Question

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.​

Answer 1

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

Answer 2

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.

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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 .

$\qquad :\implies \sf x = 2y + 5000 \:\:$

$\qquad :\implies \bf x = 2y + 5000 \:\:\:\qquad\qquad \bigg\lgroup \sf{ Eq^n \: 1 }\bigg\rgroup$

⠀⠀⠀⠀⠀⠀CASE II : The cost of 4 feature phones and two smartphones is Rs. 88,000 .

$\qquad :\implies \sf 4y + 2x = 88000 \:\:$

$\qquad :\implies \bf 4y + 2x = 88000 \:\:\:\qquad\qquad \bigg\lgroup \sf{ Eq^n \: 2}\bigg\rgroup$

⠀⠀⠀⠀⠀⠀Now , Finding the cost of Smart phone & Feature phone :

$\qquad \:\maltese\:\bf{ From \:Equation \:2 \:\::}$

$\qquad \dag\:\:\bigg\lgroup \sf{Equation \:2\:: 4y + 2x = 88000 }\bigg\rgroup$

$\qquad :\implies \sf 4y + 2x = 88000 \:\:$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: Eq^n\:1 \: : \::}}$

$\qquad \dag\:\:\bigg\lgroup \sf{Equation \:1\:: x = 2y + 5000 }\bigg\rgroup$

$\qquad :\implies \sf 4y + 2x = 88000 \:\:$

$\qquad :\implies \sf 4y +2 (2y + 5000) = 88000 \:\:$

$\qquad :\implies \sf 4y +4y + 10000 = 88000 \:\:$

$\qquad :\implies \sf 4y +4y = 88000 - 10000 \:\:$

$\qquad :\implies \sf 8y = 88000 - 10000 \:\:$

$\qquad :\implies \sf 8y = 78000 \:\:$

$\qquad :\implies \sf y = \dfrac{ 78000}{8} \:\:$

$\qquad :\implies \sf y = \cancel {\dfrac{ 78000}{8}} \:\:$

$\qquad :\implies \bf y = 9750 \:\:$

$\qquad :\implies \frak{\underline{\purple{\:y = Rs.\: 9750 }} }\:\:\bigstar$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: \: Value\:of\: y \ [ \ 9750 \ ] \:in \:Eq^n \:1 \::}}$

$\qquad \dag\:\:\bigg\lgroup \sf{Equation \:1\:: x = 2y + 5000 }\bigg\rgroup$

$\qquad :\implies \sf x = 2y + 5000 \:\:$

$\qquad :\implies \sf x = 2(9750) + 5000 \:\:$

$\qquad :\implies \sf x = 19500 + 5000 \:\:$

$\qquad :\implies \bf x = 24500 \:\:$

$\qquad :\implies \frak{\underline{\purple{\:x \:= Rs.\: 24500 }} }\:\:\bigstar$

Therefore,

• The cost of smartphone is : x = Rs. 24, 500
• The cost of feature phone: y = Rs. 9,750

Therefore,

⠀⠀⠀⠀⠀$\qquad \therefore {\underline{ \sf \:Cost \:of\:Smartphone \:and \:feature \:phone \:are\:\bf Rs. \ 24,500 \:\& \: Rs.\: 9,750\:\:\sf , \ respectively . }}$

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