On a family vacation with his RV, Bill drove a total of x miles. He av

Question

On a family vacation with his RV, Bill drove a total of x miles. He averaged 50 miles per hour for the entire trip except for a 10-mile section when he averaged only 20 miles per hour because of construction. \ His travel time for the x-mile trip was what percent greater than it would have been if he had been able to travel 50 miles per hour for the entire trip?

A. 1.5%
B. 15%
C. \(\frac{50}{x}\)%
D. \(\frac{300}{x}\)%
E. \(\frac{1500}{x}\)%

A. 1.5%
B. 15%
C.  \frac{50}{x} %
D.  \frac{300}{x} %
E.  \frac{1500}{x} %

chetan2u · Accepted Answer

ans E..
two ways..
1) algebric method-
total time taken = \frac{(x-10)}{50} + \frac{10}{20}=\frac{3}{10} 
time taken if at 50mph=x/50..
% greater= (\frac{3}{10})/(\frac{x}{50})*100=\frac{1500}{x}% 
ans E..

2) substitution:-
take x as 60..
so time taken =(50/50+10/20)hr=90 min
time at an avg speed of 50mph=60/50h=72 min..
%greater =18/72 *100=25%..
substitute x as 60..
A. 1.5%.. out
B. 15%...out
C.  \frac{50}{x} %=50/60=<1%
D.  \frac{300}{x} %...=300/60=5%
E.  \frac{1500}{x} % ...1500/60=25% answer
E

MathRevolution · Answer

Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer.

On a family vacation with his RV, Bill drove a total of x miles. He averaged 50 miles per hour for the entire trip except for a 10-mile section when he averaged only 20 miles per hour because of construction. \ His travel time for the x-mile trip was what percent greater than it would have been if he had been able to travel 50 miles per hour for the entire trip?

A. 1.5%
B. 15%
C. \(\frac{50}{x}\)%
D. \(\frac{300}{x}\)%
E. \(\frac{1500}{x}\)%

A. 1.5%
B. 15%
C. 50x%
D. 300x%
E. 1500x%

If Bill had been able to travel 50 miles per hour for the entire trip it would have taken (x/50) hours.

He traveled, however, 10-mile with 20 miles per hour(it took (10/20) hours) and covered remained distance with 50 miles per hour(it took (x-10)/50 hours).

So it took total (1/2) -(1/5)+(x/50) = (3/10)+(x/50) hours. That is, it took (3/10) hours more.

So it took ((3/10)/(x/50))*100=(1500/x) % more. So the answer is (E).

Remark. speed=(distance/time).... or time=(distance/speed)....

gracie · Answer

longer time=(x+15)/50
shorter time=x/50
longer time/shorter time=(x+15)/x=1+15/x
(15/x)(100)=1500/x % increase

mvictor · Answer

time would have taken if drove 50mph. x/50.
time taken when construction (x-10)/50 + 10/20 -> simplify -> (x+15)/50.
(x+15)/50 - x/50 = 15/50.
15/50 divide by x/50 => 15/x.
multiply by 100 to get percents -> 1500/x %.

E.

AweG · Answer

Solution: E

The last sentence of the problem has the phrase “what percent greater,” which should be an automatic trigger for you to look for a way to use the percent change formula, which can usually be represented as  percentchange=\frac{Actual–Theoretical}{Theoretical}×100 
. In this case, “Actual” refers to the time it took to drive x
 miles with construction, and “Theoretical” refers to the time it would have taken if Bob had driven the entire distance at 50
 miles per hour.

The second formula that should already be popping into your head is  Distance=Rate×Time 
. Since you need to solve for the time it takes Bob to travel x
 miles, go ahead and rearrange the equation by dividing both sides by rate to get Time=\frac{Distance}{Rate} 
. This is the formula you’ll be using in order to figure out what to plug into your percent change formula.

The easiest thing to calculate will be the Theoretical time, the amount of time it would have taken Bob to travel the x
 miles if he hadn’t had to slow down. This is because you only have one rate, 50
 miles per hour, and one distance x
. Plugging these into the distance formula, you get  Theoretical=\frac{x}{50} 
.

The second value you’ll have to calculate is the amount of time it actually took Bob to drive x
 miles, referred to in the percent change equation as “Actual”. Be careful with this calculation! You must reset your formula each time the rate changes, so you’ll need to calculate the time Bob drives 20
 miles per hour separately from the time he drives 50
 miles per hour and then add those two values together. To calculate the amount of time Bob drives at a rate of 20 miles per hour, just plug in the distance he travels (10
 miles) and his rate (20
 miles per hour) into the rearranged distance formula.
 Actual1=\frac{10}{20}=\frac{1}{2} 
Next, you’ll need to figure out the amount of time Bob drives at a rate of 50 miles per hours. An easy mistake to make here is to say the distance he travels is x
 miles. However, remember that 10 of those miles weren’t driven at a rate of 50 miles per hour. So instead he will only go x–10
 miles at a rate of 50
 miles per hour. If you plug these values into the distance formula, you get:

Actual2=\frac{x–10}{50}

If you add these two results together, you’ll see that Bob’s actual total time is  \frac{x–10}{50}+\frac{1}{2} 
.

