Maths SOS

[Casparina]

We're struggling with a maths question in our household. It doesn't look as if it should be that hard! Can anyone help?

i'm pretty sure that in this type of problem, by 'cross' they mean get completely clear of each other - i.e., from the point that the heads of each train first draw next to each other, to the point that the ends of the trains come clear. The 14 seconds is the duration for that time happen, rather than the length of time until the trains first meet (or collide, on the same track!).

The wording could be clearer, though. But if this is something like the GMAT exam, then there are enough examples that they assume that people know what they mean in the question, and don't really think of other interpretations.

Thanks goodness for that, it was getting a bit embarrassing being the first comment 😂

I’ll take my Nobel prize now thanks

The interpretation doesn't matter as in both cases the trains will cover 140m each making the total distance covered still 280m. In 14 seconds. Thus resulting in the same answer.

...and "cross in 14 seconds" in this type of problem does mean completely pass, not the point at which they would hit.

If that were what you were looking for, you can't actually get the answer from the question, as you don't know how far they are apart when you start. You can only work it out when you have two of the three variables - distance, speed, time. If you are trying to figure out when their fronts meet, you don't have distance between them - the length of the train itself would irrelevant in that calculation. So it can't be done.

No, because you don't know how far apart they are when you start. The length of the train is irrelevant if all you care about is when their fronts would meet.

@yumumsun Sorry I should have said - the questions are from "The Ultimate TSA Collection" based on past Oxford TSA papers, though we are using them to prepare for the general papers in 16+ entrance exams. Fortunately at 16+ they're given 2 hours for 50 questions rather than the 1.5 hours which is standard for the TSA.

Let's say Train A is stationary and train B is hurtling towards train A.
If we make train A stationary then we have to think about train B going at twice the speed that they would both have been doing as stated in the question.

At time t=0 imagine the front on the moving train - train B - reaching and passing the front of the stationary train - train A (on another track facing in the opposite direction)

If train B moves past stationary train A until the front of train B reaches the back of train A - then train B will have travelled 140m (the length of train A) - at this point B is alongside A - it has not completely passed train A yet.

For train B to completely pass train A it must continue for another 140m. We are told it takes 14s for the trains to pass.

so as speed (V) = distance (D) /time (T)
train B travels at speed = 2V
train B travels a distance D of 140+140 = 280m
train B takes T=14s to pass train A

so speed -
2V = 280m/14s
2V = 20m/s
so
V = 10m/s

but the answer is not given in units of m/s - it is given in units of km/hr

10m/s = 600m/min = 36000m/hr = 36km/hr
Answer D

@larkstar that’s a lovely clear explanation - thank you. DD has read the thread and now says that on her first (timed) attempt this afternoon she did it almost all right but forgot to convert m/s into km/h. The perils of multiple choice - nul points!

forgot to convert m/s into km/h.

Never forget your units!

What? These are 16+ questions? Seen similar level at 11+ test papers 😂

I was a bit surprised by it being from a TSA, though I suppose those aren't aimed at maths/physics bods.

@PreplexJ the 11 yr old (Y7) in our house wouldn’t be scoring highly on this. I don’t think he even knows V= D/T yet.

Here’s a full practice paper if your 11 yr old fancies a mental workout:

https://www.admissionstesting.org/Images/296238-tsa-oxford-specimen-test.pdf

Thanks for the paper will give it a try.

To the point V=D/T, I know it is not covered in KS2 curriculum but a lot of 11+ test actually test it.

Same for the LCM questions, not necessarily fill a swimming pool but build a fence, paint a house etc, seen it a couple of times in some 11+ papers...

In terms of the swimming pool one I'd have stuck a dummy value in for the volume of the swimming pool and worked it out like that: i.e. 6000L pool then Hose A is 1000 L per hour etc etc then add up the total volume per hour and divide by the 6000 you first thought of.

You can also do it nicely with algebra and X in place of the volume but using a dummy value sometimes makes it easier to conceptualise for me. (And to do without any paper which is what I'm currently doing!)

Trains:

Imagine a signal post at the point where the two trains start to cross. The passing finishes when the tail of each train reaches the same signal post. Each train has therefore travelled exactly its own length during the crossing. 140m in 14 seconds is 10m/second, or 3600*10 =36000 m/hour, which is 36km/h

Pool:

Flow rate is proportional to 1/time
Total flow rate is the sum of the individual flow rates
Total flow rate is therefore

1/(0.25) + 1/2 + 1/3 + 1/4 = 48/12 + 6/12 + 4/12 + 3/12 = 61/12

It will therefore take 12/61 days, or 288/61 hours.
which is 4 and 44/61 hours, close to 4 3/4hours, or 4 hours 45 mins.

Thank you @mushti.

In case anyone is not yet fed up with me, we have one we haven't nailed this evening. The answer is known to be C, but we're not getting our simultaneous equations right. Knowing it's C I can work it backwards and prove it right, but not otherwise! Thanks in advance for any help.

No need for simultaneous equations.

Consider the amount of powder in each packet is the same for 5 weeks of (m+2) washes as it is for 6 weeks of (m+1) washes.

5(m+2) = 6(m+1)

5m+10 = 6m + 6

m = 4

(Alternatively, 5(m+2) = 6(m+1) becomes 5x6 = 6x5 and is therefore trivially true when m+2 = 6 and m+1 = 5.)

Oh @mushti. Thank you. Elegantly simple when you know how!