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?


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Question Number 174613 by Mastermind last updated on 05/Aug/22

$In a mixture of Skettles and {M & M}^{'} s,$ $80 % of the pieces are {M & M}^{'} s . A fourth$ $of this mixture is replaced by a second$ $mixture, resulting in combination$ $which contain 16 % Skittles in total .$ $What was the percentage of Skittles$ $in the second mixture ?$
	Answered by nimnim2 last updated on 05/Aug/22

	$1 s t m i x t u r e = 100$ $\Rightarrow s k i t t l e s = 20, {M & M}^{'} s = 80$ $A f o u r t h o f m i x t u r e i s r e p l a c e d$ $∴ r e m a i n i n g m i x t u r e,$ $s k i t t l e s = 15, {M & M}^{'} s = 60$ $c l e a r l y 1 o f s k i t t l e s a n d 24 o f {M & M}^{'} s i s r e q u i r e d$ $∴ % o f s k i t t l e s i n t h e 2 n d m i x t u r e = \frac{1}{1 + 24} \times 100$ $= 4 %$
	Commented by Mastermind last updated on 05/Aug/22

	$Please with deep explanation, how did$ $you get skittles as 15 and {M & M}^{'} s as 60$
	Commented by Mastermind last updated on 05/Aug/22

	$The answer will be 4 % when {there}^{'} s$ $no skittles in the second replaced$ $mixture$
	Commented by nimnim2 last updated on 06/Aug/22

	$f o u r t h p a r t w a s t a k e n o u t . \Rightarrow r e m a i n i n g = \frac{3}{4} p a r t$ $∴ \frac{3}{4} o f 20 = 15 a n d \frac{3}{4} o f 80 = 60$
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