A coffee shop sells small cups of coffee and large cups of coffee.
A small cup of coffee costs $2.
A large cup of coffee costs $3.
On Tuesday, the store sold a total of 155 cups of coffee for a total of $335.
How many small cups of coffee did the store sell on Tuesday?
Conventional algebra:
x=number of small cups
y=number of large cups
form the system of equations
x+y=155
2x+3y=335
Solve the system for x and y.
Quick way by substituting directly:
Let 155-x=number of large cups
2x+3(155-x)=335
Solve the linear equation for x.
Faster way by mental calculations:
Assume all sold are small,
then proceeds
=2x155=310
Money left-over = $335-310 = 25
Exchange 25 small cups for 25 large cups for $25, so
small cups = 155-25=130
large cups = 25.
To solve this problem, we can set up a system of equations using the given information.
Let's say the number of small cups of coffee sold is "s" and the number of large cups of coffee sold is "l". We know that the total number of cups of coffee sold is 155, so we can write the equation:
s + l = 155 -- Equation 1
We also know that the total revenue from selling small cups of coffee is $2 multiplied by the number of small cups sold, and the total revenue from selling large cups of coffee is $3 multiplied by the number of large cups sold. The total revenue from both types of coffee is $335, so we can write the equation:
2s + 3l = 335 -- Equation 2
Now we can solve this system of equations to find the value of "s".
One way to solve this system is by using substitution. Solving Equation 1 for "l", we get:
l = 155 - s
Substituting this expression for "l" into Equation 2, we have:
2s + 3(155-s) = 335
Simplifying this equation:
2s + 465 - 3s = 335
-s = 335 - 465
-s = -130
Simplifying further:
s = 130
Therefore, the coffee shop sold 130 small cups of coffee on Tuesday.