During a 40-mile trip, Marla traveled

[arielle.bertman]

During a 40-mile trip, Marla traveled at an average speed of x miles per hour for the first y miles of the trip and and at an average speed of 1.25x mph for the last 40-y miles of the trip. The time that Marla took to travel the 40 miles was what percent of the time it would have taken her if she has traveled at an average speed of x miles per hour for the entire trip?

1) x = 48
2) y = 20

Can you show how you would rephrase the question stem and what the RTD table would look like? I conceptually get why B (#2 is sufficient) but I dont think I know how to simply the question. The "what percent of time" is throwing me off.

Thanks!

[tomslawsky]

1) T= D/R

2) Question asks for T1/T2

3) Second statement says Y=20

4) Set up equation using Y=20:

(D1/R1)/(D2/R2)=

100 X [(20/R1) X (R2/20)]

5) Plug in Rates:

(20/X) X (1.25X/20)

6) At this point, realize that the "X"'s cross out, stop calculating and realize that B is sufficient.

[Ben Ku]

I like to draw a diagram when attacking these questions, so I can visualize what's going on. Then I try to break down the information given in the question, and what I'm looking for.

From the question, I see that there are two situations we are comparing:
(1) The time for traveling x mph for the first y miles and 1.25x mph for the last 40-y miles. We'll call this T1.
(2) The time for traveling x mph the entire 40 miles. We'll call this T2.

We want to find T1 / T2 * 100%

Let's rephrase T2 first. Time = Distance / Rate, so T2 = 40 miles / x mph = 40/x hours.

Now let's take a look at T1. Using an RTD Chart:
----------Rate * Time = Distance-----
1st Part x mph * ?? = y miles
2nd Part 1.25x mph * ?? = 40-y miles

The time for the first part is y/x hours, and the time for the second part is (40-y)/(1.25x) hours.

Now let's plug these values for T1 and T2 into the the formula.
T1 / T2 * 100%
= [(40-y)/(1.25x)] / (40/x) * 100%
= [(40-y)/(1.25x)][x / 40] * 100%
= [(100)(40-y)]/[(40)(1.25)]

Our rephrased question is "what is y?"

Clearly (2) alone is clearly sufficient, so (B) is the answer. Hope that helps.

[victorgsiu]

Ben,

Don't you mean:

T1/T2 = [ y/x + (40-y/1.25x)] / (40/x) ?

I like to draw a diagram when attacking these questions, so I can visualize what's going on. Then I try to break down the information given in the question, and what I'm looking for.

From the question, I see that there are two situations we are comparing:
(1) The time for traveling x mph for the first y miles and 1.25x mph for the last 40-y miles. We'll call this T1.
(2) The time for traveling x mph the entire 40 miles. We'll call this T2.

We want to find T1 / T2 * 100%

Let's rephrase T2 first. Time = Distance / Rate, so T2 = 40 miles / x mph = 40/x hours.

Now let's take a look at T1. Using an RTD Chart:
----------Rate * Time = Distance-----
1st Part x mph * ?? = y miles
2nd Part 1.25x mph * ?? = 40-y miles

The time for the first part is y/x hours, and the time for the second part is (40-y)/(1.25x) hours.

Now let's plug these values for T1 and T2 into the the formula.
T1 / T2 * 100%
= [(40-y)/(1.25x)] / (40/x) * 100%
= [(40-y)/(1.25x)][x / 40] * 100%
= [(100)(40-y)]/[(40)(1.25)]

Our rephrased question is "what is y?"

Clearly (2) alone is clearly sufficient, so (B) is the answer. Hope that helps.

[Ben Ku]

Ben,

Don't you mean:

T1/T2 = [ y/x + (40-y/1.25x)] / (40/x) ?

