The following was asked on Facebook. Since it is a common doubt, I am making a post on it.
To cover a distance, if I increase the speed by 3 kmph, it would take 40 minutes less, whereas if I decrease the speed by 2 kmph, it would take 40 minutes more. Find the distance.
The student is looking for both the traditional solution and if there is any short-cut.
Traditional Solution
Two places beginners get stuck or waste time in this question.
We have two set of values for Speed & Time
(s + 3, t – 40) and (s – 2, t + 40).
It is also obvious that Distance is same. And hence one can equate the product of the values. And it does not matter that speed is in kmph and time is in minutes. It is explained below.
However this results in 2 variables but only 1 equation (with terms in st, s, t and constant).
Learning #1: We should realize that we also have a third set of values for Speed & Time i.e. (s, t). “When speed increases by 3 kmph, time decreases by 40 mins” obviously means we are comparing it with usual speed and usual time.
Thus, we have two equations …
[mathjax]
$$s \times t = \left(s+3\right)\times\left(t-\frac{40}{60}\right)$$
$$s \times t = \left(s-2\right)\times\left(t+\frac{40}{60}\right)$$
The terms in st will cancel from both sides and we will get two simultaneous equations in s and t, which will result in unique solution for s and t.
Learning #2: The above equations is very tedious process, majorly because of the conversion of mins to hours leading to fractions. A lot of students struggle with the above, when it can be made very easy.
Had we just taken time, t, ALSO in mins, we just needed to imagine the following to realize that we DONT NEED TO CONVERT MINS TO HOURS.
$$s \times \frac{t}{60} = \left(s+3\right)\times\frac{\left(t-40\right)}{60}$$
i.e. the equations would be:
$$s \times t = \left(s+3\right)\times\left(t-40\right)$$
$$s \times t = \left(s-2\right)\times\left(t+40\right)$$
which is a straightforward …
–40s + 3t = 120
40s – 2t = 80
Adding, t = 200 mins. Substituting, s = 12 kmph.
Distance = \(12 \times \frac{200}{60}\) = 40 km. (Obviously in this step we need to convert mins to hours).
Now for the short-cut
In such questions it is MOSTLY observed that for either speed or time, the decrease in one case and the increase in other case is the same numeric value. If this is the case, we can solve this question, without pencil-work.
In this example …
The decrease and increase is numerically same for Time.
Hence use ratio of (decrease in Speed) and (increase in Speed).
⇒ the two Time taken will be in ratio of (decrease in speed) to (increase in speed) i.e. in ratio 2 : 3
We also know the difference in the two Time taken is 80 mins. Knowing the ratio and the difference, we can orally find the two Time taken as 160 mins and 240.
⇒ the two Speeds will be in ratio of (decrease in speed) to (increase in speed) i.e. in ratio 2 : 3
We also know the difference in the two Speeds is 5 kmph.
Knowing the ratio and the difference, we can orally find the two Speeds as 10 kmph and 15 kmph.
Please pair the speed and time correctly ……
higher speed, 15 kmph, will take lower time, 160 mins;
or lower speed, 10 kmph, will take higher time, 240 mins i.e. 4 hours,
so distance = 40 km.
When you learn short-cuts please do pay attention to when is the short-cut application. E.g. taken below is a similar (yet different) example from Takshzila’s book (Q. 7 page 142) …
Everyday I cover the distance from home to office at a usual speed and take a certain time. If I increase my usual speed by 5 kmph, I take 10 mins less than usual. If I reduce my usual speed by 5 kmph, I take 15 mins more than usual. Find the distance from home to office.
In this example,
The increase and decrease is numerically equal for Speeds
Hence use ratio of (decrease in Time taken) to (increase in Time taken)
⇒ the two Time taken are in ratio of (decrease in Time taken) and (increase in Time taken) i.e. in ratio 10 : 15 i.e. 2 : 3. And since difference in Time taken is 25 mins, the two time taken are 50 mins and 75 mins.
⇒ the two Speeds are in ratio of (decrease in Time taken) and (increase in Time taken) i.e. in ratio 10 : 15 i.e. 2 : 3. And since difference in Speeds is 10 kmph, the two Speeds are 20 kmph and 30 kmph.
Distance = 20 kmph × 75 mins OR 30 kmph × 50 mins = 25 km.
In this question neither changes in time & speed are same. I saw the following methodology to handle this; check whether it’s logical correct or not.
Q. A man covered a certain distance at some speed, had he moved 7m/s faster, he would have taken 7sec less and had he moved 3m/s slower, he would have taken 9sec more. Find the distance.
Ans.speed F/S = 7/3( increase in speed 3m/s slower to 7m/s faster so it would have direct effect on speed) * 9/7 ( time consumed decrease from 9sec more to 7sec less so it would have inverse effect on speed)= 3/1
Time become F/S = 1/3
now equate ratio with actual number.
Difference of 2 in speed ratio = 10m/s so 3= 15m/s and 1=5m/s
Difference of 2 in time ratio= 16sec so 1=8 sec & 3=24sec
Distance= 15*8 =120mt
(15-7 )8*15(8+7)=120mt
5*24=120mt.
Can we do in this way using proportion.
what if the difference(increase or decrease) is not same in both time and speed?
Then the short-cut will not work. There was a process, alligation based, which did give the result, but I could not really figure out the logic behind it. Hence have forgotten it. Will try to figure it out again. Will get back when it clicks. Till then, it is the equation way, I guess.
Never thought of the shortcut. Thanks Sir.