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Question Number 208637 by efronzo1 last updated on 20/Jun/24

$s$
	Answered by Berbere last updated on 20/Jun/24
	$find fractin that one man can compete per day V_m and women V_w V_m =(1/(16.24)) V_w =(1/(16.48)) so it just find x ; 14.12.V_m +x(12V_m +12V_w )=1 x.((1/(32))+(1/(64)))=(9/(16))⇒x=4 Total 14+4=18$
	$f i n d f r a c t i n t h a t o n e m a n c a n c o m p e t e p e r d a y V_{m}$ $a n d w o m e n V_{w}$ $V_{m} = \frac{1}{16.24}$ $V_{w} = \frac{1}{16.48}$ $s o i t j u s t f i n d x;$ $14.12 . V_{m} + x (12 V_{m} + 12 V_{w}) = 1$ $x . (\frac{1}{32} + \frac{1}{64}) = \frac{9}{16} \Rightarrow x = 4$ $T o t a l 14 + 4 = 18$
	Commented by efronzo1 last updated on 20/Jun/24

	$x = 12$
	Answered by A5T last updated on 20/Jun/24

	$I n 1 d a y, t h e 48 w o m e n d i d \frac{1}{16} o f t h e j o b$ $\Rightarrow 1 w o m a n d i d \frac{1}{16 \times 48} o f t h e j o b p e r d a y$ $S i m i l a r l y, 1 m a n w o u l d d o \frac{1}{16 \times 24} p e r d a y$ $(12 \times \frac{1}{16 \times 24} \times 14) + x (\frac{12}{16 \times 24} + \frac{12}{16 \times 48}) = 1$ $\Rightarrow x = 12$
	Answered by MM42 last updated on 20/Jun/24

	$\frac{16 \times 48}{16 \times 24} = 2 \Rightarrow m = 2 w$ $\Rightarrow n u m b e r o f d a y s t o w i t h 12 m e n s$ $16 \times 24 = 12 \times x \Rightarrow x = 32$ $t h e r e m a i n i n g d a y s o f d o i n g w o r k w i t h 12 m e n s$ $32 - 14 = 18$ $12 m & 12 w \equiv 18 m$ $\Rightarrow 12 \times 18 = 18 \times x \Rightarrow x = 12 ✓$
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