Linear Equations in One Variable (Word Problems) interactive ... - Free Printable

Let's solve each problem step by step.

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Problem 1:

The numerator of a fraction is 6 less than the denominator. If 3 is added to the numerator, the fraction is equal to \( \frac{2}{3} \). What is the original fraction?

#### Solution:
1. Let the numerator be \( x \) and the denominator be \( y \).
2. According to the problem:
- The numerator is 6 less than the denominator: \( x = y - 6 \).
- If 3 is added to the numerator, the fraction becomes \( \frac{2}{3} \): \( \frac{x + 3}{y} = \frac{2}{3} \).

3. Substitute \( x = y - 6 \) into the second equation:
\[
\frac{(y - 6) + 3}{y} = \frac{2}{3}
\]
Simplify the numerator:
\[
\frac{y - 3}{y} = \frac{2}{3}
\]

4. Cross-multiply to solve for \( y \):
\[
3(y - 3) = 2y
\]
\[
3y - 9 = 2y
\]
\[
3y - 2y = 9
\]
\[
y = 9
\]

5. Substitute \( y = 9 \) back into \( x = y - 6 \):
\[
x = 9 - 6 = 3
\]

6. The original fraction is:
\[
\frac{x}{y} = \frac{3}{9} = \frac{1}{3}
\]

Answer:
\[
\boxed{\frac{1}{3}}
\]

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Problem 2:

A sum of Rs 800 is in the form of denominations of Rs 10 and Rs 20. If the total number of notes be 50, find the number of notes of ₹10.

#### Solution:
1. Let the number of Rs 10 notes be \( x \) and the number of Rs 20 notes be \( y \).
2. According to the problem:
- The total number of notes is 50: \( x + y = 50 \).
- The total amount is Rs 800: \( 10x + 20y = 800 \).

3. Simplify the second equation by dividing everything by 10:
\[
x + 2y = 80
\]

4. Now we have the system of equations:
\[
x + y = 50
\]
\[
x + 2y = 80
\]

5. Subtract the first equation from the second:
\[
(x + 2y) - (x + y) = 80 - 50
\]
\[
x + 2y - x - y = 30
\]
\[
y = 30
\]

6. Substitute \( y = 30 \) back into \( x + y = 50 \):
\[
x + 30 = 50
\]
\[
x = 20
\]

7. The number of Rs 10 notes is:
\[
x = 20
\]

Answer:
\[
\boxed{20}
\]

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Problem 3:

Seeta Devi has Rs 9 in fifty-paise and twenty-five-paise coins. She has twice as many twenty-five-paise coins as she has fifty-paise coins. How many coins of each kind does she have?

#### Solution:
1. Let the number of fifty-paise coins be \( x \).
2. According to the problem:
- She has twice as many twenty-five-paise coins as fifty-paise coins: the number of twenty-five-paise coins is \( 2x \).
- The total value of the coins is Rs 9. Since fifty-paise is \( \frac{1}{2} \) rupee and twenty-five-paise is \( \frac{1}{4} \) rupee, the total value equation is:
\[
0.5x + 0.25(2x) = 9
\]

3. Simplify the equation:
\[
0.5x + 0.5x = 9
\]
\[
x = 9
\]

4. The number of fifty-paise coins is \( x = 9 \).
5. The number of twenty-five-paise coins is \( 2x = 2 \times 9 = 18 \).

Answer:
\[
\boxed{9 \text{ fifty-paise coins, } 18 \text{ twenty-five-paise coins}}
\]

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Problem 4:

Sunita is twice as old as Ashima. If six years is subtracted from Ashima’s age and four years added to Sunita’s age, then Sunita will be four times Ashima’s age. How old were they two years ago?

#### Solution:
1. Let Ashima's current age be \( x \) and Sunita's current age be \( 2x \).
2. According to the problem:
- Six years subtracted from Ashima's age: \( x - 6 \).
- Four years added to Sunita's age: \( 2x + 4 \).
- At that time, Sunita's age will be four times Ashima's age:
\[
2x + 4 = 4(x - 6)
\]

3. Solve the equation:
\[
2x + 4 = 4x - 24
\]
\[
4 + 24 = 4x - 2x
\]
\[
28 = 2x
\]
\[
x = 14
\]

4. Ashima's current age is \( x = 14 \).
5. Sunita's current age is \( 2x = 2 \times 14 = 28 \).

6. Two years ago:
- Ashima's age was \( 14 - 2 = 12 \).
- Sunita's age was \( 28 - 2 = 26 \).

Answer:
\[
\boxed{26 \text{ yrs (Sunita), } 12 \text{ yrs (Ashima)}}
\]

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Problem 5:

The ages of Sonu and Monu are in the ratio 7:5. Ten years hence, the ratio of their ages will be 9:7. Find their present ages.

