Percentage Practice Test 2 with solutions

Question : 36 of 40

Marks: +1, -0

The number of grams of water needed to reduce 9 grams of shaving lotion containing 50% alcohol to a lotion containing 30% alcohol, is :

Solution:

Alcohol in 9 gms

=50/100×9=4.5\gms

Let x gm of water be added.
Then,

4.5/{9+x}×100=30

⇒ 270 +30x = 450 ⇒ 30x = 180 ⇒ x = 6 gms

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