Editorial - Matwa - TO TKA 1

Problem 1

First let's find the radius, which should be half of the diagonal d.

s=7/2⋅200=700=7

d=7/2⋅200⋅√(,2)=700√(,2)=7√(,2) (m)

r=(7√(,2))/2

Now let's subtract the area of the circle by the area of the square.

A=(49⋅2)/4⋅22/7-7⋅7=28

That's it.

Takeaway: Don't miss the trees for the forest.

Problem 2

I drew it up :)

Takeaway: Take your time to read the actual question and not fumble the symbols like me!

Subproblem 2.1

AQ=12

AQ=1/3⋅36=12

True.

Subproblem 2.2

AB=30

PB=2/3⋅27=18

AB=√(,PB2+PA2)=√(,182+242)=30

True.

Subproblem 2.3

BS=18

BS=1/3⋅27=9

False.

Subproblem 2.4

PC∶CR=1∶2

Just by observation, one can guess that PC is half of AB

PC=15

Then, let's find PR

PR=√(,272+362)=45

CR=PR-PC=45-15=30

PC∶CR=15∶30=1∶2

Correct.

Subproblem 2.5

PC:CR=2∶3

No. We just disproved this.

Problem 3

This isn't solvable. Where are the annotations

Maybe don't make graphics using Office. Could the government maybe use actual creative software, like Photoshop, PAINT.NET, or Affinity? Is all that really out of their budget of corruption? Sorry, it's probably the job of some underpaid teacher. Fuck efficiency and the education budget cuts.?

Okay I found the annotated version, along with the answers

Problem 4

Tough problem which I didn't solve at first glance.

Let C be the circumference. Also, let a be the middle element and b the difference.

(a-b)+a+(a+b)=108

3*a=108

a=36

We know that one of the sides must be 36

Let's use pythagoras' theorem to solve for the rest.

Of course, the hypotenuse is always the longest side.

362+(36-b)2=(36+b)2

1296+1296-72*b+b2=1296+72*b+b2

1296=144*b

b=9

Hey, now we know the sides are 27 36 45

Knowing the height and width, we can now find the area.

A=(27⋅36)/2=486

Easy.

Takeaway: The solution isn't always obvious

Problem 5

We can clearly see that the difference b is 50000 and the starting value is 100000

(S_24)=24/2⋅(2⋅100000+23⋅50000)=16200000

But remember that we need to subtract 2 million.

S=16200000-2000000=14200000

That's option (C)

Takeaway: Honestly, you should be skipping problems like these if they're not trivial.

Problem 6

Takeaway: Honestly, you should be skipping problems like these if they're not trivial.

Subproblem 6.1

b=50000

Yeah, we found that b=50000 earlier.

Subproblem 6.2

(S_12)=4500000

(S_12)=12/2⋅(2⋅100000+11⋅50000)=4500000

Correct.

Subproblem 6.3

(S_14)-2000000=2750000

(S_14)=14/2⋅(2⋅100000+13⋅50000)=5950000

S=5950000-2000000=3950000

False.

Problem 7

Takeaway: In the actual exam, problems like these are either timewasters or free points. Tread carefully.

Subproblem 7.1

(U_12)=650000

(U_12)=100000+11⋅50000=650000

True.

Subproblem 7.2

(U_24)-(U_12)=600000

True, as every year the difference goes up 12⋅50000=600000

Subproblem 7.3

(S_31)-4000000=26350000

(S_31)=31/2⋅(2⋅100000+30⋅50000)=26350000

Oh look, that's correct?

No, we need to subtract 4000000 as 2 years have passed.

S=26350000-4000000=22350000

So, the statement is false.

Subproblem 7.4

In July 2025, 4000000 will have been taken out from the account.

Yes. Because 2 years have passed.

Subproblem 7.5

(S_32)=1750000

Clearly false. That number is too small!

Problem 8

Already hate this problem.

Here's a tip: Use a*b*c instead of x*y*z to minimize mistakes confusing z and 2

4*a+2*b+4*c=480

6*a+b+2*c=420

6*a+4*b+6*c=800

2*a+b+2*c=240

6*a+b+2*c=420

4*a=180

∴a=45

b+2*c=420-270=150

3*b+6*c=450

4*b+6*c=800-270=530

∴b=80

2*c=150-80=70

c=35

You do the menial work. Calculate all of the options. I'm not writing all that.

Takeaway: Either do these easy but tedious problems first or last to preserve your momentum.

Problem 9

Takeaway: Nothing. This is trivial.

Subproblem 9.1

The seller can only sell 350 of type 2 cakes

Clearly false. Here's why:

1500000/5000=300

That's less than 350

Subproblem 9.2

Maximum profits are gained when the seller sells 250 cakes of the first type

There are always 3 options: Sell only type 1, sell only type 2, or sell a mix of both.

x+y≤400

3000*x+5000*y≤1500000

P=1000*x+1500*y

The case where you only sell the first type. Take the smallest x

3000*x=1500000

x=500∨ x=400

P=1000⋅400=400000

The case where you only sell the second type. Take the smallest y

5000*y=1500000

y=300∨y=400

P=1500⋅300=450000

The case where you sell a mix of both.

y=400-x

3000*x+(5000)*(400-x)=1500000

x=250

y=400-250=150

P=1000⋅250+1500⋅150=475000

The second statement is false, actually!

