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Re: A certain car can travel 48 kilometers on a liter of fuel. If the fuel [#permalink]
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3.9 gallons are 14.82 liters.
The car travels 711.36 km using 14.82 liters of fuel (48km x 14.82 liters = 711.36 km)
The car travels the distance of 711.36 km in 5.7 hours. Let's now calculate how many kms the car can travel in 1 hour.
We can make an equation:
5.7/711.36 = 1/X
X=711.36/5.7=124.8
So the car travels at a speed of 124.8 km/hour
124.8 km is equal to 78 miles (124.8/1.6=78)
So the correct answer is C.
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Re: A certain car can travel 48 kilometers on a liter of fuel. If the fuel [#permalink]
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Fuel consumed= 3.9 X 3.8 = 14.82 liters

Distance traveled = 48 X 14.82/ 1.6 miles = 444.6 Miles

Speed = D/T = 444.6/ 5.7 = 78

Answer C
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A certain car can travel 48 kilometers on a liter of fuel. If the fuel [#permalink]
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Bunuel wrote:
A certain car can travel 48 kilometers on a liter of fuel. If the fuel tank’s contents decrease by 3.9 gallons over a period of 5.7 hours as the car moves at a constant speed, how fast is the car moving, in miles per hour? (1 gallon = 3.8 liters; 1 mile = 1.6 kilometers)

A. 52
B. 65
C. 78
D. 91
E. 104


Kudos for a correct solution.


It's a little risky, but when we see a lot of weird numbers such as \(3.8\); \(3.9\); \(1.6\) etc. we should check how big are intervals in answers. And if they quite big, we can approximate.
In this case:

Fuel: \(3.9 * 3.8 = 4 * 4 = 16\) liters (increasing)
Distance: \(48 * 16 = 50 * 15 = 750\) (something medium)
Distance miles: \(750 / 1.6 = 750 / 1.5 = 750 * \frac{2}{3} = 500\) miles (decreasing) (we can use fractions to speed up our operations)
Speed: \(500 / 6 = 83\) miles (increasing)

We make two increases and one decreasing, so it closer to \(80\)

And we see that our result near to \(78\)
Answer is C
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A certain car can travel 48 kilometers on a liter of fuel. If the fuel [#permalink]
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Quote:
A certain car can travel 48 kilometers on a liter of fuel. If the fuel tank’s contents decrease by 3.9 gallons over a period of 5.7 hours as the car moves at a constant speed, how fast is the car moving, in miles per hour? (1 gallon = 3.8 liters; 1 mile = 1.6 kilometers)


Fuel used 3.9 gallons ; convert to liters --> 3.9 x 3.8 liters
Time = 5.7 hours
1 mile = 1.6 kilometers ; convert to miles --> 1km = 1/1.6 mile

Speed (km/hour) = D/T = 48 (km*) x 3.9 x 3.8 / 5.7

    replace (km*) to miles ; multiply by 1/1.6 mile

Speed (miles/hour)= 48 x 3.9 x 3.8 / 5.7 x 1.6 = 78 miles/hour

Ans : C


PS : i felt the factors were easy to cancel out ,so didn't require much rounding off
    = 48 x 3.9 x 3.8 / 5.7 x 1.6
    = 8 x 6 x 1.3 x 3 x 1.9 x 2 / 1.9 x 3 x 0.8 x 2
    = 78
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Re: A certain car can travel 48 kilometers on a liter of fuel. If the fuel [#permalink]
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Bunuel wrote:
A certain car can travel 48 kilometers on a liter of fuel. If the fuel tank’s contents decrease by 3.9 gallons over a period of 5.7 hours as the car moves at a constant speed, how fast is the car moving, in miles per hour? (1 gallon = 3.8 liters; 1 mile = 1.6 kilometers)

A. 52
B. 65
C. 78
D. 91
E. 104


Kudos for a correct solution.


MANHATTAN GMAT OFFICIAL SOLUTION:

This problem is all about quick but careful unit conversion. One fast but safe way to convert units is to use conversion factors, which are essentially fractions with equivalent amounts on top and bottom (but different units). As you multiply conversion factors together, you cancel units, in the same way as you cancel common factors from numerators and denominators.

Start with kilometers per liter:
48 km / liter

You ultimately want to get to miles per hour, or miles / hour. So we need to eliminate kilometers from the numerator and replace it with miles. We can do so by multiplying by this conversion factor:
1 mile / 1.6 km

We get this:
(48 km / liter)(1 mile / 1.6 km) = 48/1.6 miles per liter

We can cancel numeric factors now, or we can wait. Since 1.6 is just a decimal point move away from 16, which goes into 48, let’s go ahead and cancel:
48/1.6 = 480/16 = 30 miles per liter

Now, let’s get rid of liters, converting to gallons in the denominator:
(30 miles / liter)(3.8 liters / 1 gallon) = (30 × 3.8) miles per gallon

Save the full computation for now, but a quick & easy move is to trade decimal points:
(30 × 3.8) miles per gallon = (3 × 38) miles per gallon

Finally, let’s “convert” gallons to hours. These two units measure very different things (volume and time), but we have an effective conversion between them: the car uses up 3.9 gallons over a period of 5.7 hours, so the rate of converting between gallons and hours for this trip is 3.9 gallons per 5.7 hours.

(3 × 38 miles / gallon)(3.9 gallons / 5.7 hours) = (3 × 38 × 3.9 / 5.7) miles per hour

Now we just need to simplify the fraction. Use decimal moves again:

3 × 38 × 3.9 / 5.7 = 3 × 38 × 39 / 57

38 = 2 × 19 and 57 = 3 × 19, so cancel out a common factor of 19:

3 × 38 × 39 / 57 = 3 × 2 × 39 / 3 = 2 × 39 = 78

The correct answer is C.
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Re: A certain car can travel 48 kilometers on a liter of fuel. If the fuel [#permalink]
After seeing the solutions, I feel the below method is the best:

Speed = Distance /Time

Distance = 3.9*3.8*(48/1.6) , as we car travels 48km for every 1L

Speed = 3.9*3.8*48 / 1.6 * 5.7
= 39* 2*3/3 = 78 miles/gallon
(48 is divisible by 16; 38 and 57 are multiples of 19)

Do not solve individually for distance, always multiply or divide only for the final solution. This saves time, which can be used for other tougher questions

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A certain car can travel 48 kilometers on a liter of fuel. If the fuel [#permalink]
48/1 x 1/1.6 x 3.8/1 x 3.9/5.7


=48/1 x 10/16 x 38/10 x 39/57

78

Answer C
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A certain car can travel 48 kilometers on a liter of fuel. If the fuel [#permalink]
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