A rectangle was altered by increasing its length by $10$ percent and decreasing its width by $p$ percent. If these alterations decreased the area of the rectangle by $12$ percent, what is the value of $p$ ?

12

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15

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20

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22

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Solution

The correct option is B $20$
Area of a rectangle $=$ Length $\times$ breadth
Let $A_{1} = L_{1} \times B_{1}$
Given, $L_{2} = 10$ % $+ L_{1} = 1. 1 L_{1}$
And , $B_{2} = B_{1} - \frac{p}{100} = \frac{100 B_{1} - p}{100}$
$\Rightarrow A_{2} = 0.88 A_{1}$
$\Rightarrow L_{2} \times B_{2} = 0.88 (L_{1} \times B_{1})$
$\Rightarrow 1. 1 L_{1} \times \frac{100 B_{1} - p}{100} = 0.88 (L_{1} \times B_{1})$
$\Rightarrow 1.1 \times (100 B_{1} - p) = 88 B_{1}$
$\Rightarrow 110 B_{1} - 1.1 p = 88 B_{1}$
$\Rightarrow 22 B_{1} = 1.1 p$
$\Rightarrow p = 20$

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A rectangle was altered by increasing its length by 10 percent and decreasing its width by p percent. If these alterations decreased the area of the rectangle by 12 percent, what is the value of p?

The correct option is B 20Area of a rectangle = Length × breadth Let A1=L1×B1Given, L2=10 % +L1=1.1L1And , B2=B1−p100=100B1−p100⇒A2=0.88A1⇒L2×B2=0.88(L1×B1)⇒1.1L1×100B1−p100=0.88(L1×B1)⇒1.1×(100B1−p)=88B1⇒110B1−1.1p=88B1⇒22B1=1.1p⇒p=20

A rectangle was altered by increasing its length by $10$ percent and decreasing its width by $p$ percent. If these alterations decreased the area of the rectangle by $12$ percent, what is the value of $p$ ?