Subproblem 9.3

Maximum profit is 475000

Yeah, we just calculated it.

Problem 10

x+y≤25

150*x+200*y≤4200

P=50000*x+60000*y

Try optimizing for x

x=25∨x=28

P=50000⋅25=1250000

Try optimizing for y

y=25∨y=21

P=60000⋅21=1260000

Try optimizing for x and y

y=25-x

150*x+200*(25-x)=4200

x=16

y=9

P=50000⋅16+60000⋅9=1340000

So, the ideal profit is 1340000

Takeaway: This is trivial. Nothing more to say.

Problem 11

Takeaway: Check and recheck your results and process.

Subproblem 11.1

(-2,-8) is on ƒ(x)

-8=-(-2)2+6⋅(-2)-8

-8=-4-12-8

-8=-24

The statement is false.

Subproblem 11.2

The axis of symmetry is at x=3

x=-b/(2*a)=-6/(-2)=3

The statement is true.

Subproblem 11.3

ƒ(x) intersects with (0,-8)

-8=-0+0-8

-8=8

The statement is true.

Subproblem 11.4

The range of ƒ(x) is -48≤y≤0

Let's test for y at x=-4,3,4

ƒ(-4)=-16-24-8=-48

ƒ(3)=-9+18-8=1

ƒ(4)=-16+24-8=0

No, the range is -48≤y≤1

Subproblem 11.5

The domain where ƒ(x)>0 is -4≤x<2

We know this is wrong because the roots of the function are 4 and 2, not -4 and 2

I don't know why it's marked as correct.

Problem 12

I don't like doing algebra, so let's find the line directly via creating a line connecting the two points.

m=(1-(-2))/(-3-6)=-1/3

That's the slope/gradient.

y=-1/3*x+c

Plug in one point to find the c

1=-(-3)/3+c

c=0

Liar.

y=-1/3*x

x+3*y+0*c=0

1. a+b=1+3=4 Correct

2. c=0 Correct

3. (0,-6) doesn't intersect as the line intersects the origin point (0,0). ⊥

Takeaway: Sometimes the best solution is the easiest solution.

Problem 13

I wouldn't actually find the composed functions themselves, as I'm lazy.

(g∘ƒ)*(0)=g(-3)=-12+2=-10

Divisible by 5

(g∘ƒ)*(1)=g(2-1-3)=g(-2)=-8+2=-6

Not divisible by 5

(g∘ƒ)*(2)=g(8-2-3)=g(3)=12+2=14

Not divisible by 5

(ƒ∘g)*(-1)=ƒ(-2)=8+2-3=7

Not divisible by 5

(ƒ∘g)*(-2)=ƒ(-6)=72+6-3=75

Divisible by 5

Takeaway: Don't overcomplicate the problem.

Problem 14

There's probably some easier way to do this, but I'll do it the hard way to confirm.

y=(3*(3*x-2)+4)/(3*x-2-5)=(9*x-2)/(3*x-7)

3*x*y-7*y=9*x-2

3*x*y-9*x=7*y-2

x=(7*y-2)/(3*y-9)

(7*y-2)/(3*y-9)

(7⋅2-2)/(3⋅2-9)=-4

Yay.

Takeaway: If you don't know the easy solution, try the hard solution.

Problem 15

I decided to do this problem because it's fun :)

x2-5*x+6≥0

(x-3)*(x-2)≥0

x≤3∪x≥2

Not equals because you can't have 0 as divisor

2*x+4>0

2*x>-4

x>-2

So, take the union.

-2<x≤3∪x≥2

Yay!

Takeaway: Domain and range problems are trivial if you know your theory.

Problem 16

Subproblem 16.1

1of wheat will generate 690of flour

y=1-0.2=0.8

g(0.8)=0.82+0.05=0.64+0.05=0.69=690

Subproblem 16.2

100of wheat will generate 150of flour

Problem 18

This problem is trivial.

1. You can do the first statement in your head.

2. Trivial if you know that mirroring by y=-x just swaps x and y while also flipping their signs.

3. Easy if you know the formula.

4. Easy if you know the formula

5. Solving 4 also solves 5

Problem 22

This problem infuriated me at first. But, it's actually exteremely easy. Just scale the graph.

Let T be the total population, s the total students of some percentage p

s=p/100⋅T

We know that:

T=14⋅100/18

Let's substitute that.

s=p/100⋅100/18⋅14=(14⋅p)/18

The problem needs s for whatever p we're testing to be ≥40

(14⋅p)/18≥40

14⋅p≥720

We have an excellent helper function here, where we can input p as the percentage, and we can test if that percentage of the population has more than 40 students. All that's next is testing each subproblem. Neat!

Takeaway: Simplify the problem.

Problem 23

Real easy. I misread at first.

1. 3⋅10⋅9⋅8⋅5⋅4=43200

2. 3⋅1⋅5⋅4=60

3. 3⋅1⋅8⋅5⋅4=480

Takeaway: Focus.

Problem 24

That means there's 40/100⋅80=32 people in the intersection

There's 80-32=48 in the just SMA region

There's 40-32=8 in the just work region

In total, there's 32+48+8=88

Which leaves 100-88=12 left

Problem 25

1. 4/5⋅2/3=8/15

2. 4/5⋅1/3+1/5⋅2/3=2/5

3. 1/5⋅1/3=1/15