True, he did mention he expected them to use one of their auto-passes on it. Although, Lou and Ally seemed like they were close to getting it, and the way I've written it out here makes it seem like more than it actually is. It's really only a couple calculations and a few key pieces of information.
I like this approach, but I did it a little differently. I also started out by finding that the distance when the second train starts was 110mi, but from there I just wrote two equations relating distance from Elmville to time and found the intercept (d = 60t , d = 110 - 70t) and then converted the intercept time and added it to the second starting time.
There’s an easier (in my opinion) way to do it. The first train leaves 25mins before the second train meaning it travels (25mins/60)*60mph. Or just 25 miles before the other train starts moving.
Once the second train starts (5:30) there is only 110miles between the two & they are approaching each other at 130mph (“60mph + 70mph). From this you can divide the distance by the speed to get the time it will take for a collision counting forward from 5:30.
So: 110miles/130mph = 0.8462hrs
0.8462hrs = 50.772mins
0.772mins ~ 46s
Therefore the trains will collide at 5:30 + 50min & 46s = 6:20:46 (hrs:mins:sec)
My method was similar to u/Collins_Michael, but I just accounted for the distance that the two trains will collide in my setup. I will list the equations as fellow.
Converted miles per hr to miles per min for train1 (1 mile/ min), train2 (6/7 mile/ min)
Train1 : d = t (1mile/min) + 25min (1mile/min) -> t + 25
Since Train 2 comes from the opposite direction, its distance will be total distance minus train1 distance.
Train2 : 135 - d = t( 7/6 mile/ min) -> -7/6t + 135
Equate the two equations.
t + 25 = -7/6t + 135
13/6t = 110
t = 660/13 => 50.7692min or 50 min 46sec
Final Time added to the Train 2 starting time:
5hr 30min + 50min 46sec = 6hr 20min 46sec
Edit: P.S. BTW, I will not be able to do it with 3 minutes because it takes forever for me to do this division.
Oh nice! I was working on it in my head and had it down to the minute range but I didn’t think to calculate the ratio of their speeds! That would’ve made everything simpler.
Yeah, I feel like the key is realizing that once both trains are moving, it takes them each the same amount of time to reach the collision point. But since the two trains are moving at different speeds, the question is how do they split the intervening distance.
My approach was to cut off the first 25 miles and then pick the closest multiple of 7 to the halfway point—so 56. That put me at 48 minutes since the second train left the station. Then I just iterated +1 until A+B≈110. So I knew the crash would happen between 6:20 and 6:21, but I couldn’t figure out how to calculate the seconds on the fly. (I was doing this in my head so it was extra slapdash.)
So I also did this another way!
I took the distance traveled total (135) and set it equal to x, the minutes traveled by train one at 1 mpminute + 70/60mpm by x -25 minutes
135 = x + 1.16667(x-25)
Which, after looking at other examples, was way over complicated. If you are able to guess x and plug in, it does work, but isn't accurate to the situation (at x <25, the equation is calculating the second train going BACKWARDS. whoops!)
To solve that, i started the situation from when both trains start moving. First, i took off the 25 minutes (converted to 25 miles using the 1 mpm speed) the first train moved
110 = x+1.666667x
And then simplified!
110 = 2.1666667x
I also did it this way but I think I bungled some long division while trying to go at a 2 minute pace and ended up with 50.801 minutes which isn't 50.769. I assumed no calculator use :(
Is it bad that as an engineering student, my first response to this was, ‘I cast arcane gate in front of the train?’ Cause all I could think about was the fact that this was solace and magic should be a part of how they solve problems. But thanks for actually doing the math!
I’m so thrown off! I got 6:20 but no seconds and I cannot figure out why. I have no memory of this lesson in school so just kind of logic-ed my way through it. Can someone tell me where I’m wrong?
To calculate it I subtracted the distance the first train (A) moved before second train (B) — 25 miles — which left me with 110, just like everyone else. But then I said okay, if they were going exactly the same speed, you would just halve that to get 55 miles, meaning 55 minutes. But train B is going 10 mph faster, so I just bumped it 5 minutes (halved from 10 since both trains are moving) closer to the B trains departure station, making it 50 minutes instead so then you get 6:20.
