TV problem

yalanand wrote:For every x TVs the factory makes, it costs 100x + 100sqrt(x) + 9000 dollars total in production costs. The factory sells each TV for $200. How many TVs must be sold to meet production costs?

(A) 81
(B) 90
(C) 99
(D) 100
(E) 108

IMO D



if x are the number of TVs therefore from eqn

100x + 100sqrt(x) + 9000 = 200x - (1)

let sqrt(x) = y
then x = y^2

putting value in (1)

100y^2 + 100y+ 9000 = 200y^2

simplifying

100y+9000 = 100y^2

dividing by 100

y+90=y^2

y^2 - y -90 = 0

y^2 - 10y + 9y - 90 =0

y = 10, -9

cannot be -ve

therefore y =10

hence x = 100

100x + 100sqrtx +9000 = 200x
100sqrtx+9000 = 100x
100x-100sqrtx = 9000
x-sqrtx = 90

At this point, I scanned the answers to see which one worked. Only D (100) works. 100 - 10 = 90.

Answer is D.

-Carrie

gaggleofgirls wrote:100x + 100sqrtx +9000 = 200x
100sqrtx+9000 = 100x
100x-100sqrtx = 9000
x-sqrtx = 90

Continuing from above,

x - sqrt (x) = 90
or, x - 90 = sqrt (x)
Squaring both sides,
x^2 -180x +8100 = x
or, x^2 -181x +8100 = 0
or, x^2 - 100x -81x +8100 = 0
or, x (x-100) -81 (x-100) = 0
or, (x-81)(x-100)=0
=> x = 81 or 100


I know I am somewhere making a mistake. Could you please help me point out that one?

Folks, I think we can avoid quadratic equation here...

Given 100x + 100sqrt(x) + 9000 = 200x

i.e 100x - 100sqrt(x) = 9000

i.e 100sqrt(x)[sqrt(x) - 1] = 9000 ( taking 100sqrt(x) common)

i.e sqrt(x)[sqrt(x)-1] = 90.

Let k = sqrt(x)

k(k-1) = 90.

So 90 must be written as porduct of two consecutive numbers which is 10x9

Hence k = 10, i.e sqrt(x) =10.

Thus x = 100.

To be quick in answering such questions, I usually like plugging-in from the answer choices.

In this case, I quickly checked my answer choices & liked the option D to start with for easy calculation. So, if x=100, then the Production costs for 100 TVs is = 100x100 + 100xsqrt(100)+9000. This solves to $20,000.

The factory sells each TV for $200. So, the selling price is= 100x$200 ==> $20,000. BINGO !!!

D is the answer.