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?
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
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