Manik12345 wrote:
A picture of an equilateral triangle is reduced by 75 percent in a photocopier. The new picture would now have to be blown up by what percentage for its area to be twice that of the original picture?
A. 32%
B. 800%
C. 3100%
D. 3200%
E. 15,000%
Does the reduction in the photo not imply that the area would have been reduced in that proportion ?
What i was doing is let area be 'a' and according after reduction it becomes a/4 and solved the diff/original *100 which was giving 700% . But the answer is 3100% which we get after reducing the sides not the area by 75%
The question should have mentioned that the copier reduces the size such that the sides are reduced by 75% for clarity. Not given anything, I might have still assumed that the lengths of the sides reduce by 75% since if we imagine a vertical straight line on a sheet and then imagine what happens when we shrink the sheet by 75%, I would say that the length of the line gets reduced by 75%. Now imagine that the line has a semi circle such that the line is a diameter. Again, if the sheet is shrunk by 75%, the length of the line reduces by 75% but the area of the semi circle reduces by much more.
If side of the triangle is s, area is \(\sqrt{3}s^2/4 = A\).
Now the side is s/4.
We need area to be 2A i.e. the side of the triangle should be \(\sqrt{2}s\)
Percentage increase in side = \((\sqrt{2}s - s/4)/(s/4) * 100 = 466%\) approximately
I don't know why they say that the answer is 3100%. While blowing up the picture, they are assuming that the reference is area and not sides!
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