5 Marks Questions

Question 15 Marks

The perimeter of a rectangle is $240\ cm$. If its length is increased by $10\%$ and its breadth is decreased by $20\%$, we get the same perimeter. Find the length and breadth of the rectangle.

Answer

Let the weight of box $A$ be $x$ kg.
According to the question, Weight of Box $A =$ Weight of Box $\text{B}+\frac{5}{2}\text{kg}$
Weight of box $B$ $=\Big(\text{x}-\frac{5}{2}\Big)\text{kg}$ &
Weight of Box $C$ = Weight of box $\text{B}+\frac{41}{4}\text{kg}$
Weight of Box $C$ $=\text{x}-\frac{5}{2}+\frac{41}{4}\text{kg}=\text{x}+\Big(\frac{-10+41}{4}\Big)\text{kg}=\Big(\text{x}+\frac{31}{4}\Big)\text{kg}$
As, total weight of three boxes $=48\frac{3}{4}\text{kg}=\frac{195}{4}\text{kg}$
Total weight of three boxes = Weight of Box $A$ + Weight of Box $B$ + Weight of Box $C$
$\frac{195}{4}=\text{x}+\text{x}-\frac{5}{2}+\text{x}+\frac{31}{4}$
$\frac{195}{4}=3\text{x}+\frac{31-10}{4}$
$\frac{195}{4}=3\text{x}+\frac{21}{4}$
$3\text{x}=\frac{195}{4}-\frac{21}{4}$
$3\text{x}=\frac{195-21}{4}$
$3\text{x}=\frac{174}{4}$
$\text{x}=\frac{174}{3\times4}$
$\text{x}=\frac{29}{2}=14\frac{1}{2}\text{kg}$
Hence, Box $A$ weighs $14\frac{1}{2}\text{kg}$

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Question 25 Marks

The perimeter of a rectangle is $240\ cm$. If its length is increased by 10% and its breadth is decreased by $20\%$, we get the same perimeter. Find the length and breadth of the rectangle.

Answer

Given, perimeter of rectangle
$= 240\ m\ 2 \times $ (Length + Breadth)
$= 240\ m$ Length + Breadth
$= 120\ m$
Let the length of rectangle be $x\ cm$ Then, the breadth of rectangle $= (120 - x)cm$
New length of rectangle $= x + 10\%$ of $x$ $=\text{x}+\frac{10}{100}\times\text{x}$
$=\frac{110}{100}\text{x }\text{cm}$ &
new length of rectangle $= (120 - x) - 20\%$ of $(120 - x)cm$
$=(120-\text{x})-\frac{20}{200}\times(120-\text{x})$
$=(120-\text{x})\Big(1-\frac{20}{100}\Big)=\frac{80}{100}(120-\text{x})\text{cm}$
Given, perimeter of new rectangle is $240\ cm$
According to the question, $2$ $\times $ (New length + New breadth) = 240
$2\times\Big[\frac{110\text{x}}{100}+\frac{90}{100}(120-\text{x})\Big]=240$
$\frac{110\text{x}+9600-80\text{x}}{100}=\frac{240}{2}$
$30\text{x}+9600=\frac{240}{2}\times100$
$30x + 9600 = 12000$
$30x = 12000 - 9600$
$30x = 2400$
$\text{x}=\frac{2400}{30}=80$
Hence, length of rectangle $= 80\ cm$ & breadth of rectangle $= 120 - 80 = 40\ cm$

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