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# Cost of $2$ apples, $3$ bananas and one coconut is Rs. $26$. Also the cost of $3$ apples, $2$ bananas and $2$ coconuts is Rs. $35$. Then the cost of $12$ apples, $13$ bananas and $7$ coconut isA.Rs. $172$B.Rs. $148$C.Rs. $143$D.Rs. $126$

Last updated date: 20th Jun 2024
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Hint: Here, cost of bananas, apples and coconut is given in the form of sum of their prices. By assuming the values of coconut, banana and apple as some variables we can form the$3$ equation and solve them accordingly. For solving the equation we have to convert the given equation such that it resembles the required equation. Although this cannot be done everywhere, it is possible with a specific set of numbers.

Let us assume the cost of an apple as $x$, cost of a banana as $y$ and cost of a coconut as $z$.
Thus as per the first condition, that is cost of $2$ apples, $3$ bananas and one coconut is Rs. $26$ , an equation is formed as $2x + 3y + z = 26$ which will be marked as equation (1).
BY the second condition, that is cost of $3$ apples, $2$ bananas and $2$ coconut is Rs. $35$ , an equation is formed as $3x + 2y + 2z = 35$ which will be marked as equation (2).
And the required equation is $12x + 13y + 7z = ?$ which will be equation (3).
Simplest way of solving these equations is to manipulate the first two equations and convert the left hand side as similar to the left hand side of equation (3).
This can be done by multiplying the first equation with 3 and equation (2) with $2$ , which gives us
$\Rightarrow 3 \times (2x + 3y + z) = 3 \times 26 \\ \Rightarrow 6x + 9y + 3z = 78 \\$
The obtained equation can be termed as equation (4).
Now, for the second equation, we get
$\Rightarrow 2 \times (3x + 2y + 2z) = 2 \times 35 \\ \Rightarrow 6x + 4y + 4z = 70 \\$
This new equation can be termed as equation (5).
On adding the left hand side of the equation (4) and equation (5), we get
$\Rightarrow (6x + 4y + 4z) + (6x + 9y + 3z) \\ =12x + 13y + 7z \\$
This, if you notice, is the exact requirement of the question.
Since equation (4) and (5) were equality equations their respective left hand side and right hand side must be equal.
So, on adding the right hand sides if equation (4) and (5), we get
$\Rightarrow$ $70 + 78 = 148$, which is required in equation (3).
Hence, option (B) is the correct option.

Note: Remember, this way of solving is applicable for only certain values of coefficients of variables. Also for solving linear equation with $3$ unknowns, you need a minimum of $3$ equations, so if such type of question is given where there are $3$ unknowns and only $2$ equations are given, it is a clear indication that you have to manipulate the two equation and get the ${3^{rd}}$ one.