Subject :IMO Class : Class 6
Subject :IMO Class : Class 7
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Subject :IMO Class : Class 8
Ans 2:
Class : Class 9
Quantity of mixture = 400 ml.
Quantity of pure alcohol in it
= 15 % of 400
= 15/100 * 400
= 60 ml
Let's consider 'x ml' of pure alcohol was added to the mixture.
After adding,
Quantity of mixture = (400 + x) ml
Quantity of pure alcohol in it = (60 + x) ml
Strength
= (60 + x)/(400 + x) = 32/100
=> 100(60 + x) = 32(400 + x)
=> 6000 + 100x = 12800 + 32x
=> 100x - 32x = 12800 - 6000
=> 68x = 6800
=> x = 6800/68
=> x = 100
Hence answer is 100 ml
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Subject :IMO Class : Class 6
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Subject :IMO Class : Class 6
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Subject :IMO Class : Class 7
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Subject :IMO Class : Class 5
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Subject :IMO Class : Class 4