CAT 2018 Question Paper | Quants Slot 2
CAT Previous Year Paper | CAT Mensuration Questions | Question 12
CAT Questions Yet another classic question from Geometry which is pretty doable if you get your equations right. This question was rolled out in CAT Previous Year paper 2018, which can prove to be a motivating factor while you are going through a series of difficult questions. Try this question before having a look at the solution provided. Solve more such questions to ace your online CAT preparation. Visit 2IIM's CAT Question Bank for more such questions.
Question 12 : The area of a rectangle and the square of its perimeter are in the ratio 1 : 25. Then the lengths of the shorter and longer sides of the rectangle are in the ratio
- 3 : 8
- 2 : 9
- 1 : 4
- 1 : 3
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Explanatory Answer
Method of solving this CAT Question from Mensuration
Given that the area of a rectangle and the square of its perimeter are in the ratio 1: 25, we have to find the ratio of the lengths of the shorter and longer sides of the rectangle i.e. b : l
The area of the rectangle = lb
The square of its perimeter = [2(l + b)]2
⟹ lb : [2(l + b)]2 = 1 : 25
⟹ lb : 4[l2 + 2lb + b2] = 1 : 25
⟹ 25 lb = 4[l2 + 2lb + b2]
⟹ 25 lb = 4l2 + 8lb + 4b2
⟹ 4l2 - 17lb + 4b2 = 0
⟹ 4l2 - 16lb - lb + 4b2 = 0
⟹ 4l[l- 4b] – b [l- 4b] = 0
⟹ [l - 4b][4l - b] = 0
By this we can find that the ratio of the shorter and longer sides of the rectangle b : l is in the ratio 1: 4
In other way ,if we want to deal with only one variable and do not want to factorize divide this 4l2 - 17lb + 4b2 = 0 by l2 throughout
⟹ \\frac{4l^2}{l^2}) - \\frac{17lb}{l^2}) + \\frac{4b^2}{l^2})
⟹ 4 - \\frac{17b}{l}) + 4[\\frac{b}{l})]2
Since we have to find ratio of b : l, \\frac{b}{l}) can be substituted with x
4 - 17x + 4x2 = 0
Now x can be found by simplifying it
The question is "The area of a rectangle and the square of its perimeter are in the ratio 1 : 25. Then the lengths of the shorter and longer sides of the rectangle are in the ratio"
Hence, the answer is 1 : 4
Choice C is the correct answer.
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