WORD PROBLEMS ON FRACTIONS

Problem 1 :

A cookie factory uses ¼ of a barrel of oatmeal in each batch of cookies. The factory used ½ of a barrel of oatmeal yesterday. How many batches of cookies did the factory make ?

Solution :

Amount of oatmeal used in each batch of cookies is

= ¼ of a barrel

No. of cookies made yesterday is

= Oatmeal used yesterday/Oatmeal used in each batch

= ½ ÷ ¼

= ½ ⋅ ⁴⁄₁

= 2

The factory made 2 batches of cookies yesterday.

Problem 2 :

Of the students in the band, ¼ play the flute and another play the clarinet. What fraction of the students in the band play either the flute or the clarinet ?

Solution :

Fraction of the students in the band play either the flute or the clarinet :

¼ + 

Least common  multiple of the denominators (4, 10) is 20.

Make each denominator as 20 by multiplying the numerator and denominator of the first fraction by 5 and the second fraction by 2.

⁽¹ ˣ ⁵⁾⁄₍₄ ₓ ₅₎ + ⁽¹ ˣ ²⁾⁄₍₁₀ ₓ ₂₎

⁵⁄₂₀²⁄₂₀

= ⁽⁵ ⁺ ²⁾⁄₂₀

⁷⁄₂₀

Fraction of the students who play either the flute or the clarinet is ⁷⁄₂₀.

Problem 3 :

Lily rode her bicycle ⁵⁄₄ miles from her house to the park. Then she rode ³⁄₂ miles from the park to the library. How many miles did Lily ride in all ?

Solution :

Total number of miles that Lily rode in all is

= ⁵⁄₄ + ³⁄₂

Least common  multiple of the denominators (4, 2) is 4.

Make each denominator as 4 by multiplying the numerator and denominator of the first fraction by 1 and the second fraction by 2.

= ⁽⁵ ˣ ¹⁾⁄₍₄ ₓ ₁₎ + ⁽³ ˣ ²⁾⁄₍₂ ₓ ₂₎

⁵⁄₄⁶⁄₄

= ⁽⁵ ⁺ ⁶⁾⁄₄

¹¹⁄₄

= 2¾

Lily rode 2¾ miles in all.

Problem 4 :

At the dealership where Mary works, she fulfilled  of her quarterly sales goal in January and another ³⁄₁₀ of her sales goal in February. If her quarterly sales goal is $12,000, what amount of sales did she make in January and February ?

Solution :

Fraction of her sales done in January and February is

=  + ³⁄₁₀

Least common multiple of the denominators (6, 10) is 30.

make each denominator as 30 by multiplying the numerator and denominator of the first fraction by 5 and the second fraction by 3.

Then we have,

= ⁽¹ ˣ ⁵⁾⁄₍₆ ₓ ₅₎ + ⁽³ ˣ ³⁾⁄₍₁₀ ₓ ₃₎

= ⁵⁄₃₀ ⁹⁄₃₀

⁽⁵ ⁺ ⁹⁾⁄₃₀

¹⁴⁄₃₀

⁷⁄₁₅

Amount of sales she made in January and February is

⁷⁄₁₅ of quarterly sales goal 

= ⁷⁄₁₅ ⋅ 12000

= 5600

Amount of sales Mary made in January and February is $5,600.

Problem 5 :

Lucy made strawberry jam and raspberry jam. She made enough strawberry jam to fill  of a jar. If she made 2 times as much raspberry jam as strawberry jam, how many jars would the raspberry jam fill ?

Solution :

Amount of strawberry jam made by Lucy is

of a jar

2 times as much raspberry jam as strawberry jam is

= 2 ⋅  of a jar

= ⁵⁄₃ of a jar

= 1⅔ of a jar

Lucy would make enough raspberry jam to fill 1⅔ jars.

Problem 6 :

David's salary is $1800. David spent 2/3 of the total money for food. He spent 1/2 of the remaining for his kids education and saved the rest. How much did he save ?

