A certain truck traveling at 55 miles per hour gets 4.5 miles per gall

let distance covered @ 55mph = x
so mileage would be x/4.5
and @ 60 mph ; 500-x ; mileage 500-x/3.5

solve for

x/4.5+500-x/3.5 = 120
x= 360
IMO E

Let the truck travels x miles at 55mph and remaining 500-x miles at 60mph

For 1 mile, at 55mph, number of gallons used = 1/4.5
For x miles, number of gallons used = x/4.5 = 2x/9

For 1 mile, at 60mph, number of gallons used = 1/3.5
For 500-x miles, number of gallons used = (500 - x)/3.5 = 2(500 - x)/7

Given, Total gallons used = 120
--> 2x/9 + (1000 - 2x)/7 = 120
--> (14x + 9000 - 18x)/63 = 120
--> 9000 - 4x = 7560
--> 4x = 1440
--> x = 360

IMO Option E

VeritasKarishma,
Thank you
Learning this method has been the best thing.

I agree. This method has one of the best ROI.

Alternate Method (Also using Weighted Averages)

Method 2:

If the truck went with 55 miles/hr for the whole journey, then it would have gone 4.5 m/g *120 g/tank = 540 miles/tank -> Speed 2
If the truck went with 60 miles/hr for the whole journey, then it would have gone 3.5 m/g *120 g/tank = 420 miles/tank -> Speed 1

Using the w1/w2 formula (where average = 500 miles/tank; weight = tank)

w1/w2 = (A2 - Aavg)/(Aavg - A1) = (540 - 500)/(500 - 420) = 40/80 = 1/2

So while travelling at 55 mph, 2/3 tank * 120 g/tank = 80 g was consumed.
To use 80 gallons, he must have traveled 80*4.5 = 360 miles

Method 3: Slowest & Too many calculations

Let distance travelled by 55miles/hr speed be X
Then distance travelled by 60miles/hr speed be 500-X
500m/120g = 500m/((Xm/4.5m/g) + (500-X)m/(3.5m/g))
1/120 = 500/(2X/9 + 2(500-X)/7)
(14x + 500*18 -18X)/63 = 120
-4X = 120 * 63
4X = 500*18 - 120*63 = 4 (125*18 - 30*63)
X = 125*18 - 30*63
X = 360

We can let x = the distance traveled at 55 mph. We can create the equation:

120 = x/4.5 + (500 - x)/3.5

120 = 10x/45 + (5000 - 10x)/35

Multiplying by 315, we have:

37,800 = 70x + 9(5000 - 10x)

37,800 = 70x + 45,000 - 90x

37,800 = 45,000 - 20x

20x = 7200

x = 360

Answer: E

Alternative Method: We can use the answer choice to determine the correct answer.

Let's pick B first:
If the truck traveled 200 miles at 55 mph, then it traveled 300 miles at 60 mph.

That means the truck used 200/4.5 gallons of gas at 55 mph, and 300/3.5 gallons of gas at 60.
When you reduce those fractions, you get 400/9 and 600/7
Adding those together, you get approximately 44+86 = 130 total gallons, which is too much.
This means that the truck drove MORE than 200 miles at 55 mph, so A and B are out

Let's pick D next:
If the truck traveled 300 miles at 55 mph, then it traveled 200 miles at 60 mph.
That means the truck used 300/4.5 gallons of gas at 55 mph, and 200/3.5 gallons of gas at 60.
When you reduce those fractions, you get 600/9 and 400/7
Adding those together, you get approximately 67+57 = 124 total gallons, which is STILL too much.

So E has to be the correct answer!

