Asked by Anonymous

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?

Answers

Answered by mathhelper
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

<b>so Artie had 8x or 88
Renata had 3x or 33</b>

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
Answered by Bosnian
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













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