Is it possible to solve a word problem by giving values? (at least for building an equation) Below is attached a problem.

There are apples and bananas in a fridge. The sum of apples and bananas is $50$ ton.  $7\%$ of apples are going bad and $8\%$ of bananas are going bad. The sum of the apples and bananas going bad is calculated as $3,8$ ton. How many sturdy apples are there? 



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*Is it possible to solve a word problem by giving values? (at least for building an equation)

*These questions make me so confused, Can I take your tips?

*What is the best way of building an equation from word problems (including kind of every question)
Note: The problem I've attached is just for showing an example. 

EDIT: let's say $\text{bananas} =25$, $\text{apples} =25$, $25(\text{apples}) + 25(\text{bananas}) = 50$. Here we get $25 \cdot \frac{7}{100} + 25 \cdot \frac{8}{100} = 3.8$ Then our equation will be $x \cdot \frac{7}{100} + y \cdot \frac{8}{100} = 3.8$ and $x+y = 50$. What about solving these questions by this method? Is it possible?
Wishing My Kindest Regards!
 A: Generally you assign variables to the quantities of interest.  I find it helpful to really write down the definitions so I can refer to it as I do the problem.  Here let $a=$ the weight of apples in the fridge in tons and $b=$ the weight of bananas in the fridge in tons.  The sentences of your problem need to be translated into equations.  You need as many equations as you have variables, so here we need two equations.  The first sentence tells us $$a+b=50$$  The next two sentences give us another equation about the weight of fruit going bad.  Can you write that equation?  You then solve the two equations simultaneously.
A: The problem in your comment is, that you assume that $a=b=25$. But apart from that you are in the right direction. 
$a+b=50$ and $a\cdot \frac7{100}+b\cdot \frac8{100}=3.8$. 
Here you have 2 equations and 2 variables. This little equation system can be solve with various methods: $\texttt{Substitution method, Addition method, ...}$
$\texttt{Substitution method}$: You solve the first equation for $a$:
$a+b=50 \quad |-b$
$a=50-b$
Now you insert the expression for $a$ into the second equation:
$a\cdot \frac7{100}+b\cdot \frac8{100}=3.8$
$(50-b)\cdot \frac7{100}+b\cdot \frac8{100}=3.8$
Multiplying out the brackets
$50\cdot \frac7{100}-b\cdot \frac7{100}+b\cdot \frac8{100}=3.8$
Simplifiying and sammerize the term with the variable $b$.
$\frac7{2}+b\cdot \frac1{100}=3.8$
The term with b has to be alone on one side. Thus we substract $\frac7{2}=3.5$ on both sides.
$\underbrace{\frac7{2}-3.5}_{=0}+b\cdot \frac1{100}=3.8-3.5$
$b\cdot \frac1{100}=0.3 \quad |\cdot 100$
I think you can proceed. Any additional questions, comments?
