# A shopkeeper buys a number of books for Rs. 80. If he had bought 4 more for the same amount each book would have cost Rs. 1 less. How many books did he buy?

A. 8

B. 16

C. 24

D. 28

Last updated date: 24th Mar 2023

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Answer

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Hint:- We Find cost per book in both the given cases and use the given condition by subtracting and putting it equal to 1.

Complete step-by-step answer:

Let the total number of books the shopkeeper bought be x.

And it is given that the cost of all books is Rs. 80.

So, the cost of one book will be \[\dfrac{{{\text{Total cost}}}}{{{\text{Number of books}}}}\].

So, the cost of one book will be Rs. \[\dfrac{{{\text{80}}}}{{\text{x}}}\].

Number if shopkeeper buys 4 more books then,

Total number of books will become x + 4.

And the total cost of books remains the same, equal to Rs. 80.

So, now the cost of one book will be Rs. \[\dfrac{{{\text{80}}}}{{{\text{x + 4}}}}\].

But as we are given in question that cost of each book becomes Rs. 1 less if the shopkeeper buys 4 more books.

So, \[\dfrac{{{\text{80}}}}{{\text{x}}}{\text{ - }}\dfrac{{{\text{80}}}}{{{\text{x + 4}}}}\] = 1 (1)

Now we had to solve equation 1, to get the value of x.

Taking LCM in equation 1. We get,

\[\dfrac{{{\text{80(x + 4) - 80x}}}}{{{\text{x(x + 4)}}}}\] = 1

Now cross-multiplying the above equation. We get,

80x + 320 – 80x = \[{{\text{x}}^{\text{2}}}\] + 4x

\[{{\text{x}}^{\text{2}}}\] + 4x – 320 = 0 (2)

Now, splitting the middle term of equation 2, to get the required value of x.

\[{{\text{x}}^{\text{2}}}\] + 20x – 16x – 320 = 0

Taking x common from the first two terms and –16 common from last two terms of the above equation. We get,

x(x + 20) – 16(x + 20) = 0

Taking (x + 20) common from the above equation. We get,

(x + 20)(x – 16) = 0

So, from the above equation we get x = –20 or x = 16.

But the number of books cannot be negative.

So, x = 16.

Hence, the shopkeeper buys 16 books.

So, the correct option will be B.

Note:- Whenever we come up with this type of problem then first, we assume the number of books equal to a variable x. And then find the cost of each book in both the cases given. And then make a quadratic equation using given conditions. Then we had to solve the quadratic equation to get the required value of x. And neglect negative value because books cannot be negative.

Complete step-by-step answer:

Let the total number of books the shopkeeper bought be x.

And it is given that the cost of all books is Rs. 80.

So, the cost of one book will be \[\dfrac{{{\text{Total cost}}}}{{{\text{Number of books}}}}\].

So, the cost of one book will be Rs. \[\dfrac{{{\text{80}}}}{{\text{x}}}\].

Number if shopkeeper buys 4 more books then,

Total number of books will become x + 4.

And the total cost of books remains the same, equal to Rs. 80.

So, now the cost of one book will be Rs. \[\dfrac{{{\text{80}}}}{{{\text{x + 4}}}}\].

But as we are given in question that cost of each book becomes Rs. 1 less if the shopkeeper buys 4 more books.

So, \[\dfrac{{{\text{80}}}}{{\text{x}}}{\text{ - }}\dfrac{{{\text{80}}}}{{{\text{x + 4}}}}\] = 1 (1)

Now we had to solve equation 1, to get the value of x.

Taking LCM in equation 1. We get,

\[\dfrac{{{\text{80(x + 4) - 80x}}}}{{{\text{x(x + 4)}}}}\] = 1

Now cross-multiplying the above equation. We get,

80x + 320 – 80x = \[{{\text{x}}^{\text{2}}}\] + 4x

\[{{\text{x}}^{\text{2}}}\] + 4x – 320 = 0 (2)

Now, splitting the middle term of equation 2, to get the required value of x.

\[{{\text{x}}^{\text{2}}}\] + 20x – 16x – 320 = 0

Taking x common from the first two terms and –16 common from last two terms of the above equation. We get,

x(x + 20) – 16(x + 20) = 0

Taking (x + 20) common from the above equation. We get,

(x + 20)(x – 16) = 0

So, from the above equation we get x = –20 or x = 16.

But the number of books cannot be negative.

So, x = 16.

Hence, the shopkeeper buys 16 books.

So, the correct option will be B.

Note:- Whenever we come up with this type of problem then first, we assume the number of books equal to a variable x. And then find the cost of each book in both the cases given. And then make a quadratic equation using given conditions. Then we had to solve the quadratic equation to get the required value of x. And neglect negative value because books cannot be negative.

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