[Dr. Ashani Dasgupta]

Welcome to the Problem Marathon for AMC 8. You can login with google and post your solutions here. The student with most correct solutions will receive an award at the end of the problem marathon.

The question paper of AMC 8, 2024 is available here.

AMC 8, 2024 Problems, Solutions and Concepts

[vineetachoudharyakshita]

Problem 1

Solution 1-

My idea is to solve it in this manner-

 

222222-(22222+2222+222+22+2)

 

I don't care about the number in the brackets but about its unit digit. As 2 is repeating 5 times in the unit digit, it will be 2×5= 10

222222-0= 2 (B)

• This reply was modified 1 year, 8 months ago by vineetachoudharyakshita.

[vineetachoudharyakshita]

Problem 2
Solution (The typical method)

$$ \frac{44}{11}+\frac{110}{44}+\frac{44}{1100} $$

We will simplify and get-

$$\frac {4} {1}+ \frac {10} {4}+ \frac {4} {100}$$

$$ = 4+ 2.5+0.04= 6.54 $$(C)

 

• This reply was modified 1 year, 8 months ago by vineetachoudharyakshita. Reason: Wrong LaTeX code used
• This reply was modified 1 year, 8 months ago by vineetachoudharyakshita.
• This reply was modified 1 year, 8 months ago by vineetachoudharyakshita.
• This reply was modified 1 year, 8 months ago by Dr. Ashani Dasgupta.

[vineetachoudharyakshita]

Problem 3

To ease the solution, I have given names to the points

Area of Grey Region= Area of HEFGJI+ Area of AKLMCB

Area of HEFGJI= \( 7^2 \) - \( 4^2 \)
= 49-16= 33
Area of AKLMCB
= \( 10^2 \) - \( 9^2 \)
= 100-81= 19
∴ 33+19=52(E)

[vineetachoudharyakshita]

Problem 4

The sum from 1 to 9= 45
 The square numbers below 45= 1,4,9,16,25 and 36 (Excluding 0)

Now, We can say that the closest 2 square numbers from 45 are 25 and 36
The difference for each is 20 and 9 respectively.

20 is not an element in our set of numbers from 1 to 10

∴ 9 is the number excluded by Yunji (E)

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