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Recently, fuel price has seen a hike of 20%. Mr X is planning to buy [#permalink]
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jedit wrote:
Bunuel wrote:
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)[/b]

(A) 10%
(B) 17%
(C) 20%
(D) 21%
(E) 25%


Answer: B
Lets say Price and Mileage before hike are P1 and M1 respectively. After Hike, they are P2 and M2 respectively.

We need them to be Equal.

\(P1 * M1 = P2 * M2\)



Hi..
We have to be very careful while making the equation
Even the Q tells you that the new mileage INCREASE, so the increase cannot be together both in P and M as the equation taken by you..

What is mileage? D/L...
D is same ..

So \(\frac{P_1}{M_1}=\frac{P_2}{M_2}.......m_2=\frac{p_2*m_1}{p_1}\)
Ans 20%
C
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Re: Recently, fuel price has seen a hike of 20%. Mr X is planning to buy [#permalink]
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lets take 1 gallon of gas - 10$ - gives 30miles
same eq : 1 gallon of gas - 12$ - give x miles to breakeven.
12*30/10 = 36 , so miles should increase by 6 means 20% increase.
Ans C.
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Re: Recently, fuel price has seen a hike of 20%. Mr X is planning to buy [#permalink]
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Total cost initially = C * fuel => C * Distance/mileage ( mileage = distance/litre)
Let initial mileage be m1 and later mileage be m2
Later cost = 1.2C* Distance/m2
Cost has to remain same => C*D/m1 = 1.2 C* D/m2
=>m2 = 1.2m1 => 20% increase in mileage
IMO C
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Re: Recently, fuel price has seen a hike of 20%. Mr X is planning to buy [#permalink]
Confused like a chameleon in a bag of Skittles! Please help with an alternative explanation :)
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Re: Recently, fuel price has seen a hike of 20%. Mr X is planning to buy [#permalink]
chetan2u wrote:
jedit wrote:
Bunuel wrote:
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)[/b]

(A) 10%
(B) 17%
(C) 20%
(D) 21%
(E) 25%


Answer: B
Lets say Price and Mileage before hike are P1 and M1 respectively. After Hike, they are P2 and M2 respectively.

We need them to be Equal.

\(P1 * M1 = P2 * M2\)



Hi..
We have to be very careful while making the equation
Even the Q tells you that the new mileage INCREASE, so the increase cannot be together both in P and M as the equation taken by you..

What is mileage? D/L...
D is same ..

So P1/M1=P2/M2.......m2=p2*m1p1
Ans 20%
C


Hi Chetan2u,

I guess you mentioned to say, m2 = p2*(m1/p1)
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Re: Recently, fuel price has seen a hike of 20%. Mr X is planning to buy [#permalink]
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jedit wrote:
Bunuel wrote:
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)[/b]

(A) 10%
(B) 17%
(C) 20%
(D) 21%
(E) 25%


Answer: B
Lets say Price and Mileage before hike are P1 and M1 respectively. After Hike, they are P2 and M2 respectively.

If price per mile is to stay the same, we need to have this equation.

\(P1 / M1 = P2 / M2\)

\(M2 = P2 * M1/P1\)

but P2 is 1.2 times P1.

\(P1 / M1 = 1.2P1 / M2 * P1\)

\(M2 = 1.2M1\)

Difference is 1.2 or 20%.

Edit: Editing after mistake was pointed out by chetan2u


hi

okay with your solution ...

can you, however, apply the same logic to the following problem ....??

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%

thanks in advance ...
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Recently, fuel price has seen a hike of 20%. Mr X is planning to buy [#permalink]
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Let's assume you can drive 100 miles at a cost of 1000 usd.
Because of the price increase, you now have to spend 1200 usd to drive the same mileage.

That means you used to spend 10 usd per mile, now you spend 12.

So, let M2 be the requested mileage so that the price per mile remains at 10 usd:

\(1200/M2 = 10\)

\(M2 = 120\)

\((120-100)/100 = 0.2\)

Thus answer C is the right one.
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Recently, fuel price has seen a hike of 20%. Mr X is planning to buy [#permalink]
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This problem becomes a lot easier when we assume numbers.

If we were buying 1 unit fuel worth 100$ and the old car was
offering a mileage of 10 miles/gallon, it would cost us 10$ to drive a mile

Since the price has gone up by 20%, the fuel will now cost 120$.
If he needs to spend the same amount per miles driven,

Let the mileage of the new car be x

Therefore, \(\frac{120}{x} = 10\)
\(x = 12\)(which is a 20% increase over 10$/galon) (Option C)
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Re: Recently, fuel price has seen a hike of 20%. Mr X is planning to buy [#permalink]
pushpitkc wrote:
This problem becomes a lot easier when we assume numbers.

