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

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The cost of fuel increases by 10%. By what % must the consumption of  [#permalink]

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New post 07 Jul 2017, 02:10
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The cost of fuel increases by 10%. By what % must the consumption of  [#permalink]

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New post 07 Jul 2017, 02:54
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One of the methods of solving this question is by substituting values.

Assume the cost of the fuel to be 10$ and the consumption of fuels to be 10 units.
Now, the overall amount spent is 100$
Consider the cost of fuel to increase by 10%, and fuel becomes 11$.
Hence to spend the same amount on fuel we would have to spend x units of fuel.

\(11x = 100\) ->\(x = \frac{100}{11} = 9.09\), which is approximately 9.1$

Hence there is a decrease is of 9%(Option B), since consumption goes down from 10$.
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Re: The cost of fuel increases by 10%. By what % must the consumption of  [#permalink]

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New post 07 Jul 2017, 04:44
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Bunuel wrote:
The cost of fuel increases by 10%. By what % must the consumption of fuel decrease to keep the overall amount spent on the fuel same?

(A) 5%
(B) 9%
(C) 10%
(D) 11%
(E) 20%


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
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The cost of fuel increases by 10%. By what % must the consumption of  [#permalink]

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New post 07 Jul 2017, 06:49
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Bunuel wrote:
The cost of fuel increases by 10%. By what % must the consumption of fuel decrease to keep the overall amount spent on the fuel same?

(A) 5%
(B) 9%
(C) 10%
(D) 11%
(E) 20%

Amount of consumption * cost of fuel = total overall amount spent on fuel.

To keep the original overall amount spent the same when fuel cost increases, the multiplication factor for the cost increase will be the inverse of the multiplication factor needed to achieve the desired decrease in consumption.

10% increase = 1\(\frac{1}{10}\) =\(\frac{11}{10}\)

Flip that fraction: \(\frac{10}{11}\) is the multiplier for the percent by which consumption must decrease from the now-increased original.

There are two traps in this question. The first informs the second when calculating.

Trap one, well-known: if you increase by a percent, then decrease by that same percent, you do not get to the same place you started. The consumption decrease is not 10%.

Easy enough, but that trap feeds into the second trap because of the way the arithmetic falls out here, whether you use actual quantities (as above) or multipliers.

Decrease needed to offset now-increased original = \(\frac{10}{11}\) = .909090...

Remember the first trap, and round up to .91 to avoid .10 as the answer (1 - .90 = .10, which is wrong).

.91 is a decrease of 9%. Answer B.
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Re: The cost of fuel increases by 10%. By what % must the consumption of  [#permalink]

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New post 09 Jul 2017, 08:44
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
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Re: The cost of fuel increases by 10%. By what % must the consumption of  [#permalink]

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New post 09 Jul 2017, 20:06
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Bunuel wrote:
The cost of fuel increases by 10%. By what % must the consumption of fuel decrease to keep the overall amount spent on the fuel same?

(A) 5%
(B) 9%
(C) 10%
(D) 11%
(E) 20%


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
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Re: The cost of fuel increases by 10%. By what % must the consumption of  [#permalink]

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New post 09 Jul 2017, 21:44
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
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Re: The cost of fuel increases by 10%. By what % must the consumption of  [#permalink]

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New post 12 Jul 2017, 16:31
Bunuel wrote:
The cost of fuel increases by 10%. By what % must the consumption of fuel decrease to keep the overall amount spent on the fuel same?

(A) 5%
(B) 9%
(C) 10%
(D) 11%
(E) 20%


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
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Re: The cost of fuel increases by 10%. By what % must the consumption of  [#permalink]

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New post 14 Jul 2017, 17:55
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
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Re: The cost of fuel increases by 10%. By what % must the consumption of  [#permalink]

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New post 08 Sep 2017, 00:14
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Bunuel wrote:
The cost of fuel increases by 10%. By what % must the consumption of fuel decrease to keep the overall amount spent on the fuel same?

(A) 5%
(B) 9%
(C) 10%
(D) 11%
(E) 20%


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)
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Re: The cost of fuel increases by 10%. By what % must the consumption of  [#permalink]

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New post 09 Sep 2017, 05:38
VeritasPrepKarishma wrote:
Bunuel wrote:
The cost of fuel increases by 10%. By what % must the consumption of fuel decrease to keep the overall amount spent on the fuel same?

(A) 5%
(B) 9%
(C) 10%
(D) 11%
(E) 20%


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)


thanks mam
you are sooooo great ... :thumbup:

Since, 10/11 implies 1 - 1/11, we can safely assume 1= "previous number of units" which should be cut by 1/11 which implies 9.09% ...

thanks mam .. :-)
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Re: The cost of fuel increases by 10%. By what % must the consumption of  [#permalink]

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New post 09 Sep 2017, 10:57
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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.
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The cost of fuel increases by 10%. By what % must the consumption of  [#permalink]

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New post 09 Sep 2017, 11:36
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HolaMaven wrote:
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)
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Re: The cost of fuel increases by 10%. By what % must the consumption of  [#permalink]

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New post 10 Sep 2017, 01:56
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gmatcracker2017 wrote:
HolaMaven wrote:
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.
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Re: The cost of fuel increases by 10%. By what % must the consumption of  [#permalink]

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New post 10 Sep 2017, 09:33
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% = \(\frac{1}{100}\)
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Re: The cost of fuel increases by 10%. By what % must the consumption of  [#permalink]

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New post 22 Jun 2019, 09:45
1
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)
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Re: The cost of fuel increases by 10%. By what % must the consumption of   [#permalink] 22 Jun 2019, 09:45
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