View Full Version : Inverse proportion..

momoko

06-22-2009, 07:01 AM

Two quantities, x and y, are in inverse proportion. It is known that y = 120 for a particular value of x.

Find the value of y when the value of x is increased by 200%.

The steps I have done thus far:

k = xy, where k is a constant.

When y = 120,

k = 120x

x = k/120

How do I continue? Or is it a totally wrong method? Thanks in advance :)

Subhotosh Khan

06-22-2009, 07:13 AM

Two quantities, x and y, are in inverse proportion. It is known that y = 120 for a particular value of x.

Find the value of y when the value of x is increased by 200%.

The steps I have done thus far:

k = xy, where k is a constant.

When y = 120,

k = 120x

x = k/120

How do I continue? Or is it a totally wrong method? Thanks in advance :)

Inverse proportion means

x is proportional to 1/y. Then,

x = k * (1/y) ............ where k is constant

x*y = k

x[sub:2a6ouq1l]1[/sub:2a6ouq1l] * y[sub:2a6ouq1l]1[/sub:2a6ouq1l] = k

x[sub:2a6ouq1l]1[/sub:2a6ouq1l] * 120 = k .........................(1)

x[sub:2a6ouq1l]2[/sub:2a6ouq1l] * y[sub:2a6ouq1l]2[/sub:2a6ouq1l] = k ..............(2)

the value of x is increased by 200%

My interpretetion of that statement is

x[sub:2a6ouq1l]2[/sub:2a6ouq1l] = 3x[sub:2a6ouq1l]1[/sub:2a6ouq1l] .....................(3)

there is a possibility that the questioneer may have meant to imply x[sub:2a6ouq1l]2[/sub:2a6ouq1l] = 2x[sub:2a6ouq1l]1[/sub:2a6ouq1l].... but in that case the statement should have been the value of x is increased to200%of the original value

using (3) in (1)

3*x[sub:2a6ouq1l]1[/sub:2a6ouq1l] * y[sub:2a6ouq1l]2[/sub:2a6ouq1l] = k ...............(4)

using (1) in (4)

3*x[sub:2a6ouq1l]1[/sub:2a6ouq1l] * y[sub:2a6ouq1l]2[/sub:2a6ouq1l] = x[sub:2a6ouq1l]1[/sub:2a6ouq1l] * 120

y[sub:2a6ouq1l]2[/sub:2a6ouq1l] = 40

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