x-3y-3 = 0
3x-9y-2= 0
Answers
Step-by-step explanation:
Current time that is shown by watch is 10 : 00 AM & Watch is getting slow by 2 minutes for each hour in a day .
Exigency To Find : What will be the time after 12 hrs ?
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⠀⠀⠀Given that ,
⠀⠀⠀⠀⠀⠀⠀⠀▪︎ ⠀The Current time that is shown by watch is 10 : 00 AM
⠀⠀⠀⠀⠀⠀⠀⠀▪︎ ⠀Watch is getting slow by 2 minutes for each hour .
⠀⠀⠀⠀⠀Now , As we have to find time that is shown by watch after 12 hrs ,
Therefore,
⠀⠀⠀⠀⠀⠀The time that get slower ( or late ) in 12 hrs will be ⠀
\begin{gathered}\qquad \qquad \sf Time \: get \:slower \:= \: 12 \times 2 \: min \:\:\\\end{gathered}
Timegetslower=12×2min
\begin{gathered}\qquad \qquad \bf Time \: get \:slower \:= \: 24 \: min \:\:\\\end{gathered}
Timegetslower=24min
\begin{gathered}\qquad :\implies \pmb{\underline{\purple{\:Time \: get \:slower_{(12 \:hrs\:)} \:= \: 24 \: min\: }} }\:\:\bigstar \\\end{gathered}
:⟹
Timegetslower
(12hrs)
=24min
Timegetslower
(12hrs)
=24min
★
Therefore ,
\begin{gathered}\dag\:\:\sf{ As,\:We\:know\:that\::}\\\\ \qquad\bigstar\:\:\bf Time \:After \: 12 \: hrs \: : \\\end{gathered}
†As,Weknowthat:
★TimeAfter12hrs:
\begin{gathered}\qquad \dag\:\:\bigg\lgroup \sf{ 10 \:: \:00 \: AM \:+ \: 12 \: hrs \: - \: Time \: get \:slower_{(12\: hrs)} }\bigg\rgroup \\\\\end{gathered}
†
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10:00AM+12hrs−Timegetslower
(12hrs)
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\begin{gathered}\qquad:\implies \sf 10 \:: \:00 \: AM \:+ \: 12 \: hrs \: - \: Time \: get \:slower_{(12\: hrs)} \\\end{gathered}
:⟹10:00AM+12hrs−Timegetslower
(12hrs)
\begin{gathered}\qquad:\implies \sf 22 \:: \:00 \: PM \: \: - \: Time \: get \:slower_{(12\: hrs)} \\\end{gathered}
:⟹22:00PM−Timegetslower
(12hrs)
⠀⠀⠀⠀⠀⠀⠀⠀OR ,
\begin{gathered}\qquad:\implies \sf 22 \:: \:00 \: PM \: \: - \: Time \: get \:slower_{(12\: hrs)} \\\end{gathered}
:⟹22:00PM−Timegetslower
(12hrs)
\begin{gathered}\qquad:\implies \sf 10 \:: \:00 \: PM \: \: - \: Time \: get \:slower_{(12\: hrs)} \\\end{gathered}
:⟹10:00PM−Timegetslower
(12hrs)
⠀⠀⠀⠀⠀⠀\begin{gathered}\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}\\\end{gathered}
⋆NowBySubstitutingtheknownValues:
\begin{gathered}\qquad:\implies \sf 10 \:: \:00 \: PM \: \: - \: Time \: get \:slower_{(12\: hrs)} \\\end{gathered}
:⟹10:00PM−Timegetslower
(12hrs)
\begin{gathered}\qquad:\implies \sf 10 \:: \:00 \: PM \:\: - \: 24 \: min \\\end{gathered}
:⟹10:00PM−24min
\begin{gathered}\qquad:\implies \sf 10 \:: \:00 \: PM \:\: - \: 24 \: min \qquad\bigg\lgroup \sf{ 1 \:hrs \:=\: 60 \: min }\bigg\rgroup \\\end{gathered}
:⟹10:00PM−24min
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1hrs=60min
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\begin{gathered}\qquad:\implies \sf 10 \:: \:00 \: PM \:\: - \: 24 \: min \qquad\bigg\lgroup \sf{ 60 \:min \:-\: 24 \: min =\: 36 \: min \:}\bigg\rgroup \\\end{gathered}
:⟹10:00PM−24min
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60min−24min=36min
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\begin{gathered}\qquad:\implies \sf 9\:: \:36 \: PM \:\: \\\end{gathered}
:⟹9:36PM
\begin{gathered}\qquad :\implies \pmb{\underline{\purple{\:Time_{( After\:12\:hrs\:)} = \: 9\:: \:36 \: PM \:\: }} }\:\:\bigstar \\\end{gathered}
:⟹
Time
(After12hrs)
=9:36PM
Time
(After12hrs)
=9:36PM
★
Therefore,
⠀⠀⠀⠀⠀\begin{gathered}\therefore {\underline{ \mathrm {\:Time\:shown\:after\:12\:hrs \:in\:the\:watch\:will \:be\:\bf{9\:: \:36 \: PM \:\:}}}}\\\end{gathered}
∴
Timeshownafter12hrsinthewatchwillbe9:36PM