unil bought a 3-year-old machine from Anil for 7.29.000. Its value decreases
by 10 p.c.p.a. What was the price of the machine

unil bought a 3-year-old machine from Anil for 7.29.000. Its value decreases
by 10 p.c.p.a. What was the price of the machine when Anil bought it? (2019)
1) 9.00,000 (2) 10,00.000 (3) 8.00.000 (4) 12,00.000​

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2 thoughts on “unil bought a 3-year-old machine from Anil for 7.29.000. Its value decreases<br />by 10 p.c.p.a. What was the price of the machine”

  1. Given Equation

    [tex] \tt \to \: \dfrac{(x + 1)}{(x – 1)} + \dfrac{(x – 2)}{(x + 2)} = 3[/tex]

    Now Take LCM

    [tex] \tt\to \: \dfrac{(x + 1)(x + 2) + (x – 2)(x – 1)}{(x – 1)(x + 2)} = 3[/tex]

    [tex] \tt \to \: \dfrac{ {x}^{2} + 2x + x + 2 + {x}^{2} – x – 2x + 2 }{ {x}^{2} + 2x – x – 2} = 3[/tex]

    [tex] \tt \to \: \dfrac{ {x}^{2} + 3x + 2 + {x}^{2} – 3x + 2}{ {x}^{2} + x – 2 } = 3[/tex]

    [tex] \tt \to \: \dfrac{ 2{x}^{2} + 4}{ {x}^{2} + x – 2} = 3[/tex]

    Using Cross multiplication

    [tex] \tt \to \: 2{x}^{2} + 4 = 3( {x}^{2} + x – 2)[/tex]

    [tex] \tt \to \: 2{x}^{2} + 4 = 3 {x}^{2} + 3x – 6[/tex]

    [tex] \tt \to \: 3 {x}^{2} – 2{x}^{2} + 3x – 6 – 4 = 0[/tex]

    [tex] \tt \to \: {x}^{2} + 3x – 10 = 0[/tex]

    Now Using Middle term Splitting

    [tex] \tt \to \: {x}^{2} + 5x – 2x – 10 = 0[/tex]

    [tex] \tt \to \: x(x + 5) – 2(x + 5) = 0[/tex]

    [tex] \tt \to \: (x – 2)(x + 5) = 0[/tex]

    [tex] \tt \to \: (x – 2) = 0 \: and \: (x + 5) = 0[/tex]

    [tex] \tt \to \: x = 2 \: and \: x = – 5[/tex]

    Answer

    [tex]\tt \to \: x = 2 \: and \: x = – 5[/tex]

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  2. Question:-

    Unil bought a 3-year-old machine from Anil for 7.29.000. Its value decreases by 10 p.c.p.a. What was the price of the machine when Anil bought it?

    To Find:-

    Find the price of the machine when Anil bought it.

    Solution:-

    Formula to be Used:-

    [tex] {\boxed{ \sf \red{ A } = \red{{P(1 ± \dfrac{ R }{ 100 } )}^{ T } }}} [/tex]

    Now ,

    We have to find the price of the machine:-

    Let the price of machine be ‘ x ‘

    [tex]\longrightarrow\sf \: 729000 = x{( 1 – \dfrac { 10 } { 100 } ) }^{ 3 } [/tex]

    [tex]\longrightarrow\sf \: 729000 = x{ ( 0.9 ) }^{ 3 } [/tex]

    [tex]\longrightarrow\sf \: 729000 = 0.729x [/tex]

    [tex]\longrightarrow\sf \: x = \cancel\dfrac{ 729000 } { 0.729 } [/tex]

    [tex]\longrightarrow\sf \: x = 10,00,000 [/tex]

    Hence ,

    Price of machine is Rs 10,00,000

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