Re: UKMT 2023 JMC Paper

You had to do 64cm-20cm and since you got 44cm^2. You then did 44cm/4 to get 11cm. You then work out a perimeter which works with an area of 11cm. You can get 5.5cm X 2 and it will work and get the answer 15.

yes i think i got that right. the answer was 16


so the area of the larger square was 64 (the question said that)

this means the side length of the large square was 8cm

now that side is equal to one long side of rectangle (call that a) and short side (call that b)

it asked for the perimeter of one rectangle so it is 2a + 2b which is 8 x 2 which is 16.

btw i recommend looking at q 22 from the 2021 jmc as it was very similar.

https://www.ukmt.org.uk/sites/default/files/ukmt/JMC%202021%20Extended%20Solutions.pdf

Yes...I used the same method and got the same answer.
overall, I think that the questions weren't that hard for the first 1-15...maybe the boundaries are high this year?

oh so i remember the question exactly. here is the question:
Beatrix was born in this century. On her birthday this year, her age was equal to the sum of the digits of the year in which she was born. In which of these years will her age on her birthday be twice the sum of the digits of that year?

A 2027
B 2029
C 2031
D 2033
E 2035

So if she was born in 2015

She is 8 years old this year.

In 2031 she is 16 years old
And the sum of the digits of the digits is 8

So its 2031

however, it depends on how you understood the question.

The answer depends, if u interprate "that year" as the year she was born the answer would be 2031


But if u Interpret "that year" as 2023 the answer would be 2035

personally i just got 2031 but u could have got 2035.

its one of those 2

Wow - photographic memory!
You are so right about the interpretation - I went for the other understanding & got 2035 but realised I’d misunderstood in the final few seconds.
I find understanding the question a bigger challenge than the maths in a lot of these papers. It’s a hard thing to work on

I'm not sure if other schools had this but everyone at my school (we did the test online) had issues with Question 2,
It went something like this,
How many of the following are factors of 30?

A0 B1 C2 D3 E4
But it wouldn't show anyone the list of numbers, so we weren't sure whether it meant the numbers in the boxes A,B,C,D and E or if there was a glitch.

oh yeah that makes sense thank you so much! also could someone please help me with that question about the dried tea bags I was so confused with it

icl i was too but idk

i know i got it wrong for sure as i wrote 480

i asked chatgpt the question and it then said it was E - 180

but my friends said they got either 195/210 (210 being the majority of my friends)

ok I guessed and got 180

guys if you want to see the questions and answers for the jmc 2023 here they are

paper: https://www.ukmt.org.uk/sites/default/files/ukmt/Junior%20Mathematical%20Challenge%202023.pdf

solution: https://www.ukmt.org.uk/sites/default/files/ukmt/JMC%202023%20Solutions.pdf

do you guys think the boundaries will be higher or lower this year?

110-115

Yes, 110-115 feels like the right sort of range (for the Olympiad).

I should add that 115, 114 and 110 are highly improbable scores and 113 is unlikely. Which suggests 111 or 112 as the cut-off.

I tend to mess up on the first half but not the second half!
But yeah I agree with your tweak & thinking, especially since it’s across 2 school years. I got more wrong this year than last but felt this years paper was harder but then also I’ve got a years more experience this year. Overall I think I’m at ‘my level’

When do you think results will come out? It feels like this solution was published earlier than last year.