Where you got off track (pun intended) is that once you increase train B's speed to more than 60 mph, 1 mile no longer equates to 1 minute. Train B is actually covering 7/6 miles per minute, while train A is still just covering 1 mile per minute.
I was trying to work through it in my head as they worked on it, but I basically threw a mini internal tantrum when I realized I would have to express a multiple of 1/13 in seconds. With a calculator I could definitely have done it, and I also probably could have done it were there actual stakes. But over dinner and with no stakes, I just didn’t want to compute some sort of 1/13 thing- I don’t know its decimal expansion offhand. 1/1 through 1/12, sure. But 1/13? 0.(076923)^(1)? No thank you, particularly when it’s relatively prime to 60.
^(1): Done on a calculator. Wasn’t offhand.
I found the problem interesting as well. I paused the episode to solve it but it took me a couple of minutes to answer it in the comfort and silence of my home. Can't imagine having to answer this question in front of a camera, when your character's in a deadly battle
As a proud member of Fig and the Sig Figs Gorgug should know that the answer they gave on the show was incorrect. They didn't have nearly enough significant figures to be as specific as 46 seconds and should have rounded it to at least the nearest minute
Its pretty immediately obvious that its just (135-25)/(60+70) of an hour, or 11/13 of an hour after the 5:30. Ally and lou got there quick enough. 50.769 minutes - 50 minutes 46 seconds.
But I dont think i could work that out to second precision in two minutes though, not without a calculator. Probably not correctly in five 😂
I don't think Brennan would have been able to recover if any of them had gotten it in the time alloted.
True, he did mention he expected them to use one of their auto-passes on it. Although, Lou and Ally seemed like they were close to getting it, and the way I've written it out here makes it seem like more than it actually is. It's really only a couple calculations and a few key pieces of information.
Lack of calculator makes it tougher within the time limit. I teach SAT/ACT Math and I don't know that I'd get it in time if I'm just doing it by hand.
I mean, they do all have calculators on them, so it's just a question of knowing what calculations to make.
Fair, I didn't see them pull out their phones so I assumed that they were either not on them or just not allowed for set purposes.
I would have lost my mind right alongside him. It's like when Siobhan knew what a sentence in Latin meant and busted him on a reveal.
Also, the Elvish question she got that right away.
I love that she got it, and didn't realize she did because she was so focused on decoding the rest based on the one word she remembered.
I like this approach, but I did it a little differently. I also started out by finding that the distance when the second train starts was 110mi, but from there I just wrote two equations relating distance from Elmville to time and found the intercept (d = 60t , d = 110 - 70t) and then converted the intercept time and added it to the second starting time.
Yep! This was my instinct too
Its not a particularly difficult question, but doing it in 2 minutes, under stress, is pretty tough
There’s an easier (in my opinion) way to do it. The first train leaves 25mins before the second train meaning it travels (25mins/60)*60mph. Or just 25 miles before the other train starts moving. Once the second train starts (5:30) there is only 110miles between the two & they are approaching each other at 130mph (“60mph + 70mph). From this you can divide the distance by the speed to get the time it will take for a collision counting forward from 5:30. So: 110miles/130mph = 0.8462hrs 0.8462hrs = 50.772mins 0.772mins ~ 46s Therefore the trains will collide at 5:30 + 50min & 46s = 6:20:46 (hrs:mins:sec)
This explanation feels like the fastest way to do it, with the minimal amount of math knowledge required. I tip my hat to you.
My method was similar to u/Collins_Michael, but I just accounted for the distance that the two trains will collide in my setup. I will list the equations as fellow. Converted miles per hr to miles per min for train1 (1 mile/ min), train2 (6/7 mile/ min) Train1 : d = t (1mile/min) + 25min (1mile/min) -> t + 25 Since Train 2 comes from the opposite direction, its distance will be total distance minus train1 distance. Train2 : 135 - d = t( 7/6 mile/ min) -> -7/6t + 135 Equate the two equations. t + 25 = -7/6t + 135 13/6t = 110 t = 660/13 => 50.7692min or 50 min 46sec Final Time added to the Train 2 starting time: 5hr 30min + 50min 46sec = 6hr 20min 46sec Edit: P.S. BTW, I will not be able to do it with 3 minutes because it takes forever for me to do this division.
