Q30:
A rectangular photograph is surrounded by a border that is 1 inch wide on each side. The total area of the photograph and the border is M square inches. If the border had been 2 inches wide on each side, the total area would have been (M + 52) square inches. What is the perimeter of the photograph, in inches?
A. 34
B. 36
C. 38
D. 40
E. 42
solution is below
Area of photograph alone = LW
Area of photograph + 1inch border = LW + (L+2)(W+2) (We add 2 to the length and width because the 1 inch border adds 1 inch to EACH SIDE of the picture width wise and length wise!)
So M= LW + (L+2)(W+2)
2 inch border:
(We will be adding 4 to the length and width this time since the border adds 2 inches to EACH SIDE of the picture width wise and length wise)
M+52 = LW + (L+4)(W+4)
We can replace the ‘M’ above with the 1st equation, giving us:
LW + (L+2)(W+2) + 52 = LW + (L+4)(W+4) …..After canceling, we are left with 2L + 2W = 40. The equation for the perimeter of rectangle is 2L + 2W or 2(L+W)
OA is 40 I think it is 38
(l+2)(b+2)=m+52
b+l=21 so times two is total perimter minus 4 outside =38
what am i doing wrong
thanks
Khurram
rectangular photograph-Gmat Prep -OA incorrect??
This topic has expert replies
hi khurram, your name means to be happy, any way, when I solved the problem, i got it 40;
if let's say if 1 inch is added, it will be added 2 to to each side of rectangle.
so, W+2, and, L+2,
in oreder not to confuse, W+2=x and, L+2=y
here, x*y=M
if 1 inch is added again, 2 inch will be added to both side, then
(x+2)(y+2)=M+52;
xy+2x+2y+4=M+52;
xy=M, cancel them,
2x+2y=48
x+y=48/2=24
now plug them with the sides of photograph, which is:
W+2+ L+2=24
W+L=20
perimeter is 2(W+L)=20*2=40
if let's say if 1 inch is added, it will be added 2 to to each side of rectangle.
so, W+2, and, L+2,
in oreder not to confuse, W+2=x and, L+2=y
here, x*y=M
if 1 inch is added again, 2 inch will be added to both side, then
(x+2)(y+2)=M+52;
xy+2x+2y+4=M+52;
xy=M, cancel them,
2x+2y=48
x+y=48/2=24
now plug them with the sides of photograph, which is:
W+2+ L+2=24
W+L=20
perimeter is 2(W+L)=20*2=40
GMAT/MBA Expert
- Ian Stewart
- GMAT Instructor
- Posts: 2621
- Joined: Mon Jun 02, 2008 3:17 am
- Location: Montreal
- Thanked: 1090 times
- Followed by:355 members
- GMAT Score:780
It's in this step (and another place later in your solution) where there is an error. The area of the photograph+1 inch border is not LW + (L+2)(W+2). It's just (L+2)(W+2). L+2 and W+2 are the length and width of the photo and the border together, so multiplying them gives the total area; no need to add on LW again.khurram wrote:
Area of photograph + 1inch border = LW + (L+2)(W+2) (We add 2 to the length and width because the 1 inch border adds 1 inch to EACH SIDE of the picture width wise and length wise!)
what am i doing wrong
GMAT/MBA Expert
- Scott@TargetTestPrep
- GMAT Instructor
- Posts: 7243
- Joined: Sat Apr 25, 2015 10:56 am
- Location: Los Angeles, CA
- Thanked: 43 times
- Followed by:29 members
We are given that a rectangular photograph is surrounded by a border that is 1 inch wide on each side and that the total area of the photograph and border is M square inches. If we let L = the length of the photograph and W = the width of the photograph, since the border surrounds the length and width on two sides, the length of the photograph and border is L + 2 and the width of the photograph and border is W + 2.khurram wrote: A rectangular photograph is surrounded by a border that is 1 inch wide on each side. The total area of the photograph and the border is M square inches. If the border had been 2 inches wide on each side, the total area would have been (M + 52) square inches. What is the perimeter of the photograph, in inches?
A. 34
B. 36
C. 38
D. 40
E. 42
Let's represent this in a diagram:
We can now represent the area of the border and photograph:
area = length x width
M = (L + 2)(W + 2)
M = WL + 2W + 2L + 4
We are also given that if the border had been 2 inches wide on each side, the total area would have been (M + 52) square inches. Thus, the new length of the border and photograph would be L + 4 and the new width of the border and photograph would be W + 4.
We can again represent this in a diagram:
The new area of the border and photograph is:
M + 52 = (L + 4)(W + 4)
M + 52 = WL + 4W + 4L + 16
M = WL + 4W + 4L - 36
We have two equations for M. Let's equate them and simplify:
WL + 2W + 2L + 4 = WL + 4W + 4L - 36
2W + 2L + 4 = 4W + 4L - 36
2W + 2L = 4W + 4L - 40
2W + 2L = 40
Since perimeter = 2L + 2W, the perimeter of the photograph is 40 inches.
Answer: D
Scott Woodbury-Stewart
Founder and CEO
[email protected]
See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews