# help me solving this question...

• Sep 11th 2012, 09:21 AM
MyNameIsKhan
help me solving this question...
i don't know if i'm posting on right place in the forum or not.. i've a question..

A curved section of road is replaced with a straight highway. Road becomes 20% shorter and the speed can be increased from 80 km/h to 100 km/h..
how changed driving time in percent?

(answer is 36% decrese in driving time...but i don't know how it's solved...)

• Sep 11th 2012, 09:37 AM
emakarov
Re: help me solving this question...
Denote the old distance by d. Express the new distance through d. Find the old driving time (old distance divided by old speed) and the new driving time. Find the ratio of the new driving time to the old driving time (d should cancel out).
• Sep 11th 2012, 09:43 AM
MyNameIsKhan
Re: help me solving this question...
• Sep 11th 2012, 10:46 AM
HallsofIvy
Re: help me solving this question...
That was "mathematically". What you want is to translate the mathematics into specific figures. If you let the original length of the road "d" kilometers, then, because "Road becomes 20% shorter", the new road has length d- .2d= .8d kilometers. On the old road, you could do 80 kilometers/hour. How long would it take you do drive d kilometer at 80 kilometers per hour? (The answer will depend on d.) How long would it take you do drive .8d kilometers at 100 kilometers per hour? Divide the second time by the first to get a percentage. There will be a "d" in both numerator and denominator that will cancel, leaving only a number as the answer.
• Sep 11th 2012, 11:16 AM
lelbel
Re: help me solving this question...
d1=t1*80 curved
d2=t2*100 straight
d2=d1*80/100 20% less
d1*80/100=t2*100
d1=t2*100*100/80
t2*100*100/80=t1*80 t2=t1*64/100 36% less
• Sep 12th 2012, 12:39 AM
MyNameIsKhan
Re: help me solving this question...
can u explain the 3rd step .....
• Sep 12th 2012, 04:18 AM
emakarov
Re: help me solving this question...
Quote:

Originally Posted by MyNameIsKhan
can u explain the 3rd step .....

Do you mean this one?
Quote:

Originally Posted by lelbel
d2=d1*80/100 20% less

What does it mean for some quantity x to increase by n percent? We know that n percent corresponds to the fraction n/100, so n percent of x is x*n/100. The new value of x is x + x*n/100 = x(1 + n/100). For example, if x is increased by 35%, the new value of x is 1.35x. Decreasing by 20% is the same as increasing by -20%, so if d2 is the result of decreasing d1 by 20%, then d2 = d1(1 - 20/100) = 0.8*d1.

Is this what you were asking? If so, then your original post should have been not about how to solve the problem, but about the meaning of the fact that the road became 20% shorter. If you don't know the definition of what it means to increase or decrease by a certain percentage, of course you can't solve the problem. It could be added that you should have read the textbook or other source about the meaning of percents before trying to solve a concrete problem. One has to learn the theory before applying it, and this site is not intended for teaching theory. To repeat, these remarks apply if indeed you were asking why d2 = d1 * 0.8.
• Sep 12th 2012, 06:58 AM
lelbel
Re: help me solving this question...
Don't worry,some people are rude and aren't patient...
so, the 3rd step ... if d1 is %100 then d2 is %80 (%20 less)
u multiply it by 80 and divide by 100 to eliminate %
this formula help u to understand %
%P*B=A*%100 where %P is how many % u have from base B (the whole amount of smth),A is amount that represent this %,
but u have to multpl it by %100 to eliminate %
80% of d1 = d2*100%, like 20% of 100 is 20 or 20%*100=20*100%
hope u got it
• Sep 12th 2012, 07:39 AM
MyNameIsKhan
Re: help me solving this question...
thanks lelbel ... i got it ..
• Sep 12th 2012, 09:35 AM
MyNameIsKhan
Re: help me solving this question...
sorry friends i disturbed you... but problem is i'm learning it in swedish language and my swedish and english both are not so strong , that's why i got problems while solving these type of problems.....

d1 = t1 * 80 (curved road )
d2 = t2 * 100 (straight road)

as the distance decreases by 20% so,,
100%-20% = 80% or 0.8
so the new distance we get after 20% decrease is
d2 = 0.8d1
substitute the values of d1 and d2
t2*100 = 0.8d1
t2= 0.8*t1*80/100
t2/t1 = 0.8*80/100
t2/t1 = 0.64
t2/t1 = 64%

there is 64% decease from origional distance i.e 100% ,so the change in driving time is..
100%-64% = 36%

have i done by right way ....
• Sep 13th 2012, 08:07 AM
lelbel
Re: help me solving this question...
excellento !!! good luck !!!