Question

ramya bought a rectangular cardboard of length 68/5cm and breadth 45/4cm respectively she wanted to cut the rectangular cardboard into rectangles of width 2 1/5 cm and the length same as that of the cardboard how many pieces can she cut find the dimensons of the leftover cardboard piece
write amswer in step by step please

Answer 1

Given:

The dimension of the rectangular cardboard that Ramya bought:

Length = $6\frac{8}{5}\:cm = \frac{38}{5}\:cm$

Breadth = $4\frac{5}{4}\:cm = \frac{21}{4}\: cm$

Ramya wanted to cut the rectangular cardboard into rectangles of width $2\frac{1}{5} \: cm$ = $\frac{11}{5} \: cm$ and the length the same as that of the cardboard.

To find:

(i) How many pieces can she cut?

(ii) Find the dimensions of the leftover cardboard piece?

Solution:

Finding the area of the rectangular cardboard:

Area of the rectangular cardboard is,

= length × breadth

= [ $\frac{38}{5} \times \frac{21}{4}$ ] cm²

Finding the area of the rectangles that are cut from the cardboard:

Area of each piece of a rectangle cut from cardboard is,

= length × breadth

= [$\frac{38}{5} \times \frac{11}{5}$] cm²

Finding the no. of pieces that she can cut:

Let "n" represent the no. of pieces that Ramya can cut from the cardboard.

∴ n = $\frac{Area \:of\: the\: cardboard}{Area \:of \:each\:piece}$

⇒ n = $\frac{\frac{38}{5} \times \frac{21}{4}}{\frac{38}{5} \times \frac{11}{5}}$

⇒ n = $\frac{ \frac{21}{4}}{ \frac{11}{5}}$

⇒ n = $\frac{5.25}{2.2}$

⇒ n = 2.3 ≈ 2

∴ $\boxed{\bold{Ramya \:can\: cut\:\underline{2 \:pieces}\:from\:the\:cardboard.}}$

Finding the dimensions of the leftover cardboard piece:

From the above calculations, we can see that since the length of the pieces of the rectangles is the same as that of the cardboard, so, the 2 pieces are cut along the length from the cardboard.

∴ The width of the leftover cardboard piece will be,

=  $\frac{21}{4} - \frac{22}{5}$

= $\frac{105 - 88}{20}$

= $\frac{17}{20}\:cm$

∴ $\boxed{\bold{The \:dimensions\: of\: the\: leftover\: cardboard \:piece \:is\:\underline{6\frac{8}{5}\:cm\times\frac{17}{20}\:cm}. }}$

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