# WORD PROBLEMS ON RATIONAL EQUATIONS WORKSHEET

Problem 1 :

Decreasing the reciprocal of a number by 7 results ⁻²⁰⁄₃. What is the number?

Problem 2 :

7 more than three times the reciprocal of a number results ³⁸⁄₅. What is the number?

Problem 3 :

Kevin needs 150 ounces of sugar for making 50 pounds cookies. If Kevin currently has 80 ounces of sugar, how many more ounces of sugar does he need to make 30 pounds cookies?

Problem 4 :

John is able to complete a certain work in 10 days, but Peter is abobe to complete the same work in 8 days. How many days will it take them to complete the work, if they work together?

Problem 5 :

The numerator and denominator of a fraction add up to 11. If 5 be added to both numerator and denominator, the fraction becomes ¾. Find the original fraction.

Problem 6 :

Divide 15 into two parts such that both of them are positive and the difference between their reciprocals is ¼ Let x be the number.

¹⁄ₓ - 7 = ⁻²⁰⁄₃

Multiply both sides by x to get rid of the denominator x on the left side.

x(¹⁄ₓ - 7) = x(⁻²⁰⁄₃)

x(¹⁄ₓ) - x(7) = ⁻²⁰ˣ⁄₃

1 - 7x = ⁻²⁰ˣ⁄₃

Multiply both sides by 3 to get rid of the denominator 5 on the right side.

3(1 - 7x) = 3(⁻²⁰ˣ⁄₃)

3 - 21x = -20x

3 = x

The number is 3.

Let x be the number.

3(¹⁄ₓ) + 7 = ³⁸⁄₅

³⁄ₓ + 7 = ³⁸⁄₅

Multiply both sides by x to get rid of the denominator x on the left side.

x(³⁄ₓ + 7) = x(³⁸⁄₅)

x(³⁄ₓ) + x(7) = ³⁸ˣ⁄₅

3 + 7x = ³⁸ˣ⁄₅

Multiply both sides by 5 to get rid of the denominator 5 on the right side.

5(3 + 7x) = 5(³⁸ˣ⁄₅)

15 + 35x = 38x

Subtract 35x from both sides.

15 = 3x

Divide both sides by 3.

5 = x

The number is 5.

Let x be the additional ounces ofsugar needed to make 30 pounds of cookies.

150 ounces : 50 pounds = (80 + x) ounces : 30 pounds

Write each ratio in the equation above as a fraction.

¹⁵⁰⁄₅₀ = ⁽⁸⁰ ⁺ ˣ⁾⁄₃₀

3 = ⁽⁸⁰ ⁺ ˣ⁾⁄₃₀

Multiply both sides by 30 to get rid of the denominator 30 on the right side.

30(3) = 30(⁽⁸⁰ ⁺ ˣ⁾⁄₃₀)

90 = 80 + x

Subtract 80 from both sides.

10 = x

Kevin needs 10 more ounces of sugar to make 30 pounds.

Givren : John is able to complete a certain work in 10 days and Peter is abobe to complete the same work in 8 days.

Part of the work completed by John in 1 hour :

=

Part of the work completed by Peter in 1 hour :

Let x be the number of days required to complete the work, if both John and Peter work together.

Part of the work completed by both John and Peter in 1 hour :

¹⁄ₓ

¹⁄ₓ

Multiply both sides by x to get rid of the denominator on the left side.

x(¹⁄ₓ) = x( )

1 = x() + x()

1 = ˣ⁄₁₀ + ˣ⁄₈

The least common multiple of (10, 8) = 40.

Multiply both sides by 40 to get rid of the denominators 10 and 8 on the right side.

40(1) = 40(ˣ⁄₁₀ + ˣ⁄₈)

40 = 40(ˣ⁄₁₀) + 40(ˣ⁄₈)

40 = 4x + 5x

40 = 9x

Divide both sides by 9.

⁴⁰⁄₉ = x

4⁴⁄₉ = x

If both John and Peter work together, they will be able to complete the work in 4⁴⁄₉ days.

Let x be the denominator of the fraction.

Since the numerator and denominator add up to 11, the numerator is (11 - x).

Fraction = ⁽¹¹ ⁻ ˣ⁾⁄ₓ

Givren : If 5 is added to both numerator and denominator, the fraction becomes ¾.

⁽¹¹ ⁻ ˣ ⁺ ⁵⁾⁄₍ₓ ₊ ₅₎ = ¾

⁽¹⁶ ⁻ ˣ⁾⁄₍ₓ ₊ ₅₎ = ¾

By cross multiplying,

4(16 - x) = 3(x + 5)

64 - 4x = 3x + 15

64 = 7x + 15

Subtract 15 from both sides.

49 = 7x

Divide both sides by 7.

7 = x

11 - x = 11 - 7 = 4

⁽¹¹ ⁻ ˣ⁾⁄ₓ ⁴⁄₇

The required fraction is ⁴⁄₇.

Let x be one of the parts of 15.

Then the other part of 15 is (15 - x).

Given : Difference between their reciprocals of the parts of 25 is ¼.

¹⁄ₓ - ¹⁄₍₁₅ ₋ ₓ₎ = ¼

Multiply both sides by x to get rid of the denominator x on the left side.

x[¹⁄ₓ - ¹⁄₍₁₅ ₋ ₓ₎] = x(¼)

x(¹⁄ₓ) - x[¹⁄₍₁₅ ₋ ₓ₎] = ˣ⁄₄

1 - ˣ⁄₍₁₅ ₋ ₓ₎ = ˣ⁄₄

Multiply both sides by (15 - x) to get rid of the denominator (15 - x) on the left side.

(15 - x)[1 - ˣ⁄₍₁₅ ₋ ₓ₎] = (15 - x)(ˣ⁄₄)

(15 - x)(1) - (15 - x)[ˣ⁄₍₁₅ ₋ ₓ₎] = (15 - x)(ˣ⁄₄)

15 - x - x = (15 - x)(ˣ⁄₄)

15 - 2x (15 - x)(ˣ⁄₄)

Multiply both sides by 4 to get rid of the denominator 4 on the right side.

4(15 - 2x) = 4(15 - x)(ˣ⁄₄)

60 - 8x = (15 - x)(x)

60 - 8x = 15x - x2

x- 8x + 60 = 15x

Subtract 15x from both sides.

x- 23x + 60 = 0

Factor and solve.

x- 3x - 20x + 60 = 0

x(x - 3) - 20(x - 3) = 0

(x - 3)(x - 20) = 0

x - 3 = 0  or x - 20 = 0

x = 3  or  x = 20

If x = 3,

15 - x = 15 - 3

15 - x = 12

If x = 20,

15 - x = 15 - 20

15 - x = -5

When x = 20, the other part (15 - x) is -5, which is negative. So, it can be ignored.

Therefore, the two parts of 15 are 3 and 12.

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