Alexander B. answered • 09/19/22
PhD specializing in Math, Science, English, and ACT/SAT/GRE Tutoring
Photograph: length=24 cm; width=20 cm.
Area of photograph: 24x20=480 cm^2.
Width of frame: (20+x), where x presumably is x/2 to the left and x/2 to the right of the photo.
Assumption: length of the frame: (24+x), and likewise here the photograph is probably centered, but that's not what they ask for.
Given: (20+x)(24+x)=480+300 --> x^2+44x-300 = 0 --> x= 6 cm or x = -50 cm. Since width cannot be negative, we conclude that x=6, and thus we have the frame width to be 20+6=26 cm and the frame length to be 24+6=30 cm. Check: 26x30-20x24=300 cm^2.
However, the problem does not explicitly state that the length has any margin between the photo and the frame, as much as one would expect that. So technically, if all that extra area is devoted to width margin and none to length margin, we instead get the following equation to solve:
(20+x)(24)=480+300 --> 24x=300 --> x=12.5 cm. While this is a simpler, linear problem, it would really look ugly to have margins only width-wise. Nonetheless, let's check if this works:
(20+12.5)(24)-(20)(24)=300 cm^2.
So, we wither have a picture with uniform margins, in which case these are 6/2=3 cm around, which seems normal, or we have an ugly solution, where there are no margin lengthwise, so we have 0 cm margin length-wise and 12.5=6.25 cm width-wise on either side of the photo.
It helps when the professor/teacher/etc. makes the problem consistent with everyday expectations and avoids sloppy descriptions. But there you have it, the only two solutions that work; one with a small and reasonable presumption and the other with no presumption but highly unlikely in reality.
Cheers.
Alexander B.
09/20/22
Lil P.
I got two answers for this problem, and yours is different than the other one, which says 3cm. I also got this answer when I was doing my work... are you sure this is right?09/20/22