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The length of a rectangle is increased by 60%. By what percent should the width be decreased to maintain the same area?
A. 35.5
B. 36.5
C. 37.5
D. 38.5

Last updated date: 26th Feb 2024
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IVSAT 2024
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Hint: Suppose the original dimensions of rectangle be l and b. So, its area will be their product or lb. Now according to the question, if new length will let’s say ${{l}^{'}}\text{ and }{{\text{b}}^{'}}$ here ${{l}^{'}}$ will \[l+\dfrac{60l}{100}\text{ }\Rightarrow \text{ }\dfrac{160l}{100}\text{ }\Rightarrow \text{ }\dfrac{8l}{5}\] then, find the area and equate it with original one. Then find the percentage decrease by using formula,
\[\dfrac{\text{Decrease}}{\text{Original}}\times 100%\]

Complete step by step answer:
In the question, we are told about a rectangle that if its length is increased by 60. Then we have to find by what percent breadth should be decreased to maintain the same area.
So, let’s suppose that originally the length and breadth of a triangle be taken as l and b respectively. So, the area of rectangle fill product of length and breadth \[\Rightarrow l\times b\text{ }\Rightarrow \text{ lb}\text{.}\]
Now in the given question we are told that length has increased by 60%. So, the new length be let's say ${{l}^{'}}$ will be represented as \[l+\dfrac{60l}{100}\text{ or }\dfrac{160l}{100}\text{ or }\dfrac{8l}{5}\].
Suppose, the new breadth be considered as ${{b}^{'}}$ then, its area will be product of length and breadth \[\Rightarrow \dfrac{8l}{5}\times b'\Rightarrow \dfrac{8lb'}{5}\].
Now we know that area of rectangle remains unchanged, so we get,
\[lb=\dfrac{8lb'}{5}\text{ }\]
\[\Rightarrow \text{b=}\dfrac{8b'}{5}\]
We can write \[b'=\dfrac{5b}{8}\]
The original breadth was b, the new breadth should be \[\dfrac{5b}{8}\] the decrease in breadth is \[b-\dfrac{5b}{8}\text{ }\Rightarrow \dfrac{3b}{8}.\]
Now we will find the decrease in percentage using formula,
\[\dfrac{\text{Decrease}}{\text{Original}}\times 100%\]
Here, decrease is \[\dfrac{3b}{8}\] original is b. So, its percentage would be:
\[\dfrac{\dfrac{3b}{8}}{b}\times 100%\text{ =}\dfrac{3}{8}\times 100%\text{ }\]
\[\Rightarrow \text{37}\text{.5 }\!\!%\!\!\text{ }\]

So, the correct answer is “Option C”.

Note: Students can also find the decrease percent by using the unitary method, by just taking b as 100% and then finding the percentage of \[\dfrac{3b}{8}\] to get the answer.