Worksheet with 22 math word problems focusing on algebra, geometry, and number relationships.

Let’s solve each problem one by one, step by step. We’ll use simple algebra — let the width be a variable (like *w*), then express length in terms of that variable, and use area = length × width to set up an equation.

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Problem 1:
Length = 2 × width → Let width = w, so length = 2w
Area = 72 → w × 2w = 72 → 2w² = 72 → w² = 36 → w = 6
Then length = 2×6 = 12
✔ Dimensions: 6 inches by 12 inches

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Problem 2:
Length = 4 × width → Let width = w, length = 4w
Area = 144 → w × 4w = 144 → 4w² = 144 → w² = 36 → w = 6
Length = 4×6 = 24
✔ Dimensions: 6 inches by 24 inches

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Problem 3:
Length = width + 4 → Let width = w, length = w+4
Area = 60 → w(w+4) = 60 → w² + 4w - 60 = 0
Factor: (w+10)(w-6)=0 → w=6 (since width can’t be negative)
Length = 6+4 = 10
✔ Dimensions: 6 yards by 10 yards

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Problem 4:
Width = length - 11 → Let length = L, width = L-11
Area = 80 → L(L-11) = 80 → L² -11L -80 = 0
Factor: (L-16)(L+5)=0 → L=16
Width = 16-11 = 5
✔ Dimensions: 5 inches by 16 inches

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Problem 5:
Length = width + 3 → Let width = w, length = w+3
Area = 70 → w(w+3) = 70 → w² +3w -70 = 0
Factor: (w+10)(w-7)=0 → w=7
Length = 7+3 = 10
✔ Dimensions: 7 inches by 10 inches

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Problem 6:
Length = width + 3 → Let width = w, length = w+3
Area = 108 → w(w+3) = 108 → w² +3w -108 = 0
Factor: (w+12)(w-9)=0 → w=9
Length = 9+3 = 12
✔ Dimensions: 9 cm by 12 cm

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Problem 7:
Width = length - 5 → Let length = L, width = L-5
Area = 84 → L(L-5) = 84 → L² -5L -84 = 0
Factor: (L-12)(L+7)=0 → L=12
Width = 12-5 = 7
✔ Dimensions: 7 meters by 12 meters

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Problem 8:
Length = 2 × width → Let width = w, length = 2w
Area = 50 → w×2w = 50 → 2w² = 50 → w² = 25 → w=5
Length = 2×5 = 10
✔ Dimensions: 5 inches by 10 inches

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Problem 9:
Length = 2×width + 1 → Let width = w, length = 2w+1
Area = 55 → w(2w+1) = 55 → 2w² + w -55 = 0
Use quadratic formula or factor: Try factoring → (2w+11)(w-5)=0 → w=5
Length = 2×5+1 = 11
✔ Dimensions: 5 feet by 11 feet

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Problem 10:
Length = 3×width - 2 → Let width = w, length = 3w-2
Area = 65 → w(3w-2) = 65 → 3w² -2w -65 = 0
Try factoring: (3w+13)(w-5)=0 → w=5
Length = 3×5 -2 = 13
✔ Dimensions: 5 meters by 13 meters

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Problem 11:
Length = 2×width - 7 → Let width = w, length = 2w-7
Area = 60 → w(2w-7) = 60 → 2w² -7w -60 = 0
Factor: (2w+15)(w-4)=0 → w=4
Length = 2×4 -7 = 1
Wait — length = 1? That’s possible, but let’s check: 4×1=4 ≠60 → Mistake!

Actually, try quadratic formula for 2w² -7w -60 = 0
Discriminant = 49 + 480 = 529 → √529=23
w = [7 ±23]/4 → w=(30)/4=7.5 or w=(-16)/4=-4 → w=7.5
Length = 2×7.5 -7 = 15-7=8
Check: 7.5 × 8 = 60 ✔
✔ Dimensions: 7.5 meters by 8 meters

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Problem 12:
Two consecutive integers: n and n+1
Product = 56 → n(n+1)=56 → n²+n-56=0
Factor: (n+8)(n-7)=0 → n=7
Integers: 7 and 8
✔ Answer: 7 and 8

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Problem 13:
Consecutive odd integers: n and n+2 (both odd)
Product = 99 → n(n+2)=99 → n²+2n-99=0
Factor: (n+11)(n-9)=0 → n=9
Integers: 9 and 11
✔ Answer: 9 and 11

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Problem 14:
Consecutive even integers: n and n+2
Square of smaller = larger + 10 → n² = (n+2) + 10 → n² = n + 12 → n² -n -12=0
Factor: (n-4)(n+3)=0 → n=4 (even, good)
Integers: 4 and 6
Check: 4²=16, 6+10=16 ✔
✔ Answer: 4 and 6

