By | September 19, 2013

The idea of getting the least common multiple of the denominator in adding dissimilar fractions is to convert them into similar fractions or fractions whose denominators are the same. Once the fractions are similar, you only need to add the numerator and  copy the denominator.

The solutions to Fraction Addition Practice Test 1 below is divided into three parts: (1) getting the least common multiple of the denominator, (2) converting the given fractions to their equivalent fractions whose denominator is the LCM and (3) adding the converted fractions. Of course, in solving this types of problem the Civil Service Exam, you don’t need to go through all the steps. You should try developing your own short cuts to make solving faster.

Solution to Number 1

Given: $\displaystyle \frac{2}{7} + \frac{3}{7}$

$\displaystyle \frac{2}{7} + \frac{3}{7} = \frac{2 + 3}{7} = \frac{5}{7}$

Answer: $\displaystyle \frac{5}{7}$

Solution to Number 2

Given: $\displaystyle \frac{3}{5} + \frac{1}{5}$

$\displaystyle \frac{3}{5} + \frac{1}{5} = \frac{3 + 1}{5} = \frac{4}{5}$

Answer: $\displaystyle \frac{4}{5}$

Solution to Number 3

Given: $\displaystyle \frac{2}{3} + \frac{1}{8}$

A. Get the LCM of the denominators 8 and 3.

Multiples of 3: 3, 6, 9, 12, 15, 18, 24

Multiples of 8: 8, 16, 24, 32, 40

Therefore, the LCM of 8 and 3 is 24.

B. Convert the given to equivalent fractions whose denominator is 24.

$\displaystyle \frac{2}{3} = \frac{x}{24}$ ($x = (24 \div 3) \times 2$ which results to $x = 16$.)

$\displaystyle \frac{1}{8} = \frac{y}{24}$ ($y = (24 \div 8) \times 1$ which results to $y = 3$.)

So, $\displaystyle \frac{2}{3} = \frac{16}{24}$ and $\displaystyle \frac{1}{8} = \frac{3}{24}$

$\displaystyle \frac{16}{24} + \frac{3}{24} = \frac{19}{24}$

Answer: $\displaystyle \frac{19}{24}$.

Solution to Number 4

Given:  $\displaystyle \frac{1}{4} + \frac{1}{2} + \frac{2}{5}$

A. Get the LCM of 4, 2, and 5.

Multiples of 2: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22

Multiples of 4: 4, 8, 12, 16, 20, 24

Multiples of 5: 5, 10, 15, 20, 25

B. Convert the given to equivalent fractions whose denominator is 20.

$\displaystyle \frac{1}{4} = \frac{x}{20}$  ($x = (20 \div 4) \times 1$ which results to $x = 5$.)

$\displaystyle \frac{1}{2} = \frac{y}{20}$ ($y = (20 \div 2) \times 1$ which results to $y = 10$.)

$\displaystyle \frac{2}{5} = \frac{z}{20}$ ($z = (20 \div 5) \times 2$ which results to $z = 8$.)

So, $\displaystyle \frac{1}{4} = \frac{5}{20}$, $\displaystyle \frac{1}{2} = \frac{10}{20}$ and $\displaystyle \frac{2}{5} = \frac{8}{20}$.

$\displaystyle \frac{5}{20} + \frac{10}{20} + \frac{8}{20} = \frac{23}{20}$

Answer: $\displaystyle \frac{23}{20}$ or $\displaystyle 1 \frac{3}{20}$

Solution to Number 5

Given: $\displaystyle \frac{1}{6} + \frac{2}{3}$

A. Get the LCM of the denominators 6 and 3.

Since 6 is divisible by 3, the LCM of 6 and 3 is 6.

B. Convert the given to equivalent fractions whose denominator is 6.

Since $\displaystyle \frac{1}{6}$ has already denominator 6, we only need to convert $\displaystyle \frac{2}{3}$.

$\displaystyle \frac{2}{3} = \frac{x}{6}$ ($x = (6 \div 3) \times 2$ which results to $x = 4$).

So, $\displaystyle \frac{2}{3} = \frac{4}{6}$.

$\displaystyle \frac{1}{6} + \frac{4}{6} = \frac{5}{6}$

Answer: $\displaystyle \frac{5}{6}$.

Solution to Number 6

Given: $\displaystyle \frac{1}{4} + \frac{1}{8}$

A. Get the LCM of the denominators 4 and 8.

Since 8 is divisible by 4, the LCM of 4 and 8 is 8.

B. Convert the given to equivalent fractions whose denominator is 8.

Since $\displaystyle \frac{1}{8}$ has already 8 as denominator, we only need to convert $\displaystyle \frac{1}{4}$.

$\displaystyle \frac{1}{4} = \frac{x}{8}$.  ($x = (8 \div 4) \times 1$ which results to $x = 2$).

So, $\displaystyle \frac{1}{4} = \frac{2}{8}$.

$\displaystyle \frac{2}{8} + \frac{1}{8} = \frac{3}{8}$

Answer: $\displaystyle \frac{3}{8}$

Solution to Number 7

Given: $\displaystyle \frac{2}{3} + \frac{1}{2} + \frac{5}{6}$

A. Get the LCM of the denominators 2, 3, and 6.

Six are both divisible by 2 and 3, sLCM of 2, 3, and 6 is 6.

B. Convert the given to equivalent fractions whose denominator is 6.

We only need to convert $\displaystyle \frac{2}{3}$ and $\displaystyle \frac{1}{2}$.

$\displaystyle \frac{2}{3} = \frac{x}{6}$  ($x = (6 \div 3) \times 2$ which results to $x = 4$).

$\displaystyle \frac{1}{2} = \frac{y}{6}$  ($y = (6 \div 2) \times 1$ which results to $x = 3$).

So, $\displaystyle \frac{2}{3} = \frac{4}{6}$ and $\displaystyle \frac{1}{2} = \frac{3}{6}$.

$\displaystyle \frac{4}{6} + \frac{3}{6} + \frac{5}{6}= \frac{12}{6}$

Answer: $\displaystyle \frac{12}{6}$ or $2$

Solution to Number 8

Given: $\displaystyle \frac{1}{11} + \frac{2}{11} + \frac{3}{11}$

$\displaystyle \frac{1}{11} + \frac{2}{11} + \frac{3}{11} = \frac{1 + 2 + 3}{11} = \frac{6}{11}$

Answer: $\displaystyle \frac{6}{11}$

Solution to Number 9

Given: $\displaystyle \frac{1}{9} + \frac{2}{9} + \frac{1}{3}$.

A. Get the LCM of the denominators 3 and 9.

Since 9 is divisible by 3, the LCM of 3 and 9 is 9.

B. Convert the given to equivalent fractions whose denominator is 9.

We only need to convert $\displaystyle \frac{1}{3}$.

$\displaystyle \frac{1}{3} = \frac{x}{9}$ ($x = (9 \div 3) \times 1$ which results to $x = 3$).

$\displaystyle \frac{1}{9} + \frac{2}{9} + \frac{3}{9} = \frac{6}{9}$.

Answer: $\displaystyle \frac{6}{9}$ or $\displaystyle \frac{2}{3}$ in lowest terms.

Solution to Number 10

Given: $\displaystyle \frac{1}{8} + \frac{1}{2} + \frac{1}{3}$

A. Get the LCM of the denominators 8, 2,  and 3.

Multiples of 2: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26

Multiples of 3: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30

Multiples of 8: 8, 16, 24, 32, 40

So, the LCM of 8, 2 and 3 is 24.

B. Convert the given to equivalent fractions whose denominator is 8.

$\displaystyle \frac{1}{8} = \frac{x}{24}$  ($x = (24 \div 8) \times 1$ which results to $x = 3$.)

$\displaystyle \frac{1}{2} = \frac{y}{24}$ ($y = (24 \div 2) \times 1$ which results to $y = 12$.)

$\displaystyle \frac{1}{3} = \frac{z}{24}$ ($z = (24 \div 3) \times 1$ which results to $z = 8$.)

So, $\displaystyle \frac{1}{8} = \frac{3}{24}$, $\displaystyle \frac{1}{2} = \frac{12}{24}$ and $\displaystyle \frac{1}{3} = \frac{8}{24}$.

$\displaystyle \frac{3}{24} + \frac{12}{24} + \frac{8}{24} = \frac{3 + 12 + 8}{24} = \frac{23}{24}$.

Answer: $\displaystyle \frac{23}{24}$

Solution to Number 11

Given: $\displaystyle \frac{1}{7} + \frac{2}{21}$

A. Get the LCM of the denominators 7 and 21.

Since 21 is divisible by 7, the least common multiple of 7 and 21 is 21.

B. Convert the given to equivalent fractions whose denominator is 21.

We only need to convert $\displaystyle \frac{1}{7}$.

$\displaystyle \frac{1}{7} = \frac{x}{21}$: ($x = (21 \div 7) \times 1$ which gives $x = 3$.)

So, $\displaystyle \frac{1}{7} = \frac{3}{21}$.

$\displaystyle \frac{3}{21} + \frac{2}{21} = \frac{5}{21}$

Answer: $\displaystyle \frac{5}{21}$

Solution to Number 12

Given: $\displaystyle \frac{7}{10} + \frac{1}{2} + \frac{2}{3}$

A. Get the LCM of the denominators 2, 3, and 10.

Multiples of 2: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32

Multiples of 3: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33

Multiples of 10: 10, 20, 30, 40, 50

The LCM of 2, 3, and 10 is 30.

B. Convert the given to equivalent fractions whose denominator is 30.

$\displaystyle \frac{7}{10} = \frac{x}{30}$ ($x = (30 \div 10) \times 7$ which results to $x = 21$.)

$\displaystyle \frac{1}{2} = \frac{y}{30}$ ($y = (30 \div 2) \times 1$ which results to $y = 15$.)

$\displaystyle \frac{2}{3} = \frac{z}{30}$ ($z = (30 \div 3) \times 2$ which results to $z = 20$.)

So, $\displaystyle \frac{7}{10} = \frac{21}{30}$, $\displaystyle \frac{1}{2} = \frac{15}{30}$ and $\displaystyle \frac{2}{3} = \frac{20}{30}$.

$\displaystyle \frac{21}{30} + \frac{15}{30} + \frac{20}{30} = \frac{56}{30}$

Answer:  $\displaystyle \frac{56}{30}$ or $1 \displaystyle \frac{26}{30}$ or $1 \displaystyle \frac{13}{15}$ in lowest terms.

Solution to Number 13

Given: $\displaystyle \frac{1}{5} + \frac{3}{8} + \frac{1}{2}$

A. Get the LCM of the denominators 8, 5,  and 2.

Multiples of 2: 2, 4, 6, 8, 10, …, 36, 38, 40, 42

Multiples of 5: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50

Multiples of 8: 8, 16, 24, 32, 40, 48, 56

So, the LCM of 8, 5, and 2 is 40.

B. Convert the given to equivalent fractions whose denominator is 40.

$\displaystyle \frac{1}{5} = \frac{x}{40}$ ($x = (40 \div 5) \times 1$ which results to $x = 8$.)

$\displaystyle \frac{3}{8} = \frac{y}{40}$ ($y = (40 \div 8) \times 3$ which results to $y = 15$.)

$\displaystyle \frac{1}{2} = \frac{z}{40}$ ($z = (40 \div 2) \times 1$ which results to $z = 20$.)

So, $\displaystyle \frac{1}{5} = \frac{8}{40}$, $\displaystyle \frac{3}{8} = \frac{15}{40}$ and $\displaystyle \frac{1}{2} = \frac{20}{40}$.

$\displaystyle \frac{8}{40} + \frac{15}{40} + \frac{20}{40} = \frac{43}{40}$

Answer: $\displaystyle \frac{43}{40}$ or $1 \displaystyle \frac{3}{40}$

Solution to Number 14

Given: $\displaystyle \frac{1}{3} + \frac{2}{9} + \frac{1}{6}$

A. Get the LCM of the denominators 3, 9, and 6.

Multiples of 3: 3, 6, 9, 12, 18, 21

Multiples of 6: 6, 12, 18, 24, 30

Multiples of 9: 9, 18, 27, 36, 45

So, the LCM of 3, 9 and 6 is 18.

B. Convert the given to equivalent fractions whose denominator is 18.

$\displaystyle \frac{1}{3} = \frac{x}{18}$ ($x = (18 \div 3) \times 1$ which results to $x = 6$.)

$\displaystyle \frac{2}{9} = \frac{y}{18}$ ($y = (18 \div 9) \times 2$ which results to $y = 4$.)

$\displaystyle \frac{1}{6} = \frac{z}{18}$ ($z = (18 \div 6) \times 1$ which results to $z = 3$.)

So, $\displaystyle \frac{1}{3} = \frac{6}{18}$, $\displaystyle \frac{2}{9} = \frac{4}{18}$ and $\displaystyle \frac{1}{6} = \frac{3}{18}$.

14. $\displaystyle \frac{6}{18} + \frac{4}{18} + \frac{3}{18} = \frac{13}{18}$

Answer: $\displaystyle \frac{13}{18}$

Solution to Number 15

Given: $\displaystyle \frac{3}{12} + \frac{8}{12} + \frac{2}{12} = \frac{13}{12}$

A. Get the LCM of the denominators 4, 3 and 6.

Multiples of 4: 4, 8, 12, 16, 20

Multiples of 3: 3, 6, 9, 12, 15

Multiples of 6:  6, 12, 18, 24, 30

So, the LCM of 4, 3, and 6 is 12.

B. Convert the given to equivalent fractions whose denominator is 8.

$\displaystyle \frac{1}{4} = \frac{x}{12}$ ($x = (12 \div 3) \times 1$ which results to $x = 4$.)

$\displaystyle \frac{2}{3} = \frac{y}{12}$ ($y = (12 \div 3) \times 2$ which results to $y = 8$.)

$\displaystyle \frac{1}{6} = \frac{z}{12}$ ($y = (12 \div 6) \times 1$ which results to $z = 2$.)

So, $\displaystyle \frac{1}{4} = \frac{3}{12}$, $\displaystyle \frac{2}{3} = \frac{8}{12}$ and $\displaystyle \frac{1}{6} = \frac{2}{12}$.

Answer: $\displaystyle \frac{13}{12}$ or $\displaystyle 1 \frac{1}{12}$

This is a very long post so please inform me if you have spotted any errors in the solutions.

## 5 thoughts on “Fraction Addition Practice Test 1 Solutions and Answers”

1. mark

thanks a lot it’s very helpful:) do you have a vocabulary in english and filipino?

1. Civil Service Reviewer Post author

Hi Mark. I have English vocabulary, but no Filipino yet. Please, check out the English page above. 🙂

2. Michelle Hererra

Got it all!100%! Thank you for meticulously explaining the process…

3. Civil Service Reviewer Post author

Yes, the answer is 10/21. Typo error. I already changed it. Thank you.