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In-a-mixture-of-Skettles-and-M-amp-M-s-80-of-the-pieces-are-M-amp-M-s-A-fourth-of-this-mixture-is-replaced-by-a-second-mixture-resulting-in-combination-which-contain-16-Skittles-in-total-What-w




Question Number 174613 by Mastermind last updated on 05/Aug/22
In a mixture of  Skettles and M&M′s,  80% of the pieces are M&M′s. A fourth  of this mixture is replaced by a second  mixture, resulting in combination  which contain 16% Skittles in total.  What was the percentage of Skittles  in the second mixture?
$$\mathrm{In}\:\mathrm{a}\:\mathrm{mixture}\:\mathrm{of}\:\:\mathrm{Skettles}\:\mathrm{and}\:\mathrm{M\&M}'\mathrm{s}, \\ $$$$\mathrm{80\%}\:\mathrm{of}\:\mathrm{the}\:\mathrm{pieces}\:\mathrm{are}\:\mathrm{M\&M}'\mathrm{s}.\:\mathrm{A}\:\mathrm{fourth} \\ $$$$\mathrm{of}\:\mathrm{this}\:\mathrm{mixture}\:\mathrm{is}\:\mathrm{replaced}\:\mathrm{by}\:\mathrm{a}\:\mathrm{second} \\ $$$$\mathrm{mixture},\:\mathrm{resulting}\:\mathrm{in}\:\mathrm{combination} \\ $$$$\mathrm{which}\:\mathrm{contain}\:\mathrm{16\%}\:\mathrm{Skittles}\:\mathrm{in}\:\mathrm{total}. \\ $$$$\mathrm{What}\:\mathrm{was}\:\mathrm{the}\:\mathrm{percentage}\:\mathrm{of}\:\mathrm{Skittles} \\ $$$$\mathrm{in}\:\mathrm{the}\:\mathrm{second}\:\mathrm{mixture}? \\ $$
Answered by nimnim2 last updated on 05/Aug/22
1st mixture=100   ⇒skittles=20 , M&M′s=80  A fourth of mixture is replaced  ∴ remaining mixture,       skittles=15, M&M′s=60  clearly 1 of skittles and 24 of M&M′s is required  ∴ %of skittles in the 2nd mixture=(1/(1+24))×100                                                                                 =4%
$$\mathrm{1}{st}\:{mixture}=\mathrm{100}\: \\ $$$$\Rightarrow{skittles}=\mathrm{20}\:,\:{M\&M}'{s}=\mathrm{80} \\ $$$${A}\:{fourth}\:{of}\:{mixture}\:{is}\:{replaced} \\ $$$$\therefore\:{remaining}\:{mixture},\: \\ $$$$\:\:\:\:{skittles}=\mathrm{15},\:{M\&M}'{s}=\mathrm{60} \\ $$$${clearly}\:\mathrm{1}\:{of}\:{skittles}\:{and}\:\mathrm{24}\:{of}\:{M\&M}'{s}\:{is}\:{required} \\ $$$$\therefore\:\%{of}\:{skittles}\:{in}\:{the}\:\mathrm{2}{nd}\:{mixture}=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{24}}×\mathrm{100} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{4\%} \\ $$
Commented by Mastermind last updated on 05/Aug/22
Please with deep explanation, how did  you get skittles as 15 and M&M′s as 60
$$\mathrm{Please}\:\mathrm{with}\:\mathrm{deep}\:\mathrm{explanation},\:\mathrm{how}\:\mathrm{did} \\ $$$$\mathrm{you}\:\mathrm{get}\:\mathrm{skittles}\:\mathrm{as}\:\mathrm{15}\:\mathrm{and}\:\mathrm{M\&M}'\mathrm{s}\:\mathrm{as}\:\mathrm{60} \\ $$
Commented by Mastermind last updated on 05/Aug/22
The answer will be 4% when there′s  no skittles in the second replaced   mixture
$$\mathrm{The}\:\mathrm{answer}\:\mathrm{will}\:\mathrm{be}\:\mathrm{4\%}\:\mathrm{when}\:\mathrm{there}'\mathrm{s} \\ $$$$\mathrm{no}\:\mathrm{skittles}\:\mathrm{in}\:\mathrm{the}\:\mathrm{second}\:\mathrm{replaced}\: \\ $$$$\mathrm{mixture} \\ $$
Commented by nimnim2 last updated on 06/Aug/22
fourth part was taken out. ⇒ remaining=(3/4) part  ∴ (3/4) of 20=15 and (3/4) of 80=60
$${fourth}\:{part}\:{was}\:{taken}\:{out}.\:\Rightarrow\:{remaining}=\frac{\mathrm{3}}{\mathrm{4}}\:{part} \\ $$$$\therefore\:\frac{\mathrm{3}}{\mathrm{4}}\:{of}\:\mathrm{20}=\mathrm{15}\:{and}\:\frac{\mathrm{3}}{\mathrm{4}}\:{of}\:\mathrm{80}=\mathrm{60} \\ $$

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