The price of a smartphone is 5000 more
than the cost of two feature phones. If
the cost of 4 feature phones and two
smartphone is 88,000, then find the cost
of a smartphone and a feature phone.
Answers
Answer:
Cost of a smartphone = Rs. 24500
Cost of a feature phone = Rs. 9750
Step-by-step explanation:
To Find:
The cost of a smartphone and a feature phone.
Let us assume:
The cost of a smartphone be x.
The cost of a feature phone be y.
The price of a smartphone is 5000 more than the cost of two feature phones.
i.e., x = 2y + 5000 _____(i)
The cost of 4 feature phones and two smartphone is 88000.
i.e., 4y + 2x = 88000 _____(ii)
Finding the cost of a smartphone and a feature phone:
In equation (ii).
⟿ 4y + 2x = 88000
Substituting the value of x from eqⁿ(i).
⟿ 4y + 2(2y + 5000) = 88000
⟿ 4y + 4y + 10000 = 88000
⟿ 8y = 88000 - 10000
⟿ 8y = 78000
⟿ y = 78000/8
⟿ y = 9750
∴ Cost of a feature phone = Rs. 9750
In equation (i).
⟿ x = 2y + 5000
⟿ x = 2(9750) + 5000
⟿ x = 19500 + 5000
⟿ x = 24500
∴ Cost of a smartphone = Rs. 24500
Step-by-step explanation:
Given : 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.
⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀
❍ 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
⎩
⎪
⎪
⎪
⎧
Eq
n
1
⎭
⎪
⎪
⎪
⎫
⠀⠀⠀⠀⠀⠀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
⎩
⎪
⎪
⎪
⎧
Eq
n
2
⎭
⎪
⎪
⎪
⎫
⠀⠀⠀⠀⠀⠀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}
†
⎩
⎪
⎪
⎪
⎧
Equation2:4y+2x=88000
⎭
⎪
⎪
⎪
⎫
\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
n
1::
\begin{gathered}\qquad \dag\:\:\bigg\lgroup \sf{Equation \:1\:: x = 2y + 5000 }\bigg\rgroup \\\\\end{gathered}
†
⎩
⎪
⎪
⎪
⎧
Equation1:x=2y+5000
⎭
⎪
⎪
⎪
⎫
\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=
8
78000
\begin{gathered}\qquad :\implies \sf y = \cancel {\dfrac{ 78000}{8}} \:\:\\\end{gathered}
:⟹y=
8
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
★
⠀⠀⠀⠀⠀⠀\begin{gathered}\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: \: Value\:of\: y \ [ \ 9750 \ ] \:in \:Eq^n \:1 \::}}\\\end{gathered}
⋆NowBySubstitutingtheValueofy [ 9750 ]inEq
n
1:
\begin{gathered}\qquad \dag\:\:\bigg\lgroup \sf{Equation \:1\:: x = 2y + 5000 }\bigg\rgroup \\\\\end{gathered}
†
⎩
⎪
⎪
⎪
⎧
Equation1:x=2y+5000
⎭
⎪
⎪
⎪
⎫
\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}
:⟹
x=Rs.24500
★
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.
⠀⠀