [#permalink]
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pushpitkc wrote:
Let the price of a sandwich be x.
If he sold n sandwiches per day, now he sells n+5 sandwiches.
But the price would come down.
Since revenue is constant at 150
nx = 150
Also, (n+5)(x-a) = 150 where a is the cost by which the sandwich goes down.

150 has factors 15,25.30 and 50 which match with answer options 10,6,5,3
But since the number of sandwiches goes up by 5, as the price of sandwiches goes down(25,30 is our pair)
Initially sandwiches cost 6$, now they cost 5$.
Hence, the price of the sandwich is 6(Option D)


Hello Pushpitkc.. really liked your approach.

but what about pair :

n=15 p=10 and n=20 and p=7,5. Here difference in n=5 . So value of p can be 10. As its not given that current price or new price is in integer form.
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Re: At a sandwich shop, all sandwiches sell for the same price. If the sho [#permalink]
Nikkb wrote:
pushpitkc wrote:
Let the price of a sandwich be x.
If he sold n sandwiches per day, now he sells n+5 sandwiches.
But the price would come down.
Since revenue is constant at 150
nx = 150
Also, (n+5)(x-a) = 150 where a is the cost by which the sandwich goes down.

150 has factors 15,25.30 and 50 which match with answer options 10,6,5,3
But since the number of sandwiches goes up by 5, as the price of sandwiches goes down(25,30 is our pair)
Initially sandwiches cost 6$, now they cost 5$.
Hence, the price of the sandwich is 6(Option D)


Hello Pushpitkc.. really liked your approach.

but what about pair :

n=15 p=10 and n=20 and p=7.5. Here difference in n=5 . So value of p can be 10. As its not given that current price or new price is in integer form.


I agree, but from the way I looked at it,
i thought because 20 is not a factor of 150.
I went with Option D as both 25 and 30 are factors of 150!

Lets wait for the OE!
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Re: At a sandwich shop, all sandwiches sell for the same price. If the sho [#permalink]
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Bunuel wrote:
At a sandwich shop, all sandwiches sell for the same price. If the shop were to decrease the price of each sandwich, it would sell 5 more sandwiches per day its daily revenue would remain constant at $150. How much does a sandwich currently cost?

A. $3
B. $4
C. $5
D. $6
E. $10


Bunuel I think question should mention that price is an Integer

Also, language should include word "could" for price

Let price = p
No. Of sandwich​ = s

Revenue = ps= 150

150=
1*150
2*75
3*50
5*30
6*25
10*15

New revenue = (p-x)(s+5)=150

*No. Of sandwiches before and after = factors of 150 at a gap of 5*

Case 1: I.e. no of sandwiches may be 25 and 30

Or

Case 2: no of sandwiches may be 5 and 10

Case 1: price earlier and after will be *6* and 5 respectively

Case 2: price earlier and after will be 30 and 15 respectively

Hence answer option : D

Posted from my mobile device
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Re: At a sandwich shop, all sandwiches sell for the same price. If the sho [#permalink]
Bunuel wrote:
At a sandwich shop, all sandwiches sell for the same price. If the shop were to decrease the price of each sandwich, it would sell 5 more sandwiches per day its daily revenue would remain constant at $150. How much does a sandwich currently cost?

A. $3
B. $4
C. $5
D. $6
E. $10


Sandwiches sold = S
Price per SW = P

S*P = 150 ; Considering the extra 5 SWs sold,

5 * P = 150 ; P = 30 is the revenue for 5 SWs

One SW = 30/5 = 6$

Is this approach correct?
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Re: At a sandwich shop, all sandwiches sell for the same price. If the sho [#permalink]
theperfectgentleman wrote:
Bunuel wrote:
At a sandwich shop, all sandwiches sell for the same price. If the shop were to decrease the price of each sandwich, it would sell 5 more sandwiches per day its daily revenue would remain constant at $150. How much does a sandwich currently cost?

A. $3
B. $4
C. $5
D. $6
E. $10


Sandwiches sold = S
Price per SW = P

S*P = 150 ; Considering the extra 5 SWs sold,

5 * P = 150 ; P = 30 is the revenue for 5 SWs

One SW = 30/5 = 6$

Is this approach correct?


theperfectgentleman,
First things first the 5 extra sandwiches don't cost 150$, but the extra 5 will cost 150$.
So if we sold x sandwiches before, now x+5 sandwiches will sell for the same 150$
Also, if you are considering the price per sandwich as P,
how are you dividing 30 with 5 to find the price of one sandwich

So, the method you have used is not correct.