#### Solution:
1. Let Sonu's current age be \( 7x \) and Monu's current age be \( 5x \).
2. According to the problem:
- Ten years hence, Sonu's age will be \( 7x + 10 \) and Monu's age will be \( 5x + 10 \).
- The ratio of their ages ten years hence will be 9:7:
\[
\frac{7x + 10}{5x + 10} = \frac{9}{7}
\]

3. Cross-multiply to solve for \( x \):
\[
7(7x + 10) = 9(5x + 10)
\]
\[
49x + 70 = 45x + 90
\]
\[
49x - 45x = 90 - 70
\]
\[
4x = 20
\]
\[
x = 5
\]

4. Sonu's current age is \( 7x = 7 \times 5 = 35 \).
5. Monu's current age is \( 5x = 5 \times 5 = 25 \).

Answer:
\[
\boxed{35 \text{ yrs (Sonu), } 25 \text{ yrs (Monu)}}
\]

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Problem 6:

Five years ago, a man was seven times as old as his son. Five years hence, the father will be three times as old as his son. Find their present ages.

#### Solution:
1. Let the present age of the father be \( F \) and the present age of the son be \( S \).
2. According to the problem:
- Five years ago, the father was seven times as old as his son:
\[
F - 5 = 7(S - 5)
\]
- Five years hence, the father will be three times as old as his son:
\[
F + 5 = 3(S + 5)
\]

3. Solve the first equation:
\[
F - 5 = 7(S - 5)
\]
\[
F - 5 = 7S - 35
\]
\[
F = 7S - 30
\]

4. Solve the second equation:
\[
F + 5 = 3(S + 5)
\]
\[
F + 5 = 3S + 15
\]
\[
F = 3S + 10
\]

5. Equate the two expressions for \( F \):
\[
7S - 30 = 3S + 10
\]
\[
7S - 3S = 10 + 30
\]
\[
4S = 40
\]
\[
S = 10
\]

6. Substitute \( S = 10 \) back into \( F = 3S + 10 \):
\[
F = 3(10) + 10 = 30 + 10 = 40
\]

7. The present ages are:
- Father: \( F = 40 \)
- Son: \( S = 10 \)

Answer:
\[
\boxed{40 \text{ yrs (Father), } 10 \text{ yrs (Son)}}
\]

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Problem 7:

At a party, colas, squash, and fruit juice were offered to guests. A fourth of the guests drank colas, a third drank squash, two fifths drank fruit juice, and just three did not drink anything. How many guests were in all?

#### Solution:
1. Let the total number of guests be \( x \).
2. According to the problem:
- A fourth of the guests drank colas: \( \frac{x}{4} \).
- A third of the guests drank squash: \( \frac{x}{3} \).
- Two fifths of the guests drank fruit juice: \( \frac{2x}{5} \).
- Three guests did not drink anything.

3. The sum of all these fractions plus the three guests who did not drink should equal the total number of guests:
\[
\frac{x}{4} + \frac{x}{3} + \frac{2x}{5} + 3 = x
\]

4. Find a common denominator for the fractions. The least common multiple of 4, 3, and 5 is 60:
\[
\frac{x}{4} = \frac{15x}{60}, \quad \frac{x}{3} = \frac{20x}{60}, \quad \frac{2x}{5} = \frac{24x}{60}
\]

5. Substitute these into the equation:
\[
\frac{15x}{60} + \frac{20x}{60} + \frac{24x}{60} + 3 = x
\]
\[
\frac{15x + 20x + 24x}{60} + 3 = x
\]
\[
\frac{59x}{60} + 3 = x
\]

6. Subtract \( \frac{59x}{60} \) from both sides:
\[
3 = x - \frac{59x}{60}
\]
\[
3 = \frac{60x}{60} - \frac{59x}{60}
\]
\[
3 = \frac{x}{60}
\]

7. Solve for \( x \):
\[
x = 3 \times 60 = 180
\]

Answer:
\[
\boxed{180}
\]

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Final Answers:

1. \(\boxed{\frac{1}{3}}\)
2. \(\boxed{20}\)
3. \(\boxed{9 \text{ fifty-paise coins, } 18 \text{ twenty-five-paise coins}}\)
4. \(\boxed{26 \text{ yrs (Sunita), } 12 \text{ yrs (Ashima)}}\)
5. \(\boxed{35 \text{ yrs (Sonu), } 25 \text{ yrs (Monu)}}\)
6. \(\boxed{40 \text{ yrs (Father), } 10 \text{ yrs (Son)}}\)
7. \(\boxed{180}\)