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 ?

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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