Solving Equation (Application)

[Drain Brain]

a man spent 78 dollars for cigarettes. has the price per box been .50 cents less, he could have had one more box. How many boxes did he buy?

this is where I can get to

let $x=$ number of box bought

$\frac{78}{x}=$ price per box

$\frac{78}{x}-.50=(x+1)$

solving for x I get 177.27

but I know I should've got a whole number answer . please tell me my mistake.

 

[kaliprasad]

a man spent 78 dollars for cigarettes. has the price per box been .50 cents less, he could have had one more box. How many boxes did he buy?

this is where I can get to

let $x=$ number of box bought

$\frac{78}{x}=$ price per box

$\frac{78}{x}-.50=(x+1)$

solving for x I get 177.27

but I know I should've got a whole number answer . please tell me my mistake.

you are right till the point
reduced price per box = $\frac{78}{x}-.50$

then you intemixed the price and number of boxes

now the number of boxes = x + 1

so cost = $78 = (\frac{78}{x}-.50)(x+1) $

now you should be able to take on from here

 

[Drain Brain]

the answer is 12 boxes right?

 

[Deveno]

I don't understand what you did.

For one thing $\dfrac{78}{x} - 0.50$ is a price (per box), while $x + 1$ is a number of boxes.

This is what you want to do:

Let $x$ = number of boxes, with $y$ = price per box.

We know that:

$xy = 78$, so $y = \dfrac{78}{x}$, so far this agrees with what you did.

Next, we know that:

$(x + 1)(y - 0.50) \leq 78$ (we don't know that if the cigarettes were 50 cents less per box, he could have bought EXACTLY one more box with nothing left over).

This becomes:

$(x + 1)(\frac{78}{x} - \frac{1}{2}) \leq 78$.

I recommend you multiply this out, and multiply both sides by $-2x$ (this will reverse the inequality).

 

[Drain Brain]

I just followed kaplrasd's suggestion

$78=\left(\frac{78}{x}-\frac{1}{2}\right)(x+1)$
.
.
.
$-0.50x^2-0.5x+78=0$ multiply this by -10

I get

$5x^2+5x-780=0$

solving for x using quadratic formula

I get

$x=12$ discarding the negative root.

 

[Deveno]

And that is "OK" in the sense that it is barely the right answer.

As I pointed out before, this is partly because the problem is imprecisely worded.

The difference between kaliprasad's answer and mine is that with his, you eventually wind up with the quadratic equation:

$x^2 + x - 156 = 0$

and with mine, you wind up with the quadratic inequality:

$x^2 + x - 156 \geq 0$.

Either way, the LHS factors as:

$x^2 + x - 156 = (x - 12)(x + 13)$.

The negative root can clearly be discarded.

With the inequality:

$(x - 12)(x + 13) \geq 0$

we must have either:

$x \geq 12$ AND $x \geq -13$, which boils down to simply: $x \geq 12$

OR:

$x \leq 12$ AND $x \leq - 13$, which amounts to just: $x \leq -13$, which is impractical, given that our solution should be positive.

To see what I mean by using the inequality instead of the equation, suppose $x = 13$.

This means that the cigarettes cost 6 dollars a box.

If the cigarettes only cost 5.50, the man would be able to buy 14 boxes for 77 dollars, which still satisfies the conditions laid out in the original problem.

 

[kaliprasad]

I just followed kaplrasd's suggestion

$78=\left(\frac{78}{x}-\frac{1}{2}\right)(x+1)$
.
.
.
$-0.50x^2-0.5x+78=0$ multiply this by -10

I get

$5x^2+5x-780=0$

solving for x using quadratic formula

I get

$x=12$ discarding the negative root.

above is the answer.