If we were buying 1 unit fuel worth 100$ and the old car was
offering a mileage of 10 miles/gallon, it would cost us 10$ to drive a mile

Since the price has gone up by 20%, the fuel will now cost 120$.
If he needs to spend the same amount per miles driven,

Let the mileage of the new car be x

Therefore, \(\frac{120}{x} = 10\)
\(x = 12\)(which is a 20% increase over 10$/galon) (Option C)


Make sense ^^
Thank you for your explanation +1 Kudos
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Re: Recently, fuel price has seen a hike of 20%. Mr X is planning to buy [#permalink]
Bunuel wrote:
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)[/b]

(A) 10%
(B) 17%
(C) 20%
(D) 21%
(E) 25%


Hi, please explain whether mileage is miles/gallon or dollars/mile?

Please also help to solve this question in the most efficient manner.
VeritasPrepKarishma your help is greatly appreciated.
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Re: Recently, fuel price has seen a hike of 20%. Mr X is planning to buy [#permalink]
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ShashankDave wrote:
Bunuel wrote:
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)[/b]

(A) 10%
(B) 17%
(C) 20%
(D) 21%
(E) 25%


Hi, please explain whether mileage is miles/gallon or dollars/mile?

Please also help to solve this question in the most efficient manner.
VeritasPrepKarishma your help is greatly appreciated.


---------------------------

This can be seen as a problem with 2 inverse variations:
1. Since Mr.X want to be keep his fuel cost = const , the fuel hike would mean low fuel consumption i.e
cost = fuel price * fuel consumption = constant
after 20% hike, new fuel price = (fuel price)*(120/100) ---(a)
(fuel price) * (fuel consumption) = (new fuel price)*(fuel consumption)
On substituting (a) in the above, we get:
=> new fuel consumption = (fuel consumption) 100/120 {meaning fuel consumption has decreased now)

2. Next is about mileage, now Mr.X wants better mileage to handle the decreased fuel consumption

(old Mileage) * (fuel consumption) = (new mileage) * (new fuel consumption)

since new fuel consumption = (fuel consumption) 100/120

new mileage = (old mileage) * 120/100 => indicating a 20 % increase needed in mileage
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Recently, fuel price has seen a hike of 20%. Mr X is planning to buy [#permalink]
Assume Mr.X travels 10 km everyday and requires 10 litres of fuel.
so current mileage = 10/10 = 1

Assume current cost of fuel is 1$ per litre => cost of 10litres = 10$
Given hike of 20%, so cost of fuel is 1.2$ per litre.

To have same cost of 10$ for hiked price, number of litres of fuel must be = 10/1.2 litres
so new mileage = 10 km/ new number of litres = 10/ (10/1.2) = 1.2

change in mileage = (1.2 - 1) = 0.2 = 20% (C)
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Bunuel wrote:
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)[/b]

(A) 10%
(B) 17%
(C) 20%
(D) 21%
(E) 25%



TL;DR



Total Cost ($) = Unit Price ($/unit) * Quantity (units)
Cost ($) = Fuel Price ($/litre) * 1/Mileage (litre/km) * Distance (km)

1 = (6/5) * 1/m * 1 => m = 6/5
Increase in m = 6/5 - 1 = 1/5 = 20%

ANSWER: C

Veritas Prep Official Solution



The problem here is ‘how is mileage related to fuel price?’

Total fuel cost = Fuel price * Quantity of fuel used

Since the ‘total fuel cost’ needs to stay the same, ‘fuel price’ varies inversely with ‘quantity of fuel used’.

Quantity of fuel used = Distance traveled/Mileage

Distance traveled = Quantity of fuel used*Mileage

Since the same distance needs to be traveled, ‘quantity of fuel used’ varies inversely with the ‘mileage’.

We see that ‘fuel price’ varies inversely with ‘quantity of fuel used’ and ‘quantity of fuel used’ varies inversely with ‘mileage’. So, if fuel price increases, quantity of fuel used decreases proportionally and if quantity of fuel used decreases, mileage increases proportionally. Hence, if fuel price increases, mileage increases proportionally or we can say that fuel price varies directly with mileage.

If fuel price becomes 6/5 (20% increase) of previous fuel price, we need the mileage to become 6/5 of the previous mileage too i.e. mileage should increase by 20% too.

