Question

A borrowed amount of 4000 amounts to 5400 in 5 years .how much will 5600amount to in 3 years at the same rate?

Answer 1

S.i =P×R×T÷100
4000×R×3÷100
1300/1×R/3
R×1=1300×3
R=3900

Answer 2

Answer:

$\small\underline\textsf{1 \:case \:= \:7\: Rate \:of \:interest }$

$\small\underline\textsf{ 2\: case\: = \:6776 \:Amount }$

$\bf\red{Given}\begin{cases}\tt\blue{Principal= 4000} \\ \tt\orange{Amount=5400} \\\tt\green{Time=5\:years}\\\tt\orange{Rate\: of \:interest = ?}\end{cases}$

$\huge\underline\textsf{Explantion:- }$

$\rightarrow\bf S.I.1 = A1 - P1$

$\rightarrow\bf(5400 - 4000) \\ \rightarrow\bf1440$

$\large\underline\textsf{Formula \:Used }$

$\rightarrow\boxed{\bf S.I = \frac{P \times R \times N}{100}}$

$\large\underline\textsf{A.T.Q }$

$\rightarrow\bf 1400 = \frac{4000 \times R \times 5}{100}$

$\rightarrow\bf r = \frac{\cancel{140000}}{\cancel{20000}}$

$\rightarrow\bf r = 7$

$\large\underline\textsf{Rate \: of \:interest \:=7}$

$\rule{300}{2}$$\large\underline\textbf{Second\: case }$

$\huge\underline\textsf{Explantion:- 2 }$

$\tt\red{Given}\begin{cases}\sf\green{Principal=5600} \\ \sf\purple{Time=3 \:years} \\\bf\pink{Rate\: of \:interest\: =7} \\ \tt{Amount=?}\end{cases}$

$\boxed{\bf S.I.2 = P \times R \times N}$

$\boxed{\bf \frac{56\cancel0\cancel0 \times 7 \times 3}{1\cancel0\cancel0} } \\ \boxed{\bf{\red{= 1176}}}$$\rule{300}{2}$

$\large\underline\textsf{Amount:- }$

$\rightarrow\bf p2 + s.i.2$

$\rightarrow\bf (5600 + 1176)$

$\rightarrow\bf amount = 6776$

$\small\underline\textsf{1 \:case \:= \:7\: Rate \:of \:interest }$

$\small\underline\textsf{ 2\: case\: = \:6776 \:Amount }$