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Linear-Equations-in-One-Variable-Word-Problems-Workbook-1 - Free Printable

Linear-Equations-in-One-Variable-Word-Problems-Workbook-1

Educational worksheet: Linear-Equations-in-One-Variable-Word-Problems-Workbook-1. Download and print for classroom or home learning activities.

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Show Answer Key & Explanations Step-by-step solution for: Linear-Equations-in-One-Variable-Word-Problems-Workbook-1
Let's solve each problem step-by-step using linear equations. We'll set up an equation for each, solve it, and explain the reasoning.

---

(1) For cleaning the car Jay was given Rs. 20. Now he has Rs. 53. How much money did he have before?



Let $ x $ = amount of money Jay had before.

He received Rs. 20, so:
$$
x + 20 = 53
$$

Subtract 20 from both sides:
$$
x = 53 - 20 = 33
$$

Answer: Jay had Rs. 33 before.

---

(2) Leena won 50 pieces of gum playing hoops. After giving some away she only has 30 remaining. How many did she give away?



Let $ x $ = number of gums given away.

She started with 50, gave away $ x $, left with 30:
$$
50 - x = 30
$$

Solve:
$$
x = 50 - 30 = 20
$$

Answer: She gave away 20 pieces.

---

(3) A recipe calls for $ 4\frac{1}{10} $ cups of milk. Ravi put in $ 5\frac{2}{9} $ cups. How many extra cups did he put in?



Convert mixed numbers to improper fractions:

- $ 4\frac{1}{10} = \frac{41}{10} $
- $ 5\frac{2}{9} = \frac{47}{9} $

Extra milk = $ \frac{47}{9} - \frac{41}{10} $

Find LCM of 9 and 10 = 90

$$
\frac{47}{9} = \frac{47 \times 10}{90} = \frac{470}{90},\quad \frac{41}{10} = \frac{41 \times 9}{90} = \frac{369}{90}
$$

$$
\text{Extra} = \frac{470 - 369}{90} = \frac{101}{90} = 1\frac{11}{90}
$$

Answer: He put in $ 1\frac{11}{90} $ extra cups.

---

(4) Last week Anand ran 19 miles less than Deepak. Anand ran 14 miles. How many miles did Deepak run?



Let $ x $ = miles Deepak ran.

Anand ran 19 less:
$$
x - 19 = 14
$$

Add 19 to both sides:
$$
x = 14 + 19 = 33
$$

Answer: Deepak ran 33 miles.

---

(5) Asha was 43 years old thirteen years ago. How old is she now?



Let $ x $ = her current age.

Thirteen years ago: $ x - 13 = 43 $

$$
x = 43 + 13 = 56
$$

Answer: Asha is 56 years old now.

---

(6) Envelopes cost Rs. 5 per piece. How many pieces did Khushi buy if she spent Rs. 50?



Let $ x $ = number of envelopes.

Total cost: $ 5x = 50 $

Divide both sides by 5:
$$
x = \frac{50}{5} = 10
$$

Answer: She bought 10 envelopes.

---

(7) Ayan and his best friend split money evenly, each getting Rs. 30. How much did they find?



Each got Rs. 30 → total money found:
$$
30 + 30 = 60
$$

Or: Let $ x $ = total money. Then:
$$
\frac{x}{2} = 30 \Rightarrow x = 60
$$

Answer: They found Rs. 60.

---

(8) The wind blew away 14 muffins, which was $ \frac{2}{5} $ of all. How many did you start with?



Let $ x $ = total muffins.

$$
\frac{2}{5}x = 14
$$

Multiply both sides by 5:
$$
2x = 70 \Rightarrow x = 35
$$

Answer: You started with 35 muffins.

---

(9) Ten markers cost Rs. 80. Find price of one marker.



Let $ x $ = price of one marker.

$$
10x = 80 \Rightarrow x = \frac{80}{10} = 8
$$

Answer: One marker costs Rs. 8.

---

(10) If the weight of a package multiplied by $ \frac{2}{3} $ is 8 pounds, find the weight.



Let $ x $ = weight of package.

$$
\frac{2}{3}x = 8
$$

Multiply both sides by $ \frac{3}{2} $:
$$
x = 8 \times \frac{3}{2} = 12
$$

Answer: The package weighs 12 pounds.

---

(11) Dilip and ten friends went out to eat. They split the bill evenly. Each paid Rs. 19. What was the total bill?



Number of people = 1 (Dilip) + 10 friends = 11

Each paid Rs. 19 → total bill:
$$
11 \times 19 = 209
$$

Or: Let $ x $ = total bill.
$$
\frac{x}{11} = 19 \Rightarrow x = 11 \times 19 = 209
$$

Answer: Total bill was Rs. 209.

---

Final Answers Summary:



| Q | Answer |
|---|--------|
| 1 | Rs. 33 |
| 2 | 20 pieces |
| 3 | $ 1\frac{11}{90} $ cups |
| 4 | 33 miles |
| 5 | 56 years old |
| 6 | 10 envelopes |
| 7 | Rs. 60 |
| 8 | 35 muffins |
| 9 | Rs. 8 |
|10 | 12 pounds |
|11 | Rs. 209 |

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