two equal sums of money were lent at s.i 11%p.a for 3½ years and 4½ years respectively. if the differences in interests for two perioeds was rs412.50 find each sum
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Answers
Explanation:
AnswEr :
Let the Equal Sum of money be Rs. x
\begin{gathered}\bold{First \: Part} \begin{cases} \sf{Principal=Rs. x} \\ \sf{Rate=11\% \: p.a.} \\ \sf{Time=4 \dfrac{1}{2} \: = 4.5 \: Yr. }\end{cases}\end{gathered}
FirstPart
⎩
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎧
Principal=Rs.x
Rate=11%p.a.
Time=4
2
1
=4.5Yr.
\begin{gathered}\bold{Second \: Part} \begin{cases} \sf{Principal=Rs. x} \\ \sf{Rate=11\% \: p.a.} \\ \sf{Time=3 \dfrac{1}{2} \: =3.5 \: Yr. }\end{cases}\end{gathered}
SecondPart
⎩
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎧
Principal=Rs.x
Rate=11%p.a.
Time=3
2
1
=3.5Yr.
_________________________________
\longrightarrow \large \sf{Difference = SI_1 - SI_2}⟶Difference=SI
1
−SI
2
\longrightarrow \sf{Diff. = \dfrac{PRT_1}{100} - \dfrac{PRT_2}{100}}⟶Diff.=
100
PRT
1
−
100
PRT
2
\longrightarrow \sf{41250 = \dfrac{x \times 11 \times 4.5}{100} - \dfrac{x \times 11 \times 3.5}{100}}⟶41250=
100
x×11×4.5
−
100
x×11×3.5
\longrightarrow \sf{41250 = \dfrac{49.5x}{100} - \dfrac{38.5x}{100} }⟶41250=
100
49.5x
−
100
38.5x
\longrightarrow \sf{41250 = \dfrac{(49.5x - 38.5x)}{100}}⟶41250=
100
(49.5x−38.5x)
\longrightarrow \sf{ \cancel{41250} = \dfrac{ \cancel{11}x}{100}}⟶
41250
=
100
11
x
\longrightarrow \sf{3750 = \dfrac{x}{100}}⟶3750=
100
x
\longrightarrow \sf{3750 \times 100 = x}⟶3750×100=x
\longrightarrow \large \sf{x =Rs. 375000}⟶x=Rs.375000
⠀
\therefore∴ Equal Sum is Rs. 3,75,000.
Answer: