**Problem 1 :**

The denominator of a fraction exceeds the numerator by 5. If 3 be added to both, the fraction becomes 3/4. Find the fraction.

**Solution :**

Let x/y be the required fraction.

The denominator of the fraction exceeds the numerator.

Then,

y = x + 5 -----(1)

If 3 be added to both, the fraction becomes 3 / 4

From the above information, we have

(x + 3) / (y + 3) = 3 / 4

(x + 3) / (x + 5 + 3) = 3 / 4

Simplify.

(x + 3) / (x + 8) = 3/4

4(x + 3) = 3(x + 8)

4x + 12 = 3x + 24

x = 12

Plug x = 12 in (1)

(1)-----> y = 12 + 5 = 17

So, the required fraction is 12/17.

**Problem 2 :**

The denominator of the fraction is 5 more than its numerator. The sum of the fraction and its reciprocal is 73/24. Find the fraction.

**Solution :**

Let x/y be the required fraction.

The denominator of the fraction is 5 more than its numerator.

Then,

y = x + 5

x / y = x / (x + 5)

The sum of the fraction and its reciprocal is 73/24.

x / (x + 5) + (x + 5) / x = 73 / 24

x^{2} + (x + 5)^{2} / x(x + 5) = 73/24

(2x^{2} + 10x + 25)/(x^{2 }+ 5x) = 73/24

24(2x^{2} + 10x + 25) = 73(x^{2} + 5x)

48x^{2} + 240x + 600 = 73x^{2} + 365x

73x² – 48x² + 365x – 240x – 600 = 0

25x^{2} + 125x – 600 = 0

x^{2} + 5x –
24 = 0

(x + 8) (x – 3) = 0

x + 8 = 0 x = -8 |
x - 3 = 0 x = 3 |

By applying the value of x in (1), we get

y = x + 5

y = 3 + 5

y = 8

So, the required fraction is 3/8.

**Problem 3 :**

The denominator of the fraction is one more than twice the numerator. If the sum of the fraction and its reciprocal is 58/21, find the fraction.

**Solution :**

Let x/y be the required fraction.

The denominator is one more than twice the numerator.

Then,

y = 2x + 1

x/y = x/(2 x + 1)

The sum of the fraction and its reciprocal is 58/21

x/(2x + 1) + (2x + 1)/x = 58/21

x^{2} + (2x + 1)^{2} / x(2x + 1) = 58/21

Simplifying the numerator :

x^{2} + (2x + 1)^{2 } = x^{2 }+ 4x^{2} + 4x + 1

x^{2} + (2x + 1)^{2} = 5x^{2} + 4x + 1

Simplifying the numerator :

x(2x + 1) = 2x^{2 }+ x

(5x^{2} + 4x + 1) / (2x^{2 }+ x) = 58/21

21(5x^{2} + 4x + 1) = 58(2x^{2 }+ x)

105x^{2} + 84x + 21 = 116x^{2} + 58x

116x^{2} - 105x^{2 }+ 58x - 84x - 21 = 0

11x^{2} - 26x - 21 = 0

(x – 3) (11 x + 7) = 0

x - 3 = 0 x = 3 |
11x + 7 = 0 x = -7/11 |

When x = 3,

y = 2x + 1

y = 2(3) + 1

y = 7

So, the required fraction is 3/7

**Problem 4 : **

If the numerator of a fraction is increased by 2 and the denominator by 1, it becomes 1. In case, the numerator is decreased by 4 and the denominator by 2, it becomes 1/2. Find the fraction.

**Solution :**

Let x/y be the required fraction.

If the numerator is increased by 2 and the denominator by 1, the fraction becomes 1.

Then,

(x + 2) / (y + 1) = 1

Simplify.

(x + 2) / (y + 1) = 1

x + 2 = y + 1

x - y = - 1 ------(1)

In case the numerator is decreased by 4 and the denominator by 2, the fraction becomes 1/2.

Then,

(x - 4) / (y - 2) = 1 / 2

Simplify.

2(x - 4) = 1(y - 2)

2x - 8 = y - 2

2x - y = 6 ------(2)

Solving (1) and (2), we get

x = 7 and y = 8

So, the required fraction is 7/8.

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