# Some T-shirts are purchased for \$1000. If 5 more had been purchased for the same amount they would have cost 12 cents less per shirt. How many shirts were purchased.

justaguide | Certified Educator

N shirts were purchased for \$1000. This gives the price per shirt as `1000/N` . If 5 more shirts are bought for \$1000 the price per shirt is `1000/(N + 5)` which is 12 cents less than `1000/N` . This gives:

`1000/N - 0.12 = 1000/(N + 5) `

=> `(1000 - 0.12N)/N = 1000/(N + 5)`

=> `(N + 5)*(1000 - 0.12N) = 1000*N `

=> `1000N + 5000 - 0.12N^2 - 0.6N = 1000*N`

=> `0.12N^2 + 0.6N - 5000 = 0`

This gives `N = -(5*sqrt(60009)+15)/6` and `N = (5*sqrt(60009)-15)/6`

As the number of shirts should be a rational number there is no solution for the given problem.

vaaruni | Student

Let number of shirt purchased be x. Since total cost of 'x' number of shirt = 1000\$ therefore cost of one shirt=(1000/x)\$. If 5 more shirts are bought the, number of shirt=(x+5) and cost of each shirt = 1000/(x+5) . since cost of each reduced by 12cents(0.12\$) therefore, cost of each shirt=(1000/x)-0.12.                                                         Or, 1000/x - 0.12= 1000/(x+5)                                                    Or, 1000/x - 1000/(x+5) = 0.12                                                  0r, 1000(x+5) - 1000x=0.12*x*(x+5)                                          or, 1000x +5000-1000x=0.12x^2+0.6x                                        Or, 0.12x^2+0.6x-5000=0                                                            Or, 12x^2+60x-500000 =0                                                           Or, 3x^2 +15x+125000=0                                                                              Or, x= (-15+sqrt((15^2)-4*3*125000))/(2*3) [Using formula for solution of quadratic equation  x=sqrt((b^2-4*a*c))                      and x=(-15- sqrt((15)^2-4*3*125000))/(2*3) [ Using the same above formula ]                                                                             In either case value of 'x' is imaginary [ since b^2 < 4*a*c]

Hence the above priblem has no solution <---Answer

vaaruni | Student

Let number of shirt purchased be x. Since total cost of 'x' number of shirt = 1000\$    therefore cost of one shirt=(1000/x)\$. If 5 more shirts are bought the, number of shirt=(x+5)     and  cost of each shirt = 1000/(x+5) . since cost of each reduced by 12cents(0.12\$)  therefore,  cost of each shirt=(1000/x)-0.12.                        Or, 1000/x - 0.12= 1000/(x+5)                                                   Or, 1000/x - 1000/(x+5) = 0.12                                                      0r, 1000(x+5) - 1000x=0.12*x*(x+5)                                              or,  1000x +5000- 1000x=0.12x^2+0.6x                                         Or, 0.12x^2+0.6x-5000=0                                                              Or, 12x^2+60x-500000 =0                                                             Or, 3x^2 +15x+125000=0                                                              Or, x= (-15+sqrt((15^2)-4*3*125000))/(2*3) [Using formula for solution of quadratic equation ,x=sqrt((b^2-4*a*c)]                      and x=(-15- sqrt((15)^2-4*3*125000))/(2*3)  [ Using the same above     formula ]                                                                        In either case value of  'x'  is imaginary [ since  b^2 > 4*a*c]       Hence, the above problem has no solution <--Answer