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Let one man alone can finish the work in x days

and one woman alone can finish the work in y days

Step 1:

One day work of a man = $\frac{1}{x}$

One day work of a woman = $\frac{1}{y}$

Case 1:

8 men and 12 women can finish a piece of work in 10 days

=> $\frac{8}{x} + \frac{12}{y} = \frac{1}{10}$ .......................(1)

Case 2:

6 men and 8 women can finish a piece of work in 14 days

=> $\frac{6}{x} + \frac{8}{y} = \frac{1}{14}$ .........................(2)

Step 2:

Put $\frac{1}{x}$ = u and $\frac{1}{y}$ = v

(1) => 8u + 12v = $\frac{1}{10}$

=> 10(8u + 12v) = 1

=> 80u + 120v = 1 ....................................(3)

and equation (2) implies

6u + 8v = $\frac{1}{14}$

=> 14(6u + 8v) = 1

=> 84u + 112v = 1 ...................................(4)

Step 3:

Subtract equation (3) from equation (4)

=> 84u + 112v - (80u + 120v) = 1 - 1

=> 84u + 112v - 80u - 120v = 0

=> 4u - 8v = 0

=> u - 2v = 0

=> u = 2v ........................................(5)

Step 4:

Put u = 2v in equation (3)

80 * 2v + 120v = 1

=> 160v + 120v = 1

=> 280v = 1

=> v = $\frac{1}{280}$, put in equation (5)

u = 2 * $\frac{1}{280}$

or u = $\frac{1}{140}$

Step 5:

Since, $\frac{1}{x}$ = u and $\frac{1}{y}$ = v

=> $\frac{1}{x}$ = $\frac{1}{140}$

=>** x = 140**

and $\frac{1}{y}$ = $\frac{1}{280}$

=>** y = 280**

**Answer:** One can finish a piece of work in 140 days and one woman in 280 days.

and one woman alone can finish the work in y days

Step 1:

One day work of a man = $\frac{1}{x}$

One day work of a woman = $\frac{1}{y}$

Case 1:

8 men and 12 women can finish a piece of work in 10 days

=> $\frac{8}{x} + \frac{12}{y} = \frac{1}{10}$ .......................(1)

Case 2:

6 men and 8 women can finish a piece of work in 14 days

=> $\frac{6}{x} + \frac{8}{y} = \frac{1}{14}$ .........................(2)

Step 2:

Put $\frac{1}{x}$ = u and $\frac{1}{y}$ = v

(1) => 8u + 12v = $\frac{1}{10}$

=> 10(8u + 12v) = 1

=> 80u + 120v = 1 ....................................(3)

and equation (2) implies

6u + 8v = $\frac{1}{14}$

=> 14(6u + 8v) = 1

=> 84u + 112v = 1 ...................................(4)

Step 3:

Subtract equation (3) from equation (4)

=> 84u + 112v - (80u + 120v) = 1 - 1

=> 84u + 112v - 80u - 120v = 0

=> 4u - 8v = 0

=> u - 2v = 0

=> u = 2v ........................................(5)

Step 4:

Put u = 2v in equation (3)

80 * 2v + 120v = 1

=> 160v + 120v = 1

=> 280v = 1

=> v = $\frac{1}{280}$, put in equation (5)

u = 2 * $\frac{1}{280}$

or u = $\frac{1}{140}$

Step 5:

Since, $\frac{1}{x}$ = u and $\frac{1}{y}$ = v

=> $\frac{1}{x}$ = $\frac{1}{140}$

=>

and $\frac{1}{y}$ = $\frac{1}{280}$

=>

Step 1:

Let base (width) of the cube = x

Then length of the box = 2x

Height of the box = y

[Since the base of the box is a rectangle whose length is twice its width]

Let S be the surface area of the box

=> S = sum of the areas of its six sides ...................(1)

Step 2:

Area of the top and bottom of the box = 2x^{2} (each)

Area of the front and back of the box = 2xy (each)

Area of the left and right sides of the box = xy (each)

(1) => S = 2x^{2} + 2x^{2} + 2xy + 2xy + xy + xy

=> S = 4x^{2} + 6xy .........................(2)

Step 3:

Volume of the box = length * width * height

=> 124 = 2x^{2} y

=> y = $\frac{124}{2x^2}$ = $\frac{62}{x^2}$

(2) => S = 4x^{2} + 6xy

=> S = 4x^{2} + 6x * $\frac{62}{x^2}$

= 4x^{2} + 6 * $\frac{62}{x}$

=> S = $\frac{4x^3 + 372}{x}$ surface area of the box.

Let base (width) of the cube = x

Then length of the box = 2x

Height of the box = y

[Since the base of the box is a rectangle whose length is twice its width]

Let S be the surface area of the box

=> S = sum of the areas of its six sides ...................(1)

Step 2:

Area of the top and bottom of the box = 2x

Area of the front and back of the box = 2xy (each)

Area of the left and right sides of the box = xy (each)

(1) => S = 2x

=> S = 4x

Step 3:

Volume of the box = length * width * height

=> 124 = 2x

=> y = $\frac{124}{2x^2}$ = $\frac{62}{x^2}$

(2) => S = 4x

=> S = 4x

= 4x

=> S = $\frac{4x^3 + 372}{x}$ surface area of the box.