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Problem 15:
Consecutive odd integers: n and n+2
Product = 2×sum -1 → n(n+2) = 2(n + n+2) -1 → n²+2n = 2(2n+2)-1 → n²+2n = 4n+4-1 → n²+2n = 4n+3
→ n² -2n -3 =0 → (n-3)(n+1)=0 → n=3
Integers: 3 and 5
Check: product=15, sum=8, 2×8-1=15 ✔
✔ Answer: 3 and 5

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Problem 16:
Consecutive integers: n and n+1
Product = 3×sum -3 → n(n+1) = 3(n + n+1) -3 → n²+n = 3(2n+1)-3 → n²+n = 6n+3-3 → n²+n = 6n
→ n² -5n =0 → n(n-5)=0 → n=5
Integers: 5 and 6
Check: product=30, sum=11, 3×11-3=30 ✔
✔ Answer: 5 and 6

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Problem 17:
Consecutive even integers: n and n+2
Product = 3×sum +6 → n(n+2) = 3(n + n+2) +6 → n²+2n = 3(2n+2)+6 → n²+2n = 6n+6+6 → n²+2n = 6n+12
→ n² -4n -12=0 → (n-6)(n+2)=0 → n=6
Integers: 6 and 8
Check: product=48, sum=14, 3×14+6=48 ✔
✔ Answer: 6 and 8

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Problem 18:
Consecutive odd integers: n and n+2
Product = 2×larger +77 → n(n+2) = 2(n+2) +77 → n²+2n = 2n+4+77 → n²+2n = 2n+81
→ n² = 81 → n=9 (odd, good)
Integers: 9 and 11
Check: product=99, 2×11+77=22+77=99 ✔
✔ Answer: 9 and 11

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Problem 19:
Consecutive integers: n and n+1
Product = 3×larger +5 → n(n+1) = 3(n+1) +5 → n²+n = 3n+3+5 → n²+n = 3n+8
→ n² -2n -8=0 → (n-4)(n+2)=0 → n=4
Integers: 4 and 5
Check: product=20, 3×5+5=20 ✔
✔ Answer: 4 and 5

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Problem 20:
Three consecutive integers: n, n+1, n+2
Four times sum of all three = 2 × product of larger two
Sum = n + (n+1) + (n+2) = 3n+3
Larger two: (n+1) and (n+2), product = (n+1)(n+2)
Equation: 4(3n+3) = 2(n+1)(n+2)
→ 12n + 12 = 2(n² + 3n + 2) → 12n + 12 = 2n² + 6n + 4
→ 0 = 2n² -6n -8 → divide by 2: n² -3n -4=0 → (n-4)(n+1)=0 → n=4
Integers: 4,5,6
Check: sum=15, 4×15=60; product of larger two=5×6=30, 2×30=60 ✔
✔ Answer: 4, 5, 6

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Problem 21:
Three consecutive integers: n, n+1, n+2
Three times sum = product of larger two
Sum = 3n+3
Larger two: (n+1)(n+2)
Equation: 3(3n+3) = (n+1)(n+2)
→ 9n + 9 = n² + 3n + 2
→ 0 = n² -6n -7 → (n-7)(n+1)=0 → n=7
Integers: 7,8,9
Check: sum=24, 3×24=72; product of larger two=8×9=72 ✔
✔ Answer: 7, 8, 9

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Problem 22:
Right triangle sides: shortest = x
Medium = x + 7
Longest = 3x - 7
Pythagorean theorem: (shortest)² + (medium)² = (longest)²
→ x² + (x+7)² = (3x-7)²
Expand:
x² + x² +14x +49 = 9x² -42x +49
→ 2x² +14x +49 = 9x² -42x +49
Subtract 49 from both sides:
2x² +14x = 9x² -42x
Bring all to one side:
0 = 7x² -56x → 7x(x - 8) = 0 → x=8 (x=0 not valid)
Shortest side = 8
Check: medium=15, longest=3×8-7=17 → 8²+15²=64+225=289=17² ✔
✔ Answer: 8

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Final Answer:
1. 6 in × 12 in
2. 6 in × 24 in
3. 6 yd × 10 yd
4. 5 in × 16 in
5. 7 in × 10 in
6. 9 cm × 12 cm
7. 7 m × 12 m
8. 5 in × 10 in
9. 5 ft × 11 ft
10. 5 m × 13 m
11. 7.5 m × 8 m
12. 7 and 8
13. 9 and 11
14. 4 and 6
15. 3 and 5
16. 5 and 6
17. 6 and 8
18. 9 and 11
19. 4 and 5
20. 4, 5, 6
21. 7, 8, 9
22. 8