The value of \(\left( {\frac{7}{5} \div \frac{7}{{10}}of\frac{3}{4}} \right) \div \frac{4}{9} + \left( {\frac{7}{{16}} \div 10\frac{1}{2} \times 7\frac{1}{5}} \right) \times \frac{5}{{12}} \) is:
To find the value of the given expression, we need to follow the order of operations, commonly known as BODMAS or PEMDAS. This rule dictates the sequence in which operations should be performed:
In the given expression, \(\left( {\frac{7}{5} \div \frac{7}{{10}}of\frac{3}{4}} \right) \div \frac{4}{9} + \left( {\frac{7}{{16}} \div 10\frac{1}{2} \times 7\frac{1}{5}} \right) \times \frac{5}{{12}}\), we have brackets, 'of', division, multiplication, and addition.
The first bracket is \(\left( {\frac{7}{5} \div \frac{7}{{10}}of\frac{3}{4}} \right)\). Inside this bracket, we first perform the 'of' operation, which is multiplication:
\(\frac{7}{{10}}of\frac{3}{4} = \frac{7}{{10}} \times \frac{3}{4} = \frac{7 \times 3}{10 \times 4} = \frac{21}{40}\)
Now, the expression inside the first bracket becomes \(\frac{7}{5} \div \frac{21}{40}\). We perform the division by multiplying by the reciprocal of the second fraction:
\(\frac{7}{5} \div \frac{21}{40} = \frac{7}{5} \times \frac{40}{21}\)
Simplify before multiplying:
\(= \frac{\cancel{7}^1}{5} \times \frac{40}{\cancel{21}^3} = \frac{1}{5} \times \frac{40}{3}\)
\(= \frac{1}{\cancel{5}^1} \times \frac{\cancel{40}^8}{3} = \frac{1 \times 8}{1 \times 3} = \frac{8}{3}\)
So, the value of the first bracket is \(\frac{8}{3}\).
The second bracket is \(\left( {\frac{7}{{16}} \div 10\frac{1}{2} \times 7\frac{1}{5}} \right)\). First, convert the mixed numbers to improper fractions:
\(10\frac{1}{2} = \frac{(10 \times 2) + 1}{2} = \frac{21}{2}\)
\(7\frac{1}{5} = \frac{(7 \times 5) + 1}{5} = \frac{36}{5}\)
The expression inside the second bracket becomes \(\left( {\frac{7}{{16}} \div \frac{21}{2} \times \frac{36}{5}} \right)\). Within the bracket, perform division and multiplication from left to right.
First, perform the division:
\(\frac{7}{{16}} \div \frac{21}{2} = \frac{7}{{16}} \times \frac{2}{{21}}\)
Simplify before multiplying:
\(= \frac{\cancel{7}^1}{16} \times \frac{2}{\cancel{21}^3} = \frac{1}{16} \times \frac{2}{3}\)
\(= \frac{1}{\cancel{16}^8} \times \frac{\cancel{2}^1}{3} = \frac{1 \times 1}{8 \times 3} = \frac{1}{24}\)
Now, perform the multiplication with the result:
\(\frac{1}{24} \times \frac{36}{5}\)
Simplify before multiplying:
\(= \frac{1}{\cancel{24}^2} \times \frac{\cancel{36}^3}{5} = \frac{1 \times 3}{2 \times 5} = \frac{3}{10}\)
So, the value of the second bracket is \(\frac{3}{10}\).
Substitute the simplified bracket values back into the original expression. The expression now looks like:
\(\frac{8}{3} \div \frac{4}{9} + \frac{3}{10} \times \frac{5}{{12}}\)
According to BODMAS, we perform division and multiplication next, from left to right.
First, the division:
\(\frac{8}{3} \div \frac{4}{9} = \frac{8}{3} \times \frac{9}{4}\)
Simplify before multiplying:
\(= \frac{\cancel{8}^2}{\cancel{3}^1} \times \frac{\cancel{9}^3}{\cancel{4}^1} = \frac{2 \times 3}{1 \times 1} = 6\)
Next, the multiplication:
\(\frac{3}{10} \times \frac{5}{{12}}\)
Simplify before multiplying:
\(= \frac{\cancel{3}^1}{\cancel{10}^2} \times \frac{\cancel{5}^1}{\cancel{12}^4} = \frac{1 \times 1}{2 \times 4} = \frac{1}{8}\)
The expression is now simplified to:
\(6 + \frac{1}{8}\)
Finally, perform the addition:
\(6 + \frac{1}{8} = \frac{6 \times 8}{8} + \frac{1}{8} = \frac{48}{8} + \frac{1}{8} = \frac{48 + 1}{8} = \frac{49}{8}\)
The value of the expression is \(\frac{49}{8}\).
Comparing this with the given options, we find that the value matches option 1.
| Rule | Operation | Notes |
|---|---|---|
| B / P | Brackets / Parentheses | Evaluate expressions inside brackets first. |
| O / E | Orders / Exponents | Calculate powers and roots. |
| D / M | Division / Multiplication | Perform from left to right. 'Of' is treated as multiplication and is usually done before division/multiplication. |
| A / S | Addition / Subtraction | Perform from left to right. |
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