price of 4 apples and 3 mango is 34 price of 2 mango and 5 orange is 49 price o...

Questions

June 2021 1 23 Report

price of 4 apples and 3 mango is 34
price of 2 mango and 5 orange is 49
price of 3 orange and 6 apples is 54
find the price of 1 apples , 1 mango , 1 orange

Answers & Comments

Avengers00

Verified answer

$\underline{\underline{\Huge{\textbf{Solution:}}}}$

Given,

price of 4 apples and 3 mango is 34

price of 2 mango and 5 orange is 49

price of 3 orange and 6 apples is 54

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$\sf\textsf{Assume variables for price of one apple,}\\\sf\textsf{one mango, one orange.}$

$\sf\textsf{Let price of 1 Apple be \textbf{x Rs.}}\\\sf\textsf{Let price of 1 Mango be \textbf{y Rs.}}\\\sf\textsf{Let price of 1 Orange be \textbf{z Rs.}}$

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$\sf\textsf{Rewrite the given data using considered}\\\sf\textsf{variables to form System of equations.}$

$\begin{alignedat}{34}\sf 4x+3y&= \sf 34&\quad \cdots\cdots&\sf (1)\\\sf 2y+5z&=\sf49 &\quad\cdots\cdots&\sf(2)\\\sf3z+6x&=\sf54&\quad \cdots\cdots&\sf(3) \end{alignedat}$

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$\sf\textsf{Do 3$\times$[2] + 5$\times$[3]}$

$\begin{alignedat}{147}\sf 6y+15z&=\sf 147 &\quad \cdots\cdots&\sf(3)\\\sf15z+30x&=\sf270&\quad\cdots\cdots&\sf(4)\end{alignedat}$

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$\sf\textsf{Do [4]-[3]}$

$\begin{array}{ccccccccr}&30x&+&&15z&=&&270&\\&6y&-&&15z&=&&147&\\-&&&-&&&-&\\\\&&30x&-&6y&=&&123&\cdots\cdots[5]\quad\end{array}$

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$\sf\textsf{Do 2$\times$[1]}$

$\begin{alignedat}{68}8x+6y&= 68&\quad \cdots\cdots&(6)\end{alignedat}$

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$\sf\textsf{Do [5] + [6] to get value of x}$

$\begin{array}{cccccccc}&30x&-&&6y&=&&123\\&8x&+&&6y&=&&68\\\\&38x&&&&=&&191\end{array}$

$\mathsf{\implies x = \dfrac{191}{38}\quad\cdots\cdots[7]}$

$\begin{array}{r|l} &\bf 5.02 \\\cline{1-2} \sf 38 & 1\: 9\: 1\\ &1\: 9\: 0\: \\ \cline{2-2} & \quad\: 1\: 0\\&\quad \quad0\\\cline{2-2} &\quad\: 1\: 0\: 0\\&\quad\quad 7\: 6\\\cline{2-2}&\quad \quad \boxed{2\: 4 }\\ \end{array}$

$\therefore\mathsf{x = \dfrac{191}{38} = 5.02}$

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$\sf\textsf{Substitute [7] in [1] to get value of y}$

$\begin{aligned}\implies 4\left(\dfrac{191}{38}\right)+3y&= 34\\\\\implies \dfrac{382}{19}+3y &= 34 \\\\\implies 3y &= 34 - \dfrac{382}{19}\\\\\implies 3y &= \dfrac{646-382}{19}\\\\\implies 3y &= \dfrac{264}{19}\\\\\implies y &= \dfrac{88}{19}\end{aligned}$

$\begin{array}{r|l} &\bf 4.63 \\\cline{1-2} \sf 19 & 8\: 8\\ &7\: 6\\\cline{2-2} &1\: 2\: 0\\&1\: 1\: 4\\\cline{2-2} &\quad\: \: 6\: 0\\&\quad\: \: 5\: 7\\\cline{2-2}&\quad\: \boxed{\: 3}\\ \end{array}$

$\therefore\mathsf{y = \dfrac{88}{19} = 4.63}$

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$\sf\textsf{Substitute [7] in [3] to get value of z}$

$\begin{aligned}\implies 3z+ 6\left(\dfrac{191}{38}\right) &= 54\\\\\implies 3z+ \dfrac{573}{19}&= 54\\\\\implies 3z &= 54-\dfrac{573}{19}\\\\\implies 3z &= \dfrac{1026-573}{19}\\\\\implies 3z &= \dfrac{451}{19}\\\\\implies z &=\dfrac{151}{19} \end{aligned}$

$\begin{array}{r|l} &\bf 7.94 \\\cline{1-2} \sf 19 & 1\: 5\: 1\\&1\: 3\: 3\\\cline{2-2} &\quad \! 1\: 8\: 0\\&\quad\! 1\: 7\: 1\\\cline{2-2} &\quad\quad\! \! 9\: 0\\&\quad\quad\!\! 7\: 6\\\cline{2-2}&\quad\quad\! \!\boxed{1\: 4}\\ \end{array}$

$\therefore\mathsf{z = \dfrac{151}{19} = 7.94}$

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Swarnimkumar22 When i saw your answers I am very happy

Avengers00 Thank you (:

Swarnimkumar22 :-)

seemajhafea17 thankyou so much but i also got same answer but when i put values in question i didn't get exact value

seemajhafea17 i think we can just get approximate answer

Avengers00 Yes! It i due to the fact that the denominator obtained in all the results contain 19(which doesn't contain either 2 or 5 or both), will be a non terminating non repeating decimals.

Avengers00 Substituting fraction will satisfy the given data accordingly.

Avengers00 :) Hope this clarifies your doubts

seemajhafea17 oh yes

seemajhafea17 i got same answer but not sure with it