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If the difference between the compound interest andsimple interest for 2 years at 12%p.a. compounded annually isRs108, find the sum borrowed. |
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Answer» <P>Rs. 7500 Step-by-step explanation: Let the sum borrowed was P Here Time = 2 years and Rate = 12% PER ANNUM COMPOUNDED annually Lets Find Compound Interest We know that CI=P[1+\frac{12}{100}]^2-PCI=P[1+ 100 12
] 2 −P CI=P[\frac{112}{100}]^2-PCI=P[ 100 112
] 2 −P CI=P[\frac{28}{25}]^2-PCI=P[ 25 28
] 2 −P CI=P[\frac{784}{625}]-PCI=P[ 625 784
]−P CI=[\frac{159P}{625}]CI=[ 625 159P
] ------- ( i ) Now Simple Interest SI = (P × R × n)/100 SI = ( P × 12 × 2)/100 SI = 6P/25 ---- (ii ) According to Question CI - SI = 108 Putting values of CI and SI from ( i ) and ( ii ) [\frac{159P}{625}]-\frac{6P}{25}=108[TEX] < /p > < p > [tex]\frac{9P}{625} = 108[ 625 159P
]− 25 6P
=108[tex] [tex] 625 9P
=108 P=\frac{108\times625}{9}P= 9 108×625
P = 7500 ∴ Sum borrowed was Rs. 7500 |
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