L\] and b respectively. Final length\[=l\TIMES \frac{160}{100}\] Final width\[={{b}_{1}}\](SAY) Initial AREA = final area \[l\times b=l\times \frac{160}{100}\times {{b}_{1}}\Rightarrow {{b}_{1}}=\frac{100}{160}=\frac{5}{8}b\] Percentage decrement in width \[=\frac{b-{{b}_{1}}}{{{b}_{1}}}\times 100=\frac{b-\frac{5}{8}b}{b}\times 100%\] \[=37.5%\]

"> L\] and b respectively. Final length\[=l\TIMES \frac{160}{100}\] Final width\[={{b}_{1}}\](SAY) Initial AREA = final area \[l\times b=l\times \frac{160}{100}\times {{b}_{1}}\Rightarrow {{b}_{1}}=\frac{100}{160}=\frac{5}{8}b\] Percentage decrement in width \[=\frac{b-{{b}_{1}}}{{{b}_{1}}}\times 100=\frac{b-\frac{5}{8}b}{b}\times 100%\] \[=37.5%\]

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The length a rectangle is increased by 60%. By what percentage would the width have to be decreased to maintain the same area

7th Class Mathematics in 7th Class 11 months ago

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Let the initial length and width are \[L\] and b respectively. Final length\[=l\TIMES \frac{160}{100}\] Final width\[={{b}_{1}}\](SAY) Initial AREA = final area \[l\times b=l\times \frac{160}{100}\times {{b}_{1}}\Rightarrow {{b}_{1}}=\frac{100}{160}=\frac{5}{8}b\] Percentage decrement in width \[=\frac{b-{{b}_{1}}}{{{b}_{1}}}\times 100=\frac{b-\frac{5}{8}b}{b}\times 100%\] \[=37.5%\]

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