Question

A full water bucket is \[32{\text{ kg}}\]. When the water is half, the bucket weight is \[20{\text{ kg}}\]. What was the only bucket weight?

Answer 1

Hint: We will assume the only bucket weight as \[x{\text{ kg}}\] and weight of only water in the full bucket as \[y{\text{ kg}}\]. As given, the full water bucket is \[32{\text{ kg}}\] i.e., \[x + y = 32\]. Also, when the water is half, the bucket weight is \[20{\text{ kg}}\] i.e., \[x + \dfrac{y}{2} = 20\]. We will solve these two linear equations to find the only bucket weight.

Complete step by step answer:
Let the only bucket weight be \[x{\text{ kg}}\] and the weight of only water in the full bucket be \[y{\text{ kg}}\].
Given in the question that the full water bucket is \[32{\text{ kg}}\]. So, we can write
\[ \Rightarrow x + y = 32\]
\[ \Rightarrow y = 32 - x - - - (1)\]
Also given, when the water is half, the bucket weight is \[20{\text{ kg}}\]. So, we can write
\[ \Rightarrow x + \dfrac{y}{2} = 20\]
Taking LCM, we get
\[ \Rightarrow 2x + y = 40 - - - (2)\]
Now, we have obtained two linear equations in two variables. An equation is said to be linear equation in two variables if is it is of the form \[ax + by + c = 0\], where \[a\] and \[b\] is not equal to zero and \[a\], \[b\] and \[c\] are real numbers.
Now, putting \[(1)\] in \[(2)\], we get
\[ \Rightarrow 2x + 32 - x = 40\]
\[ \Rightarrow x + 32 = 40\]
Subtracting \[32\] from both the sides, we get
\[ \Rightarrow x = 40 - 32\]
\[ \Rightarrow x = 8\]
Therefore, the only bucket weight was \[8{\text{ kg}}\].

Note:
Here, we have two variables i.e., \[x\] and \[y\]. In these types of questions if there are \[n\] variables in an equation then there should be a minimum of \[n\] different equations, to get the value of all the variables. By substituting the values of variables in different equations we can get the values of different variables.