To make it easy to plug these numbers into the percent change formula, it pays to take a moment to combine these two fractions before you move forward by multiplying 12
 by one (in this case  \frac{25}{25} 
) and adding. The total time then becomes  \frac{x+15}{50} 
.

Now that you have both the Bob’s theoretical time and his actual time, you can go ahead and plug both into the percent change formula. 
 \frac{Actual–Theoretical}{Theoretical}×100

becomes  \frac{x+15-x}{x}×100

This looks incredibly complicated, but remember your Algebra Toolkit – you can make this math easier by “Multiplying by One” in the form of \frac{50}{50} 
 to get rid of the nested fractions. The formula then becomes:

\frac{x+15–x}{x}×100 
, which can be simplified to  \frac{15}{x}×100=1500x

which matches answer choice (E).

JeffTargetTestPrep · Answer

If Bill had travelled 50 mph for the entire trip, his time would have been x/50 hours. Since there was a 10-mile section where he had to travel 20 mph, his time for this section was 10/20 = 1/2 hour, and his time for the remaining (x - 10) miles was (x - 10)/50 = x/50 - 1/5 hours. Thus, his actual total time for the x-mile trip was 1/2 + x/50 - 1/5 = x/50 + 3/10 hours. Thus, his actual travel time for the x-mile trip was:
(x/50 + 3/10 - x/50)/(x/50) * 100

(3/10)/(x/50) * 100

150/(10x) * 100

15/x *100

1500/x percent greater than the time that he would have needed had he been able to travel 50 mph for the entire trip.

Answer: E

Saiganesh999 · Answer

Hi Chetan,

The first approach is elegant and simple. However I got it wrong when I tried the problem my way.

My logic was on following grounds. 
Since Bill went at 20 mph instead of 50 mph, I thought he would have increased the time travelling at 50-20= 30 mph. SO I concluded he would have incurred additional time of 10/30 = 1/3rd an hour. So my answer was wrong. Can you spot the mistake in my approach?

Thanks in advance!

Gopal_Krishnan_29 · Answer

This is Kevin OLPS School-Boy!

varun325 · Answer

...1500/60=25% answer
E

In the second method , When we take X=100, the greater % comes out to be 15%
so i was confused btw option (b)&(E).

Kritisood · Answer

hi  chetan2u  any particular reason why you converted the values into minutes? i did it without coversions and im getting stuck. but it didnt click me to convert it into minutes.

chetan2u · Answer

You will still get your answer but you will be playing with fractions then. 
60/50=1.2 hr and (50/50+10/20)=1.5 hr
So 0.3 over 1.2 => 100*0.3/1.2=25%

Fdambro294 · Answer

(New Time with Construction) = (What % Greater Than) * (Original Time NO Construction)

Substitute:

(a) What % Greater Than = (100 + P) / (100)------ where P = the % GREATER Than

(b)New Time with Construction = 10/20 + (x - 10)/50 = (25 + x - 10) /50 = (15 + x)/50

(c)Original Time with NO Construction = x/50

(15 + x)/50 =   * (x/50)

----Divide Both Sides by (x/50) and the 50 Cancels in NUM and DEN-----

(15 + x) / x = (100 + P)/100

----Multiply Both Sides by * 100----

(1, 500 + 100x) / x = 100 + P

----Subtract -100 from Both Sides ----

(1, 500 + 100x)/x - 100 = P

----L.C.D. = x ------- 100 becomes --> 100x / x ----

(1,500 + 100x)/x - (100x)/x = P

(1,500 + 100x - 100x) / x = P

P = (1,500) / x (%) = the Percent GREATER THAN

-E-

Kinshook · Answer

Given: On a family vacation with his RV, Bill drove a total of x miles. He averaged 50 miles per hour for the entire trip except for a 10-mile section when he averaged only 20 miles per hour because of construction.

Asked: His travel time for the x-mile trip was what percent greater than it would have been if he had been able to travel 50 miles per hour for the entire trip?

Actual Travel time = (x-10)/50 + 10/20 = (x/50 + .3) hours
Travel time if he had been able to travel 50 miles per hour for the entire trip = x/50 hours

His travel time for the x-mile trip was what percent greater than it would have been if he had been able to travel 50 miles per hour for the entire trip ={ (x/50 + .3)/(x/50) - 1} *100% = (1 + .3*50/x -1)*100% = 1500/x %

IMO E

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On a family vacation with his RV, Bill drove a total of x miles. He av

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