Aptitude Discussion

Q. |
Two numbers $A$ and $B$ are such that the sum of 5% of $A$ and 4% of $B$ is two-third of the sum of 6% of $A$ and 8% of $B$. Find the ratio of $A : B.$ |

✖ A. |
2 : 3 |

✖ B. |
1 : 1 |

✖ C. |
3 : 4 |

✔ D. |
4 : 3 |

**Solution:**

Option(**D**) is correct

$5\% \text{ of } A + 4\% \text{ of } B = \dfrac{2}{3}(6\% \text{ of } A + 8\% \text{ of } B)$

$\Rightarrow \left(\dfrac{5}{100}\right)×A + \left(\dfrac{4}{100}\right)×B = \dfrac{2}{3}\left(\dfrac{6}{100}×A + \dfrac{8}{100}×B\right)$

$\Rightarrow \left(\dfrac{1}{20}\right)×A + \left(\dfrac{1}{25}\right)×B = \left(\dfrac{1}{25}\right)×A + \left(\dfrac{4}{75}\right)×B$

$\Rightarrow \left(\dfrac{1}{2} - \dfrac{1}{25}\right)×A = \left(\dfrac{4}{75} - \dfrac{1}{25}\right)×B$

$\Rightarrow \left(\dfrac{1}{100}\right)×A = \left(\dfrac{1}{75}\right)×B $

$\Rightarrow \dfrac{A}{B} = \dfrac{100}{75} = \dfrac{4}{3}$

So required ratio = **4:3**

**Edit:** For easier calculation, check out **KARTIK's observation.**

**Aman Sharma**

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**Zeeshan**

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Karthic ur answer is 3:4 (answer c)

**KARTIK**

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you could cancel off the % on both side by taking \dfrac{1}{100}$ common. that way it would be easy to solve the equation.

$5A+4B=(2/3)(6A+8B)$

$15A+12B = 12A + 16B$

$3A=4B$

$A:B = 4:3$

Yo bro, that's a cool way man.

Thank you kartik, really very helpful :)

nyc trick bro..givemore trick of short type