Aptitude Discussion

Q. |
A student multiplied a number by 3/5 instead of 5/3. What is the percentage error in the calculation? |

✖ A. |
$34\%$ |

✖ B. |
$44\%$ |

✖ C. |
$54\%$ |

✔ D. |
$64\%$ |

**Solution:**

Option(**D**) is correct

Let the number be $x$.

Then, error

\(=\dfrac{5x}{3}-\dfrac{3x}{5}\)

\(=\dfrac{16x}{5}\)

Error $\%$

\(=\dfrac{16x/15}{5x/3}\times 100\)

\(=64\%\)

It is not a bad idea to assume some number and then proceed with the given instructions. Let's assume the number to be $15$.

Now, if the student had multiplied it correctly by $\left(\dfrac{5}{3}\right)$, he/she would have got the result as $=\left(\dfrac{5}{3}\right)\times 15=25$. But instead the student ended up multiplying by $=\left(\dfrac{3}{5}\right)$ and got $=\left(\dfrac{3}{5}\right) \times 15 = 9$ as a result.

Here, we can see that error percentage,

$\text{% error}=\left(\dfrac{25-9}{25}\right)$ $=0.64$ or $\textbf{64%}$

**Edit:** For yet another alternative solution, check comment by **Chirag Goyal.**

**ABHIJEET**

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**Chirag Goyal**

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Shortcut,

$$\text{Error %} = \left(1 - \dfrac{ \text{New fraction Multiplied}}{\text{Original fraction was to be Multiplied}}\right)100$$

Great one (y). Is it general enough or valid only for this question?

It depends on whether the new fraction is smaller than the original one or not.

If new fraction (actually multiplied) is greater than original fraction

Then it would be,

$\text{Error %} = \left( \dfrac{ \text{New fraction Multiplied}}{\text{Original fraction was to be Multiplied}}-1\right)100$

Got it, Thank you :)

ye student pakka sindhi h .. :P :P