Relating increament of speed to time

[MoniMini]

Aditya rides a bicycle from town P to town Q with a constant speed. If he increases his speed by 3 m/s, he will arrive to town Q three times faster. How many times faster will Aditya arrive to town Q, if he increases his speed by 6 m/s



s=$\frac{d}{t}$



Let his constant speed be x
x= d/t
Therefore x+3 = (d/t) +3
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C'est tout! What next?

 

[tiny-tim]

Hi MoniMini!

Let his constant speed be x
x= d/t
Therefore x+3 = (d/t) +3

ok so far!

now translate "he will arrive to town Q three times faster"

 

[MoniMini]

Hi MoniMini!


ok so far!

now translate "he will arrive to town Q three times faster"

Umm...That's the problem...I'm not being able to do that :(

 

[tiny-tim]

if the new speed, x+3, is 3 times faster,

that means it must be 3x

 

[Nickie]

If Aditya increases his speed by 3 m/s, he will arrive at town Q in 1/3 of the original time. This means that his new speed is 3 times faster than his original speed.

If he increases his speed by 6 m/s, he will arrive at town Q in 1/6 of the original time. This means that his new speed is 6 times faster than his original speed.

Therefore, if Aditya increases his speed by 6 m/s, he will arrive at town Q 6 times faster than his original speed. This can also be calculated by dividing the original time by the new time (1/6 divided by 1/3), which gives us 2. This means that Aditya will arrive at town Q 2 times faster if he increases his speed by 6 m/s compared to increasing it by 3 m/s.