Question

answer this question

Answer 1

Answer:

Question :

${\sf{\pink{10}. \: Find \: the \: value \: of \: x :}}$

$\sf{:\implies\dfrac{9x + 7}{2} - {\Bigg[x - {\bigg\lgroup \dfrac{x - 2}{7} \bigg\rgroup}\Bigg]} = 36}$

(A) 9 ⠀⠀⠀⠀(B) 18

(C) 5 ⠀⠀⠀⠀(D) 4

$\begin{gathered}\end{gathered}$

Solution :

$\sf{:\implies\dfrac{9x + 7}{2} - {\Bigg[x - {\bigg\lgroup \dfrac{x - 2}{7} \bigg\rgroup}\Bigg]} = 36}$

Opening round brackets

$\sf{:\implies\dfrac{9x + 7}{2} - {\Bigg[x - \dfrac{x - 2}{7}\Bigg]} = 36}$

we can write x as x/1

$\sf{:\implies\dfrac{9x + 7}{2} - {\Bigg[\dfrac{x}{1} - \dfrac{x - 2}{7}\Bigg]} = 36}$

Now, taking LCM of denominators

$\sf{:\implies\dfrac{9x + 7}{2} - {\Bigg[\dfrac{(x \times 7) - (x \times 1 - 2 \times 1)}{7}\Bigg]} = 36}$

$\sf{:\implies\dfrac{9x + 7}{2} - {\Bigg[\dfrac{7x- x -2}{7}\Bigg]} = 36}$

Subtracting x from 7x

$\sf{:\implies\dfrac{9x + 7}{2} - {\Bigg[\dfrac{6x -2}{7}\Bigg]} = 36}$

Now, opening square brackets

$\sf{:\implies\dfrac{9x + 7}{2} - \dfrac{6x -2}{7}= 36}$

Now, again taking LCM of denominators 2 and 7

$\sf{:\implies\dfrac{7(9x + 7) - 2(6x -2)}{14}= 36}$

Multiple 7 into 9x+7 and 2 into 6x-2

$\sf{:\implies\dfrac{63x + 49- 12x - 4}{14}= 36}$

Subtracting 12x from 63x and 4 from 49

$\sf{:\implies\dfrac{51x + 45}{14}= 36}$

Taking LHS (14) to RHS (36)

$\sf{:\implies{51x + 45= 36 \times 14}}$

Multiplying 36 into 14

$\sf{:\implies{51x + 45= 504}}$

Taking LHS (39) to RHS (504)

$\sf{:\implies{51x= 504 - 45}}$

$\sf{:\implies{51x= 459}}$

Now, taking LHS (53) to RHS (459)

$\sf{:\implies{x= \dfrac{459}{51} }}$

Cutting the values

$\sf{:\implies{x= \cancel{\dfrac{459}{51}}}}$

$\sf{:\implies{x= 9}}$

$\large\bigstar \: \red{\underline{\boxed{\sf{x = 9}}}}$

Hence, the value of x is 9.

So, the option (A) is the correct answer.

$\begin{gathered}\end{gathered}$

Verification :

$\sf{:\implies\dfrac{9x + 7}{2} - {\Bigg[x - {\bigg\lgroup \dfrac{x - 2}{7} \bigg\rgroup}\Bigg]} = 36}$

Substituting the value of x (9)

$\sf{:\implies\dfrac{9 \times 9 + 7}{2} - {\Bigg[9 - {\bigg\lgroup \dfrac{9 - 2}{7} \bigg\rgroup}\Bigg]} = 36}$

$\sf{:\implies\dfrac{81 + 7}{2} - {\Bigg[9 - {\bigg\lgroup \dfrac{9 - 2}{7} \bigg\rgroup}\Bigg]} = 36}$

Subtracting 2 from 9

$\sf{:\implies\dfrac{81 + 7}{2} - {\Bigg[9 - {\bigg\lgroup \dfrac{7}{7} \bigg\rgroup}\Bigg]} = 36}$

Simplifying 7/7

$\sf{:\implies\dfrac{81 + 7}{2} - {\Bigg[9 - {\bigg\lgroup \cancel{\dfrac{7}{7}} \bigg\rgroup}\Bigg]} = 36}$

$\sf{:\implies\dfrac{81 + 7}{2} - {\Bigg[9 - {\bigg\lgroup \: \: 1 \: \: \bigg\rgroup}\Bigg]} = 36}$

Opening round brackets

$\sf{:\implies\dfrac{88}{2} - {\Bigg[9 - 1\Bigg]} = 36}$

Subtracting 1 fron 9

$\sf{:\implies\dfrac{88}{2} - {\Bigg[ \: \: 8 \: \: \Bigg]} = 36}$

Opening square brackets

$\sf{:\implies\dfrac{88}{2} -8 = 36}$

Simplifying 88/2

$\sf{:\implies \cancel{\dfrac{88}{2}} -8 = 36}$

$\sf{:\implies{ 44 -8 = 36}}$

$\sf{:\implies{ 36 = 36}}$

$\large\bigstar \: \red{\underline{\boxed{\sf{LHS = RHS}}}}$

Hence Verified!

$\begin{gathered}\end{gathered}$

Learn More :

☼ BODMAS :

↝ BODMAS rule is an acronym used to remember the order of operations to be followed while solving expressions in mathematics.

It stands for :-

• ↠ B - Brackets,
• ↠ O - Order of powers or roots,
• ↠ D - Division,
• ↠ M - Multiplication 
• ↠ A - Addition
• ↠ S - Subtraction.

↝ It means that expressions having multiple operators need to be simplified from left to right in this order only.

☼ BODMAS RULE :

↝ First, we solve brackets, then powers or roots, then division or multiplication (whatever comes first from the left side of the expression), and then at last subtraction or addition.

• ↠ Addition (+)
• ↠ Subtraction (-)
• ↠ Multiplication (×)
• ↠ Division (÷)
• ↠ Brackets ( )

☼ EXPONENT :

↝ The exponent of a number says how many times to use the number in a multiplication.

☼ LAW OF EXPONENT :

The important laws of exponents are given below:

• ↠ ${\rm{{a}^{m} \times {a}^{n} = {a}^{m + n}}}$
• ↠ ${\rm{{a}^{m}/{a}^{n} = {a}^{m - n}}}$
• ↠ ${\rm{({a}^{m})^{n} = {a}^{mn}}}$
• ↠ ${\rm{{a}^{n}/{b}^{n} = ({a/b})^{n} }}$
• ↠ ${\rm{{a}^{0} = 1}}$
• ↠ ${\rm{{a}^{ - m} = {1/a}^{m}}}$
• ↠ ${\rm{{a}^{\frac{1}{n} } = \sqrt[n]{a}}}$

$\rule{220pt}{3pt}$