Question

Solve \[16 - 2 \div 7 + 6 \times 2 = \] ?

Answer 1

Hint: : Here we are asked to solve the given problem \[16 - 2 \div 7 + 6 \times 2\]. This can be done by the BODMAS rule. The abbreviation of BODMAS is brackets, order, division, multiplication, addition, and subtraction. According to this rule, we have to solve the given problem in the BODMAS order.
Formula: The formula that we need to know before solving this problem:
BODMAS rule – when there are a lot of operations that come together in a single problem, we have to solve that problem in a certain order that is brackets, order, division, multiplication, addition, and subtraction.

Complete step by step answer:
It is given that \[16 - 2 \div 7 + 6 \times 2\], we aim to solve this problem.
We know that when there are a lot of different operations that come in a single problem then we have to follow a rule called BODMAS.
BODMAS – brackets, order, division, multiplication, addition, and subtraction
The problem should be solved in this order only.
According to this rule, we have to solve the division first, since we don’t have any brackets, powers, and roots.
\[16 - 2 \div 7 + 6 \times 2 = 16 - \dfrac{2}{7} + 6 \times 2\]
Then by the rule, we have to solve the multiplication.
\[ = 16 - \dfrac{2}{7} + 12\]
After that, we have to solve the additional part.
\[ = 16 - \dfrac{{2 + 84}}{7}\]
On further simplification, we get
\[ = 16 - \dfrac{{86}}{7}\]
At last, we have to solve the subtraction part
\[ = \dfrac{{112 - 86}}{7}\]
On simplifying the above expression, we get
\[16 - 2 \div 7 + 6 \times 2\]\[ = \dfrac{{26}}{7}\]
Thus, we got the value of the given expression that is \[16 - 2 \div 7 + 6 \times 2\]\[ = \dfrac{{26}}{7}\].

Note:
 In the above problem, while doing the division part we didn’t solve \[\dfrac{2}{7}\] since it will give us a decimal value, this is totally up to the students if one wants to find the final answer in decimal then they can go for solving the division part \[\dfrac{2}{7}\]. Then in the addition and subtraction part, we had a rational number a whole number so we took LCM to solve that part.