# 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.

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### 3 Answers

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.**

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**

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**