Aptitude Discussion

Q. |
If the price of petrol increases by $25%$, by how much must a user cut down his consumption so that his expenditure on petrol remains constant? |

✖ A. |
$25\%$ |

✖ B. |
$16.67\%$ |

✔ C. |
$20\%$ |

✖ D. |
$33.33\%$ |

**Solution:**

Option(**C**) is correct

Let the price of petrol be Rs.100 per litre. Let the user use 1 litre of petrol. Therefore, his expense on petrol = $100\times 1$ = Rs.100

Now, the price of petrol increases by $25\%$. Therefore, the new price of petrol = Rs.125.

As he has to maintain his expenditure on petrol constant, he will be spending only Rs.100 on petrol.

Let ‘$x$’ be the number of litres of petrol he will use at the new price.

Therefore,

\(125\times x=100\)

\(\Rightarrow x=\dfrac{100}{125}\)

\(=\dfrac{4}{5}=0.8\)litres

He has cut down his petrol consumption by 0.2 litres

\(=\dfrac{0.2}{1}\times 100\)

\(=20\%\) reduction.

There is a **shortcut for solving this problem**.

If the price of petrol has increased by $25\%$, it has gone up $1/4^{th}$ of its earlier price.

Therefore, the $\%$ of reduction in petrol that will maintain the amount of money spent on petrol constant

\(=\dfrac{1}{4+1}\)

\(=\dfrac{1}{5}=20\%\)

i.e. Express the percentage as a fraction. Then add the numerator of the fraction to the denominator to obtain a new fraction. Convert it to percentage - that is the answer.

**Edit:** For direct formula involving such questions, check comment by **Rushi** and **Chirag Goyal.**

**Edit: **For more insights on the question, check comment by **Priya.**

**Vijay**

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**Param**

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Simplest of all . No Need to remember formula. No advance calculation. Based on basic calculation.

Let assume original price be 100. Increased by 25% become Rs125.

Logic : Now we have to keep the consumption same,so we have to bring Rs 125 back to Rs 100. How much we have to decrease so that Rs 125 become Rs 100.

(125-100)125*100 % = 20%

Done . No need to go to consumption. You can calculate using one variable only.

**Priya**

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25% means 1/4 and 20% means 1/5. If the question is given as "increased by 25" then $\dfrac{1}{(4+1)}=\dfrac{1}{5}=20\%$

If the question is given as "decreased by 25" then $\dfrac{1}{(4-1)}=\dfrac{1}{3}=33\dfrac{1}{3}\%$

i.e., the value of $33\dfrac{1}{3}\%$ is $\dfrac{1}{3}$

It is easy if you know the values for all the percentages!

**Rushi**

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There is shortcut formula for this kind of sums which is $\left(\dfrac{R}{100+R}\right) \times 100$ in the case of rising and $\left(\dfrac{R}{100-R}\right) \times 100$ in the case of fall.

But this formula is applicable only when exp. have to remain constant.

Thanks $Rushi$

I'm Rewriting the same Shortcut using TeX commands for better understanding.

$\text{Consumption Should be Decrease or Increase}$

$\text=\dfrac{R}{100\pm R}\times{100}$

(R*100)/(100+R)..... this formula we will use here.

25*100/(125)=20

so 20% is anwser.