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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