A shop owner spent $540 to purchase a stock of computer keyboards. If the price of each keyboard had been reduced by $2, he could have bought 3 more keyboards. Find the price of one keyboard.

Answer:

Price before discount = $20 per keyboard

Price after discount = $18 per keyboard

Before the discount, you can buy 27 keyboards. After the discount, you can buy 30 keyboards.

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Work Shown:

k = cost of one keyboard before the price reduction

540/k = amount of keyboards purchased before the price reduction

k-2 = cost of one keyboard after the price reduction

540/(k-2) = amount of keyboards purchased after the price reduction

540/(k-2) = (540/k) + 3

————–

If you multiply both sides by the LCM k(k-2), then you’ll clear out the fractions and we can solve for k like so

540/(k-2) = (540/k) + 3

540k = 540(k-2) + 3k(k-2)

540k = 540k – 1080 + 3k^2 – 6k

0 = 540k – 1080 + 3k^2 – 6k – 540k

0 = 3k^2 – 6k – 1080

3k^2 – 6k – 1080 = 0

3(k^2 – 2k – 360) = 0

k^2 – 2k – 360 = 0

(k – 20)(k + 18) = 0

k-20 = 0 or k+18 = 0

k = 20 or k = -18

We ignore the negative result because a negative price doesn’t make sense.

————–

If k = 20, then

540/k = 540/20 = 27

Meaning that you can buy 27 keyboards before the price reduction

In other words, (27 keyboards)*(20 dollars per keyboard) = 540 dollars total.

After the price reduction, the cost per keyboard is now k-2 = 20-2 = 18

We can now buy 540/(k-2) = 540/18 = 30 keyboards, which is an increase of 30-27 = 3 extra keyboards. This helps confirm we have the correct answer.

Answer:Price before discount = $20 per keyboardPrice after discount = $18 per keyboard

Before the discount, you can buy 27 keyboards. After the discount, you can buy 30 keyboards.

===============================================

Work Shown:

k = cost of one keyboard before the price reduction

540/k = amount of keyboards purchased before the price reduction

k-2 = cost of one keyboard after the price reduction

540/(k-2) = amount of keyboards purchased after the price reduction

540/(k-2) = (540/k) + 3

————–

If you multiply both sides by the LCM k(k-2), then you’ll clear out the fractions and we can solve for k like so

540/(k-2) = (540/k) + 3

540k = 540(k-2) + 3k(k-2)

540k = 540k – 1080 + 3k^2 – 6k

0 = 540k – 1080 + 3k^2 – 6k – 540k

0 = 3k^2 – 6k – 1080

3k^2 – 6k – 1080 = 0

3(k^2 – 2k – 360) = 0

k^2 – 2k – 360 = 0

(k – 20)(k + 18) = 0

k-20 = 0 or k+18 = 0

k = 20 or k = -18

We ignore the negative result because a negative price doesn’t make sense.

————–

If k = 20, then

540/k = 540/20 = 27

Meaning that you can buy 27 keyboards before the price reduction

In other words, (27 keyboards)*(20 dollars per keyboard) = 540 dollars total.

After the price reduction, the cost per keyboard is now k-2 = 20-2 = 18

We can now buy 540/(k-2) = 540/18 = 30 keyboards, which is an increase of 30-27 = 3 extra keyboards. This helps confirm we have the correct answer.

Answer:$20

Step-by-step explanation:X: the price of a keyboard

Y: number of keyboards to buy (original)

The shop owner spent $540 to purchase a stock of computer keyboards, so:

XY=540

⇒X=540/Y

If the price of each keyboard had been reduced by $2, he could have bought 3 more keyboards:

(X-2)(Y+3)=540

⇒3X-2Y=6

⇒3.540/Y – 2Y=6

⇒Y=27

⇒X=20

⇒ the price of one keyboard: $20