# CAT 2018 Quant Questions

Question:
Points E, F, G, H lie on the sides AB, BC, CD, and DA, respectively, of a square ABCD. If EFGH is also a square whose area is 62.5% of that of ABCD and CG is longer than EB, then the ratio of length of EB to that of CG is

 2 : 5 4 : 9 3 : 8 1 : 3
Let the area of ABCD be 100. Side of ABCD = 10 Area of EFGH is 62.5 => Side of EFGH = √62.5
Triangles AEH, BFE, CGF and DHG are congruent by ASA.
Let AE = BF = CG = DH = x; EB = FC = DG = AH = 10 -xx
$\mathrm{AE}^{2}+\mathrm{AH}^{2}=\mathrm{EH}^{2}$
$\mathrm{x}^{2}+(10-\mathrm{x})^{2}=(\sqrt{62.5})^{2}$
Solving, x = 2.5 or 7.5
Since it’s given that CG is longer than EB, CG = 7.5 and EB = 2.5.
Therefore, EB : CG = 1 : 3

Get one day FREE Trial of CAT online Full course
Also Check: 841+ CAT Quant Questions with Solutions

#### CAT Quant Online Course

• 1000+ Practice Problems
• Detailed Theory of Every Topics
• Online Live Sessions for Doubt Clearing
• All Problems with Video Solutions

CAT 2018 Questions Paper with Solution PDF

## Try CAT 2020 online course for 1 day for FREE

Please provide your details to get FREE Trial of Bodhee Prep's Online CAT Course for one day. We will inform you about the trial on your whatsApp number with the activation code.