SOLUTION: Erin wants to reduce both the length and width of a picture by 1/3. If the original picture is 12 inches by 6 inches, what will be the dimension, in inches, of the reduces picture

Algebra ->  Percentage-and-ratio-word-problems -> SOLUTION: Erin wants to reduce both the length and width of a picture by 1/3. If the original picture is 12 inches by 6 inches, what will be the dimension, in inches, of the reduces picture      Log On


   

Question 35584: Erin wants to reduce both the length and width of a picture by 1/3. If the original picture is 12 inches by 6 inches, what will be the dimension, in inches, of the reduces picture?
a. 18 by 9
b. 12 by 4
c. 9 by 4
d. 8 by 6
e. 8 by 4
Found 2 solutions by Earlsdon, stanbon:
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Erin wants to reduce the original length (L) by one third, so we can write the new length (nL) as:
nL+=+L+-+L%2F3%29
The original length, L = 12 inches.
nL+=+12+-+12%2F3
nL+=+12+-+4
nL+=+8inches.
Likewise for the original width (W) which is 6 inches. Let the new width be nW.
nW+=+W+-+W%2F3
nW+=+6+-+6%2F3
nW+=+6+-+2
nW+=+4inches.
The dimensions of the reduced picture are:
8 inches by 4 inches.

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Erin wants to reduce both the length and width of a picture by 1/3. If the original picture is 12 inches by 6 inches, what will be the dimension, in inches, of the reduces picture?
a. 18 by 9
b. 12 by 4
c. 9 by 4
d. 8 by 6
e. 8 by 4
Erin reducing the length and the width by 1/3.
So she will still have 2/3 of the length and 2/3 of the width.
(2/3)12= 8 inches
(2/3)6 = 4 inches
Cheers,
Stan H.