Knewton Challenge: Win A Knerd Shirt and GMAT Club Tests!

Question

Hey guys,

Great thinking on these last few challenge questions. Take a stab at this data sufficiency problem. The winner will be selected randomly from the correct answer submissions and announced before the end of the day on Friday. As always, only answers with explanations will be considered in the drawing. Good luck!

Uswetha · Answer

Answer is option E. With total amount being same in both cases, it is enough if we prove that amount after 2 years in special account is greater than basic By solving the equations for two years, Total Amount in SI (basic) : (P+(2*P*b)/100) = 2k*b+100k (k=1000) Total amount in CI(Special): (P(1+(s/100))^2)=100k+2ks+s^2 Consider A: b<1.05s. If b=s substituting 1, A(for SI)=102k, A(for CI)=102.001k hence CI>SI if b=1.02s, Substituing s=1, A(for SI)=102.40k, A(for CI)= 102.001k hence SI>CI. Cannot be determined with A alone. Consider B: Given interest after one year, solving the equation, b>11%.Nothing is mentioned about s. Cannot be determined with B alone Consider A and B: solving the PI for one year 100k*b/100>11k, b>11%. s>11*20/21. If b<=s, Special is greater If b>s, Basic is greater than special. Hence cannot be determined. Posted from my mobile device

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Knewton Challenge: Win A Knerd Shirt and GMAT Club Tests!

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