Linear Equations in One Variable (Word Problems) activity | Live ... - Free Printable

Let's solve each problem step by step.

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Problem 1:

Statement: I am currently 5 times as old as my son. In 6 years time, I will be three times as old as he will be then. What are our ages now?

#### Solution:
Let the son's current age be \( x \) years.
Then, the father's current age is \( 5x \) years.

In 6 years:
- Son's age will be \( x + 6 \).
- Father's age will be \( 5x + 6 \).

According to the problem, in 6 years, the father will be three times as old as the son:
\[
5x + 6 = 3(x + 6)
\]

Solve for \( x \):
\[
5x + 6 = 3x + 18
\]
\[
5x - 3x = 18 - 6
\]
\[
2x = 12
\]
\[
x = 6
\]

So, the son's current age is \( x = 6 \) years.
The father's current age is \( 5x = 5 \times 6 = 30 \) years.

#### Answer:
Father: \( 30 \) years, Son: \( 6 \) years

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Problem 2:

Statement: I have Rs 1000 in ten and five rupee notes. If the number of ten rupee notes that I have is ten more than the number of five rupee notes, how many notes do I have in each denomination?

#### Solution:
Let the number of five rupee notes be \( y \).
Then, the number of ten rupee notes is \( y + 10 \).

The total value of the notes is Rs 1000:
\[
5y + 10(y + 10) = 1000
\]

Simplify and solve for \( y \):
\[
5y + 10y + 100 = 1000
\]
\[
15y + 100 = 1000
\]
\[
15y = 900
\]
\[
y = 60
\]

So, the number of five rupee notes is \( y = 60 \).
The number of ten rupee notes is \( y + 10 = 60 + 10 = 70 \).

#### Answer:
Five rupee notes: \( 60 \), Ten rupee notes: \( 70 \)

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Problem 3:

Statement: There are 180 multiple-choice questions in a test. If a candidate gets 4 marks for every correct answer and for every unattempted or wrongly answered question, one mark is deducted from the total score of correct answers. If a candidate scored 450 marks in the test, how many questions did he answer correctly?

#### Solution:
Let the number of correct answers be \( c \).
The total number of questions is 180, so the number of unattempted or wrongly answered questions is \( 180 - c \).

The scoring system is:
- \( 4c \) marks for correct answers.
- \( -(180 - c) \) marks for unattempted or wrongly answered questions.

The total score is given as 450:
\[
4c - (180 - c) = 450
\]

Simplify and solve for \( c \):
\[
4c - 180 + c = 450
\]
\[
5c - 180 = 450
\]
\[
5c = 630
\]
\[
c = 126
\]

#### Answer:
The candidate answered \( 126 \) questions correctly.

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Problem 4:

Statement: A labourer is engaged for 20 days on the condition that he will receive Rs 60 for each day he works and he will be fined Rs 5 for each day he is absent. If he receives Rs 745 in all, for how many days he remained absent?

#### Solution:
Let the number of days the labourer worked be \( w \).
Then, the number of days he was absent is \( 20 - w \).

The total earnings are calculated as:
\[
60w - 5(20 - w) = 745
\]

Simplify and solve for \( w \):
\[
60w - 100 + 5w = 745
\]
\[
65w - 100 = 745
\]
\[
65w = 845
\]
\[
w = 13
\]

So, the labourer worked for 13 days.
The number of days he was absent is:
\[
20 - w = 20 - 13 = 7
\]

#### Answer:
The labourer remained absent for \( 7 \) days.

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Problem 5:

Statement: Ravish has three boxes whose total weight is \( 60 \frac{1}{2} \) kg. Box B weighs \( 3 \frac{1}{2} \) kg more than box A, and box C weighs \( 5 \frac{1}{3} \) kg more than box B. Find the weight of box A.

#### Solution:
Let the weight of box A be \( a \) kg.
Then, the weight of box B is \( a + 3 \frac{1}{2} = a + \frac{7}{2} \) kg.
The weight of box C is \( (a + \frac{7}{2}) + 5 \frac{1}{3} = a + \frac{7}{2} + \frac{16}{3} \) kg.

