Which is the easiest and hardest language ??
Answers
Question ❔
Which is the easiest and hardest language ??
Answer;)
Easiest Language--English
Hardest Language--Mandarin
.
Explanation:
Hope it helps uh!!
#ammuqueen
Answer:
The price of a smartphone is 5000 more than the cost of two feature phones &
the cost of 4 feature phones and two smartphone is 88,000 .
Exigency To Find : The cost of a smartphone and a feature phone.
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❍ Let's Consider the cost price of smart phone and feature phone be Rs. x & Rs. y , respectively .
⠀⠀⠀⠀⠀⠀CASE I : The price of a smartphone is Rs. 5000 more than the cost of two feature phones .
\begin{gathered}\qquad :\implies \sf x = 2y + 5000 \:\:\\\end{gathered}
:⟹x=2y+5000
\begin{gathered}\qquad :\implies \bf x = 2y + 5000 \:\:\:\qquad\qquad \bigg\lgroup \sf{ Eq^n \: 1 }\bigg\rgroup\\\end{gathered}
:⟹x=2y+5000
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Eq
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1
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⠀⠀⠀⠀⠀⠀CASE II : The cost of 4 feature phones and two smartphones is Rs. 88,000 .
\begin{gathered}\qquad :\implies \sf 4y + 2x = 88000 \:\:\\\end{gathered}
:⟹4y+2x=88000
\begin{gathered}\qquad :\implies \bf 4y + 2x = 88000 \:\:\:\qquad\qquad \bigg\lgroup \sf{ Eq^n \: 2}\bigg\rgroup\\\end{gathered}
:⟹4y+2x=88000
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Eq
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2
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⠀⠀⠀⠀⠀⠀Now , Finding the cost of Smart phone & Feature phone :
\begin{gathered}\qquad \:\maltese\:\bf{ From \:Equation \:2 \:\::}\\\end{gathered}
✠FromEquation2:
\begin{gathered}\qquad \dag\:\:\bigg\lgroup \sf{Equation \:2\:: 4y + 2x = 88000 }\bigg\rgroup \\\\\end{gathered}
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Equation2:4y+2x=88000
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\begin{gathered}\qquad :\implies \sf 4y + 2x = 88000 \:\:\\\end{gathered}
:⟹4y+2x=88000
⠀⠀⠀⠀⠀⠀\begin{gathered}\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: Eq^n\:1 \: : \::}}\\\end{gathered}
⋆NowBySubstitutingtheEq
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\begin{gathered}\qquad \dag\:\:\bigg\lgroup \sf{Equation \:1\:: x = 2y + 5000 }\bigg\rgroup \\\\\end{gathered}
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Equation1:x=2y+5000
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\begin{gathered}\qquad :\implies \sf 4y + 2x = 88000 \:\:\\\end{gathered}
:⟹4y+2x=88000
\begin{gathered}\qquad :\implies \sf 4y +2 (2y + 5000) = 88000 \:\:\\\end{gathered}
:⟹4y+2(2y+5000)=88000
\begin{gathered}\qquad :\implies \sf 4y +4y + 10000 = 88000 \:\:\\\end{gathered}
:⟹4y+4y+10000=88000
\begin{gathered}\qquad :\implies \sf 4y +4y = 88000 - 10000 \:\:\\\end{gathered}
:⟹4y+4y=88000−10000
\begin{gathered}\qquad :\implies \sf 8y = 88000 - 10000 \:\:\\\end{gathered}
:⟹8y=88000−10000
\begin{gathered}\qquad :\implies \sf 8y = 78000 \:\:\\\end{gathered}
:⟹8y=78000
\begin{gathered}\qquad :\implies \sf y = \dfrac{ 78000}{8} \:\:\\\end{gathered}
:⟹y=
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78000
\begin{gathered}\qquad :\implies \sf y = \cancel {\dfrac{ 78000}{8}} \:\:\\\end{gathered}
:⟹y=
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78000
\begin{gathered}\qquad :\implies \bf y = 9750 \:\:\\\end{gathered}
:⟹y=9750
\begin{gathered}\qquad :\implies \frak{\underline{\purple{\:y = Rs.\: 9750 }} }\:\:\bigstar \\\end{gathered}
:⟹
y=Rs.9750
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⠀⠀⠀⠀⠀⠀\begin{gathered}\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: \: Value\:of\: y \ [ \ 9750 \ ] \:in \:Eq^n \:1 \::}}\\\end{gathered}
⋆NowBySubstitutingtheValueofy [ 9750 ]inEq
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\begin{gathered}\qquad \dag\:\:\bigg\lgroup \sf{Equation \:1\:: x = 2y + 5000 }\bigg\rgroup \\\\\end{gathered}
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Equation1:x=2y+5000
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\begin{gathered}\qquad :\implies \sf x = 2y + 5000 \:\:\\\end{gathered}
:⟹x=2y+5000
\begin{gathered}\qquad :\implies \sf x = 2(9750) + 5000 \:\:\\\end{gathered}
:⟹x=2(9750)+5000
\begin{gathered}\qquad :\implies \sf x = 19500 + 5000 \:\:\\\end{gathered}
:⟹x=19500+5000
\begin{gathered}\qquad :\implies \bf x = 24500 \:\:\\\end{gathered}
:⟹x=24500
\begin{gathered}\qquad :\implies \frak{\underline{\purple{\:x \:= Rs.\: 24500 }} }\:\:\bigstar \\\end{gathered}
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x=Rs.24500
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Therefore,
The cost of smartphone is : x = Rs. 24, 500
The cost of feature phone: y = Rs. 9,750
Therefore,
⠀⠀⠀⠀⠀\begin{gathered}\qquad \therefore {\underline{ \sf \:Cost \:of\:Smartphone \:and \:feature \:phone \:are\:\bf Rs. \ 24,500 \:\& \: Rs.\: 9,750\:\:\sf , \ respectively . }}\\\end{gathered}
∴
CostofSmartphoneandfeaturephoneareRs. 24,500&Rs.9,750, respectively.
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