STEP maths I, II, III 1992 solutions

I/1

Today is June 26th 1992
We want to know April 3rd 1905
This was 86 years ago, plus a year and a bit more.
Those 86 years, starting in 1905, will have 21 leap years.
A leap year has 366 days. 366 mod7 shows this adds two days every leap year.
A non-leap year has 365 days. 365 mod7 shows this adds one day.
So, from 1905, we are adding 21*2 + 65*1 = 42 + 65 = 107 days.
107 mod7 = 2.
So we currently have a shift of 2 day, up until April 3rd 1991.
If June 26th 1992 was a Friday, and 1992 is a leap year, then June 26th 1991 will be a Wednesday.
This makes June 5th 1991 a Wednesday.
This makes June 1st 1991 a Sat.
This makes May 31st 1991 a Friday.
Between April 3rd and May 31st there are 27 + 31 = 58 days. 58 mod7 = 2
So April 3rd of 1991 was a Wednesday.
Remembering the two days shift to go back to 1905 gives us Monday

Done.

Reply 22

Looking at an online calendar confirms your answer Lusus btw

Reply 23

There are some quick "trick" methods of finding the weekday of any date. I wonder if they would credit those methods in the exam, if used without justification...

Reply 24

generalebriety

16

Do you expect the first question on the 'easy' paper to be difficult? :wink:

Ok, doing II/3.

Edit: will type it up tomorrow. Tired.

Reply 25

What are they, Lusus?

(doing I/4)

Reply 26

Speleo

What are they, Lusus?

This page explains a method well.

(I guess I'll try another STEP question as well, to help get through 1992)

Reply 27

Speleo

II/1

Someone might want to check v.) and vi.), v.) because it seems so easy, vi.) because I'm not certain I can neglect the terms I did... (limit's right though).

Think I'll attempt Lusus' challenge now, I expect to fail miserably :tongue:

v) is definitely right, it must be that easy... at least I agree :tongue:

vi) Seems okay, don't know what else they could expect...

I'll start typing up the matrix one now (it's finished) but it will take time! Latexing matrices is not nice... (and I feel I took a shortcut to the proof but I think it is okay...)

Reply 28

Thanks Lusus and nota :smile:

Here's I/4.

STEP I Q14 is, ah, different, to say the least. :s-smilie:

Reply 30

O_o
I can see that hypothesis tests come in a bit but mostly...

Anyway, what's the verdict: pull an all-nighter doing STEP or go to sleep now?

Reply 31

There are some nice-looking Mechanics questions on a couple of the papers but I do not have the perseverance to get them out tonight, so I shall say goodnight. Sleep well, Speleo. :smile:

Reply 32

Good night Lusus :smile:

I'm thinking of an all-nighter or possibly going to bed around 7 :p:

(good thing to be finished with exams is I can turn around the days how I want...)

Reply 33

EDIT: Can't be bothered, going to go sleep :tongue:

Reply 34

TheDuck

2

Q15, probability and statistics... of STEP I
Sorry if you wanted the other bits doing! - I'm just in a probability mood right now.

The arrival of passengers is uniform. Let train A be the train which takes 60 minutes and train B the one which takes 75.

If a passenger catches A, then his/her journey time is uniformly distributed over
(60,84), while if he/she catches B, then the journey is similarly distributed over (75,111).

So quite simply the average journey time =
average for A * P(catching A) + average for B * P(catching B)
=72*(24/60)+93*(36/60)
=84.6 as required.

For the second part, the journey time for catching A will be uniformly distributed over (60, 72+x), while for B, it will be (75,123-x)

average journey time =
average for A * P(catching A) + average for B * P(catching B)
=(60+72+x)/2*(x+12)/60+(75+123-x)/2*(48-x)/60
=-(17/20)*x+462/5+(1/60)*x²

differentiating this and setting it to zero gives...
0=-17/20+(1/30)*x
so x=20.5
giving an average journey time of 82.0 minutes

Reply 35

Okay, STEP I Question 2

(David, hope I've made all cases clear etc. this looks like one of those tedious ones again...)

To start with I'll list all possible cases by brute force (this was how I solved it...). Therefore I find the whole layout of the question a bit weird as a) doesn't lead to b) etc. with my method...

Important to keep in mind is the criterion (which means the sum has to be possible to compose in 8 ways with the digits 1-9):
a+b+c=d+e+f=g+h+k=a+d+g=b+e+h=c+f+k=a+e+k=c+e+g

SUM
24: 789 = 1 way
23: 986 = 1 way
22: 985, 876 = 2 ways
21: 984, 876 = 2 ways
20: 983, 974, 965, 875 = 4 ways
19: 982, 973, 964, 874, 865 = 5 ways
18: 981, 972, 963, 954, 873, 864, 765 = 7 ways
17: 971, 962, 953, 872, 854, 863, 764 = 7 ways
16: 961, 952, 943, 871, 862, 853, 763, 754 = 8 ways
15: 951, 942, 861, 852, 843, 762, 753, 654 = 8 ways
14: 941, 932, 851, 842, 761, 752, 743, 653 = 8 ways
13: 193, 184, 175, 283, 274, 265, 364 = 7 ways
12: 129, 138, 147, 156, 273, 264, 345 = 7 ways
11: 128, 137, 146, 236, 245 = 5 ways
10: 127, 136, 145, 235 = 4 ways
9: 135, 234 = 2 ways
8: 125, 134 = 2 ways
7: 124 = 1 way
6: 123 = 1 way

From this we can see that there are 3 possible ways of getting a sum that has eight ways of being combined. However, for this type of array we are trying to create 'e' will appear in four sums (d+e+f and b+e+h and a+e+k and c+e+g). For the sum 14 the max number of times a digit appears is three (for 1, 2, 3, 4, 5 and 7) and for 15 the max number of times a digit appears is four (for 5), lastly for 16 the max number of times a digit appears is three (for 9, 8, 7, 6, 5 and 3). This means to create an array of this type the combinations that will be present are the ones presented in the emboldened line in the list of possibilities.

Therefore follows the answers to the questions
i) Shown 15 above.
ii) Shown 5 above.
iii) Implicitly shown as we only have two sums with 9 involved when creating the sum 15. This means that the 9 must be located at a position where it only appears in two summations (therefore b, d, f or h). (On a sidenote e appears in four sums as earlier mentioned and a, c, g and k appear in three summations)
iv) Setting b=9 (knowing e=5) gives h=1 and a+c= 4 and 2 (from our table of possibilities) we get two possible arrays (one with first line 492 and one with 294).
v) Calculating these possibilities I think is just to multiply 4 by 2 (because there are four possible positions of the 9, b, d, f and h) to get 8 because once you fix one number all the other will fall out with two possibilities.

Hope this is correct :redface:

Reply 36

Swayum

18

Is it me or is STEP I going beyond the C4 syllabus? I swear I'm not meant to know about complex numbers yet. Also, what should I have studied to be able to do mechanics and statistics questions in STEP I, II and III?

Reply 37