percentage problem (powerprep)

[ma200]

The original value of machine X is V dollars, while the original value of machine Y is 2V dollars. Both machines depreciate in value at a constant rate of 10 percent of their original value per year.

A: value of machine X after 3 years
B: value of machine Y after 6 years

OA = B

is there a quicker way to do this question?

[walt526]

Let V=100

X after 3 years
Year 0: 100
Year 1: 90
Year 2: 81
Year 3: ~73

Y after 6 years
Year 0: 200
Year 1: 180
Year 2: 162
Year 3: ~146
Year 4: ~131
Year 5: ~118
Year 6: ~106

That took less than 1 minute (they key is to not worry too much about precise values). Of course, another way to do it would be to just look at Y after 3 years and realize that at 3 years to go (~146) it is much larger than the initial value of X...

[ved_theone]

Originally Posted by walt526

Let V=100

X after 3 years
Year 0: 100
Year 1: 90
Year 2: 81
Year 3: ~73

Y after 6 years
Year 0: 200
Year 1: 180
Year 2: 162
Year 3: ~146
Year 4: ~131
Year 5: ~118
Year 6: ~106

That took less than 1 minute (they key is to not worry too much about precise values). Of course, another way to do it would be to just look at Y after 3 years and realize that at 3 years to go (~146) it is much larger than the initial value of X...

+1, Cant get simpler than tha!!!

[nicrey]

Originally Posted by ma200

The original value of machine X is V dollars, while the original value of machine Y is 2V dollars. Both machines depreciate in value at a constant rate of 10 percent of their original value per year.

A: value of machine X after 3 years
B: value of machine Y after 6 years

OA = B

is there a quicker way to do this question?

X after 3 years = 0.7 V

Y after 6 years = 0.4* 2V = 0.8V

they "depreciate in value at a constant rate of 10 percent of their original value".

They do not depreciate 10% with respect to the value at the beginning of the year.

[munimM]

after 3 years x=v(.9)^3
after 6 years y=2v(.9)^6....now X=1,Y=2(.9)^3=2*(.729)=1.4