Let’s start with the applications of ratios today. Before you go on to an actual question, there is a relation between variables that you need to understand:

Distance = Speed × Time

Now let’s say my driving speed is a cool 100 mph. If I have to travel 100 miles, how much time will it take me? An hour, simple! Alright. If I have to travel 200 miles, how long will it take me? 2 hrs, you say? That is correct. What if I have to travel 500 miles? How long will it take me? 5 hrs, of course.. (now I am wasting your time, I know, but bear with me) When I hold my speed at a steady 100 mph, do you see a relation between Distance and Time? Can I say that if my distance doubles, my time taken doubles too? Can I say that if the distance that I have to travel on two different days is in the ratio 1: 5 (100 miles and 500 miles respectively), then the time I take on these two days will also be in the ratio 1:5 (1 hr and 5 hrs respectively)?

Yes, I can. Same is the relation of Distance with Speed keeping time constant. If this makes sense, you are half way there. We say Distance varies directly with Time (when Speed is kept constant) and Distance varies directly with Speed (when Time is kept constant). Anyway, we will take variation some other day.

Now, let me take another scenario.

Say, I have to travel 200 miles. If I keep a speed of 100 mph, I reach in 2 hrs. But what if the maximum speed allowed on the entire stretch is just 50 mph (and I am afraid of the law)? How long will it take me? Since my speed has reduced to half, it takes me double the time i.e. 4 hrs to cover 200 mph. So can I say that if there are two routes, both of 200 miles, and if my** speed on the routes is in the ratio 2:1 **(100 mph and 50 mph respectively), then my **time taken will be in the ratio 1:2** (2 hrs and 4 hrs respectively)? We say that Speed varies inversely with Time (when Distance is kept constant). So if speed in two cases is in the ratio 3:7, time taken in the two cases will be in the ratio 7:3 (keeping the Distance same in both the cases)

Now let’s look at the question I posted in the previous post:

Two trains, A and B, traveling toward each other on parallel tracks, started simultaneously from opposite ends of a 250 mile route. A takes a total of 3 hours to reach the opposite end while B takes a total of 2 hours to reach the opposite end. When train A meets train B during the journey, how far is train A from its starting point?

Both A and B travel a 250 mile route i.e. they travel the same distance. So the ratio of their speeds must be inverse of the ratio of time taken. (Say if train A’s speed is 250 mph, it will take 1 hr and if train B’s speed is 125 mph, it will take 2 hrs – ratio of speeds is 2:1, ratio of time taken is 1:2). Now we know that A takes 3 hrs to cover 250 miles and B taken 2 hrs to cover the same distance so ratio of time taken by A:B is 3:2. Then, ratio of their speeds must be 2:3. So for every 2 miles that A covers, B covers 3 miles in the same time.

As shown in the previous ratios post, the total distance in ratio terms will be 5 which is equal to 250 miles (this is the distance the trains have covered together when they meet). The multiplier is 50. So A covers a distance of 2*50 = 100 miles and B covers a distance of 3*50 = 150 miles. So train A is 100 miles away from its starting point! No equations needed.

Now, let’s take another example.

The speed of bus A is 20% more than the speed of bus B. Bus B takes 2 hours longer than bus A to travel 600 miles. What is the speed of bus A?

Ordinarily, people would make equations and solve them to get to the answer. But we can do it quickly and orally.

Speed of bus A is 20% more than speed of bus B. This means that speed of bus A : speed of bus B is 120:100 i.e. 6:5. To travel the same distance, time taken by bus A: time taken by bus B will be 5:6. This difference of 1 in the ratio of time taken is actually given to be 2 hours. Hence, the multiplier is 2. Time taken by bus A to travel the 600 miles must be 5*2 = 10 hrs and time taken by bus B to travel the 600 miles must be 6*2 = 12 hrs.

Speed of bus A = 600/10 = 60 mph

Try this logic on the next Time-Speed-Distance question you come across. Next week, we will look at the application of Ratios in Work problems.

*Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep in Detroit, Michigan, and regularly participates in content development projects such as this blog!*

## 17 thoughts on “Quarter Wit, Quarter Wisdom: Applications of Ratios in TSD”

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I really want to congratulate you for all the amazing posts that have been published. Great Work!