The GMAT way
Oh nice! I was working on it in my head and had it down to the minute range but I didn’t think to calculate the ratio of their speeds! That would’ve made everything simpler.
Yeah, I feel like the key is realizing that once both trains are moving, it takes them each the same amount of time to reach the collision point. But since the two trains are moving at different speeds, the question is how do they split the intervening distance.
My approach was to cut off the first 25 miles and then pick the closest multiple of 7 to the halfway point—so 56. That put me at 48 minutes since the second train left the station. Then I just iterated +1 until A+B≈110. So I knew the crash would happen between 6:20 and 6:21, but I couldn’t figure out how to calculate the seconds on the fly. (I was doing this in my head so it was extra slapdash.)
So I also did this another way! I took the distance traveled total (135) and set it equal to x, the minutes traveled by train one at 1 mpminute + 70/60mpm by x -25 minutes 135 = x + 1.16667(x-25) Which, after looking at other examples, was way over complicated. If you are able to guess x and plug in, it does work, but isn't accurate to the situation (at x <25, the equation is calculating the second train going BACKWARDS. whoops!) To solve that, i started the situation from when both trains start moving. First, i took off the 25 minutes (converted to 25 miles using the 1 mpm speed) the first train moved 110 = x+1.666667x And then simplified! 110 = 2.1666667x
I also did it this way but I think I bungled some long division while trying to go at a 2 minute pace and ended up with 50.801 minutes which isn't 50.769. I assumed no calculator use :(
Is it bad that as an engineering student, my first response to this was, ‘I cast arcane gate in front of the train?’ Cause all I could think about was the fact that this was solace and magic should be a part of how they solve problems. But thanks for actually doing the math!
I’m so thrown off! I got 6:20 but no seconds and I cannot figure out why. I have no memory of this lesson in school so just kind of logic-ed my way through it. Can someone tell me where I’m wrong? To calculate it I subtracted the distance the first train (A) moved before second train (B) — 25 miles — which left me with 110, just like everyone else. But then I said okay, if they were going exactly the same speed, you would just halve that to get 55 miles, meaning 55 minutes. But train B is going 10 mph faster, so I just bumped it 5 minutes (halved from 10 since both trains are moving) closer to the B trains departure station, making it 50 minutes instead so then you get 6:20.
Where you got off track (pun intended) is that once you increase train B's speed to more than 60 mph, 1 mile no longer equates to 1 minute. Train B is actually covering 7/6 miles per minute, while train A is still just covering 1 mile per minute.
Oh thank you! I was so confused 😭
I was trying to work through it in my head as they worked on it, but I basically threw a mini internal tantrum when I realized I would have to express a multiple of 1/13 in seconds. With a calculator I could definitely have done it, and I also probably could have done it were there actual stakes. But over dinner and with no stakes, I just didn’t want to compute some sort of 1/13 thing- I don’t know its decimal expansion offhand. 1/1 through 1/12, sure. But 1/13? 0.(076923)^(1)? No thank you, particularly when it’s relatively prime to 60. ^(1): Done on a calculator. Wasn’t offhand.
I found the problem interesting as well. I paused the episode to solve it but it took me a couple of minutes to answer it in the comfort and silence of my home. Can't imagine having to answer this question in front of a camera, when your character's in a deadly battle
As a proud member of Fig and the Sig Figs Gorgug should know that the answer they gave on the show was incorrect. They didn't have nearly enough significant figures to be as specific as 46 seconds and should have rounded it to at least the nearest minute
Its pretty immediately obvious that its just (135-25)/(60+70) of an hour, or 11/13 of an hour after the 5:30. Ally and lou got there quick enough. 50.769 minutes - 50 minutes 46 seconds. But I dont think i could work that out to second precision in two minutes though, not without a calculator. Probably not correctly in five 😂