What is (3 2/5+1 7/8)+2 1/5 ?

The solution to (3 2/5+1 7/8)+2 1/5 is 7 19/40

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Popular >

(3 2/5+1 7/8)+2 1/5

Solution

(3 \frac{2}{5} + 1 \frac{7}{8}) + 2 \frac{1}{5}

Solution

7 \frac{19}{40}

Solution steps

(3 \frac{2}{5} + 1 \frac{7}{8}) + 2 \frac{1}{5}

Convert mixed numbers to improper fractions:

2 \frac{1}{5} = \frac{11}{5}

2 \frac{1}{5}

Convert mixed numbers to improper fraction:

a \frac{b}{c} = \frac{a \cdot c + b}{c}

2 \frac{1}{5} = \frac{2 \cdot 5 + 1}{5}

= \frac{11}{5}

= (3 \frac{2}{5} + 1 \frac{7}{8}) + \frac{11}{5}

Convert mixed numbers to improper fractions:

3 \frac{2}{5} = \frac{17}{5}

3 \frac{2}{5}

Convert mixed numbers to improper fraction:

a \frac{b}{c} = \frac{a \cdot c + b}{c}

3 \frac{2}{5} = \frac{3 \cdot 5 + 2}{5}

= \frac{17}{5}

= (\frac{17}{5} + 1 \frac{7}{8}) + \frac{11}{5}

Convert mixed numbers to improper fractions:

1 \frac{7}{8} = \frac{15}{8}

1 \frac{7}{8}

Convert mixed numbers to improper fraction:

a \frac{b}{c} = \frac{a \cdot c + b}{c}

1 \frac{7}{8} = \frac{1 \cdot 8 + 7}{8}

= \frac{15}{8}

= (\frac{17}{5} + \frac{15}{8}) + \frac{11}{5}

Follow the PEMDAS order of operations

Calculate within parentheses

(\frac{17}{5} + \frac{15}{8}) : \frac{211}{40}

\frac{17}{5} + \frac{15}{8}

\frac{17}{5} + \frac{15}{8} = \frac{211}{40}

\frac{17}{5} + \frac{15}{8}

Least Common Multiplier of

5, 8 : 40

5, 8

Least Common Multiplier (LCM)

Prime factorization of

5 : 5

5

5

is a prime number, therefore no factorization is possible

= 5

Prime factorization of

8 : 2 \cdot 2 \cdot 2

8

8

divides by

28 = 4 \cdot 2

= 2 \cdot 4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2

Multiply each factor the greatest number of times it occurs in either

5

8

= 5 \cdot 2 \cdot 2 \cdot 2

Multiply the numbers:

5 \cdot 2 \cdot 2 \cdot 2 = 40

= 40

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

40

For

\frac{17}{5} :

multiply the denominator and numerator by

8

\frac{17}{5} = \frac{17 \cdot 8}{5 \cdot 8} = \frac{136}{40}

For

\frac{15}{8} :

multiply the denominator and numerator by

5

\frac{15}{8} = \frac{15 \cdot 5}{8 \cdot 5} = \frac{75}{40}

= \frac{136}{40} + \frac{75}{40}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{136 + 75}{40}

Add the numbers:

136 + 75 = 211

= \frac{211}{40}

= \frac{211}{40}

= \frac{211}{40} + \frac{11}{5}

Add and subtract (left to right)

\frac{211}{40} + \frac{11}{5} : \frac{299}{40}

\frac{211}{40} + \frac{11}{5}

\frac{211}{40} + \frac{11}{5} = \frac{299}{40}

\frac{211}{40} + \frac{11}{5}

Least Common Multiplier of

40, 5 : 40

40, 5

Least Common Multiplier (LCM)

Prime factorization of

40 : 2 \cdot 2 \cdot 2 \cdot 5

40

40

divides by

240 = 20 \cdot 2

= 2 \cdot 20

20

divides by

220 = 10 \cdot 2

= 2 \cdot 2 \cdot 10

10

divides by

210 = 5 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 5

2, 5

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2 \cdot 5

Prime factorization of

5 : 5

5

5

is a prime number, therefore no factorization is possible

= 5

Multiply each factor the greatest number of times it occurs in either

40

5

= 2 \cdot 2 \cdot 2 \cdot 5

Multiply the numbers:

2 \cdot 2 \cdot 2 \cdot 5 = 40

= 40

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

40

For

\frac{11}{5} :

multiply the denominator and numerator by

8

\frac{11}{5} = \frac{11 \cdot 8}{5 \cdot 8} = \frac{88}{40}

= \frac{211}{40} + \frac{88}{40}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{211 + 88}{40}

Add the numbers:

211 + 88 = 299

= \frac{299}{40}

= \frac{299}{40}

= \frac{299}{40}

Convert improper fractions to mixed numbers:

\frac{299}{40} = 7 \frac{19}{40}

\frac{299}{40} = 7

Remainder

19

\frac{299}{40}

Write the problem in long division format

40 \overline{∣ 299}

Divide

299

40

to get

7

Divide

299

40

to get

7

7 40 \overline{∣ 299}

Multiply the quotient digit

(7)

by the divisor

40

7 40 \overline{∣ 299} \underline{280}

Subtract

280

from

299

7 40 \overline{∣ 299} \underline{280} 19

7 40 \overline{∣ 299} \underline{280} 19

The solution for Long Division of

\frac{299}{40}

7

with remainder of

19

7 Remainder 19

Convert to mixed number: Quotient

\frac{Remainder}{Divisor}

\frac{299}{40} = 7 \frac{19}{40}

= 7 \frac{19}{40}

= 7 \frac{19}{40}

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Frequently Asked Questions (FAQ)

What is (3 2/5+1 7/8)+2 1/5 ?
The solution to (3 2/5+1 7/8)+2 1/5 is 7 19/40