Question

a retailer planned to buy some computers from a total of1800000.before the retailer could buy the computer the price was reduced by 4000.this reduction enabled the retailer to buy 5 more computers using the same amount of money originally planned.find number ofcomputers the retailer bought

Answer 1

Let the original price of each computer be \( P \). The retailer planned to buy \( N \) computers.

The total amount the retailer planned to spend is \( 1,800,000 \), so we have the equation:

\[
N \times P = 1,800,000
\]

After the price reduction of \( 4,000 \), the new price of each computer becomes \( P - 4,000 \). With this new price, the retailer is able to buy \( N + 5 \) computers for the same total budget. So we can set up a second equation:

\[
(N + 5) \times (P - 4,000) = 1,800,000
\]

Now we can simplify the first equation to express \( P \):

\[
P = \frac{1,800,000}{N}
\]

Substituting \( P \) in the second equation, we get:

\[
(N + 5) \left( \frac{1,800,000}{N} - 4,000 \right) = 1,800,000
\]

Expanding this, we have:

\[
(N + 5) \left( \frac{1,800,000 - 4,000N}{N} \right) = 1,800,000
\]

Multiplying both sides by \( N \) to eliminate the fraction:

\[
(N + 5)(1,800,000 - 4,000N) = 1,800,000N
\]

Expanding the left side:

\[
1,800,000N + 9,000,000 - 4,000N^2 - 20,000N = 1,800,000N
\]

Now, we can subtract \( 1,800,000N \) from both sides:

\[
9,000,000 - 4,000N^2 - 20,000N = 0
\]

Rearranging gives us a standard quadratic form:

\[
4,000N^2 + 20,000N - 9,000,000 = 0
\]

To simplify, we can divide the entire equation by 4,000:

\[
N^2 + 5N - 2,250 = 0
\]

Next, we can use the quadratic formula \( N = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Here, \( a = 1 \), \( b = 5 \), and \( c = -2,250 \):

First calculate the discriminant:

\[
b^2 - 4ac = 5^2 - 4 \times 1 \times (-2,250) = 25 + 9,000 = 9,025
\]

Now, taking the square root:

\[
\sqrt{9,025} = 95
\]

Now, substituting back into the quadratic formula:

\[
N = \frac{-5 \pm 95}{2}
\]

Calculating the two possible values:

1. \( N = \frac{90}{2} = 45 \)
2. \( N = \frac{-100}{2} = -50 \) (not a valid solution)

Thus, the retailer originally planned to buy \( N = 45 \) computers.

Finally, the number of computers the retailer actually bought after the price reduction is:

\[
N + 5 = 45 + 5 = 50
\]

So, the retailer bought **50 computers**.