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805+ (Hard)|   Word Problems|      
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sujoykrdatta
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At a sandwich shop, all sandwiches sell for the same price. If the sho 04 Jan 2025, 05:59
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AvidDreamer09
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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It doesn't say that the price has reduced by $1
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At a sandwich shop, all sandwiches sell for the same price. If the sho 04 Jan 2025, 06:01
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jm2k5
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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x^2 - ax + 30a = 0

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

This is not correct - If you had 30a^2 in the first step, then it would have been fine.
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At a sandwich shop, all sandwiches sell for the same price. If the sho 20 Apr 2026, 05:47
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