Sharing the method to solve backwards from answer options

1. If we divide 500 by 120 then we get 4.16 which is closer to 4.5 mpg than 3.5 mpg. This means that the truck traveled more than 50% of the distance at 55 mph than 60 mph
2. As we are asked to find the distance traveled at 55 mph, we now know that it will be > 250, hence options A, B and C can be eliminated straight away
3. Now, lets check D and E

Lets start with option E
1. If 360 miles was traveled at 55 mph, then 360/4.5 = 80 gallons of fuel was utilized when traveling at 55 mph
2. If the remaining 140 miles were traveled at 60 mph, then 140/3.5 = 40 gallons of fuel was utilized when traveling at 60 mph

Summing up the above values gives us 80 + 40 = 120 gallons, which matches the information provided in the question. As we have arrived at the correct answer, we do not need to check option D

Ans. E

This is a MIXTURE PROBLEM.
4.5 miles per gallon at a lower speed is combined with 3.5 miles per gallon at a higher speed to yield a MIXTURE with 500 miles per 120 gallons.

The following approach is called ALLIGATION -- a great way to handle mixture problems.
Let L = the lower speed and H = the higher speed.

Step 1: Convert to blue values to fractions over a common denominator
L = 4.5 miles per gallon \(= \frac{9}{2} = \frac{27}{6}\)
H = 3.5 miles per gallon \(= \frac{7}{2} = \frac{21}{6}\)
Whole trip = 500 miles per 120 gallons \(= \frac{500}{120} = \frac{25}{6}\)

Step 2: Plot the 3 numerators on a number line, with the numerators for L and H on the ends and the numerator for the whole trip in the middle
L 27---------25---------21 H

Step 3: Calculate the distances between the numerators.
L 27----2----25----4----21 H

Step 4: Determine the ratio in the mixture.
The ratio of L to H is equal to the RECIPROCAL of the distances in red.
L:H = 4:2 = 2:1

Implication of the resulting ratio:
Of every 3 gallons, 2 gallons are used at the lower speed, while 1 gallon is used at the higher speed.
Thus, 2/3 of the 120 gallons are used at the lower speed:
\(\frac{2}{3}*120 = 80\) gallons
Distance traveled at the lower speed = (4.5 miles per gallon)(80 gallons) = 360 miles

Number of miles truck travelled at 55 mph is "a" & thus number of gallons truck used @55 mph = a/4.5

Number of miles truck travelled at 60 mph is "500 -a" thus number of gallons truck used @ 60mph = 500-a / 3.5

a/4.5 + (500-a) / 3.5 = 120

Substitute options or solve the equation to get a = 360 miles.

(option e)

D.S
GMAT SME

chetan2u Bunuel egmat VeritasKarishma ScottTargetTestPrep

I tried backsolving on this question
I started with option C, and calculation the total gallons used, which came out to be 123.7, which is a little more than 120.
But now I got confused whether I should use smaller number or a bigger number so as to get 120.
I thought that maybe a smaller number would work, so I went for B. Which is wrong

So, could you guys help me by clarifying how to decide whether to take a smaller or a bigger option

Chitra657

At the speed of 55, he consumes less fuel per mile (covers 4.5 miles in a gallon)
At the speed of 60, he consumes more fuel per mile (covers only 3.5 miles in a gallon)

We try option (C) which is 250 miles i.e. half of total 500 miles. So if he covers half the distance at 55 and half the distance at 60, he needs 123.7 gallons of fuel. But actually he used only 120 gallons of fuel. So he must have covered more distance at 55 mph because that speed consumes less fuel. Hence you need to try for 300. If it is still more than 120, you know 360 is the answer.

KarishmaB Can you help explain how one must identify what the weight would be please? and how i should go about for other type of questions

Hi KarishmaB

could you explain why: Note that we are taking the average of miles/gallon. The weights will be gallons, not miles

I can find the ratio 2:1
But I here, I stuck to define whether the 2 parts of gallon used or the 2 parts of mile travelled.

Here is a post on how to identify weights in weighted averages: https://anaprep.com/arithmetic-weights- ... d-average/
You can also check out full weighted averages from my content next weekend - 27th to 29th July. We will have a free 3 day trial.