its vaaste
dono ki problem_____xD
Answers
Answer:
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Explanation:
Given : Simple interest on a certain sum for 4 years at 7% p.a. is more than simple interest on the same sum for 2.5 years at the same rate by 840.
Exigency To Find : The Principal amount.
⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀
❍ Let's Consider the Principal amount be P .
\begin{gathered}\dag\:\:\sf{ As,\:We\:know\:that\::}\\\\ \qquad\maltese\:\:\bf \:Formula\:for\:Simple\:Interest\:\::\\\end{gathered}
†As,Weknowthat:
✠FormulaforSimpleInterest:
\begin{gathered}\qquad \dag\:\:\bigg\lgroup \sf{Simple \:Interest \:: \dfrac{P \times R \times T}{100} }\bigg\rgroup \\\\\end{gathered}
†
⎩
⎪
⎪
⎪
⎧
SimpleInterest:
100
P×R×T
⎭
⎪
⎪
⎪
⎫
⠀⠀⠀⠀⠀Here , P is the Principal, R is the Rate of Interest & T is the Time.
⠀⠀⠀⠀⠀⠀\begin{gathered}\underline {\boldsymbol{\star\: According \:to\: the \:Question \::}}\\\end{gathered}
⋆AccordingtotheQuestion:
⠀⠀⠀⠀⠀━━━ Simple interest on a certain sum for 4 years at 7% p.a. is more than simple interest on the same sum for 2.5 years at the same rate by 840.
\begin{gathered}\qquad:\implies \sf 1^{st} \: Simple \:Interest \: = \: 2^{nd} \: Simple \:Interest\:\:+ 840 \\\end{gathered}
:⟹1
st
SimpleInterest=2
nd
SimpleInterest+840
\begin{gathered}\qquad:\implies \sf \dfrac{ P \times 7 \:\times 4 }{100} = \dfrac{ P \times 7 \:\times 2.5 }{100} + 840 \\\end{gathered}
:⟹
100
P×7×4
=
100
P×7×2.5
+840
\begin{gathered}\qquad:\implies \sf \dfrac{ P \times 28 }{100} = \dfrac{ P \times 17.5 }{100} + 840 \\\end{gathered}
:⟹
100
P×28
=
100
P×17.5
+840
\begin{gathered}\qquad:\implies \sf \dfrac{ 28P }{100} = \dfrac{ 17.5P }{100} + 840 \\\end{gathered}
:⟹
100
28P
=
100
17.5P
+840
\begin{gathered}\qquad:\implies \sf \cancel {\dfrac{ 28P }{100}} = \dfrac{ 17.5P }{100} + 840 \\\end{gathered}
:⟹
100
28P
=
100
17.5P
+840
\begin{gathered}\qquad:\implies \sf 0.28 \:P = \dfrac{ 17.5P }{100} + 840 \\\end{gathered}
:⟹0.28P=
100
17.5P
+840
\begin{gathered}\qquad:\implies \sf 0.28 \:P = \cancel {\dfrac{ 17.5P }{100}} + 840 \\\end{gathered}
:⟹0.28P=
100
17.5P
+840
\begin{gathered}\qquad:\implies \sf 0.28 \:P = 0.175P + 840 \\\end{gathered}
:⟹0.28P=0.175P+840
\begin{gathered}\qquad:\implies \sf 0.28 \:P- 0.175 P = 840 \\\end{gathered}
:⟹0.28P−0.175P=840
\begin{gathered}\qquad:\implies \sf 0.105 P = 840 \\\end{gathered}
:⟹0.105P=840
\begin{gathered}\qquad:\implies \sf P = \dfrac{840}{0.015} \\\end{gathered}
:⟹P=
0.015
840
\begin{gathered}\qquad:\implies \sf P = \cancel {\dfrac{840}{0.015}} \\\end{gathered}
:⟹P=
0.015
840
\begin{gathered}\qquad:\implies \bf P = 8000 \\\end{gathered}
:⟹P=8000
\begin{gathered}\qquad:\implies \frak{\underline{\purple{\:P = Rs.8000 }} }\:\:\bigstar \\\end{gathered}
:⟹
P=Rs.8000
★
Therefore,
⠀⠀⠀⠀⠀\begin{gathered}\therefore {\underline{ \mathrm {\:Principal \:amount\:\:is\:\bf{Rs.8000}}}}\\\end{gathered}
∴
PrincipalamountisRs.8000