The owner of an art shop conducts his business in the following manner. Every once in a while he raises his prices fo x%, then a while later he reduces all the new prices by x%. After one such up-down cycle the price of a painting decreased by $ 441. After a second up-down cycle the painting was sold for $ 1944.81. What was the original price of the painting in $?
Answers
Answer:
$ 2756.25
Step-by-step explanation:
Let say Original Price of Painting = $ P
Price after x % increase = P + (x/100)P = P (1 + 0.01x) $
Price after then x % decrease = P (1 + 0.01x) (1 - 0.01x) $
= P ( 1 - 0.0001x) $
Net decrease in Price = P - P ( 1 - 0.0001x) = 0.0001Px $
0.0001Px = 441
=> Px = 4410000
After Second up down cycle Price will be (P - 441)(1 - 0.0001x)
P - 0.0001Px - 441 + 0.0441x = 1944.81
=> P - 441 - 441 + 0.0441x = 1944.81
=> P + 0.0441x = 2826.81
=> P + 441²/P = 2826.81
=> P² - 2826.81 P + 441² = 0
=> P² - 2756.25P - 70.56P + 441² = 0
=> P(P - 2756.25) - 70.56(P - 2756.25) = 0
=> (P - 70.56)(P - 2756.25) = 0
=> P = 2756.25 ( p = 70.56 not possible as can not be less than 1944.81)
Other way to solve
Let say after one up down of price price become a times the original price
Original Price = P
after 1 st up down Price = aP
after 2nd up down price = a²P
P - aP = 441
P(1-a) = 441 - eq 1
a²P = 1944.82 - eq2
eq2/eq1
=> a²/(1-a) = 4.41
=> a² = 4.41 - 4.41a
=> a² + 4.41a - 4.41 = 0
=> a² - 0.84a + 5.25a - 4.41 = 0
=> a(a -0.84) + 5.25(a - 0.84) = 0
=> (a - 0.84)(a + 5.25) = 0
=> a = 0.84 a = -5.25 not possible as a can not be -ve
P(1-a) = 441
=> P(1 - 0.84) = 441
=> P = 441/0.16
=> P = 2756.25
Answer:
Step-by-step explanation:
x% increase and decrease so by using successive % change
x-x-(x*x)/100= (-x^2/100)%
Let initial price be P
(-x^2/100)*(1/100)*P= 441 .....(1)
Let P1 be price after first cycle
P1 = P-441
FOR second up/down as well
Change =( -x^2/100)%
Now,
P-441 + (-x^2/100)*(1/100)*(P-441) = 1944.81
Putting values from (1) we get quadratic equation
P^2 -2826.81 P + 194481 =0 (solving it is
cumbersome)
We get P= 2756.25