The ratio of Artie’s CDs to Renato’s CDs was 8 to 3. Artie gave 1/2 of his CDs to Renato. He counted his remaining CDs and found that he now had 33 fewer CDs than Renato. How many CDs did Artie have at first?

Answer 1

At start:
A ---- 8x
R ---- 3x

after the gifting:
A ---- 8x - (1/2)(8x) = 4x
R ----- 3x + 4x = 7x

R - A = 33
7x - 4x = 33
x = 11

so Artie had 8x or 88
Renata had 3x or 33

check: A gives away 44, so has 44 left
R had 33 , gets 44, so now has 77 , which is 33 more than A
My answer is correct

Answer 2

A = Artie’s CDs

R = Renato’s CDs

The ratio of Artie’s CDs to Renato’s CDs was 8 to 3 means:

A / R = 8 / 3

Cross multiply

3 A = 8 R

A = 8 / 3 A

R = 3 / 8 A

After Artie gave 1 / 2 of his CDs to Renato he have A / 2 Cds.

His remaining = A / 2

Renato have R + A / 2 Cds

Artie counted his remaining CDs and found that he now had 33 fewer CDs than Renato means:

A / 2 = R + A / 2 - 33

Subtract A / 2 to both sides

0 = R - 33

R = 33

A = 8 / 3 R = 8 • 33 / 8 = 88

Renato at first have 33 CDs

Artie at first have 88 CDs

Check result:

A / R = 88 / 33 = 11 • 8 / 11 • 3 = 8 / 3

After Artie gave 1 / 2 of his CDs to Renato he have A / 2 Cds = 88 / 2 = 44 CDs.

Renato have R + A / 2 = 33 + 44 = 77 CDs

77 - 44 = 33