Algebra and equations · Ages 10–11

Equations and setting them up worksheet

Part of: First algebra and simple equations

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Algebra and equations · Ages 10–11

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Write each answer in the space. For word problems, show your working and write a sentence answer.

1.Solve 84 = r − 55: r =
Answer:
2.Ka-yee had some stickers and buys 5 packs of 12, making 115 in all. With x for the stickers at first, the equation is x + 5 × 12 = 115. What is x?
Answer:
3.May had some stickers and buys 6 packs of 9, making 85 in all. With x for the stickers at first, the equation is x + 6 × 9 = 85. What is x?
Answer:
4.Solve 32 = r − 28: r =
Answer:
5.Given k + 3 = 17, the value of 8k is
Answer:
6.Given k + 10 = 30, the value of 9k is
Answer:

13.A box of cookies is shared equally among 2 children. After Tsz-hin gets a share, Mum gives Tsz-hin 6 more, and Tsz-hin now has 11. How many cookies were in the box?
Working:
Answer:
14.Dad is 55, which is 7 more than 4 times May's age. With x for May's age, which equation is right?
- A. x + 4 + 7 = 55
- B. 4x − 7 = 55
- C. 4(x + 7) = 55
- D. 4x + 7 = 55
Choose the right answer
Working:
Answer:
15.Ming buys 3 croissants at the same price, pays $44 and gets $14 change. How many dollars is each?
Working:
Answer:
16.A rectangle's perimeter (all four sides added) is 44 cm and its length 13 cm. With x cm for the width, which equation is right?
- A. 13 × x = 44
- B. 2 × (13 + x) = 44
- C. 2 × 13 + x = 44
- D. 13 + x = 44
Choose the right answer
Working:
Answer:
17.Ka-yan cuts a ribbon into 3 equal pieces, then cuts 15 cm off one piece, leaving 24 cm of it. With x cm for the ribbon's length, which equation is right?
- A. 3x − 15 = 24
- B. x − 15 = 24
- C. x3 + 15 = 24
- D. x3 − 15 = 24
Choose the right answer
Working:
Answer:
18.Ka-lok buys 5 boxes of the same chocolates, eats 18 and has 77 left. How many chocolates are in each box?
Working:
Answer:

Show the answers and working

Equations and setting them up worksheet: Answers and working

1. 139
The r being on the right doesn't matter: r minus 55 is 84. Add 55 to both sides: what do you get?
With the unknown on the right it works the same: add 55 to both sides, 84 + 55 = r, so r = 139. Check: 139 − 55 = 84.
2. 55
Work out the stickers bought first: 5 × 12. x plus that makes 115, so how do you get x?
x = 115 − 5 × 12 = 55; check: 55 + 5 × 12 = 115.
3. 31
Work out the stickers bought first: 6 × 9. x plus that makes 85, so how do you get x?
x = 85 − 6 × 9 = 31; check: 31 + 6 × 9 = 85.
4. 60
The r being on the right doesn't matter: r minus 28 is 32. Add 28 to both sides: what do you get?
With the unknown on the right it works the same: add 28 to both sides, 32 + 28 = r, so r = 60. Check: 60 − 28 = 32.
5. 112
The question asks for 8k, not k. Find k from k + 3 = 17 first, then multiply by 8.
Solve the equation first: take 3 from both sides, k = 17 − 3 = 14; then substitute: 8k = 8 × 14 = 112.
6. 180
The question asks for 9k, not k. Find k from k + 10 = 30 first, then multiply by 9.
Solve the equation first: take 10 from both sides, k = 30 − 10 = 20; then substitute: 9k = 9 × 20 = 180.

7. True
Substitute x = 8: work out 3 × 8 + 10, then see whether it makes 34.
Check by substituting: 3 × 8 + 10, is it 34? The solution is x = 8.
8. B. 2x + 6 = 16
An equation needs two things at once: an unknown (a letter) and an equals sign. Check each one.
An equation is an equality with an unknown: it needs a letter and an equals sign.
9. False
Substitute x = 3: work out 7 × 3 + 10, then see whether it makes 38.
Check by substituting: 7 × 3 + 10, is it 38? The solution is x = 4.
10. C. 4x + 18 = 38
First think how these fit together: the cost of 4 notebooks, the $18 change and the $38 paid.
The notebooks plus the change make what was paid: 4x + 18 = 38.
11. D. x + 6 = 21
Put x = 15 into each equation in turn and see which one balances.
Substitute x = 15 into each: only 15 + 6 = 21 holds.
12. D. x + 5 = 11
Put x = 6 into each equation in turn and see which one balances.
Substitute x = 6 into each: only 6 + 5 = 11 holds.

Marking: 2 for the working, 1 for the answer, 1 for the sentence.

13. 10 cookies
Let the box hold x. Shared among 2, each gets x2; plus 6 more makes 11.
Take 6 from both sides first, then multiply both sides by 2.
Let the box hold x, so each share is x2: x2 + 6 = 11. Take 6 from both sides: x2 = 5; multiply both sides by 2: x = 10. Check: 10 ÷ 2 + 6 = 11.
14. D. 4x + 7 = 55
Use x to write “7 more than 4 times May's age” first. Whose age should that equal?
4 times May's age plus 7 is Dad's age: 4x + 7 = 55, x = 12.
15. 10 dollars
Let each cost x dollars. The 3 croissants plus the $14 change make the $44 paid.
Let each be x: 3x + 14 = 44, 3x = 44 − 14, x = 10. Check: 3 × 10 + 14 = 44.
16. B. 2 × (13 + x) = 44
Perimeter = 2 × (length + width). Put in the length 13 and the width x, and see what it must equal.
Perimeter = 2 × (length + width): 2 × (13 + x) = 44, so x = 9.
17. D. x3 − 15 = 24
First write one piece's length with x: cut into 3 equal pieces, how long is each? Then think what cutting 15 cm off does.
Each piece is x3 cm; cutting 15 cm off leaves 24 cm: x3 − 15 = 24, so x = 117.
18. 19 chocolates
Let each box hold x. 5 boxes hold 5x; after eating 18 there are 77: write the equation first.
A two-step equation: undo the subtraction of 18 first, then divide by 5.
Let each box hold x: 5x − 18 = 77. Add 18 to both sides: 5x = 95; divide both sides by 5: x = 19. Check: 5 × 19 − 18 = 77.

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