HAST Exam Page (3 Viewers)

[dem()sthenes]

No lol, that question isn't hard, it's just badly written.

I agree.

 

[ADHERETOLOGIC]

One of each shape works, but you could argue that the red heart is not an allowed shape in the first place. Your mirrored statement is just wrong. Take the middle column for instance.

Oh yeah. But it applies for every other column. And the shape can't "not be allowed" tbh.

 

[hstb_]

View attachment 28604

Guys this stumped me

C

 

[hstb_]

Anyone else who just doesn't know wtf is going on and just worrying about trials and whatnot

lmaoo me

 

[SaltSolutionPromo]

Where can I find example HAST questions and/or past papers?

 

[dem()sthenes]

Where can I find example HAST questions and/or past papers?

You can find it in their website. It's either pay or pirate.

 

[Pahulbir Thind]

POST MORE QUESTIONS PLEASE, ESPECIALLY MATHEMATICAL REASONING

 

[hifum]

 

[Velocifire]

1) Find the degrees of each one

so u get 135 120 and 105 for each segment
Double check by adding to 360

then see if u can find the common factor between the three
seems 15 can fit perfectly

135/15 = 9
120/15 = 8
105/15 = 7

add them and u get 24 (a)
Fits the criteria of under 17 as well

 

[D-BOSS]

a

 

[D-BOSS]

This is AMC right?

 

[hifum]

 

[Velocifire]

Angle sum to 360, that bit ain’t hard innit

just divide and use the 90 quarter as a reference point and add/subtract

@hifum also lucked out by using the 15 30 45 as Ages 4 8 and 12 which also got the correct answer but that’s just more like a guess and check rather than certainty with the angle proportion and taking a hcf or 15

 

[Velocifire]

Also @hifum new problem the answer is 41 I think

 

[Velocifire]

For a number to be divisible by 3 (multiple of 3), its digits must add to a multiple of 3.
I count

sets of 3 digits with all digits are odd, and they add to
a multiple of 3 (3, 6, 9, 12, 15, 18, 21, 24, or 27).
Those sets are:
{1,1,1} , {1,1,7} ,
{1,3,5} , {1,5,9} ,
{1,7,7} ,
{3,3,3} , {3,3,9} ,
{3,5,7} ,
{3,9,9} ,
{5,5,5} ,
{5,7,9} ,
{7,7,7} , and
{9,9,9} .

Among those

sets,

sets have 3 different digits,
Each of those

sets can be arranged in

different arrangements/permutations,
making

different three-digit numbers whose digits are all odd.

Among those

sets listed above,
there are also

sets made of just one single repeated digit,
and each one of those sets can be arranged just one way,
to form just one three-digit number,
so from them we can get another

three-digit numbers whose digits are all odd.

The remaining

of the

sets listed above contain only two different digits, one of them repeated.
Form each of those

sets, we can make

different three-digit numbers,
because there are 3 positions to place the unrepeated digit,
and that gives us another

three-digit numbers whose digits are all odd.

That makes a total of

three-digit numbers divisible by 3, whose digits are all odd.

 

[D-BOSS]

For a number to be divisible by 3 (multiple of 3), its digits must add to a multiple of 3.
I count

sets of 3 digits with all digits are odd, and they add to
a multiple of 3 (3, 6, 9, 12, 15, 18, 21, 24, or 27).
Those sets are:
{1,1,1} , {1,1,7} ,
{1,3,5} , {1,5,9} ,
{1,7,7} ,
{3,3,3} , {3,3,9} ,
{3,5,7} ,
{3,9,9} ,
{5,5,5} ,
{5,7,9} ,
{7,7,7} , and
{9,9,9} .

Among those

sets,

sets have 3 different digits,
Each of those

sets can be arranged in

different arrangements/permutations,
making

different three-digit numbers whose digits are all odd.

Among those

sets listed above,
there are also

sets made of just one single repeated digit,
and each one of those sets can be arranged just one way,
to form just one three-digit number,
so from them we can get another

three-digit numbers whose digits are all odd.

The remaining

of the

sets listed above contain only two different digits, one of them repeated.
Form each of those

sets, we can make

different three-digit numbers,
because there are 3 positions to place the unrepeated digit,
and that gives us another

three-digit numbers whose digits are all odd.

That makes a total of

three-digit numbers divisible by 3, whose digits are all odd.

Wait I found that online lmao: https://www.algebra.com/algebra/homework/word/numbers/Numbers_Word_Problems.faq.question.978706.html

 

[Velocifire]

Yeah that I did lol the other I didn’t
do you think I would have time to do that?

 

[hifum]

nice guys!!!

 

[hifum]

this took me a while but i got the answer!

 

[D-BOSS]

156