Question 409077
I think they are saying that the area of the pipe opening has been
reduced by 20%.
Let {{{A}}} = original area
Let {{{r}}} = original radius
----------------------
If the diameter is reduced by 1 mm, then the 
radius is reduced by .5 mm.
given:
(1) {{{A = pi*r^2}}}
(2) {{{A - .2A = pi*(r - .5)^2}}}
------------------------
(2) {{{.8A = pi*(r - .5)^2}}}
substituting from (1):
(2) {{{.8*pi*r^2 = pi*(r^2 - r + .25)}}}
(2) {{{.8*r^2 = r^2 - r + .25}}}
(2) {{{.2r^2 - r + .25 = 0}}}
(2) {{{20r^2 - 100r + 25 = 0}}}
(2) {{{4r^2 - 20r + 5 = 0}}}
Use quadratic formula
{{{r = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
{{{a = 4}}}
{{{b = -20}}}
{{{c = 5}}}
{{{r = (-(-20) +- sqrt( (-20)^2-4*4*5 ))/(2*4) }}} 
{{{r = (20 +- sqrt( 400 - 80 ))/8 }}} 
{{{r = (20 +- sqrt( 320 ))/8 }}} 
{{{r = (20 + 8*sqrt(5))/8}}}
{{{r = 5/2 + sqrt(5)}}}
{{{r = 4.736}}} mm
The original diameter = 2r, or 9.472 mm
check answer:
(2) {{{A - .2A = pi*(r - .5)^2}}}
(2) {{{.8A = pi*(4.736 - .5)^2}}}
(2) {{{.8A = pi*4.236^2}}}
(2) {{{.8A = 3.141*17.944}}}
(2) {{{.8A = 56.361}}}
(2) {{{A = 70.451}}} mm2
and
(1) {{{A = pi*r^2}}}
(1) {{{A = 3.141*4.736^2}}}
(1) {{{A = 70.451}}} mm2
OK
Hope I got it