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

Answer:

The original duration of the tour  is 18 days.

Step-by-step explanation:

Let

the number of days spent be T

Daily expenses be D

Initially ,

total tour expenses was 360

$D \times T=360$

$D = \frac{360}{T}$-------------(1)

If he extends his tour for 4 days, he has to cut down his daily expenses by dollar 3

$(D + 4) \cdot(T -3) = 360$---------------(2)

Expanding the eqaution(2)

DT-3D + 4T -12 = 360

Substituting the value of D

$(\frac{360}{T})T -3( \frac{360}{T}) + 4T -12 = 360$

$(\frac{360}{T})T -3( \frac{360}{T}) + 4T= 360 + 12$

$(\frac{360}{T})T -3( \frac{360}{T}) + 4T= 372$

$360 - ( \frac{1080}{T}) + 4T= 372$

$( \frac{1080}{T}) + 4T= 372 - 360$

$( \frac{1080}{T}) + 4T= 12$

$1080 + 4T^2= 12T$

Solving the quadratic equation we get

T = 18  and T = -15

T is the number of gays and it cannot be negative , hence T is 18

Substituting in equation (1) we get

$D = \frac{360}{18}$

D =  20