The cost of fuel increases by 10%. By what % must the consumption of

Let Initial cost per consumed unit of fuel = 100
Let Initial Consumption = 100

Initial Total cost = 100*100


New cost per consumed unit of fuel = 110
Let New Consumption = C

New Total cost = 110*C

now 100*100 = 110*C
i.e. C = 90.90 = 91 (approx)

% Decrease = 9%

ANswer: Option B

assuming numbers will be best strategy to tackle this kind of problems...

let the cost be 10$ and the consumption be 10 litres,,
total cost is 100..
after the increase, the cost is 11$.. in 100$ we can purchase slightly more than 9 litres of fuel,,
hence required decrease in consumption shud be 9%

ans B

Let original cost of fuel = 100
total consumption = 10

New cost of fuel = 110
total consumption post fuel price increase = x
100*10 = 110*x
=> x = 1000/110 = 100/11 = 9 % approx

Answer B

Amount spent = Cost * Consumption = P*Q (say)

Here we have to keep amount spent (the product of two quantities) same. Question is that if one quantity increases by 10%, then by how much the other must decrease so as to keep the product same (as P*Q only)

So if P increases by 10% (or 1/10) new value of P is = P + P/10 = 11P/10
To keep the product same, new value of Q must be = 10Q/11 (because then only 11P/10 * 10Q/11 = p*Q only)

Thus, Q needs to decrease by 1/11 or by 9.09%. Closest option is B, hence B answer

We can let the original cost of fuel = x and the original consumption = y. Thus, the original cost = xy. If cost increases 10%, the new cost is 1.1x. We can let n = the percentage decrease and create the following equation:

xy = (1.1x)(y(1-n/100))

1 = (1.1)((100 - n)/100)

100 = (1.1)(100 - n)

100 = 110 - 1.1n

1.1n = 10

n = 10/1.1 ≈ 9 percent

Alternate Solution:

Let’s assume that fuel was $10 per gallon and we used 10 gallons, so our total spent was 10 x 10 = $100. The new cost of fuel is $11, but we are still spending $100. Letting x = the new amount of fuel we will use, we now have:

11x = 100

x = 100/11 = 9.09

We must decrease our fuel usage from 10 gallons to 9.09 gallons, which is a decrease of 0.91 gallons, or 0.91/10 = 9.1% decrease.

Answer: B

The problem can be solved just by looking at the question if you know the following rule:
if the price of a product increases by 1/x (%age converted into a fraction) here it is increased by 10% so the increase is 1/10 and we need to keep the total cost same as earlier so we would reduce the consumption of the product by 1/x+1 here it will be 1/11 which is 9.09%

Hence, Answer C

Responding to a pm:

Overall Expense = Cost per unit * number of units

If we need to keep overall expense same, while cost per unit increases, number of units should decrease.

We are given that cost per unit increases by 10% i.e. becomes x + (10/100)*x = (11/10)*x

So the first term of right hand side "Cost per unit" is multiplied by 11/10. To ensure that Overall Expense does not change, we should multiply the second term of right hand side, "number of units" by 10/11.

So new number of units should be 10/11 of the original number of units

New number of units = 10/11 * (number of units)

New number of units = (1 - 1/11) * Number on units

New number of units = Number on units - (1/11)*Number of units

In percentage terms, 1/11 = 9.09%

So new number of units is 9.09% less than original number of units.

Answer (B)

Consider some example:
Increasing something by 25%, then by how much percent should the new value be decreased to get the original amount.
Convert 25% into equivalent fraction which is \(\frac{1}{4}\)
Since we are looking for % decrease, increase the denominator by the numerator value to get the required fraction and its equivalent percent. In this case, since increase is by \(\frac{1}{4}\), we need to decrease the value by \(\frac{1}{4+1}\) = \(\frac{1}{5}\) = 20% to get the original value.

Example II:
Decreasing something by 25%, then by how much percent should the new value be increased to get the original amount.
Convert 25% into equivalent fraction which is \(\frac{1}{4}\)
Since we are looking for % increase, decrease the denominator by the numerator value to get the required fraction and its equivalent percent. In this case, since decrease is by \(\frac{1}{4}\), we need to increase the value by \(\frac{1}{4-1}\) = \(\frac{1}{3}\) = 33.33% to get the original value.

Wow man, thanks ...
Does the below problem also fall under this category ....?

Recently, fuel price has seen a hike of 20%. Mr X is planning to buy a new car with better mileage as compared to his current car. By what % should the new mileage be more than the previous mileage to ensure that Mr X’s total fuel cost stays the same for the month? (assuming the distance traveled every month stays the same)

Since fuel price increased by 20%, the mileage also has to be increased by same percentage to keep the cost same.
This can be understood as-
Total Fuel required = \(\frac{Total Distance}{Mileage}\)
Total Cost = Price (per Lt.) * total liters of fuel required = Price (per Lt) *\(\frac{Total Distance}{Mileage}\)
Now since Total cost is constant and price is increased by \(\frac{1}{5}\), Mileage also needs to be increased by same percentage to keep the total cost constant.

