
Solve the following:
A. $({4^0} + {5^{ - 1}}) \times {5^2} \times \dfrac{1}{3}$
B. $({2^{ - 1}} + {3^{ - 1}} + {4^{ - 1}}) \times \dfrac{3}{4}$
Answer
500.7k+ views
Hint: When the equation has more than one algebraic function, use the BODMAS rule. BODMAS rules says that firstly solve the brackets, then division, multiplication, addition and subtraction, in this specified manner itself or else there will be wrong answers.
${x^0} = 1$, here x can be any natural number.
${x^{ - 1}} = \dfrac{1}{x}$, this is an inverse function of x, here x can be any natural number.
Complete step-by-step answer:
A. $({4^0} + {5^{ - 1}}) \times {5^2} \times \dfrac{1}{3}$
Equating the equation in the question to x:
$({4^0} + {5^{ - 1}}) \times {5^2} \times \dfrac{1}{3} = x$
To find- x
Substituting in LHS, Using equations 1, 2 and ${5^2} = 25$, we get
$x = (1 + \dfrac{1}{5}) \times 25 \times \dfrac{1}{3}$
Solving the bracket 1st, we get
$x = \dfrac{6}{5} \times 25 \times \dfrac{1}{3}$
On multiplying all the terms, we get,
$x = 10$
Therefore, $({4^0} + {5^{ - 1}}) \times {5^2} \times \dfrac{1}{3} = 10$
B. $({2^{ - 1}} + {3^{ - 1}} + {4^{ - 1}}) \times \dfrac{3}{4}$
Equating the equation in the question to y:
$({2^{ - 1}} + {3^{ - 1}} + {4^{ - 1}}) \times \dfrac{3}{4} = y$
To find: y
Substituting in LHS, Using equation 2, we get
$y = (\dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4}) \times \dfrac{3}{4}$
Solving the bracket 1st, we get,
Here, the base of all the fractions is not the same, therefore, it has to be made the same and then we can add.
For making the base the same, take the LCM. LCM comes as 12.
Therefore, $y = (\dfrac{{1 \times 6}}{{2 \times 6}} + \dfrac{{1 \times 4}}{{3 \times 4}} + \dfrac{{1 \times 3}}{{4 \times 3}}) \times \dfrac{3}{4}$
$y = \dfrac{{13}}{{12}} \times \dfrac{3}{4}$
On multiplying all the terms, we get,
$y = \dfrac{{13}}{{16}}$
Therefore, $({2^{ - 1}} + {3^{ - 1}} + {4^{ - 1}}) \times \dfrac{3}{4} = \dfrac{{13}}{{16}}$
Note: While solving such sums, BODMAS rules are very important to use. Brackets are given the 1st priority. In this sum, even when there is an operation of multiplication, we don’t solve that first, instead we go for the brackets and solve them First. While solving the addition and subtraction of fractions, bases should be the same or the operation becomes invalid to perform. Read the question very carefully. Anything raised to zero is one and not zero, many mistakes happen here.
${x^0} = 1$, here x can be any natural number.
${x^{ - 1}} = \dfrac{1}{x}$, this is an inverse function of x, here x can be any natural number.
Complete step-by-step answer:
A. $({4^0} + {5^{ - 1}}) \times {5^2} \times \dfrac{1}{3}$
Equating the equation in the question to x:
$({4^0} + {5^{ - 1}}) \times {5^2} \times \dfrac{1}{3} = x$
To find- x
Substituting in LHS, Using equations 1, 2 and ${5^2} = 25$, we get
$x = (1 + \dfrac{1}{5}) \times 25 \times \dfrac{1}{3}$
Solving the bracket 1st, we get
$x = \dfrac{6}{5} \times 25 \times \dfrac{1}{3}$
On multiplying all the terms, we get,
$x = 10$
Therefore, $({4^0} + {5^{ - 1}}) \times {5^2} \times \dfrac{1}{3} = 10$
B. $({2^{ - 1}} + {3^{ - 1}} + {4^{ - 1}}) \times \dfrac{3}{4}$
Equating the equation in the question to y:
$({2^{ - 1}} + {3^{ - 1}} + {4^{ - 1}}) \times \dfrac{3}{4} = y$
To find: y
Substituting in LHS, Using equation 2, we get
$y = (\dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4}) \times \dfrac{3}{4}$
Solving the bracket 1st, we get,
Here, the base of all the fractions is not the same, therefore, it has to be made the same and then we can add.
For making the base the same, take the LCM. LCM comes as 12.
Therefore, $y = (\dfrac{{1 \times 6}}{{2 \times 6}} + \dfrac{{1 \times 4}}{{3 \times 4}} + \dfrac{{1 \times 3}}{{4 \times 3}}) \times \dfrac{3}{4}$
$y = \dfrac{{13}}{{12}} \times \dfrac{3}{4}$
On multiplying all the terms, we get,
$y = \dfrac{{13}}{{16}}$
Therefore, $({2^{ - 1}} + {3^{ - 1}} + {4^{ - 1}}) \times \dfrac{3}{4} = \dfrac{{13}}{{16}}$
Note: While solving such sums, BODMAS rules are very important to use. Brackets are given the 1st priority. In this sum, even when there is an operation of multiplication, we don’t solve that first, instead we go for the brackets and solve them First. While solving the addition and subtraction of fractions, bases should be the same or the operation becomes invalid to perform. Read the question very carefully. Anything raised to zero is one and not zero, many mistakes happen here.
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