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Pritam

· started a discussion

· 1 Months ago

Simple interest (Question: 118068.0 )

Plz anybody provide short solution...thanx

Question:
A certain sum amount to \(\unicode{x20B9} \)5,184 in 2 years and \(\unicode{x20B9} \)5,832 in 3 years, then what is its rate?
Options:
A) \(16\cfrac{1}{3}\)%
B) \(15\cfrac{2}{3}\)%
C) \(15\cfrac{3}{5}\)%
D) \(16\cfrac{2}{3}\)%
Solution:
Let the sum be x

The sum amount to \(\unicode{x20B9} \)5184 in 2 years

\(\Rightarrow\)Interest + Principal = 5184

Let   Rate = R

therefore,  \({x*R*2 \over 100}+x = 5184\)\(\rightarrow\)1

the amount to \(\unicode{x20B9} \)5832 in 3 years

\(\Rightarrow\)\({x*R*3 \over 100}+x = 5832\)\(\rightarrow\)2

Dividing 2 by 1

\({x*R*3 \over 100}+x\over{x*R*2 \over 100}+x\)=\({5832 \over 5184}\)


\({3R \over 100}+1\over{2R \over 100}+1\)=\({9\over 8}\)


\({24R \over 100}+8 = {18R \over 100}+9\)


\({6R \over 100}=1\)


\(R = {100\over 6}={50\over 3} = 16{2\over 3}\)%


therefore,   Rate = \(16{2\over 3}\)%

kesi

· commented

· 1 Months ago

1 year intest 5832-5184=648 648*2=1296

5184-1296=3888


648/3888*100=50/3
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