A bookstore can purchase several calculators for a total cost of $350. If each calculator cost $2 less, the bookstore could purchase 20 additional calculators at the same total cost. How many calculators can be purchased at the regular price?
Sorry, made an error.
once you complete the square:
(y+10)^2 = 3600
solving for y gives you 50
so 50 calculators.
you could also used the quadratic formula to solve for y and get the same answer.
x=cost of calculator
y=number of calculators
350=(x-2)*(20+y) <-- equation 1
we know that
total cost = cost per calc * number of calc bought
350 = x*y
so
350/y = x <-- equation 2
we can substitute x from equation 2 into equation 1
350=((350/y)-2)*(20+y)
solving this you get:
y^2+20y = 3500
by completing the square, add 100 to each side:
y^2+20y+100 = 3600
factoring this gives:
(y^2+10)(y^2+10)=3600
y^2+10 = 3600
y = 59.91
which rounds to 60 calculators
Let's assume that the regular price of each calculator is "x" dollars.
According to the given information, the bookstore can purchase several calculators for a total cost of $350. This can be expressed as:
x * n = 350 (where n is the number of calculators purchased at the regular price)
If each calculator cost $2 less, the bookstore could purchase 20 additional calculators at the same total cost. This can be expressed as:
(x - 2) * (n + 20) = 350
Now we have two equations:
x * n = 350 --> Equation 1
(x - 2) * (n + 20) = 350 --> Equation 2
To solve these equations, let's simplify Equation 2:
x * n + 20x - 2n - 40 = 350
Now we can substitute Equation 1 into Equation 2:
350 + 20x - 2n - 40 = 350
Simplifying further:
20x - 2n - 40 = 0
20x - 2n = 40
Now let's rewrite Equation 1 as:
n = 350 / x --> Equation 3
Substituting Equation 3 into Equation 2:
20x - 2(350 / x) = 40
Simplifying further:
20x - 700 / x = 40
Multiplying throughout by x to eliminate the fraction:
20x^2 - 700 = 40x
Rearranging the terms:
20x^2 - 40x - 700 = 0
Dividing throughout by 20 to simplify:
x^2 - 2x - 35 = 0
Now we can solve this quadratic equation to find the value of x (the regular price of each calculator).
Using factoring or the quadratic formula, we find that x = 7 or x = -5.
Since the price cannot be negative, the regular price of each calculator is $7.
Now let's substitute x = 7 into Equation 1 to find the number of calculators purchased at the regular price:
7 * n = 350
Dividing both sides by 7:
n = 50
Therefore, the bookstore can purchase 50 calculators at the regular price of $7 each.
To solve this problem, let's break it down into steps:
Step 1: Let's assume the number of calculators that can be purchased at the regular price is "x".
Step 2: The cost per calculator at the regular price is "y".
Step 3: Therefore, the total cost of purchasing "x" calculators at the regular price would be "x * y".
According to the given information:
1. A bookstore can purchase several calculators for a total cost of $350. So, we have the equation: x * y = 350.
Next, let's consider the second piece of information:
2. If each calculator cost $2 less, the bookstore could purchase 20 additional calculators at the same total cost. This means, the cost per calculator at the reduced price would be "y - 2", and the number of calculators that can be purchased would be "x + 20". So, we have the equation: (x + 20) * (y - 2) = 350.
Now, we can solve these two equations to find the values of "x" and "y" and determine the number of calculators that can be purchased at the regular price.
Equation 1: x * y = 350
Equation 2: (x + 20) * (y - 2) = 350
Solving these two equations will give us the values of "x" and "y", and thus the number of calculators that can be purchased at the regular price.