Help with simple math question?

Question

I have these two questions from an ATB practice test on line. Although I got the answers I did not know how to set it up algebraically, I would really appreciate the help, thanks.

4. After having to pay increased income taxes this year, Edmond has to sell his BMW. Edmond bought the car for $49,000, but he sold it for a 20% loss. What did Edmond sell the car for?

A. $24,200

B. $28,900

C. $35,600

D. $37,300

E. $39,200

12. A computer is on sale for $1600, which is a 20% discount off the regular price. What is the regular price?

A. $1800

B. $1900

C. $2000

D. $2100

E. $2200

Answer 1

4.

$49,000 * .20 = 9,800

$49,000 - 9,800 = $39,200

Answer: E

12.

x = original cost

100% - 20% = 80%

You have to think of it as 1600 is 80% of the original cost

$1,600 = .80x

x = $2,000

Answer: C

Answer 2

E. $39200.

(.80 * $49000.)

C.$2000.

($1600/.8 )

Answer 3

4. 49000-0.2(49000)= 39,200

12. 1600 = X - 0.2(X)

      Solving for X,

      you get X = 2000

thats it.

Answer 4

is/of = %/100

N/49000=20/100   (cross multiply)

100/980000=9800  (divide)

now subtract

49000-9800=39200

is/of=%/100

is always how you set it up when doing percents.

Hope this helps

Answer 5

A. e

B. c

Answer 6

4)  car price = 49,000

20% in car price = 49000*0.2 = 9800

so edmond sell the car for = 49000-9800= 39200 dollars

so E) is the answer

12)  1600 is after discount

if you take x as regular price

orginal price-discount =1600

x-0.2x=1600

so orignal or regular price is (x-0.2x) = 1600

0.8x = 1600

x= 1600/0.8 = 2000 dollars

so C) is correct

Answer 7

4. x = 49000*(80)

12. 0.8x = 1600

      x = 1600/0.8