Question

pls help me by solving the above problem​

Answer 1

Solution :

This is a problem based on the commutative properties of rational numbers. The only method to solve this is by calculating the values -_-

• ¼ - ⅙ - 1/48

> Take the LCM as 48

> [ 12 - 8 - 1 ]/48

> 3/48

> 1/16 . . . (1)

• ¼ - ( ⅙ - 1/48 )

> ¼ - ⅙ + 1/48

> Take the LCM as 48

> [ 12 - 8 + 1 ]/48

> 5/48 . . . (2)

Dividing (1) by (2)

>3/48 × 48/5

> 5/3 .

• ¼ × ⅙ - 1/48

> 1/24 - 1/48

> 2/48 - 48

> 1/48 . . . (3)

¼ × ( ⅙ - 1/48 )

> ¼ × ( 7/48 )

> 7/192 . . (4)

Dividing (3) by (4)

>( 1/48 ) × ( 48 × 4)/7

> 4/7

5/3 × 4/7

> 20/21.

This is the required answer.

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Additional Information :

$\boxed{\begin{minipage}{6 cm}\bf{\dag}\:\:\underline{\textsf{Fraction Rules :}}\\\\\bigstar\:\:\sf\dfrac{A}{C} + \dfrac{B}{C} = \dfrac{A+B}{C} \\\\\bigstar\:\:\sf{\dfrac{A}{C} - \dfrac{B}{C} = \dfrac{A-B}{C}}\\\\\bigstar\:\:\sf\dfrac{A}{B} \times \dfrac{C}{D} = \dfrac{AC}{BD}\\\\\bigstar\:\:\sf\dfrac{A}{B} + \dfrac{C}{D} = \dfrac{AD}{BD} + \dfrac{BC}{BD} = \dfrac{AD+BC}{BD} \\\\\bigstar\:\:\sf\dfrac{A}{B} - \dfrac{C}{D} = \dfrac{AD}{BD} - \dfrac{BC}{BD} = \dfrac{AD-BC}{BD}\\\\\bigstar \:\:\sf \dfrac{A}{B} \div \dfrac{C}{D} = \dfrac{A}{B} \times \dfrac{D}{C} = \dfrac{AD}{BC}\end{minipage}}$

Answer 2

Answer:

Solution :-

At first let's do subtraction

$\sf \: \dfrac{1}{4} - \dfrac{1}{6} - \dfrac{1}{48}$

$\sf \: \dfrac{(1 \times 12) - (1 \times 8) - 1}{48} = \dfrac{12 - 8 - 1}{48}$

$\sf \dfrac{3}{48} = \bold{\dfrac{1}{16} (EQ 1)}$

¼ - ( ⅙ - 1/48 )

¼ - ⅙ + 1/48

[ 12 - 8 + 1 ]/48

5/48 . (2)

3/48 × 48/5

5/3 .

• ¼ × ⅙ - 1/48

1/24 - 1/48

2/48 - 48

2-1/48

1/48 . . . (3)

¼ × ( ⅙ - 1/48 )

¼ × ( 7/48 )

¼ × 7/48

> 7/192 . (4)

• Divided Eq 3 and 4

( 1/48 ) × ( 48 × 4)/7

1/48 × 192/7

4/1 × 1/7

4/7

5/3 × 4/7

20/21.