Question 175135
{{{drawing( 400, 300, -10, 10, -10, 10,
  circle( -3, 0, 4 ),
  circle( 2, 0, 6 ),
  rectangle( -7, -8, 8, 8 ),
  locate( -5, 0, M[p] ), locate( 4, 0, S[p] ), locate( -2, 0, SM[p] ), 
locate( -5, 7, SM[f] )
  )}}}

In the above figure, {{{S[p]}}} represents the set of all students who passed in only Science, {{{M[p]}}} represents the set of all students who passed in only Mathematics, {{{SM[p]}}} represents the set of all students who passed in both Science and Mathematics and {{{SM[f]}}} represents the set of all students who failed in both Science and Mathematics.


Let the total no. of students = {{{N}}}.
So we can write {{{S[p] + M[p] + SM[p] + SM[f] = N }}} ______ (1)


It is given that {{{SM[p] = 475}}}.
Again, {{{S[p] + SM[p] = 0.75N}}} i.e. {{{S[p] + 475 = 0.75N}}}.
Also, {{{M[p] + SM[p] = 0.65N}}} i.e. {{{M[p] + 475 = 0.65N}}}.
Further, {{{SM[f] = 0.15N}}}.


Substituting these into equation (1)
{{{0.75N - 475 + 0.65N - 475 + 475 + 0.15N = N }}}
{{{1.55N - 475 = N }}}
{{{1.55N - N = 475 }}}
{{{0.55N = 475 }}}
{{{N = 475/0.55 }}}
{{{N = 95}}}

Thus, a total of 95 students appeared for the examination.