One Bad Mango Spoils The Bunch

Let the total number of mangoes be denoted as $x$ . From the start, there are 80% of $x$ number of high-quality mangoes and 20% of $x$ number of low-quality mangoes.

Then when 10 low-quality mangoes are removed, the new total number is $x - 10$ .

And we know that 96% of this new total have high-quality is equal to $96\% \cdot (x-10) = 80\% \cdot x$ as the number of high-quality ones doesn't change.

Then solving for $x$ :

$16\% \cdot x - 9.6 \Rightarrow x = \dfrac{9.6\times100}{16} = \boxed{60} \text{ mangoes}.$

Nice question. ;)

80℅ of a number would become 96℅, if the number is reduced to 80/96=5/6th of the original number. In other words, if the number is reduced by 1/6th. Here the number is reduced by 10 which is 1/6th of 60, therefore, the answer is 60 mangoes.

Let $x$ be the total number of mangoes.

Given that 10 low-quality mangoes are removed, then, $0.2x - 10$ mangoes comprises 4% of the remaining number of mangoes represented by $x - 10$ .

$0.2x - 10 = 0.04(x-10)$

$0.2x - 0.04x = 10 - 0.4$

$0.16x = 9.6$

$\boxed{x = 60}$

we let $x$ be the number of mangoes from the start

$0.8x$ = number of high quality mangoes

$0.2x$ = number of low quality mangoes

Now we remove $10$ pieces low quality mangoes, the total number of mangoes becomes $x-10$ and the percentage of high quality mangoes becomes $96$ %. So we set up our equation,

$\frac{0.8x}{x-10}$ $=0.96$

$0.8x=0.96x-9.6$

$x=60$

Let "L" denote low quality mangoes and let "H" denote high quality mangoes. Then: $\frac{L}{H}=\frac{20}{80}$ If then 10 low-quality mangoes are subtracted, the percentage of high-quality mangoes is equal to 96% $\Rightarrow\frac{L-10}{H}=\frac{4}{96}$ We can now split up the fraction $\frac{L-10}{H}$ into $\frac{L}{H}-\frac{10}{H}$ Since $\frac{L}{H}=\frac{20}{80}$ ; $\Rightarrow\frac{20}{80}-\frac{10}{H}=\frac{4}{96}$ After cancelling down the fractions we get $\frac{1}{4}-\frac{10}{H}=\frac{1}{24}$ Thus $\frac{10}{H}=\frac{1}{4}-\frac{1}{24}$ which equals $\frac{5}{24}$

Hence $5\times H=240$

Therefore, there are 48 high-quality mangoes. Initially, these 48 mangoes were 80% of all the mangoes $\Rightarrow$ Hence 100% of the mangoes is equal to $\boxed{60}$ .

Superior quality mangoes = 80%, Inferior quality mangoes = 20%, 10 mangoes = 16%, 100/16 = 62.5

closest value is 60.