Easy Percentages Solved QuestionAptitude 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.

(6) Comment(s)

Aman Sharma
()

nyc trick bro..givemore trick of short type

Zeeshan
()

ABHIJEET
()

zeeshan.. his name is kartik not karthick

KARTIK
()

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 + 16B3A=4BA:B = 4:3\$

Jessy
()

Yo bro, that's a cool way man.

Naaz
()

Thank you kartik, really very helpful :)