Question

Find the value of ‘f’ for the expression $0.25\left[ {4f - 3} \right] = 0.05\left[ {10f - 9} \right]$

Answer 1

Hint: In this question, we need to evaluate the value of ‘f’ such that $0.25\left[ {4f - 3} \right] = 0.05\left[ {10f - 9} \right]$. For this, we will try to get all the terms including ‘f’ on either side of the equation and then using cross multiplication method, determine the value of ‘f’.

Complete step by step answer:
The given expression is $0.25\left[ {4f - 3} \right] = 0.05\left[ {10f - 9} \right]$
Solving the above expression for ‘f’:
We will take 0.05 from the right hand side of the equation to the denominator of the left hand side of the equation, we get
$
  0.25\left[ {4f - 3} \right] = 0.05\left[ {10f - 9} \right] \\
   \Rightarrow \dfrac{{0.25}}{{0.05}}\left[ {4f - 3} \right] = 10f - 9 \\
   \Rightarrow 5\left[ {4f - 3} \right] = 10f - 9 \\
 $
Now, we will multiply the terms in the left hand side of the equation, we get
$20f - 15 = 10f - 9$
Taking all the terms of ‘f’ on either side of the equation and the constant terms of the other side, we get
$
  20f - 10f = - 9 + 15 \\
   \Rightarrow 10f = 6 \\
 $
Now, using cross multiplication method, taking 10 from the left hand side of the equation to the denominator of the right hand side, we get
$
  10f = 6 \\
   \Rightarrow f = \dfrac{6}{{10}} \\
 $
Taking 2 common from the numerator and the denominator of the above equation, we get
$
  f = \dfrac{{2 \times 3}}{{2 \times 5}} \\
   \Rightarrow f = \dfrac{3}{5} \\
 $
Hence, the value of ‘f’ is $\dfrac{3}{5}$ in the equation $0.25\left[ {4f - 3} \right] = 0.05\left[ {10f - 9} \right]$.

Note: The problem involves one variable which has to be evaluated from the adding all the like terms together. Whenever a term in addition or subtraction is shifted from one side of the equation to the other side of the equation, then there is a sign change while a term in the multiplication or division is shifted from one side of the equation to the other side, then multiplication will change to division and vice versa.