Sample page from a Grade 3 Math League competition featuring six mixed math problems ranging from basic arithmetic to logic puzzles.

Let's solve each problem step by step.

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1) If 7 + 6 = 20 - □, find the number that belongs in the box.

First, compute the left side:
- 7 + 6 = 13

So:
- 13 = 20 - □
- □ = 20 - 13 = 7

✔ Answer: 7

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2) In the two problems at the right, the larger answer is ____ more than the smaller answer.

We are given:

```
133
+148
-----
281
```

and

```
321
- 73
-----
248
```

Now compare:
- Larger answer: 281
- Smaller answer: 248
- Difference: 281 - 248 = 33

✔ Answer: 33

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3) If Mr. O'Brien was able to cut a board of wood into exactly eight 3-foot pieces, then he could have cut the same board into ____ 4-foot pieces.

First, find the total length of the board:
- 8 pieces × 3 feet = 24 feet

Now divide by 4 feet per piece:
- 24 ÷ 4 = 6 pieces

✔ Answer: 6

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4) There are 17 people meeting in Room 101. There are 7 people meeting in Room 201. If ____ people were moved from Room 101 to Room 201, there would be twice as many people in Room 201 as in Room 101.

Let’s say x people are moved from Room 101 to Room 201.

After moving:
- Room 101: 17 - x
- Room 201: 7 + x

We want:
> Room 201 = 2 × Room 101
So:
> 7 + x = 2(17 - x)

Solve:
- 7 + x = 34 - 2x
- x + 2x = 34 - 7
- 3x = 27
- x = 9

Check:
- Room 101: 17 - 9 = 8
- Room 201: 7 + 9 = 16
- Is 16 = 2 × 8? Yes!

✔ Answer: 9

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5) A, B and C represent different digits. AB represents a 2-digit number. AB + C = 50; BC + A = 41. In the answer column, place the values for A, B and C.

Let’s interpret:
- AB means 10A + B (a two-digit number)
- BC means 10B + C

Given:
1) AB + C = 50 → (10A + B) + C = 50
2) BC + A = 41 → (10B + C) + A = 41

So we have:
- (1) 10A + B + C = 50
- (2) 10B + C + A = 41

Let’s write both equations:
- Eq1: 10A + B + C = 50
- Eq2: A + 10B + C = 41

Subtract Eq2 from Eq1:
- (10A + B + C) - (A + 10B + C) = 50 - 41
- 9A - 9B = 9
- A - B = 1 → So A = B + 1

Now substitute A = B + 1 into Eq1:
- 10(B + 1) + B + C = 50
- 10B + 10 + B + C = 50
- 11B + C = 40 → C = 40 - 11B

Try integer values of B (digit from 0–9):

Try B = 3:
- C = 40 - 33 = 7
- A = B + 1 = 4
- Check: A=4, B=3, C=7 — all different ✔️

Check original equations:
- AB = 43, AB + C = 43 + 7 = 50 ✔️
- BC = 37, BC + A = 37 + 4 = 41 ✔️

Perfect!

✔ Answer:
- A = 4
- B = 3
- C = 7

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6) In a bank, Mrs. Wallace, Mrs. Thomas and Mrs. Ramariz held the positions of bank teller, loan officer and branch manager, but not necessarily in that order. The teller, who just began working that year, earned the least. Mrs. Thomas and Mrs. Ramariz worked for the bank for many years. Mrs. Thomas earned more than the loan officer. Who was the loan officer?

Let’s analyze:

#### Step 1: Identify who is the teller.
- Teller is the new employee, just began that year.
- Mrs. Thomas and Mrs. Ramariz have worked for many years → so they are not the teller.
- Therefore, Mrs. Wallace must be the teller.

#### Step 2: Teller earns the least.
- So Mrs. Wallace earns the least.

#### Step 3: Mrs. Thomas earned more than the loan officer.
- So Mrs. Thomas ≠ loan officer (because she earns more than the loan officer).
- Therefore, Mrs. Thomas is not the loan officer.

So far:
- Teller: Mrs. Wallace
- Loan officer: Not Mrs. Thomas, not Mrs. Wallace → Must be Mrs. Ramariz

Wait! Let’s double-check.

We have:
- Positions: Teller, Loan Officer, Branch Manager
- People: Wallace, Thomas, Ramariz

We know:
- Teller = Wallace (only one who is new)
- So remaining: Thomas and Ramariz → Loan Officer and Branch Manager

But: Mrs. Thomas earned more than the loan officer → So Mrs. Thomas cannot be the loan officer (she earns more), so she must be the branch manager.

Therefore:
- Loan officer = Mrs. Ramariz

✔ Answer: Mrs. Ramariz

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✔ Final Answers:

1) 7
2) 33
3) 6
4) 9
5) A = 4, B = 3, C = 7
6) Mrs. Ramariz

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