Question

QUESTION 11.1

Answer 1

The group paid $ 5250 at first city and $ 6250 at second city

Solution:

Let x = the charge in 1st city before taxes

Let y = the charge in 2nd city before taxes

The hotel charge before tax in the  second city was $1000 higher than in the first

Then the charge at the second hotel before tax will be x + 1000

y = x + 1000 ----- eqn 1

The tax in the first city was 8.5% and the  tax in the second city was 5.5%

The total hotel tax paid for the two cities was $790

Therefore, a equation is framed as:

8.5 % of x + 5.5 % of y = 790

$\frac{8.5}{100} \times x + \frac{5.5}{100} \times y = 790$

0.085x + 0.055y = 790 ------- eqn 2

Let us solve eqn 1 and eqn 2

Substitute eqn 1 in eqn 2

0.085x + 0.055(x + 1000) = 790

0.085x + 0.055x + 55 = 790

0.14x = 790 - 55

0.14x = 735

x = 5250

Substitute x = 5250 in eqn 1

y = 5250 + 1000

y = 6250

Thus the group paid $ 5250 at first city and $ 6250 at second city