Alice had twice as much money as Peter. After Alice spent 1/6 of her money

Question

and Peter spent
2/3 of his money, they had a total of $720 left.
(a) How much money did Alice and Peter have altogether at first?

(b) Peter spent 3/4 of his remaining money on a bag. What fraction of his
original amount of money did he spend on the bag?
(Give your answer in its simplest form.)

Answer 1

I will solve part (a) only.
Let x be the amount Peter had at first, in dollars.

Then Alice had 2x dollars.

Write equation for the total remaining money after spending, as you read the problem

(5/6)(2x) + (1/3)x = 720 dollars.

To solve, multiply both sides of the equation by 6 and simplify

10x + 2x = 6*720

12x = 6*720

x = (6*720)/12 = 6*60 = 360.

Peter had $360 at first; Alice had twice of it, i.e. $720 dollars.

Altogether, they had 360 + 720 = 1080 dollars, at first.

Answer 2

(a)
Let the money Peter had be y, therefore, Alice had 2y.
Total money= (y+2y)=3y

Alice spent (1/6 of 2y)=1/3y
Peter spent (2/3 of y)=2/3y
Total money spent= (1/3y+2/3y)=y

Total money left= (3y-y)=2y

2y=$720 (divide both sides by 2)
y=$360

Peter had $360
Alice had (2×360)=$ 720

Hence, total money is (360+720) =$1080

(b)
After Peter spent 2/3, the remaining fraction of money is (1-2/3)=1/3
(1/3 of $360) =$120

Money spent on the bag =(3/4 of $120)=$90

Therefore, fraction of the amount spent on bag over the original amount
=90/360
=1/4