
Fill in the blank. \[23 \times 56 = 20 \times 56 + \_\_\_\_\_\_ \times 56\] ?
Answer
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Hint: In this simplification problem, we have to find the missing values in the question is \[23 \times 56 = 20 \times 56 + \_\_\_\_\_\_ \times 56\]
First, assume the missing value as x, then we will use the BODMAS concept to solve this problem.
It stands for B -Bracket, O - Of , D - Division , M - Multiplication , A – Addition , S- Subtraction.
Complete step by step solution:
Let us suppose the missing value be \[x\].
Then the given question can be written as :
\[ \Rightarrow 23 \times 56 = 20 \times 56 + x \times 56\]
First , we transfer \[20 \times 56\]from RHS to LHS side .
Hence we have
\[ \Rightarrow 23 \times 56 - 20 \times 56 = x \times 56\] .
Taking common 56 on LHS. We get
\[ \Rightarrow \left( {23 - 20} \right) \times 56 = x \times 56\].
On simplifying
\[ \Rightarrow \left( 3 \right) \times 56 = x \times 56\]
Dividing both side by 56 we get
\[ \Rightarrow x = 3\]
Hence \[x = 3\] is the required solution.
Therefore, \[23 \times 56 = 20 \times 56 + 3 \times 56\].
Note:
The BODMAS rule is used to solve simplification problems. Some examples are given as follows.
While moving numbers from LHS TO RHS or vice versa, sign changes.
For example: \[a + b = c + d\]. If we move ‘c’ from LHS To RHS Then sign of ‘c‘ will change . i.e.\[d = a + b - c\].
Also while dividing and multiplying the sign does not change. But, we have to multiply and divide both sides .
For example: \[a \times b = c \times b\]. we divide both sides by b, thus the sign does not change . hence, we get \[a = c\].
Similarly, we can multiply any number to both sides without altering the question.
For example if \[a = b\] then we can multiply both sides by ‘c’ to get \[a \times c = b \times c\].
One more example is \[5-2 + 6 \div 3\]: The division must be completed first \[6 \div 3 = 2\] which then leaves addition and subtraction; As both are of the same importance, we can then work from left to right \[5-2 + 2 = 5\]. since, \[6 \div 3 = 2\]
This may be commonly miscalculated as either 3 by working from left to right, or as 1 by wrongly assuming that addition should be completed before subtraction. Here, The correct answer is \[5\].
First, assume the missing value as x, then we will use the BODMAS concept to solve this problem.
It stands for B -Bracket, O - Of , D - Division , M - Multiplication , A – Addition , S- Subtraction.
Complete step by step solution:
Let us suppose the missing value be \[x\].
Then the given question can be written as :
\[ \Rightarrow 23 \times 56 = 20 \times 56 + x \times 56\]
First , we transfer \[20 \times 56\]from RHS to LHS side .
Hence we have
\[ \Rightarrow 23 \times 56 - 20 \times 56 = x \times 56\] .
Taking common 56 on LHS. We get
\[ \Rightarrow \left( {23 - 20} \right) \times 56 = x \times 56\].
On simplifying
\[ \Rightarrow \left( 3 \right) \times 56 = x \times 56\]
Dividing both side by 56 we get
\[ \Rightarrow x = 3\]
Hence \[x = 3\] is the required solution.
Therefore, \[23 \times 56 = 20 \times 56 + 3 \times 56\].
Note:
The BODMAS rule is used to solve simplification problems. Some examples are given as follows.
While moving numbers from LHS TO RHS or vice versa, sign changes.
For example: \[a + b = c + d\]. If we move ‘c’ from LHS To RHS Then sign of ‘c‘ will change . i.e.\[d = a + b - c\].
Also while dividing and multiplying the sign does not change. But, we have to multiply and divide both sides .
For example: \[a \times b = c \times b\]. we divide both sides by b, thus the sign does not change . hence, we get \[a = c\].
Similarly, we can multiply any number to both sides without altering the question.
For example if \[a = b\] then we can multiply both sides by ‘c’ to get \[a \times c = b \times c\].
One more example is \[5-2 + 6 \div 3\]: The division must be completed first \[6 \div 3 = 2\] which then leaves addition and subtraction; As both are of the same importance, we can then work from left to right \[5-2 + 2 = 5\]. since, \[6 \div 3 = 2\]
This may be commonly miscalculated as either 3 by working from left to right, or as 1 by wrongly assuming that addition should be completed before subtraction. Here, The correct answer is \[5\].
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