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 ⅒play 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 + ⅒
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₎
= ⁵⁄₂₀ + ²⁄₂₀
= ⁽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₎
= ⁵⁄₄ + ⁶⁄₄
= ⁽⁵ ⁺ ⁶⁾⁄₄
= ¹¹⁄₄
= 2¾
Lily rode 2¾ miles in all.
Problem 4 :
At the dealership where Mary works, she fulfilled ⅙ 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
= ⅙ + 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 ⅚ 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
= ⅚ of a jar
2 times as much raspberry jam as strawberry jam is
= 2 ⋅ ⅚ 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
= ⅔ ⋅ 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 = (⅓)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
= (⅓) ⋅ 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 = 40°
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 ⅓. 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 ⅓.
⁽ˣ ⁺ ²⁾⁄₍₄ₓ ₊ ₁ ₊ ₂₎ = ⅓
⁽ˣ ⁺ ²⁾⁄₍₄ₓ ₊ ₃₎ = ⅓
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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