Divide 5,000 into two parts such that the
simple interest on the first part for two years
at 9% per annum is equal to the simple
interest on the second part for 4 years at 8%
per annum.
Answers
Answer:
Let 5000 divided into two parts x and y.
According to the question
⇒ x × (21/5) × (20/3) = 2 × y × (11/4) × 4
⇒ x/y = 22/28
⇒ x/y = 11/14
Suppose the ratio of x ∶ y = 11k ∶ 14k
⇒ 11k + 14k = 5000
⇒ 25k = 5000
⇒ k = 200
The difference between x and y = 14k – 11k
⇒ 3k
⇒ 3 × 200
⇒ 600
Answer:
Let the First Part be Rs. x and Second Part be Rs. (5,000 - x).
\begin{gathered}\bold{First \: Part} \begin{cases} \sf{Principal=Rs. x} \\ \sf{Rate=10\% \: p.a.} \\ \sf{Time=2 \: Yr.}\end{cases}\end{gathered}
FirstPart
⎩
⎪
⎪
⎨
⎪
⎪
⎧
Principal=Rs.x
Rate=10%p.a.
Time=2Yr.
\begin{gathered}\bold{Second \: Part} \begin{cases} \sf{Principal=Rs. (5,000 -x)} \\ \sf{Rate=20\% \: p.a.} \\ \sf{Time=3 \: Yr.}\end{cases}\end{gathered}
SecondPart
⎩
⎪
⎪
⎨
⎪
⎪
⎧
Principal=Rs.(5,000−x)
Rate=20%p.a.
Time=3Yr.
• According to Question Now ;
\leadsto\sf{SI_1 = SI_2}⇝SI
1
=SI
2
\leadsto\sf{\dfrac{PRT_1}{\cancel{100}}=\dfrac{PRT_2}{\cancel{100}}}⇝
100
PRT
1
=
100
PRT
2
Using the Values Now
\leadsto \sf{x \times 10 \times 2 = (5000 - x) \times 20 \times 3}⇝x×10×2=(5000−x)×20×3
\leadsto \sf{20x = 300000 - 60x}⇝20x=300000−60x
\leadsto \sf{20x + 60x= 300000}⇝20x+60x=300000
\leadsto \sf{80x = 300000 }⇝80x=300000
\leadsto \sf{x = \cancel\dfrac{300000}{80} }⇝x=
80
300000
\leadsto⇝ Rs. 3,750
• Second Part be :
⇒ Rs. (5,000 - x)
⇒ Rs. (5,000 - 3,750)
⇒ Rs. 1,250
⠀
\therefore∴ Rs. 3750 and Rs. 1250 is two different Part on which Simple Interest will be Equal.