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secondary 2 | Maths
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Date Posted: 3 years ago
Views: 311
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p is directly proportional to √q
So p = k√q , k is a constant.
When q is increased by h%, p increases by 200%
Now 200% = 200/100 = 2
So p increased by 2 times. It is now 3 times the original value, or 3p.
q is now (100+h)% or (1 + h/100) of original.
So, 3p = k√((1 + h/100)q)
But remember that p = k√q. So we substitute it into the new equation.
3k√q = k√((1+ h/100)q)
3√q = √((1+ h/100)q)
3√q = √(1 + h/100) √q
3 = √(1 + h/100)
3² = 1 + h/100
9 = 1 + h/100
h/100 = 8
h = 800
So p = k√q , k is a constant.
When q is increased by h%, p increases by 200%
Now 200% = 200/100 = 2
So p increased by 2 times. It is now 3 times the original value, or 3p.
q is now (100+h)% or (1 + h/100) of original.
So, 3p = k√((1 + h/100)q)
But remember that p = k√q. So we substitute it into the new equation.
3k√q = k√((1+ h/100)q)
3√q = √((1+ h/100)q)
3√q = √(1 + h/100) √q
3 = √(1 + h/100)
3² = 1 + h/100
9 = 1 + h/100
h/100 = 8
h = 800
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