If the magic square
given below is filled,
all the numbers will
be
consecutive
integers.
positive
What is the sum of
the
smallest
and
largest number of
the given magic
square?
Answers
SUM OF SMALLEST AND LARGEST NUMBER
= 65 + 71
= 136
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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 ?
⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀
⠀⠀⠀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}
†
⎩
⎪
⎪
⎪
⎧
10:00AM+12hrs−Timegetslower
(12hrs)
⎭
⎪
⎪
⎪
⎫
\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
⎩
⎪
⎪
⎪
⎧
1hrs=60min
⎭
⎪
⎪
⎪
⎫
\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
⎩
⎪
⎪
⎪
⎧
60min−24min=36min
⎭
⎪
⎪
⎪
⎫
\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
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