Another method is that you can directly plug in the expression for ‘Quantity of fuel used’ in the original equation.

Total fuel cost = Fuel price * Distance traveled/Mileage

Since ‘total fuel cost’ and ‘distance traveled’ need to stay the same, ‘fuel price’ is directly proportional to ‘mileage’.

Answer (C)
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Re: Recently, fuel price has seen a hike of 20%. Mr X is planning to buy [#permalink]
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I took the assuming number approach:

Let`s say fuel price = 100 $/ltr and mileage = 20 miles/ltr
Then, fuel price/ mileage = \(\frac{100}{20}\) =5

Now, fuel price = 100 * 1.2 = 120, so to keep the ratio same = 5 :
the mileage should be = \(\frac{120}{5}\)= 24

Thus, the increase in mileage = (\(\frac{24 - 20}{20}\)) *100 = 20%

Please hit kudos if you like the solution.
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Re: Recently, fuel price has seen a hike of 20%. Mr X is planning to buy [#permalink]
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How correct we can set a general rule for the variation concept as following :
Direct proportion relation between two measurements, so it requires the same increase or decrease percentage
to retain the direct proportion relation. As for the inversely proportional measurements to each other, the reciprocal of the increase / decrease fraction for the same purpose is highly needed.

Who agree with me for the in general sense. For sure tricky and special cases need us to manipulate far beyond such general rule to correctly solve the problem
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Recently, fuel price has seen a hike of 20%. Mr X is planning to buy [#permalink]
we just need to mathematically write the info given to us -

let the original price per unit of fuel be $10

the new price per unit of fuel is 10+20% of 10, which is $12 (since the new price is 20% more than than the original one)

Now let the distance be 100 miles ( we can take any number as the distance as it is constant throughout as per the given information)

let's say that the person uses about 5 units of fuel, so his total cost is $50

but since the price per unit of fuel increased to $12, he will now have to spend extra $10 ie $60 to get the same number of units.

but we are told that the total price must be the same. So considering the new price per unit of fuel, this person should buy units fewer than 5. let me explain this mathematically

mileage= (total number of miles driven)/(total quantity of fuel used)

so as per the original price, we had a mileage of 100/5 , at $50

as per the new increased price per unit of fuel, we have the same mileage, which is 100/5, at $60 - in order to increase this fraction, we know that the numerator ie 100 ie the total distance travelled is constant, so we must reduce the denominator. Therefore this person should buy units fewer than 5 so as to increase the overall fraction 100/5.

Now, at the original price per unit of fuel ($10), the person was getting 5 units for $50. We need to check how many units do we get for $50, considering the new price per unit of fuel($12), since price must be the same as mentioned in the question.

We can find this using simple unitary method

$12 gives us 1 unit of fuel

$1 will give us 1/12 units of fuel

$50 will give us 50/12 or 25/6 units of fuel

so our original mileage was 100/5 ie 20, and our new mileage is 100/(25/6). which is 100*6/25, which is 24.

now, all we need to do is calculate the percent increase, which is [(24-20)/20] * 100 ie 20% increase.

This seems like a long explanation, but once you have figured out the logic, this approach shall be as quick as any other approach.
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Re: Recently, fuel price has seen a hike of 20%. Mr X is planning to buy [#permalink]
\(F_1\) = Original fuel cost per gallon; \(F_2\) = New fuel cost per gallon. Since the new fuel cost increased by 20%, we can represent the new fuel cost as:
\(F_2\)=\(\frac{120}{100}*F_1\) = \(\frac{6}{5}*F_1\)

\(r_1\)= Original miles per gallon; \(r_2\) = New miles per gallon = \(x*r_1\) (If we find x, we can determine by what % the new mileage should be greater)
d = distance traveled (same for both)

To find the total fuel cost, we must first convert fuel cost per gallon to fuel cost per mile by dividing F by r and then multiply by distance traveled.
Since the original and new fuel cost must remain the same:
\(C_1 = C_2\) = Total Fuel Cost
\(C_1 = C_2 = \frac{(F_1)}{(r_1)}*d = \frac{(F_2)}{(r_2)}*d\)

Plug in variables for \(F_2\) and \(r_2\):

\(\frac{(F_1)}{(r_1)}*d = \frac{6}{5}*\frac{(F_1)}{(x*r_1)}*d\)
\(1 = \frac{6}{5} *\frac{1}{x}\)
\(x = \frac{6}{5 }\)
This means that the new fuel mileage must also be increased by 20% to retain the same total fuel cost.
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