Tim and Joe had 354 marbles altogether. Tim lost 84 marbles to Joe in the first game. Joe lost 52 marbles to Tim in the second game. Joe then had twice as many marbles as Tim. How many marbles did Joe have at first?

Answer 1

t+j = 354
2(t-84+52) = j+84-52
solve for j

Answer 2

t = Tim's marbles at the beginning

j = Joe 's marbles at the beginning

Tim and Joe had 354 marbles altogether, means:

t + j = 354

Subtract j to both sides

t = 354 - j

When Tim lost 84 marbles, Tim has:

t - 84 marbles

Replace t by 354 - j

354 - j - 84 = 270 - j

Tim has 270 - j marbles

Joe has:

j + 84 marbles

When Joe lost 52 marbles Joe has:

j + 84 - 52 = j + 32 marbles

Tim has:

270 - j + 52 = 322 - j marbles

Joe then had twice as many marbles as Tim, means:

( j + 32 ) / ( 322 - j ) = 2

Multiply both sides by 322 - j

j + 32 = 2 ( 322 - j )

j + 32 = 644 - 2 j

Add 2 j to both sides.

3 j + 32 = 644

Subtract 32 to both sides.

3 j = 612

j = 612 / 3

j = 204

Joe have 204 marbles at first.

Check result.

t = 354 - j = 354 - 204 = 150

Tim have 150 marbles at first.

When Tim lost 84 marbles to Joe, Tim has:

150 - 84 = 66 marbles

Joe has:

204 + 84 = 288 marbles

When Joe lost 52 marbles Joe has:

288 - 52 = 236 marbles

Tim has:

66 + 52 = 118 marbles

236 / 118 = 2