Calculate the compound interest on a of Rs- 20000 the end of 3 years the rate of 10- p-a- compounded annually-6620562034102386

Question

Toppr Admin · Accepted Answer

Given- P-20000-r-10-n-3 yearsA-P-1-r100-n-A-20000-1-10100-3-A-20000-xD7-1110-xD7-1110-xD7-1110A-Rs-26620Compound interest - Amount -x2212- PrincipalC-I-26620-x2212-20000C-I-Rs-6620

Calculate the compound interest on a sum of Rs. $20000$ at the end of $3$ years at the rate of $10 %$ p.a. compounded annually.
$6620$
$5620$
$3410$
$2386$

Given, $P = 20000, r = 10 %, n = 3$ years
$A = P [{(1 + \frac{r}{100})}^{n}]$
$A = 20000 [{(1 + \frac{10}{100})}^{3}]$
$A = 20000 \times \frac{11}{10} \times \frac{11}{10} \times \frac{11}{10}$
$A = Rs. 26620$
Compound interest $=$ Amount $-$ Principal
$C . I . = 26620 - 20000$
$C . I . = Rs. 6620$

Calculate the compound interest on a sum of Rs. 20000 at the end of 3 years at the rate of 10% p.a. compounded annually.6620562034102386

Given, P=20000,r=10%,n=3 yearsA=P[(1+r100)n]A=20000[(1+10100)3]A=20000×1110×1110×1110A=Rs.26620Compound interest = Amount − PrincipalC.I.=26620−20000C.I.=Rs.6620

Calculate the compound interest on a sum of Rs. $20000$ at the end of $3$ years at the rate of $10 %$ p.a. compounded annually.
$6620$
$5620$
$3410$
$2386$

Given, $P = 20000, r = 10 %, n = 3$ years
$A = P [{(1 + \frac{r}{100})}^{n}]$
$A = 20000 [{(1 + \frac{10}{100})}^{3}]$
$A = 20000 \times \frac{11}{10} \times \frac{11}{10} \times \frac{11}{10}$
$A = Rs. 26620$
Compound interest $=$ Amount $-$ Principal
$C . I . = 26620 - 20000$
$C . I . = Rs. 6620$