Banking Sample Paper SBI Junior Associates (Main) Sample Paper Test-2

  • question_answer
    The compound interest (compounded annually) on Rs.9100 for 2 years @ R% pa is Rs.6279. Had the rate of interest been (R - 15)%, what would have been the compound interest on the same sum of money for the same time?

    A)  Rs.6554.35    

    B)       Rs.2132.50

    C)  Rs.2934.75    

    D)       Rs.3105.25

    E)  Rs.2748.65

    Correct Answer: C

    Solution :

    Logical Method: \[\text{CI = 6279}\] Now, ? % of 9100 = 6279 \[\therefore ?=\frac{6279}{9100}\times 100=69%\] Now, there is compound interest for 2 years. So rate of interest will be \[\left( 30+30+\frac{30\times 30}{100} \right)=69%\] Hence, R = 30% Now, new rate = 30 - 15 = 15% Equivalent rate of 15% compound rate of interest for 2 years \[=15+15+\frac{15\times 15}{100}\] \[=30+2.25=32.25%\] Now, 32.25% of 9100 \[=\frac{129}{400}\times 9100=\text{Rs}\text{. 2934}\text{.75}\] Another Method: \[\text{CI=P}\left[ {{\left( \text{1+}\frac{\text{R}}{\text{100}} \right)}^{\text{n}}}\text{-1} \right]\] Or, \[\text{I+}\frac{\text{CI}}{\text{P}}\text{=}{{\left( \text{1+}\frac{\text{R}}{\text{100}} \right)}^{\text{n}}}\] Now, \[\text{1+}\frac{\text{6279}}{\text{9100}}\text{=}{{\left( \text{1+}\frac{\text{R}}{\text{100}} \right)}^{2}}\] Or, \[\text{1+0}\text{.69=}{{\left( \text{1+}\frac{\text{R}}{\text{100}} \right)}^{\text{2}}}\] Or, \[\text{1}\text{.3=I+}\frac{\text{R}}{\text{100}}\] \[\therefore \text{R=30 }\!\!%\!\!\text{ }\] Again, new rate \[=(30-15)%=15%\] \[\therefore \text{CI=2years15 }\!\!%\!\!\text{ per}\,\,\text{annum}\] \[=15+15+\frac{15\times 15}{100}=32.25%\] \[\therefore 32.25%\] of \[\text{9100=Rs}\text{. 2934}\text{.75}\]


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