Quote:
Originally Posted by ashishjha100 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
Pl help..... |
@ashishjha100 -
I hope this will help -
Avg (Arithmetic mean) cost for x calculators = 300/x
Selling price for 1 calculator = (300/x) + 5 (5 more than avg price of calculator)
No. of calculators sold by merchant = x-2 (2 less than no. of calculators bought)
Total revenue from the sale = 300 + 120 = 420
we know that - Total revenue = Selling price for 1 calculator * No. of calculators sold
=> [(300/x) + 5] * [x-2] = 420
After simplifying quadratic equation, you'll get 30 as the answer. Also, you can try putting 30 directly in the above equation.
P.S - I read somewhere that when you need to chose and eliminate the choices, start from E to A.