Question

A person on tour has dollar 360 for his daily expenses. If he extends his tour for 4 days, he has to cut down his daily expenses by dollar 3. find the original duration of the tour. (Hint use quadratic equation)

Answer 1

The original duration of the tour = 20 days

Explanation:

Solution:

Total expenses for the tour = $360

Let the original tour duration be for
`x` days.

So, for
`x` days the total expense = $360

Thus the daily expense in dollars can be given by =
`(360)/(x)`

Tour extension and effect on daily expenses.

The tour is extended by 4 days.

Tour duration now =
`(x+4)` days

On extension, his daily expense is cut by $3

New daily expense in dollars =
`((360)/(x)-3)`

Total expense in dollars can now be given as:
`(x+4)((360)/(x)-3)`

Simplifying by using distribution (FOIL).

`(x.(360)/(x))+(x(-3)+(4.(360)/(x))+(4(-3))`

`360-3x+(1440)/(x)-12`

`348-3x+(1440)/(x)`

We know total expense remains the same which is = $360.

So, we have the equation as:

`348-3x+(1440)/(x)=360`

Multiplying each term with
`x` to remove fractions.

`348x-3x^2+1440=360x`

Subtracting
`348x` both sides

`348x-348x-3x^2+1440=360x-348x`

`-3x^2+1440=12x`

Dividing each term with -3.

`(-3x^2)/(-3)+(1440)/(-3)=(12x)/(-3)`

`x^2-480=-4x`

Adding
`4x` both sides.

`x^2+4x-480=-4x+4x`

`x^2+4x-480=0`

Solving using quadratic formula.

For a quadratic equation:
`ax^2+bx+c=0`

`x=(-b\pm√(b^2-4ac))/(2a)`

Plugging in values from the equation we got.

`x=(-4\pm√((4)^2-4(1)(-480)))/(2(1))`

`x=(-4\pm√(16+1920))/(2)`

`x=(-4\pm√(1936))/(2)`

`x=(-4\pm44)/(2)`

So, we have

`x=(-4+44)/(2)` and
`x=(-4-44)/(2)`

`x=(40)/(2)` and
`x=(-48)/(2)`

∴
`x=20` and
`x=-24`

Since number of days cannot be negative, so we take
`x=20` as the solution for the equation.

Thus, the original duration of the tour = 20 days