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# Elemntary Problem Solving Question

A shopkeeper buys some books for £80. If he had bought 4 more books for the same amount, each book would have cost £1 less. Find the number of books he bought?

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### 1 Answer by Expert Tutors

John M. | Engineering manager professional, proficient in all levels of MathEngineering manager professional, profic...
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1. Let b = the number of books he bought for £80
2. Let c = the cost of each book
3. Then, b*c = £80    {Eqn 1}
4. And (b+4) * (c-£1) = £80    {Eqn 2}
5. Two equations, two unknowns.  So you can find a unique solution.  And you can do this several ways.  I'll use substitution.  For convenience, I'll drop the £ sign.
6. Rewrite Eqn 1 as b = 80/c
7. Then substitute into Eqn 2:   (80/c + 4) * (c- 1) = 80
8. Now multiply out using FOIL method:  (80/c * c) + (80/c * -1) + (4c) + (4 * -1) = 80
9. Rewrite:  80 - 80/c + 4c - 4 = 80
10. Rearrange:  -80/c + 4c = 4
11. Divide both sides of equation by 4:   -20/c + c = 1
12. Rearrange:  -20/c + c - 1 = 0  {Eqn 3}
13. Get a common denominator of c:   (-20 + c2 - c) / c = 0  {Eqn 4}
14. Notice in Eqn 4 that the numerator must be equal to zero in order for the equation to hold.  So, we can ignore the c in the denominator
15. Rewrite:  c2-c-20 = 0   {Eqn 5}
16. Now factor Eqn 5:  (c-5)(c+4) = 0.  The solution is c = 5 or c = -4.  Since c is a cost, it cannot be negative, so the only solution is 5
17. The cost of each book originally is £5
18. Substituting back into Eqn 1, the number of books bought by the shopkeeper is 16