300, gm of salt solution has 40 % salt in it. | vedclass

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(B) Initial amount of salt in $300\, gm$ solution $= 300 	imes 0.40 = 120\, gm$.Let $x\, gm$ of salt be added.New amount of salt $= 120 + x$.New tota

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(B) Initial amount of salt in $300\, gm$ solution $= 300 	imes 0.40 = 120\, gm$.Let $x\, gm$ of salt be added.New amount of salt $= 120 + x$.New total weight of solution $= 300 + x$.According to the problem,the new concentration is $50 \%$,so:$\frac{120 + x}{300 + x} = \frac{50}{100} = \frac{1}{2}$.$2(120 + x) = 300 + x$.$240 + 2x = 300 + x$.$x = 300 - 240 = 60\, gm$.Alternatively,using the alligation method:Initial concentration $= 40 \%$,Added salt concentration $= 100 \%$,Target concentration $= 50 \%$.The ratio of initial solution to added salt is $(100 - 50) : (50 - 40) = 50 : 10 = 5 : 1$.Since the initial solution is $300\, gm$,the added salt $= \frac{300}{5} 	imes 1 = 60\, gm$.

$300\, gm$ of salt solution has $40 \%$ salt in it. How much salt should be added to make it $50 \%$ in the solution? (in $gm$)

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