ps 500 test21 #19

[dunkin77]

Hi,

the answer is E) but could not get to it..... can anyone pls help?

19. A merchant paid $300 for a shipment of x identical calculators. The merchant used 2 of the calculators as
demonstrators and sold each of the others for $5 more than the average (arithmetic mean) cost of the x calculators. If the total revenue from the sale of the calculators was $120 more than the cost of the shipment, how many calculators were in the shipment?
(A) 24
(B) 25
(C) 26
(D) 28
(E) 30

[jayhawk2001]

Total num of calc sold = x-2
Arith mean = 300/x

(x-2)(300/x +5) = 120 + 300
5x -600/x = 130

x^2 - 26x - 120 = 0
(x-30)(x+4) = 0
so, x = 30 or x = -4

Hence E

[Cybermusings]

x identical calculators for a total of $300
Hence price of each = 300/x
Total revenue = 120 + 300 or 420
Price at which each is sold = 300/x + 5
(300/x + 5) (x - 2) - 300 = 120
(300/x + 5) (x - 2) = 420
x^2 - 26x - 120 = 0
x^2 - 30x + 4x - 120 = 0
x (x - 30) + 4 (x-30)
Hence x = -4 and x = 30
Since x can'e be negative x = 30