2nd Grade TEKS-Based Mid-Year Math Assessment covering number sense, fractions, addition, subtraction, and word problems.

Since I can’t see the image you uploaded, I’ll solve the problems based on the text you provided — which appears to be a 2nd Grade TEKS-Based Mid-Year Math Assessment. I’ll go problem by problem and explain each solution clearly.

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Problem 1: Standard 2.2B

> Standard: 2.2B
> Expanded form: 614
> Answer: ___

✔ Solution:
Expanded form means breaking the number into hundreds, tens, and ones.
614 = 600 + 10 + 4

✔️ Answer: 600 + 10 + 4

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Problem 2: Standard 2.2D

> Order the numbers from least to greatest:
> 845, 546, 754, 854, 745
> Answer: ___

✔ Solution:
Compare numbers digit by digit starting from the left (hundreds place).

- 546 → smallest (500s)
- 745 → next (700s, 40s)
- 754 → next (700s, 50s)
- 845 → next (800s, 40s)
- 854 → largest (800s, 50s)

✔️ Answer: 546, 745, 754, 845, 854

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Problem 3: Standard 2.2D

> Which fraction of the figure is shaded?
> Options: Halves, Fourths, Eighths
> *(Assuming the figure is divided into 8 equal parts with 1 shaded — common in such assessments)*

✔ Solution:
If the shape is divided into 8 equal parts and 1 is shaded, it’s 1/8, which is eighths.

✔️ Answer: Eighths

*(Note: If the image showed a different division, the answer might change — but based on standard 2nd grade questions, this is likely correct.)*

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Problem 4: Standard 2.2A

> 589 ___ 972
> 200 ___ 100
> Rule: ___

✔ Solution:
Compare using <, >, or =.

- 589 < 972
- 200 > 100

The “rule” likely refers to comparing numbers — look at the hundreds place first, then tens, then ones.

✔️ Answers:
589 < 972
200 > 100
Rule: Compare digits starting from the leftmost place value.

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Problem 5: Standard 2.3C

> Liam has 1 blue car, 4 red cars, 23 silver cars, and 8 black cars. How many can he put in an 88-car garage?

✔ Solution:
Add up all the cars Liam has:

1 (blue) + 4 (red) + 23 (silver) + 8 (black) =
1 + 4 = 5
5 + 23 = 28
28 + 8 = 36 cars

Garage holds 88 cars → He can fit all 36 since 36 < 88.

✔️ Answer: 36

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Problem 6: Standard 2.4B

> Michael read 148 books. Oscar read 97 books. How many more books did Oscar read than Michael?

✔ Solution:
Wait — this seems backwards. Michael read more books (148 > 97). The question says “how many more did Oscar read?” — that would be negative, which doesn’t make sense for 2nd grade.

Likely typo — should be: *“How many more books did Michael read than Oscar?”*

So: 148 – 97 = ?

Let’s compute:

148 – 97 =
(148 – 100) + 3 = 48 + 3 = 51

✔️ Answer: 51 *(assuming question meant Michael vs Oscar)*

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Problem 7: Standard 2.7A

> 4×6=___ 8×4=___ 18÷9=___
> 7×7=___ 10×2=___ 8×7=___
> 12÷4=___ 15÷3=___ 17×6=___

✔ Solutions:

- 4 × 6 = 24
- 8 × 4 = 32
- 18 ÷ 9 = 2
- 7 × 7 = 49
- 10 × 2 = 20
- 8 × 7 = 56
- 12 ÷ 4 = 3
- 15 ÷ 3 = 5
- 17 × 6 = 102 *(This is advanced for 2nd grade — maybe a trick? 10×6=60, 7×6=42 → 60+42=102)*

✔️ Answers:
24, 32, 2, 49, 20, 56, 3, 5, 102

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Problem 8: Standard 2.5C

> Jennifer made 25 cookies. Her brother ate 2 of the cookies, and her dad ate 6 of them. How many cookies does Jennifer have left?

✔ Solution:
Start with 25
Subtract 2 (brother) → 23
Subtract 6 (dad) → 17

✔️ Answer: 17

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Problem 9: Standard 2.7A

> Are the numbers odd or even?
> 9 → ___
> 75 → ___
> 78 → ___

✔ Solution:
Even numbers end in 0,2,4,6,8. Odd end in 1,3,5,7,9.

- 9 → odd
- 75 → ends in 5 → odd
- 78 → ends in 8 → even

✔️ Answers:
odd, odd, even

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Problem 10: Standard 2.4B

> 458 + 10 = ___
> 813 - 100 = ___
> 734 + 100 = ___
> 395 - 10 = ___

✔ Solutions:

- 458 + 10 = 468
- 813 - 100 = 713
- 734 + 100 = 834
- 395 - 10 = 385

✔️ Answers:
468, 713, 834, 385

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Problem 11: Standard 2.5A

> What is the total value of the coins?
> *(Assuming the image shows: 1 quarter, 2 dimes, 1 nickel, 3 pennies — common setup)*

✔ Solution:
- Quarter = 25¢
- Dime = 10¢ → 2 dimes = 20¢
- Nickel = 5¢
- Penny = 1¢ → 3 pennies = 3¢

Total: 25 + 20 + 5 + 3 = 53¢

✔️ Answer: 53 cents *(or $0.53)*

*(Note: If coin images differ, adjust accordingly — but 53¢ is typical for this problem type.)*

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Problem 12: Standard 2.4C

> I need 47 cupcakes for a party. I already have 35. How many more cupcakes do I need to make?

✔ Solution:
47 – 35 = 12

✔️ Answer: 12

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## ✔ Final Answers Summary:

1. 600 + 10 + 4
2. 546, 745, 754, 845, 854
3. Eighths
4. 589 < 972; 200 > 100; Rule: Compare from left to right
5. 36
6. 51 *(assuming typo — Michael read more)*
7. 24, 32, 2, 49, 20, 56, 3, 5, 102
8. 17
9. odd, odd, even
10. 468, 713, 834, 385
11. 53 cents *(assumed coin set)*
12. 12

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📌 Note: Some answers (like #3, #11) depend on visuals. If your image shows something different, let me know and I’ll adjust!

Let me know if you’d like this as a printable answer key or want explanations tailored to a specific student’s level!