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In an examination, the score of $\text{A}$ was $10\%$ less than that of $\text{B}$, the score of $\text{B}$ was $25\%$ more than that of $\text{C}$, and the score of $\text{C}$ was $20\%$ less than that of $\text{D}$. If $\text{A}$ scored $72$, then the score of $\text{D}$ was _______
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Let the score of $B=100\%$

Then, the score of $A = 100\%-10\times \left(\frac{100}{100}\right)\% = 90\%$

The ratio of $A$ and $B=90\%:100\%$

  • $A:B=9:10 \quad \longrightarrow (1)$

Let the score of $C=100\%$

Then the score of $B=100\%+100\times \left(\frac{25}{100}\right)\% = 125\%$

The ratio of $B$ and $C=125\%:100\%$

  • $B: C=5: 4 \quad \longrightarrow (2)$

Let the score of $D=100\%$

Then the score of $C =100\%-100\times \left(\frac{20}{100}\right)\% = 80\%$

The ratio of $C$ and $D=80\%:100\%$

  • $C: D=4: 5 \quad  \longrightarrow (3)$

Now, from the eq$^{n} \; (2)$ and $(3)$, combining the ratios, we get 

  • $B: C: D = 5: 4: 5$
  • $B: C: D = 10: 8: 10 \quad [\because \text{Multiply by ‘2’, for combining the ratios}]\quad \longrightarrow (4)$

Now, from the eq$^{n} \; (1)$ and $(4)$, combining the ratios, we get

  • $A : B : C : D = 9 : 10 : 8 : 10$

Given that, the score of $A=72$

Let $x$ be a score of $D$. Then,

  • $9 \longrightarrow 72$
  • $10 \longrightarrow x$

 $ \implies 9x=72\times10$

 $ \implies x=80$

$\textbf{Short Method:}$

The score of $D=72 \times \left(\frac{100}{90}\right) \times \left(\frac{100}{125}\right) \times \left(\frac{100}{80}\right) = 80$

$\therefore$ The score of $D$ was $80.$

Correct Answer $:80$

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