a sum of money amounts to 2100 in a year and 2205 in two years . find the rate of interest and sum.
Year 1 "=" 2100
Year 2 = 2205
r "=" ?
p "=" ?
Difference "=" 2205 "-" 2100
"=" 105
I "=" p "\\times" R "\\times" T/100
105"=" "\\frac {2100\\times R\\times1}{100}"
R"=" 5%
Part b
Let the principal for first year be x
So we'll use this formula;
I "=" P "\\times" R"\\times" T/100
Whereby
"\\frac{5x}{100}" "=" "\\frac{x}{20}"
Amount for first year "=" x"+\\frac{x}{20}" "=\\frac{21x}{20}"
For second year p "=" "\\frac{21x}{20}"
R= 5%p.a T = 1 year
So we solve for this;
I "=" "[\\frac{21x}{20}\\times5\\times1 ]\/" 100
"=" "\\frac{21x}{4}\/" 100
From the equation above we'll get
"\\frac{21x}{4}" /100
=2100
400"\\times" "\\frac{21x}{400}=" 2100"\\times" 400
"\\frac{21x}{21}" "=" "\\frac{840000}{21}"
X"=" 40000
So the original sum of money is
40,000
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