Question 1100944
ONE APPROACH:
{{{N}}}= number of bikes in the shop waiting to be repaired.
Of those,
25% have flat tires only,
5% have bent handlebars only, and
10% have ripped seats only.
That adds up to 25%+5%+10%=40% .
So, the number of bikes needing only 1 repair is
{{{(40/100)N=(2/5)N}}} .
The number of bikes needing 3 repairs is {{{(1/12)N}}} .
The rest of the bikes is the fraction of all bikes that need exactly 2 repairs.
As a fraction of all the bikes, that is
{{{1-2/5-1/12=60/60-24/60-5/60=31/60}}} .
As a number of bikes, it is
{{{N-(2/5)N-(1/12)N=(1-2/5/1/12)N=(31/60)N}}} .
Counting the number of repairs needed, we have
{{{(2/5)N}}} repairs for the {{{(2/5)N}}} bikes that need only 1 repair,
{{{2(31/60)N=(31/30)N}}} repairs for the {{{(31/60)N}}} bikes that need exactly 2 repairs,
and
{{{3(1/12)N=(1/4)N}}} repairs for the {{{(1/12)N}}} bikes that need 3 repairs.
The number of repairs adds up to
{{{(2/5)N+(31/60)N+(1/4)N=101}}}
{{{(2/5+31/60+1/4)N=101}}}
{{{(24/60+31/60+15/60)N=101}}}
{{{(101/60)N=101}}}
{{{(60/101)(101/60)N=(30/101)101}}}
{{{highlight(N=60)}}}
There are {{{highlight(60)}}} bikes in the repair department.
 
Of those {{{60}}} bikes,
{{{(31/60)60=highlight(31)}}} need exactly two repairs,
{{{1/12}}} , or {{{5}}} bikes, need all 3 repairs including their ripped seats,
25% , or {{{15}}} bikes, only need flat tires fixed,
5% , or {{{3}}} bikes, only need bent handlebars fixed, and
10%, or {{{6}}} bikes only need their ripped seats fixed.
In pictures, we could represent is as
{{{drawing(400,400,-2.6,2.6,-1.75,3.45,
red(circle(0,1.732,1.49)),red(circle(0,1.732,1.51)),
red(arrow(2,3.031,0.75,3.031)),locate(2.05,3.15,red(R)),
locate(1.88,2.97,red("( ripped")),locate(1.9,2.75,red("seats )")),
locate(0,2,6),locate(0,0.77,5),
circle(1,0,1.5),circle(-1,0,1.5),
locate(-1.2,0,15),locate(1.2,0,3),
arrow(-2.06,1.9,-2.06,1.06),locate(-2.3,2.1,tires),
arrow(2.06,1.9,2.06,1.06),locate(1.7,2.1,handlebars),
locate(-0.8,1,"?"),locate(0.75,1,"?"),locate(0,-0.1,x)
)}}} .

The last question asks for the range of values for value of {{{x}}} ,
The number of "bikes
that need both the tires and handlebars repaired without needing to fix the seat".
 
From the picture,
if {{{"?"+"?"=0}}} , {{{x=60-6-5-15-3=31}}} is the greatest possible {{{x}}} ;
If {{{6+5+"?"+"?"=29}}} , {{{x=60-29-15-3=13}}} is the least possible {{{x}}} .
 
Without a picture:
If {{{R}}} is the total number of ripped seat repairs,
we do not know {{{R}}} exactly, but
it includes at least the {{{6}}} that need only the seat repaired,
and the {{{5}}} that need all 3 repairs, so
{{{R>=6+5}}} or {{{11<=R}}} .
If less than half the bikes have a ripped seat
(meaning {{{29}}} or less bikes),
we also know that {{{R<=29}}} , so
{{{11<=R<=29}}} .
 
If we exclude 
those {{{R}}} bikes with ripped seats,
the {{{15}}} bikes that only need flat tires fixed,
and the {{{3}}} bikes that only need bent handlebars fixed,
we are excluding {{{R+15+3=R+18}}} ,
what is left is the group of bikes that needs
"both the tires and handlebars repaired without needing to fix the seat."
That number is {{{x=60-(R+18)=60-R-18=42-R}}} .
As {{{11<=R<=29}}} ,
{{{42-11>=42-R>=42-29}}}
{{{31>=x>=13}}} .
There are at least {{{13}}} and at most {{{31}}} bikes that need 
"both the tires and handlebars repaired without needing to fix the seat."
The range can be expressed as [13,31], or as {{{13<=x<=31}}}