Vintage 5th-grade math competition worksheet featuring a mix of arithmetic, geometry, and logic problems.

Let’s solve each problem one by one, carefully and step by step.

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Problem 1:
If a Δ b means a + a - b, find the value of (5 Δ 3) Δ 2 in simplest form.

First, understand the operation:
a Δ b = a + a - b → that’s the same as 2a - b

So, first compute (5 Δ 3):
= 2×5 - 3 = 10 - 3 = 7

Now take that result and do Δ with 2:
(5 Δ 3) Δ 2 = 7 Δ 2
= 2×7 - 2 = 14 - 2 = 12

✔ Final answer for #1: 12

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Problem 2:
100,000 ÷ 1,000 = 10,000 ÷ ___

Left side: 100,000 ÷ 1,000 = 100

So we have:
100 = 10,000 ÷ ?

That means: ? = 10,000 ÷ 100 = 100

✔ Final answer for #2: 100

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Problem 3:
What number results when the product of 8 and 8 is divided by the sum of 8 and 8?

Product of 8 and 8 = 8 × 8 = 64
Sum of 8 and 8 = 8 + 8 = 16

Now divide: 64 ÷ 16 = 4

✔ Final answer for #3: 4

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Problem 4:
A one-foot cube is to be divided completely into 6" cubes. All these 6" cubes are to be placed one on top of another. How many feet high is it from the bottom to the top?

First, note: 1 foot = 12 inches. So a 1-foot cube is 12" x 12" x 12".

Each small cube is 6" on each side.

How many 6" cubes fit along one edge of the big cube?
12" ÷ 6" = 2 cubes per edge

So total small cubes = 2 × 2 × 2 = 8 cubes

Now, stack all 8 cubes one on top of another. Each cube is 6" tall.

Total height = 8 × 6" = 48 inches

Convert to feet: 48 ÷ 12 = 4 feet

✔ Final answer for #4: 4

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Problem 5:
A 12 gallon tank is filled with water at the rate of ¼ gallon in 5 seconds. It will take ___ min. ___ sec. to fill ⅔ of the tank.

First, find how much water is ⅔ of 12 gallons:
⅔ × 12 = 8 gallons

Rate: ¼ gallon every 5 seconds.

How many ¼ gallons are in 8 gallons?
8 ÷ (¼) = 8 × 4 = 32 intervals

Each interval takes 5 seconds → total time = 32 × 5 = 160 seconds

Convert 160 seconds to minutes and seconds:
160 ÷ 60 = 2 minutes and 40 seconds (since 2×60=120, 160-120=40)

✔ Final answer for #5: 2 min. 40 sec.

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Problem 6:
In the addition problem:

```
2BA
+ C6D
------
8AD
```

Different letters represent different digits. Same letter = same digit.

We need to find B + C.

Let’s write it vertically:

```
2 B A
+ C 6 D
---------
8 A D
```

Start from the rightmost column (units place):

Column 1 (units): A + D = D or ends in D → so A must be 0? But wait — if A + D = D, then A = 0. BUT look at the tens place — there might be a carry.

Actually, let’s think again.

Units column: A + D = D (mod 10), possibly with a carry.

So: A + D ≡ D mod 10 → subtract D from both sides → A ≡ 0 mod 10 → so A = 0 or 10, but digit → A = 0.

But if A = 0, then in the tens column: B + 6 + carry_from_units = A (which is 0) or ends in 0.

Wait — let’s try assuming no carry first.

Assume A + D = D → implies A = 0. Then units digit is D, which matches.

Tens column: B + 6 = A → but A = 0 → B + 6 = 0 or 10? Since B is digit 0-9, B+6 can’t be 0 → must be 10 → so B = 4, and carry 1 to hundreds.

Hundreds column: 2 + C + carry(1) = 8 → 2 + C + 1 = 8 → C = 5

Check if all letters are different:
A = 0, B = 4, C = 5, D = ? We haven’t used D yet.

From units column: A + D = D → 0 + D = D → always true, so D can be any digit except 0,4,5. Let’s pick D=1 for example — doesn’t affect B+C.

So B = 4, C = 5 → B + C = 9

Wait — let’s verify the whole thing:

```
2 4 0
+ 5 6 D
-------
8 0 D
```

Add: 240 + 56D = 80D?

Let’s plug D=1: 240 + 561 = 801 → yes! 240+561=801 → correct.

Digits: A=0, B=4, C=5, D=1 → all different → good.

So B + C = 4 + 5 = 9

✔ Final answer for #6: 9

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Final Answer:
1) 12
2) 100
3) 4
4) 4
5) 2 min. 40 sec.
6) 9