You're right! Thanks for the catch. Victor's correct expression can be simplified:

T1/T2 = [ y/x + (40-y/1.25x)] / (40/x)
= (y + (40-y)/1.25)/40
= (1.25y + 40 - y) / 50
= (0.25 y + 40)/50

In other words, the rephrasing is still "what is y?"

[vishalsahdev03]

"what percent of the time it would have taken her if she has traveled at an average speed of x miles per hour for the entire trip? "

I had problems trying to understand, of what value is the question asking the percentage of !!
can someone pls break this statement me !

Thanks in advance !

[Ben Ku]

During a 40-mile trip, Marla traveled at an average speed of x miles per hour for the first y miles of the trip and and at an average speed of 1.25x mph for the last 40-y miles of the trip. The time that Marla took to travel the 40 miles was what percent of the time it would have taken her if she has traveled at an average speed of x miles per hour for the entire trip?

The actual time Marta spent, let's call that T1, is the combination of the first y miles at x mph and the second 40-y miles at 1.25x mph.

If she had traveled x mph the whole time, her new time T2 = 40/x.

The question is asking T1 is what percent of T2? In other words, T1 = (P/100)*T2.

Hope that makes sense.

[NNadjmabadi1]

I'm having trouble seeing why the x's cancel out. Can you explain that part?

[messi10]

Hi NNadjmabadi1,

I am assuming that you are ok up to this part:

T1/T2 = [ y/x + (40-y/1.25x)] / (40/x)

After this, you can take 1/x out common as a factor from both numerator and denominator. To better visualize this, I will use a new variable z, which is equal to 1/x. Also, I will treat numerator (T1) and denominator (T2) separately. Lets take the numerator and replace 1/x with z

T1 = yz + z(40-y)/1.25
= z [y + (40-y)/1.25]

Do the same with the denominator

T2 = 40z
= z(40)

Now you can see that z is common between Numerator and Denominator and can be cancelled out.

Regards

Sunil

[RonPurewal]

yep

[virginia.c.chan]

Hello,

I understand the explanations given (thank you), but I had a question. I initially grouped the two rates together so then tried to solve by dividing total distance over total rate to get the total time.

Total distance: y + (40-y) = 40

Total rate: x + 1.25x = 2.25x

Obviously that cancelled out the y's and I didn't get to the right answer, but I was just wondering why I can't do it this way (I thought it was like the average rate thing where the formula is total distance/time).

Thanks in advance.

[RonPurewal]

Hello,

I understand the explanations given (thank you), but I had a question. I initially grouped the two rates together so then tried to solve by dividing total distance over total rate to get the total time.

Total distance: y + (40-y) = 40

Total rate: x + 1.25x = 2.25x

Obviously that cancelled out the y's and I didn't get to the right answer, but I was just wondering why I can't do it this way (I thought it was like the average rate thing where the formula is total distance/time).

Thanks in advance.

if you use a certain fact to generate variable expressions, you can't plug those expressions right back into the same fact -- you have to combine them with something else.
if you plug back into the same fact that you used to generate the variable expressions in the first place, then you will *always* get a tautological expression in which everything cancels out everything else.

[AishwaryaK584]

Hello

I have a question on this. Sorry this is a basic question. I don't understand the simplification for [y/x +(40-y)/ 1.25x] * (x/40) How are u cancelling out all the x's? You can only cancel out 2x's here. How area cancelling all out?

[tim]

Try using the distributive property first. Then you'll have two fractions added together, but both of them have an x on top and an x on bottom, which can be canceled out. Let us know if you need any further help with this one.

[RonPurewal]

Hello

I have a question on this. Sorry this is a basic question. I don't understand the simplification for [y/x +(40-y)/ 1.25x] * (x/40) How are u cancelling out all the x's? You can only cancel out 2x's here. How area cancelling all out?

You probably know that, for instance, 3(x + y) = 3x + 3y. Note that the 3 is multiplied by both of the terms in the addition.

The x/40 does the same thing here, allowing the x to "cancel" the x's in both denominators.