## Challenge:

A length of a rectangle is increased by \(20\%\). What percent of its width should be decreased to keep the area of the rectangle the same?

**A-** \(15\%\)

**B-** \(16\frac{2}{3}\%\)

**C-** \(20\%\)

**D-** \(25\%\)

**E-** \(30\%\)

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The correct answer is B.

The area of a rectangle is the product of its width and length. Let x be the area of the rectangle.

Area of a rectangle = Width × Length = x

Length → 1.2Length → Area of a rectangle = Width × 1.2Length = x

Let W1 be the width of the first rectangle and W2 be the width of the second rectangle.

W1 × 1.2L = W2 × L → W1 × 1.2 = W2 → W1 \(= \frac{1}{1.2}\)W2 →

W1 \(= \frac{5}{6 }\)W2

To keep the area of the rectangle the same \(\frac{1}{6 }\) or \(16\frac{2}{3}\) percent of the width should be decreased.

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