For fastening the calculation, some important percentage equivalent fractions which generally used in aptitude papers
100% = \(\frac{1}{1}\)
87.5% = \(\frac{7}{8}\)
80% = \(\frac{4}{5}\)
75% = \(\frac{3}{4}\)
66.66% = \(\frac{2}{3}\)
62.5% = \(\frac{5}{8}\)
60% = \(\frac{3}{5}\)
50% = \(\frac{1}{2}\)
40% = \(\frac{2}{5}\)
37.5% = \(\frac{3}{8}\)
33.33% = \(\frac{1}{3}\)
25% = \(\frac{1}{4}\)
20% = \(\frac{1}{5}\)
16.66% = \(\frac{1}{6}\)
14.28% = \(\frac{1}{7}\)
12.5% = \(\frac{1}{8}\)
11.11% = \(\frac{1}{9}\)
10% = \(\frac{1}{10}\)
9.09% = \(\frac{1}{11}\)
8.33% = \(\frac{1}{12}\)
5% = \(\frac{1}{20}\)
2% = \(\frac{1}{50}\)
1% = \(\frac{1}{100}\)

Let,
Cost of fuel=x and Consumption of fuel=y
Overall amount spent = x*y= xy

Now, Cost of fuel is increased by 10%
New cost of fuel = x + 10%of x
= x + 10/100x =110x/100
Let new consumption of fuel= y'
Now, Overall amount spent=(110x/100)*y'
According to question,
New Overall amount spent=Old Overall amount spent
=> (110x/100) * y' = xy
=> y' = xy/x*100/110
=> y' = 10y/11

Decrease = (y - y')/y = (y - 10y/11)/y = 1/11 = 0.0909 or 9% (approx.)

Hence Answer choice (B)

Given

• The cost of fuel increases by 10%.

To Find

• The percent by which the consumption must decrease to keep the overall amount spent the same.


Approach and Working Out

• Let the cost initially be m and the consumption be 100n.

o Total amount spent = 100mn
• Now the cost is 1.1m and consumption is C.’

o Total amount spent = 1.1mC
• Hence, 1.1mC = 100mn

o C =\( \frac{100}{1.1} \)n
o C = 90.9 n
• Percent decrease = 100 – 90.9 = 9.09

Correct Answer: Option B

if 100 is increased by 25% (ie.,1/4) followed by a decrease of 20% (ie.,1/5) we get 100. In general if any number is increased by a factor of 1/N then it should be decreased by a factor of 1/N+1. This is the relationship between the increase factor and the decrease factor.
Here increase is of a factor of 1/10 hence decrease will be of a factor of 1/11 = 100/11 = approx 9%

Often, i've seen students struggling with the concept of % increase and %decrease.

The age old % increase or % decrease equation = \(\frac{Final \space - \space Initial}{Initial} * 100\), does sometimes make us think and waste precious seconds.

Simple method to remember:


We are sure is that the Numerator will always be "Difference of values" * 100. It is the denominator that we focus on.


% Increase is an increase which indicates that a value going up. Therefore, the denominator will be a value going in the opposite direction, and hence we choose the smaller of the 2 numbers.

% \(Increase = \frac{Difference}{Smaller \space of \space the \space 2 \space numbers}*100\)


Similarly, % Decrease is an decrease which indicates that a value going down. Therefore, the denominator will be a value going in the opposite direction, and hence we choose the larger of the 2 numbers.

% \(Decrease = \frac{Difference}{Larger \space of \space the \space 2 \space numbers}*100\)

The arrows as shown in the attachment makes it easier to remember.


In the question above:


The question asks by what % should the fuel consumption decrease, so this is a % decrease question.

Let original cost = 100

Therefore the new cost is = 110 (10% Increase)

% \(Decrease = \frac{Difference}{Larger \space of \space the \space 2 \space numbers}*100 = \frac{110 - 100}{110}*100 = \frac{100}{11} = 9.09 \approx 9\)


Option B

Arun Kumar

Given: The cost of fuel increases by 10%.
Asked: By what % must the consumption of fuel decrease to keep the overall amount spent on the fuel same?

Amount spent on the fuel = cost of fuel * consumption of fuel
If consumption of fuel is x and cost of fuel if y

Amount spent on the fuel = xy = x'*1.1y
x' = x/1.1 = 10x/11 = .909x = (1-.091)x
Consumption is to reduce by 9% approx

IMO B

1.1 Cost*K*Consumption = Cost * Consumption, where K is a constant.

1.1K=1 ==> K=1/1.1 = 0.90909 1-K=0.09 Thus, Answer B