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A taqueria has exactly two items on its menu, a taco and a burrito. On a particular day, it sold 600 items, exactly half of which were tacos. What was the taqueria’s revenue on that day?

Number of taco sold = 300 & number of burrito = 300
Let the price of each taco = t & that of burrito = b

(1) The average price of an item on the menu is $4.50.
--> (300t + 300b)/600 = 4.5
--> 300(t + b) = 2700
--> t + b = 9
Total revenue = 300t + 300b = 300(t + b) = 300*9 = $ 2,700 --> Sufficient

(2) A burrito costs twice as much as a taco.
Given, b = 2t
Total revenue = 300t + 300b = 300(t + b) = 300*3t = $ 900t --> Insufficient

Option A
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Quote:

A taqueria has exactly two items on its menu, a taco and a burrito. On a particular day, it sold 600 items, exactly half of which were tacos. What was the taqueria’s revenue on that day?

(1) The average price of an item on the menu is $4.50.
(2) A burrito costs twice as much as a taco.

300t+300b=300(t+b)=?

(1) sufic
t+b/2=4.5, t+b=9
300(t+b)=300(9)=$2700

(2) insufic

Ans (A)
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taco= t , burrito = b
and price of taco= x and price of burrito = y
we have t= b=300
need to find t*x
#1
The average price of an item on the menu is $4.50.
t*x+b*y = 4.5*600
t*x+b*y= 2700
insufficient
#2
A burrito costs twice as much as a taco.
y=2x
insufficient

from 1 &2
t*x+b*y= 2700
300*(x+2x)=2700
3x= 9
x= 3
so total revenue = 300*3 ; 900$
IMO C sufficient

A taqueria has exactly two items on its menu, a taco and a burrito. On a particular day, it sold 600 items, exactly half of which were tacos. What was the taqueria’s revenue on that day?

(1) The average price of an item on the menu is $4.50.
(2) A burrito costs twice as much as a taco.
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Quote:
A taqueria has exactly two items on its menu, a taco and a burrito. On a particular day, it sold 600 items, exactly half of which were tacos. What was the taqueria’s revenue on that day?

(1) The average price of an item on the menu is $4.50.
(2) A burrito costs twice as much as a taco.

number of Tacos = number of Burritos = 300
let the cost of a Tacos be x and Burrito be y.

statement 1:
4.50 = \(\frac{300(x+y)}{600}\)
x+y = 9
not sufficient

statement 2:
y = 2*x
not sufficient

combining both statements,
3*x = 9; x = 3 and y = 6
taqueria’s revenue = 300(3+6)= 2700
Ans: C
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300 Tacos and 300 Burritos
--> The price of Taco - x
--> The price of Burrito - y
300x + 300y = 300(x+y)=???

(Statement1): The average price of an item on the menu is $4.50.
--> (x+ y)/2 = 4.5 --> x+y = 9
300(x+y)= 300*9 = 2700
Sufficient

(Statement2): A burrito costs twice as much as a taco.
--> y =2x
Nothing tells us about the prices of items
Insufficient

Answer(A).
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Answer is : A
We need the total revenue
we know, half of total (600 items) are tacos, i.e. 300 tacos and 300 burritos were sold
lets assume tacos each cost a$ and burritos each cost b$
we need to find (300a+300b) which is the total revenue
option 1:-
since there are only two items on the menu,
the avg price of an item on the menu will be=(a+b)/2
so we are given this as 4.5$
thus we know a+b= 2*4.5=9
thus we can caluclate total revenue as 300*(a+b)=2700$(you don't need to make the calculations)
therefore option 1 is sufficient to answer
Option2:-
we know that b=2a,
but we don't know anything else about a or b to calculate either values.
Thus this info isn't sufficient to calculate the total revenue ,300(a+b).
therefore answer is A
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