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A man spends 75% of his income. If his income is increased by 20% and he increased his expenditure by 10%. By what percent will saving increased

seo-qna
Last updated date: 22nd Mar 2024
Total views: 393.9k
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MVSAT 2024
Answer
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Hint: The problem deals with the concept of income, expenditure and saving. The concept of the percentage is also used as some part of the income is saved and some is spent.

Complete step by step solution: Let us assume the Initial income of the person, $I = 100$ Rs.
Initial Expenditure of the person is 75% of the income ,
$
  {E_i} = 75\% \times 100 \\
  {E_i} = \dfrac{{75}}{{100}} \times 100 \\
  {E_i} = 75 \\
 $
His expenditure is 75 Rs.
Savings of the person is given by,
${S_i} = {I_i} - {E_i}$
Substitute the value of Income and Expenditure in equation (1),
$
  {S_i} = 100 - 75 \\
  {S_i} = 25 \\
 $
The savings of the person is 25 Rs.
Now, the income has increased by 20%. The increased income is given by,
$
  {I_f} = 100 + \dfrac{{20}}{{100}} \times 100 \\
  {I_f} = 100 + 20 \\
  {I_f} = 120 \\
 $
New income is 120 Rs.
New Expenditure is 10% more than the 75 %. The increased expenditure by,
$
  {E_f} = 75 + \dfrac{{10}}{{100}} \times 75 \\
  {E_f} = 75 + 7.5 \\
  {E_f} = 82.5 \\
 $
The new expenditure is 82.5 Rs.
The new savings is,
$
  {S_i} = 120 - 82.5 \\
  {S_i} = 37.5 \\
 $
The new savings is 37.5 Rs.
The percent increase in savings is given by
$\% P = \dfrac{{{S_f} - {S_i}}}{{{S_i}}} \times 100$
$
  \% P = \dfrac{{37.5 - 25}}{{25}} \times 100 \\
  \% P = 50 \\
 $
Therefore, the percent increase in the savings of the person is 50%.

Note: The concept of percentage is used here, so it is the important step and should be very clear.
If something is $x\% $ of something, then it is written as $\dfrac{x}{{100}}$ .
If A is 10% of B , then it is written as , $A = \dfrac{{10}}{{100}}B = 0.1B$
If A is 10% more than B , then it is written as ,
$
  A = B + \dfrac{{10}}{{100}}B \\
  A = \dfrac{{110B}}{{100}} \\
  A = 1.1B \\
 $
Similarly, if A is 5% more than B, then it is written as, $A = 1.05B$.