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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 byMastermind 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 byMastermind 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 bynimnim2 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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