I have a question regarding the aforementioned issue, and I would really want to understand the fundamentals behind it.

I did read the previous post mentioned by I still couldn’t understand how from having the ratio of speeds (2:3 in the example mentioned) we can say that the total distance in ratio terms will be 5. From my understanding the total speed in ratio terms will be 5.

Moreover, I do understand the proportion and inversely proportional relations that can derive from u*t=S. For example if S=constant then if u doubles t must halved.

What I can’t understand is how the ratios work. For example if we know that s1=s2=same number then if the ratio of t1/t2 is 2/1 then u1/u2 must be 1/2. Why is that?

I know that if I draw a table with the 2 cases mentioned and divide them then: s1/s2=1 and if t1/t2=2/1 then u1/u2 must be 1/2 because their product must equal to 1. Is this how someone must think about that?

So the logic is that in a RTD table when we have 2 elements that are the same, for example s1=s2 or u1=u2 or t1=t2 then we can use the ratios to find a solution faster?

Sorry for the really long post.

Congrats again for your effort and that you make our life’s way easier.

I apologize; I just missed responding to your post. Thanks for reminding.

We have Distance = Speed * Time

Say, I drive from my home to office everyday i.e. the distance I travel is the same every day. Yesterday I traveled at 100 mph and took 10 mins to reach. If today I travel at only 50 mph, I will take 20 mins to reach, right? This is how we use proportion and inversely proportion relations, right? Ratio is exactly the same concept.

If, the ratio of my speed yesterday : today = 2:1 (100 mph:50 mph), then ratio of time taken yesterday : today = 1:2 (assuming that the distance I travel is the same in both cases i.e. the distance is kept constant). This is the inversely proportional relation we were talking about above

If distance constant, Speed ? 1/Time

If speed halves i.e. speed becomes 1 if it was 2 previously, time doubles i.e. it becomes 2 if it was 1 previously. So ratio of old speed : new speed = 2:1, then ratio of old time : new time = 1:2

“when we have 2 elements that are the same, for example s1=s2 or u1=u2 or t1=t2 then we can use the ratios to find a solution faster”

– Yes. You will almost always find that one of the elements is constant. Say the speed of two people is the same. Their time taken will be in the same ratio as distance traveled. Say if P1 travels 100 miles in 1 hr and P2 travels 200 miles, he will take 2 hrs. If ratio of distance covered = 1:2, then ratio of time taken = 1:2. It is obvious with simple examples, right?

“I still couldn’t understand how from having the ratio of speeds (2:3 in the example mentioned) we can say that the total distance in ratio terms will be 5. From my understanding the total speed in ratio terms will be 5.”

Sometimes, it is not obvious that an element is constant. e.g. if two trains start simultaneously from opposite ends and meet, starting from the time they started till the time they meet, they have traveled for the same time. So time in both cases is the same. Therefore, ratio of distances covered will be same as ratio of speeds. Say speed of T1: speed of T2 = 2:3, then distance covered by T1: distance covered by T2 = 2:3. So total distance in ratio terms is also 5.

In most questions, all you need to do is identify the element that is constant, recall the relation between the other two elements (e.g. Distance is directly proportional to Speed OR Speed is inversely proportional to Time etc) and hence establish the ratios (e.g. if ratio of distances covered = 4:7, ratio of speeds must be 4:7 too OR if ratio of speeds = 5:6, then ratio of time taken = 6:5 etc. The ratio flips in case of inversely proportional. Take a few numbers to convince yourself about this.)

Thanks so much! :))))

You are a genius, Karishma, explaining complicated concepts in the most subtle way. Thank you very much.

I have definitely improved my overall sense of perception towards a problem.

I bet these wisdom snippets are the best !! Keep it up..Cheers

Karishma? an input please? :)

Hi Karishma,

Thanks for all your wonderful posts. I have a question regarding the second example – I’m a bit confused as to how you got that 2 hours somehow equated to the multiplier for the 5:6 ratio. Thanks so much!!

Thanks Lisa. Glad you like them!

