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(Maths) Can someone help please??

X is proprtional to the square root of y where y >0
Y is increased by 44%
Work out the percentage increase in x
Original post by Akhtee3
X is proportional to the square root of y where y >0
Y is increased by 44%
Work out the percentage increase in x

There a few different ways of working this out.
1) x is proportional to the square root of y. So, if y doubles, x increases by 2\sqrt{2}. So an increase in y of 44% leads to an increase in x of 1.44=1.2\sqrt{1.44}=1.2, or 20%.

2) (A bit more formal) X is proportional to square root of y, so x=kyx=k\sqrt{y}. Rearranging for the constant, k:
k=xyk=\frac{x}{\sqrt{y}}.
This constant is the same for all x and y. Let x1x_1 and y1y_1 be the original values and x2x_2 and y2y_2 be the new values, where y2=1.44y1y_2 = 1.44y_1.

x1y1=x2y2\frac{x_1}{\sqrt{y_1}}=\frac{x_2}{\sqrt{y_2}}

The fraction x2x1\frac{x_2}{x_1} is the number of times bigger x2x_2 is than x1x_1.

x2x1=y2y1=y2y1=1.44y1y1=1.44=1.2\frac{x_2}{x_1}=\frac{\sqrt{y_2}}{\sqrt{y_1}}=\sqrt{\frac{y_2}{y_1}}=\sqrt{\frac{1.44y_1}{y_1}}=\sqrt{1.44}=1.2. Therefore a 20% increase.

3) You could pick actual numbers and see what happens.
(edited 2 years ago)

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