Algebra (Word Problems) worksheet - Free Printable

Let's solve each problem step by step:

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Problem 1:

Think of a number, add 2 to it and then multiply the sum by 6, the result is 42. Number = ?

#### Solution:
Let the number be \( x \).

1. Add 2 to the number: \( x + 2 \).
2. Multiply the sum by 6: \( 6(x + 2) \).
3. The result is 42: \( 6(x + 2) = 42 \).

Solve for \( x \):
\[
6(x + 2) = 42
\]
Divide both sides by 6:
\[
x + 2 = 7
\]
Subtract 2 from both sides:
\[
x = 5
\]

Answer: \( \boxed{5} \)

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Problem 2:

The side of a regular hexagon is \( s \) cm. Find its perimeter.

#### Solution:
A regular hexagon has 6 equal sides. If each side is \( s \) cm, the perimeter is:
\[
\text{Perimeter} = 6s
\]

Answer: \( \boxed{6s} \)

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Problem 3:

The number of bacteria in a culture is \( x \) now. It becomes the square of itself after one week. What will be its number after two weeks?

#### Solution:
1. After one week, the number of bacteria becomes \( x^2 \).
2. After another week, the number of bacteria becomes the square of \( x^2 \):
\[
(x^2)^2 = x^4
\]

Answer: \( \boxed{x^4} \)

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Problem 4:

If there are \( x \) rows of chairs and each row contains \( x^2 \) chairs, determine the total number of chairs.

#### Solution:
The total number of chairs is the product of the number of rows and the number of chairs per row:
\[
\text{Total chairs} = x \times x^2 = x^3
\]

Answer: \( \boxed{x^3} \)

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Problem 5:

The length of a rectangle is \( y \) times its breadth \( x \). The area of the rectangle is \( \square \).

#### Solution:
1. Let the breadth be \( x \).
2. The length is \( y \times x = yx \).
3. The area of the rectangle is:
\[
\text{Area} = \text{length} \times \text{breadth} = yx \times x = yx^2
\]

Answer: \( \boxed{yx^2} \)

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Problem 6:

The length and breadth of a room are \( 3x^2y^3 \) and \( 6x^3y^2 \). Find its area.

#### Solution:
The area of the room is the product of its length and breadth:
\[
\text{Area} = (3x^2y^3) \times (6x^3y^2)
\]
Multiply the coefficients and the variables separately:
\[
\text{Coefficients: } 3 \times 6 = 18
\]
\[
\text{Variables: } x^2 \times x^3 = x^{2+3} = x^5
\]
\[
y^3 \times y^2 = y^{3+2} = y^5
\]
Thus, the area is:
\[
\text{Area} = 18x^5y^5
\]

Answer: \( \boxed{18x^5y^5} \)

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Problem 7:

What is the value of \( x + x + x + \ldots \) (\( y \) times)?

#### Solution:
Adding \( x \) \( y \) times is equivalent to multiplying \( x \) by \( y \):
\[
x + x + x + \ldots \text{ (y times)} = y \times x = xy
\]

Answer: \( \boxed{xy} \)

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Problem 8:

A chair costs ₹\( x \). The cost of \( x^2y \) chairs is = ₹ \( \square \).

#### Solution:
The cost of one chair is ₹\( x \). Therefore, the cost of \( x^2y \) chairs is:
\[
\text{Cost} = x \times x^2y = x^3y
\]

Answer: \( \boxed{x^3y} \)

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Problem 9:

A man spends ₹\( x \) per week. The total money spent by him in \( xy^2 \) weeks is = ₹ \( \square \).

#### Solution:
If the man spends ₹\( x \) per week, the total money spent in \( xy^2 \) weeks is:
\[
\text{Total money} = x \times xy^2 = x^2y^2
\]

Answer: \( \boxed{x^2y^2} \)

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Problem 10:

There are 47 boys in the class. This is three more than four times the number of girls. How many girls are there in the class?

#### Solution:
Let the number of girls be \( g \).

1. Four times the number of girls is \( 4g \).
2. Three more than four times the number of girls is \( 4g + 3 \).
3. This is equal to the number of boys, which is 47:
\[
4g + 3 = 47
\]

Solve for \( g \):
\[
4g + 3 = 47
\]
Subtract 3 from both sides:
\[
4g = 44
\]
Divide both sides by 4:
\[
g = 11
\]

Answer: \( \boxed{11} \)

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Final Answers:

1. \( \boxed{5} \)
2. \( \boxed{6s} \)
3. \( \boxed{x^4} \)
4. \( \boxed{x^3} \)
5. \( \boxed{yx^2} \)
6. \( \boxed{18x^5y^5} \)
7. \( \boxed{xy} \)
8. \( \boxed{x^3y} \)
9. \( \boxed{x^2y^2} \)
10. \( \boxed{11} \)