Question

points for joint family is better than nuclear family

Answer 1

Answer:

Let assume that ABCD be the sheet whose

Length, AB = 40 cm

Width, BC = 30 cm

Now, Further given that a border of 3 cm width is printed on a sheet of paper.

Let assume that PQRS be the area of sheet after the 3 cm border.

So,

Length, PQ = 40 - 3 × 2 = 40 - 6 = 34 cm

Breadth, QR = 30 - 3 × 2 = 30 - 6 = 24 cm

Now,

\begin{gathered} \red{\rm \: Area_{(Border)} = Area_{(Rectangle \: ABCD)} - Area_{(Rectangle \: PQRS)}} \\ \end{gathered}

Area

(Border)

=Area

(RectangleABCD)

−Area

(RectanglePQRS)

\begin{gathered}\rm \: = \: AB \times BC - PQ \times QR \\ \end{gathered}

=AB×BC−PQ×QR

\begin{gathered}\rm \: = \: 40 \times 30 - 34 \times 24 \\ \end{gathered}

=40×30−34×24

\begin{gathered}\rm \: = \: 1200 - 816 \\ \end{gathered}

=1200−816

\begin{gathered}\rm \: = \: 384 \: {cm}^{2} \\ \\ \end{gathered}

=384cm

2

So,

\begin{gathered}\bf \: Area_{(Border \: of \: 500 \: sheets)} = 500 \times 384 = 192000 \: {cm}^{2} \\ \\ \end{gathered}

Area

(Borderof500sheets)

=500×384=192000cm

2

Further given that,

\begin{gathered}\rm \: Cost \: of \: 100 \: {cm}^{2} \: = \: 50 \: paisa \\ \end{gathered}

Costof100cm

2

=50paisa

\begin{gathered}\rm \: Cost \: of \: 192000 \:{cm}^{2} = 192000 \times \frac{50}{100} = 96000 \: paisa \\ \end{gathered}

Costof192000cm

2

=192000×

100

50

=96000paisa

\begin{gathered}\bf\implies \: Cost \: of \: 192000 \:{cm}^{2} \: = \: Rs \: 960 \\ \\ \end{gathered}

⟹Costof192000cm

2

=Rs960