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Driving 1.5 times slower, Bill was late for school today. What is the usual time it takes Bill to drive to school? (Assume that each day Bill takes the same route).

1) It took Bill 15 more minutes to drive to school today than usually 2) The distance between home and school is 15 miles

S1 tells us that on average Bill drives for 15 minutes less. Since he drove 1.5 times slower, the 15 minutes account for the difference. Therefore, it is possible to find the difference. (Can Someone Please Explain to Me Here In an Equation HOw we Can come up with that Answer if We Needed To. Thank you.

S2 does not tell us much; we don't really need to know the distance and by itself it is insufficeint.

Distance=Speed*time Here distance is constant, so Original Speed*Original time=New Speed*New time Acc. to stmt 1 s*t=(s/1.5)*(t+15) 1.5t= t+15 0.5t=15 t=30 min.

Ouch!! It is the first time I am hearing the phrase "times slower", in this case "1.5 times slower". I am not sure if this automatically means Velocity/1.5 just as"1.5 times faster" would mean Velocity*1.5. Can someone clarify this ? This seems pretty basic but I have difficulty with it.

Driving 1.5 times slower, Bill was late for school today. What is the usual time it takes Bill to drive to school? (Assume that each day Bill takes the same route).

1) It took Bill 15 more minutes to drive to school today than usually 2) The distance between home and school is 15 miles

S1 tells us that on average Bill drives for 15 minutes less. Since he drove 1.5 times slower, the 15 minutes account for the difference. Therefore, it is possible to find the difference. (Can Someone Please Explain to Me Here In an Equation HOw we Can come up with that Answer if We Needed To. Thank you.

S2 does not tell us much; we don't really need to know the distance and by itself it is insufficeint.

BELOW IS REVISED VERSION OF THIS QUESTION:

On Tuesday it took Bill 1.5 times as long to drive from home to school as it took him the day before on Monday. How long did it take Bill to drive from home to school on Monday? (Assume that each day Bill takes the same route).

(1) On Tuesday it took Bill 15 minutes longer to drive from home to school than on Monday. (2) The distance between home and school is 15 miles.

Suppose it took Bill \(t\) minutes to drive from home to school on Monday, then on Tuesday, it would took him \(1.5t\) minutes.

(1) On Tuesday it took Bill 15 minutes longer to drive from home to school than on Monday --> \(1.5t=t+15\) --> \(t=30\) minutes. Sufficient.

(2) The distance between home and school is 15 miles. Useless info. Not sufficient.

Driving 1.5 times slower, Bill was late for school today. What is the usual time it takes Bill to drive to school? (Assume that each day Bill takes the same route).

1) It took Bill 15 more minutes to drive to school today than usually 2) The distance between home and school is 15 miles

S1 tells us that on average Bill drives for 15 minutes less. Since he drove 1.5 times slower, the 15 minutes account for the difference. Therefore, it is possible to find the difference. (Can Someone Please Explain to Me Here In an Equation HOw we Can come up with that Answer if We Needed To. Thank you.

S2 does not tell us much; we don't really need to know the distance and by itself it is insufficeint.

When I saw this question today, I knew there was something wrong with it. How can you drive 1.5 times slower? That is 150% slower and more like a negative movement. The revised version makes more sense.
_________________

Almost the same solution we'll get if we denote the usual time by t and distance by d. Then the equation looks like d/t=1.5d/(t+15), which doesn't depend on d and finally t+15=1.5t. A is suff alone.

Ouch!! It is the first time I am hearing the phrase "times slower", in this case "1.5 times slower". I am not sure if this automatically means Velocity/1.5 just as"1.5 times faster" would mean Velocity*1.5. Can someone clarify this ? This seems pretty basic but I have difficulty with it.

Ouch!! It is the first time I am hearing the phrase "times slower", in this case "1.5 times slower". I am not sure if this automatically means Velocity/1.5 just as"1.5 times faster" would mean Velocity*1.5. Can someone clarify this ? This seems pretty basic but I have difficulty with it.

Thanks

Thanks dzyubam. I sometimes have problems with the language in the questions.Making an equation itself becomes hard. My english skills are't bad and I have gone through Manhattan Word Translations as well. Still there is a gap as long as interpreting the questions go. I get trapped in 600 level questions because of this. Any suggestions ?

An example of the question type I just talked on is the one below. Took lots of time to formulate the equation. Picking numbers gets confusing as well:-

"I am twice as old as you were when I was as old as you are.The sum of the ages is 64.Find the ages"

For a problem like this you might want to consider picking numbers instead of actually solving the problem. The wording of this problem is indeed tricky.

An example of the question type I just talked on is the one below. Took lots of time to formulate the equation. Picking numbers gets confusing as well:-

"I am twice as old as you were when I was as old as you are.The sum of the ages is 64.Find the ages"

kaptain wrote: An example of the question type I just talked on is the one below. Took lots of time to formulate the equation. Picking numbers gets confusing as well:-

"I am twice as old as you were when I was as old as you are.The sum of the ages is 64.Find the ages"

First, the sum S of the ages must divisible by 7. If S=63 (instead of the given 64), the ages are 36 and 27 (9 years ago: 27 and 18). if S=70, then 40 and 30 (30 and 20), if S=7 then 4 and 3. If x and y are the ages, then till the difference between the ages x-y=y-x/2, or 3x=4y. In general, the ages are 4S/7 and 3S/7. And the difference between the ages will be S/7. If S is not a multiple of 7 you'll get confusing fractions as ages.

Let T(1) be the usual time Let T(2) be todays time then the problem reads: T(2)=1,5 * T(1) Solve for T(1) Data 1 reads: T(2)=T (1)+15 min ==> 2 equations with 2 variables ==> sufficient Data 2 reads: Distance = 15 miles ==> no information about T(1) ==> not necessary and insufficient

k..1.5 times slower means = if original speed is 6x then 1.5 times slower speed is =4x

a) t=d/s so for original thing t= d/6x and with reduced speed t+15=d/4x so one can calculate frm here two equations , two unknowns , and no quadratic so u have the ans!

B) how late u cannot say

ans A!
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" What [i] do is not beyond anybody else's competence"- warren buffett My Gmat experience -http://gmatclub.com/forum/gmat-710-q-47-v-41-tips-for-non-natives-107086.html