As for your question, actually this step is based on a concept we discussed in a previous post: http://www.veritasprep.com/blog/2011/03/quarter-wit-quarter-wisdom-the-sorcery-of-ratios/

Get back to me if there are any further queries.

Consider the below text that i copied from Karishma’s previous post. This might help you understand why she multiplied the ration with 2.

Given: A:B=4:5. B scored 10 points more than A. How many points did A score?

B:A is 5:4. For 4 points of A, B scored 1 extra. If B actually scored 10 points more than A, A must have scored 10 times 4 i.e. 40 points (and B must have scored 50 points)

In each of the cases above, you can think of 10 as the common multiplier. It helps you arrive at all the actual values. When you multiply the numbers in ratio terms with the multiplier, you get the actual values.

In our case, the A:B=5:6 and B takes 2 hrs more than A. So for evry 5 hrs of A, B takes an extra hour. And it is given that B takes 2 hrs extra. SO A must have taken 10hours.

“Now we know that A takes 3 hrs to cover 250 miles and B taken 2 hrs to cover the same distance so ratio of time taken by A:B is 3:2. Then, ratio of their speeds must be 2:3. So for every 2 miles that A covers, B covers 3 miles in the same time”

I didn’t get the last sentence that “when the ratio is between speeds is given ,, we are taking distances and considering time as the same..

Please explain.

We know that the ratio of speeds of A and B is 2:3. What does this mean? It means that if A’s speed is 2 mph, B’s speed is 3 mph. If A’s speed is 4 mph, B’s speed is 6 mph. This means that if A covers 2 miles in 1 hr, B covers 3 miles in 1 hr. If A covers 4 miles in 1 hr, B covers 6 miles in 1 hr. If A covers 8 miles in 2 hrs, B covers 12 miles in 2 hrs and so on… So for every 2 miles that A covers, B covers 3 miles in the same time.

Karishma ,

I have difficulty in applying this concept on non -simultaneous train problems.

e.g A train leaves chennai for mumbai travelling 240 miles per hour at 12 Noon/PM.Ten minutes later a train leaves mumbai for chennai travelling 160 miles per hour.If chennai & mumbai are 300 miles apart ,at what time will the trains pass each other ?

Can you pls help ?

When you encounter a non-simultaneous problem, just convert it to simultaneous (since you know how to deal with that!) and get going. This is how you can do it:

A train leaves Chennai at 12. It covers (240/60) * 10 = 30 miles in 10 minutes.

At 12:10, the train leaves a point 30 miles ahead of Chennai for Mumbai at a speed of 240 mph and at the SAME TIME, a train leaves Mumbai for Chennai at 160 mph. The distance between the two trains is 270 miles (300 – 30 miles). Now it is a problem you know how to solve!

Honestly, thank you so much.

I found you on Gmatclub!! and I have just bought the VeritasPrep Combinatorics book.

It is amazing!!

Really thanks :D

We are glad you liked it!

A car traveling at a certain constant speed takes 2 seconds longer to travel 1 kilometer than it would take to travel 1 kilometer at 75 kilometers per hour. At what speed, in kilometers per hour, is the car traveling?

A. 71.5

B. 72

C. 72.5

D. 73

E. 73.5

B is the correct answer.

Hi Karishma,

I came across the above question in the GMAT Prep Question Pack #1 and I tried to used the concept that you taught in this blog. Since the distance is constant (1km for both cases), the rate will be inversely related to the time. Then, I was stuck because the question did not give us the ratio between the two rates nor the two time. The question only states the differences between two time. Thus, I end up doing the “textbook” way, by setting up two RT=D equations and substituting one equation into the other equation. Did I overlook anything in the problem that would have allow me to use the strategy that you discussed in this blog? Thanks.

Actually, you do have the ratio of time taken.

Time taken when speed is 75 kmph = (1km/75kmph)*60*60 secs = 48 secs

Time taken at current speed = 48 + 2 = 50 secs

So 48/50 = x/75 (here, units don’t matter since it is a ratio. You could convert both 48 and 50 into hrs and the ratio would still remain 48/50)

x = 72 kmph

Great explanations, Karsihma.

I think I’m better equipped to handle distance rate problems now.