Question

Both questions please

Answer 1

Let there be
`h` 50 cent coins (call them halfs),
`d` 10 cent coins (dimes) and
`n` 5 cent coins (nickles).

81 coins:


`h+d+n=81`

Three more dimes than halfs:


`d=h+3`

Nickels are twice halfs plus dimes:


`n=2(h+d)`

Three equations in three unknowns. We'll eliminate
`n` first:


`n=81-(h+d)=2(h+d)`


`81=3(h+d)`


`h+d=27`

Substituting
`d=h+3`


`h+h+3=27`


`2h=24`


`h=12`


`d=h+3=15`


`n=2(h+d)=2(27)=54`

Check we have the right number of coins:


`h+d+n=12+15+54=81 \quad\checkmark`

Good. The amount of money is


`50h+10d+5n=0.50(12)+0.10(15)+0.05(54)=\$10.20`

Part 2.

a for amount Tim gave cousin. Initially Tim has $146.65, cousin $10.55

After he gave the money Tim had twice as much:


`146.65 - a = 2(10.55 + a)`


`146.65 - a = 21.10 + 2a`


`146.65 - 21.10 = 3a`


`a = (146.65 - 21.10)/(3)= 41.85`

That's the answer to b. After the transfer the cousin has
`10.55+41.85=\$52.40`