Solution :

Money spent on food  is

= 1800 ⋅ 

= 1200

Remaining = 1800 - 1200

= 600 ----(1)

Money spent on kids education is

= 1/2 of remaining

= 1/2 ⋅ 600

= 300 ----(2)

Then, his savings is

= (1) - (2)

= 600 - 300

= 300

David saved $300.

Problem 7 :

A, B and C are friends. A has one-third of the money that B has. C has one-half of the money that A has. If they all together have  $450. How much money do A, B and C have separately? 

Solution :

Let us assume that B has the money y.

Then,

A = ()y = ʸ⁄₃

C = ½ of A

C = ½  ʸ⁄₃

C = ʸ⁄₆

Given : A, B and C together have $450.

A + B + C = 450

ʸ⁄₃ + y + ʸ⁄₆ = 450

Least common multiple of the denominators (3, 6) is 6.

Multiply both sides of the equation by 6 to get rid of the denominators 3 and 6.

6(ʸ⁄₃ + y + ʸ⁄₆) = 6(450)

6(ʸ⁄₃) + 6(y) + 6(ʸ⁄₆) = 2700

2y + 6y + y = 2700

9y = 2700

y = 300

ʸ⁄₃ = ³⁰⁰⁄₃ = 100

ʸ⁄₆ ³⁰⁰⁄₆ = 50 

A has $100, B has $300 and C has $50.

Problem 8 :

John's present age is one-third of David's age 5 years back. If David is 20 years old now, find the present age of John.

Solution :

Present age of David = 20 years.

David's age 5 years back = 20 - 5 = 15 years

Given : John's present age is 1/3 of David's age 5 years back.

John's present age :

= ⅓ of 15 years

= (⋅ 15

= 5 years

Problem 9 :

In a triangle, the first angle is one-half of the third angle and the second angle is three-fourth of the third angle. Find the three angles of the triangle.

Solution :

In the triangle, since both the first and second angles are linked to the third angle, introduce a variable for the third angle and  write the first and second angles in terms of that variable and solve for it.

Let x be the measure of the third angle.

First angle = (½)x = ˣ⁄₂

Second angle = (¾)x = ³ˣ⁄₄

In a triangle, sum of the three angles is equal to 180°.

first angle + second angle + third angle = 180°

ˣ⁄₂ + ³ˣ⁄₄ + x = 180°

Least commpon  multiple of the denominator (2, 4) is 4.

Multiply both sides of the equation by 4 to get rid of the denominators 2 and 4.

4(ˣ⁄₂ + ³ˣ⁄₄ + x) = 4(180°)

4(ˣ⁄₂) + 4(³ˣ⁄₄) + 4(x) = 720°

2x + 3x + 4x = 720°

9x = 720°

x = 80°

ˣ⁄₂ = ⁸⁰⁄₂ = 4

³ˣ⁄₄ = ⁽³ ˣ ⁸⁰⁾⁄₄ = 60°

The three angles of the triangle are 40°, 60° and 80°.

Problem 10 :

The denominator of a fraction is 1 more than thrice the numerarotor. If 2 be added to both numerator and denominator, the fraction would become . Find the fraction.

Solution :

Let x be the numerator.

Then, the denominator is (4x + 1).

Fraction = ˣ⁄₍₄ₓ ₊ ₁₎

Given : Adding 2 to both numerator and denominator of the fraction results .

⁽ˣ ⁺ ²⁾⁄₍₄ₓ ₊ ₁ ₊ ₂₎ =

⁽ˣ ⁺ ²⁾⁄₍₄ₓ ₊ ₃₎ =

3(x + 2) = 1(4x + 3)

3x + 6 = 4x + 3

6 = x + 3

3 = x

4x + 1 = 4(3) + 1

= 12 + 1

= 13

Sustitute x = 3 and 4x + 1 = 13 in (1).

Fraction = ³⁄₁₃

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