Simple Math Question (: (aM Ii right?)?

Question

Here's the question that i cannot form and equation with.

Paul have some marbles. He gave 1/4 of them to Peter and 3/7 of the remainder to Mary. If he had 312 marbles left, how many marbles did he have at first?

Please give me the answer using algebra or workings. Please do not give model methods cause my school dosen't allow. But i tried using the model method and i'm positive the answer is 728. Am i right?

Answer 1

Let: x = number of marbles Paul had at first.

(1/4)x = number of marbles given to Peter

(3/7)[x - (1/4)x] = number of marbles given to Mary

Equation: x - (1/4)x - (3/7)[x - (1/4)x] = 312

Solution: x - (1/4)x - (3/7)x + (3/28)x = 312

Multiply each term by 28 to remove the fractions, then solve for x:

28x - 7x - 12x + 3x = 312(28)

12x= 8736

x = 8736/12

x = 728

(1/4)x = (1/4)728= 182 = number of marbles given to Peter

728 - 182 = 546 = remainder

(3/7)546 = 234 = number of marbles given to Mary

To check:

728 - 182 - 234 = 312

312 = 312 Yes!

Therefore, Paul had 728 marbles at first.

YOU'RE RIGHT. CONGRATULATIONS!

Answer 2

Yes. You are correct.

If you set X as number of marbles that Paul originally has.

312 = (1 - (1/4) - (3/4)*(3/7))*X

312 = (3/7) * X

X = 312 / (3/7)

You will get X = 728

Answer 3

Let Paul have x marbles

Peter gets x/4 marbles

Remainder = 3x/4 marbles

Mary gets (3/7) (3x/4) = 9x / 28 marbles

Number given away = x/4 + 9x/28 = 7x/28 + 9x/28 = 16x/28

Remainder = x - 16x/28 = 12x/28

12x/28 = 312

x = 312 x 28 / 12

x = 728

Answer 4

let initial amt of marble = x

amount he gave to peter = x/4

amt left = x(1-(1/4)) = 3x/4

amt he gave to mary = (3/7)*(3x/4) = 9x/28

amt left = (3x/4)*(1-(3/7)) = (3x/4)*(4/7) = 3x/7

3x/7= 312

x = 312*7/3 = 728 <--     seems like you're right