Hope its clear!
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Re: At a sandwich shop, all sandwiches sell for the same price. If the sho [#permalink]
[quote="Bunuel"]At a sandwich shop, all sandwiches sell for the same price. If the shop were to decrease the price of each sandwich, it would sell 5 more sandwiches per day its daily revenue would remain constant at $150. How much does a sandwich currently cost?

A. $3
B. $4
C. $5
D. $6
E. $10

[/here is my own solution to this problem but its kind of long :) can anybody recommend the way to solve it in two minutes ? :)
let x be cost per sandwich
let n be the number of sandwiches
than revenue= x*n= 150
to find number of sandwiches x = 150/n
If the shop were to decrease the price of each sandwich, it would sell 5 more sandwiches per day its daily revenue would remain constant at $150

then (x-5)(n+5)=150
xn+x5-25-5 =150
xn+x5-30=150
xn+x5=180
divide by X and plug in instead of x this = 150/n
150+750/n=180
750n=180-150
750n=30
n = 25

now plug in n=25 into x*n= 150
25x=150
x=6

Answer is D :)
]
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At a sandwich shop, all sandwiches sell for the same price. If the sho [#permalink]
dave13 wrote:
Bunuel wrote:
At a sandwich shop, all sandwiches sell for the same price. If the shop were to decrease the price of each sandwich, it would sell 5 more sandwiches per day its daily revenue would remain constant at $150. How much does a sandwich currently cost?

A. $3
B. $4
C. $5
D. $6
E. $10

[/here is my own solution to this problem but its kind of long :) can anybody recommend the way to solve it in two minutes ? :)
let x be cost per sandwich
let n be the number of sandwiches
than revenue= x*n= 150
to find number of sandwiches x = 150/n
If the shop were to decrease the price of each sandwich, it would sell 5 more sandwiches per day its daily revenue would remain constant at $150

then (x-5)(n+5)=150
xn+x5-25-5 =150
xn+x5-30=150
xn+x5=180
divide by X and plug in instead of x this = 150/n
150+750/n=180
750n=180-150
750n=30
n = 25

now plug in n=25 into x*n= 150
25x=150
x=6

Answer is D :)
]


dave13 Few points:

after the reduction in price ur equation is (x-5)(n+5)= 150 How?
it just says number of sandwich increased by 5, and we don't know new price of sandwich. So you cannot subtract 5 from x
also ur solved equation should be : xn-5n+5x-25=150 -> How u got xn+5x-25-5=150 ? here u have considered n=1 ?

Please recheck your equations or plz correct me if m wrong :)
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Re: At a sandwich shop, all sandwiches sell for the same price. If the sho [#permalink]
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a requirement for the price to be an integer is missing. Otherwise both d and e are possible and viable
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Re: At a sandwich shop, all sandwiches sell for the same price. If the sho [#permalink]
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Bunuel wrote:
At a sandwich shop, all sandwiches sell for the same price. If the shop were to decrease the price of each sandwich, it would sell 5 more sandwiches per day its daily revenue would remain constant at $150. How much does a sandwich currently cost?

A. $3
B. $4
C. $5
D. $6
E. $10


We know that the quantity sold is a whole number, and if we are also assuming that the price is a whole number (which is suggested by the answer choices), we can write 150 as the product of two positive integers as follows (where the first integer represents price and the second represents quantity):

150 = 1 x 150 = 2 x 75 = 3 x 50 = 5 x 30 = 6 x 25 = 10 x 15

We see that if the price is $6, then 25 sandwiches will be sold, and if we reduce the price by $1, then 30 sandwiches will be sold, which is exactly 5 more sandwiches than before. So, the current price must be $6.

Alternate Solution:

We will test each answer choice, except for choice B (since 150 is not divisible by 4).

If the original prices of the sandwiches are 3, 5, 6, or 10 dollars, then the shop originally sells 50, 30, 25, or 15 sandwiches, respectively.

We are given that the shop sells 5 more sandwiches by reducing the price, which means it will sell 55, 35, 30, or 20 sandwiches. Of these prices, 150 is divisible only by 30; therefore, 25 sandwiches must have been sold originally, which corresponds to a price of 6 dollars.

Answer: D
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Re: At a sandwich shop, all sandwiches sell for the same price. If the sho [#permalink]
Hi all,

How do we know that the new price has to be an integer?
If it does not have to be an interger, the right answer could also be E (new price 7.5$)?
Am I missing something here?