The total weight of the three boxes is \( 60 \frac{1}{2} = \frac{121}{2} \) kg:
\[
a + \left(a + \frac{7}{2}\right) + \left(a + \frac{7}{2} + \frac{16}{3}\right) = \frac{121}{2}
\]

Combine like terms:
\[
a + a + \frac{7}{2} + a + \frac{7}{2} + \frac{16}{3} = \frac{121}{2}
\]
\[
3a + \frac{7}{2} + \frac{7}{2} + \frac{16}{3} = \frac{121}{2}
\]
\[
3a + \frac{14}{2} + \frac{16}{3} = \frac{121}{2}
\]
\[
3a + 7 + \frac{16}{3} = \frac{121}{2}
\]

Convert 7 to a fraction with a denominator of 3:
\[
7 = \frac{21}{3}
\]
\[
3a + \frac{21}{3} + \frac{16}{3} = \frac{121}{2}
\]
\[
3a + \frac{37}{3} = \frac{121}{2}
\]

Eliminate the fractions by finding a common denominator (6):
\[
3a + \frac{37}{3} = \frac{121}{2}
\]
\[
3a + \frac{37 \times 2}{6} = \frac{121 \times 3}{6}
\]
\[
3a + \frac{74}{6} = \frac{363}{6}
\]
\[
3a = \frac{363}{6} - \frac{74}{6}
\]
\[
3a = \frac{289}{6}
\]
\[
a = \frac{289}{18}
\]

#### Answer:
The weight of box A is \( \frac{289}{18} \) kg.

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Problem 6:

Statement: The numerator of a rational number is 3 less than the denominator. If the denominator is increased by 5 and the numerator by 2, we get the rational number \( \frac{1}{2} \). Find the rational number.

#### Solution:
Let the denominator be \( d \).
Then, the numerator is \( d - 3 \).
The rational number is \( \frac{d - 3}{d} \).

According to the problem, if the denominator is increased by 5 and the numerator by 2, the new rational number is \( \frac{1}{2} \):
\[
\frac{(d - 3) + 2}{d + 5} = \frac{1}{2}
\]
\[
\frac{d - 1}{d + 5} = \frac{1}{2}
\]

Cross-multiply to solve for \( d \):
\[
2(d - 1) = 1(d + 5)
\]
\[
2d - 2 = d + 5
\]
\[
2d - d = 5 + 2
\]
\[
d = 7
\]

So, the denominator is \( d = 7 \).
The numerator is \( d - 3 = 7 - 3 = 4 \).

The rational number is:
\[
\frac{4}{7}
\]

#### Answer:
The rational number is \( \frac{4}{7} \).

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Problem 7:

Statement: The sum of a two-digit number and the number obtained by reversing its digits is 121. Find the number if its unit place digit is 5.

#### Solution:
Let the two-digit number be \( 10x + 5 \), where \( x \) is the tens digit and 5 is the units digit.
The number obtained by reversing the digits is \( 50 + x \).

According to the problem:
\[
(10x + 5) + (50 + x) = 121
\]

Simplify and solve for \( x \):
\[
10x + 5 + 50 + x = 121
\]
\[
11x + 55 = 121
\]
\[
11x = 66
\]
\[
x = 6
\]

So, the tens digit is \( x = 6 \).
The number is:
\[
10x + 5 = 10 \times 6 + 5 = 65
\]

#### Answer:
The number is \( 65 \).

---

Problem 8:

Statement: The denominator of a fraction is greater than the numerator by 8. If the numerator is increased by 17 and the denominator is decreased by 1, the number obtained is \( \frac{3}{2} \). Find the fraction.

#### Solution:
Let the numerator be \( n \).
Then, the denominator is \( n + 8 \).
The fraction is \( \frac{n}{n + 8} \).

According to the problem, if the numerator is increased by 17 and the denominator is decreased by 1, the new fraction is \( \frac{3}{2} \):
\[
\frac{n + 17}{n + 8 - 1} = \frac{3}{2}
\]
\[
\frac{n + 17}{n + 7} = \frac{3}{2}
\]

Cross-multiply to solve for \( n \):
\[
2(n + 17) = 3(n + 7)
\]
\[
2n + 34 = 3n + 21
\]
\[
34 - 21 = 3n - 2n
\]
\[
n = 13
\]

So, the numerator is \( n = 13 \).
The denominator is \( n + 8 = 13 + 8 = 21 \).

The fraction is:
\[
\frac{13}{21}
\]

#### Answer:
The fraction is \( \frac{13}{21} \).

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Final Answers:

1. Father: \( 30 \) years, Son: \( 6 \) years
2. Five rupee notes: \( 60 \), Ten rupee notes: \( 70 \)
3. Correct answers: \( 126 \)
4. Absent days: \( 7 \)
5. Weight of box A: \( \frac{289}{18} \) kg
6. Rational number: \( \frac{4}{7} \)
7. Number: \( 65 \)
8. Fraction: \( \frac{13}{21} \)

\[
\boxed{30, 6, 60, 70, 126, 7, \frac{289}{18}, \frac{4}{7}, 65, \frac{13}{21}}
\]