Thanks!
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Re: At a sandwich shop, all sandwiches sell for the same price. If the sho [#permalink]
Hi, I am not a native English speaker so I'm struggling to understand the question. If the question reads "How much does a sandwich currently COST"? why most of the answers given indicates that the price must be 6?
From the question we know that the Price of all sandwiches is the same and that if we increase the PRICE in $1 then the amount of total sold sandwiches increases in 5. Finally it is stated that the REVENUE will remain the same at $150.
How can we determine the COST of any of the sandwiches if we do not have gross income, variable costs, fixed costs, etc?
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Re: At a sandwich shop, all sandwiches sell for the same price. If the sho [#permalink]
lescudero wrote:
Hi, I am not a native English speaker so I'm struggling to understand the question. If the question reads "How much does a sandwich currently COST"? why most of the answers given indicates that the price must be 6?
From the question we know that the Price of all sandwiches is the same and that if we increase the PRICE in $1 then the amount of total sold sandwiches increases in 5. Finally it is stated that the REVENUE will remain the same at $150.
How can we determine the COST of any of the sandwiches if we do not have gross income, variable costs, fixed costs, etc?


Hi ,
Your Question is not very clear, however let me try and help.

The question says all sandwiches sell for the same price however when the cost of each sandwich is reduced , 5 more sandwiches are sold to get the same revenue of 150. ( Note : The reduced price of each sandwich must be an integer , this is an assumption in the question )

Now we pick from the answer choices

If the initial SP of each sandwich ( sw ) is 3 then there are 50 sws. ( 150/3)
after price reduction we sell 55 sws and we get same revenue so cost of each sw is 150/55 = not an integer .

if the initial SP of each Sw is 4 then we sell 150/4 = sandwiches , this is not an integer number of sandwich , hence we can stop here.

if the initial SP of each Sw is 5 then we sell 150/5 = 30 SW, after reduction we sell 35 , so price for each after reduction 150/35 = not an integer
if the initial SP of each Sw is 6 then we sell 150/6 = 25 SW, after reduction we sell 30 , so price for each after reduction 150/30 = 5 an integer.

similarly option E is rejected
hence only D gives an integer for the reduced price .
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At a sandwich shop, all sandwiches sell for the same price. If the sho [#permalink]
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At a sandwich shop, all sandwiches sell for the same price. If the shop were to decrease the price of each sandwich, it would sell 5 more sandwiches per day its daily revenue would remain constant at $150. How much does a sandwich currently cost?

Price of one sandwich x Original number of sandwiches= NP=150
(Price - $1) x (Original number + 5)= (P-1)(N+5)=150

(P-1)(N+5)=PN-N+5P-5=150 (get rid of the N variable since question asks for P)
Since NP=150 and then N=150/P

P(150/P)-150/P+5P-5=150
150-150/P-5=150
\(P^2-P-30=0\)
\(P^2-6P+5P-30=0\)
(P-6)(P+5)=0

Price cannot be Negative so P = $6
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Re: At a sandwich shop, all sandwiches sell for the same price. If the sho [#permalink]
Let the cost of the sandwich be x.
No. of sandwiches sold = n

nx=150

Let the decrease in price be a.

(n+5)(x-a)=150
Putting n=150/x:

(150/x + 5)(x-a)=150
150-150a/x+5x-5a = 150
150a/x + 5x -5a= 0
30a/x + x - a = 0

x^2 - ax + 30a = 0

(x-6a)(x+5a) = 0

Therefore x = 6a (x cannot be negative)

Putting it in nx=150:

na = 25

Possible solutions: (n,a) = (1,25),(5,5),(25,1)

Case 1: (n,a) = (1,25):
x= 6a = 150
n=1

Plugging the values in our equations:
nx=150 => 1*150 = 150 (satisfied)
(n+5)(x-a)=150 => (1+5)(150-25) = 6*125 (not satisfied)

Case 2: (n,a) = (5,5):
x= 6a = 30
n=5

Plugging the values in our equations:
nx=150 => 5*30 = 150 (satisfied)
(n+5)(x-a)=150 => (5+5)(30-5) = 10*25 (not satisfied)

Case 3: (n,a) = (25,1):
x= 6a = 6
n=25

Plugging the values in our equations:
nx=150 => 25*6 = 150 (satisfied)
(n+5)(x-a)=150 => (25+5)(6-1) = 30*5 (satisfied)

So, the cost of the sandwich was 6$.
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