the shopkeeper buys some books for ₹80 if he had bought 4 more books for the same amount each book would have cost rupees ₹1 less. find the number of book he bought.
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Let the no. of books bought by the shopkeeper be x.
Price of x books = Rs. 80
=> Price of 1 book = rs 80/x ...........(i)
Now, if the no. of books had been (x+4)
Price of x+4 books = Rs. 80
=> Price of 1 book = Rs 80/ (x+4)........(ii)
From the question, (i) - (ii) = 1
=> 80/x - 80/ (x+4) = 1
By solving we'll get
=> (x-16) (x+20) = 0
=> x= 16 or x= -20.
But number of books can never be -ve.
So, number of books he bought = 16
Price of x books = Rs. 80
=> Price of 1 book = rs 80/x ...........(i)
Now, if the no. of books had been (x+4)
Price of x+4 books = Rs. 80
=> Price of 1 book = Rs 80/ (x+4)........(ii)
From the question, (i) - (ii) = 1
=> 80/x - 80/ (x+4) = 1
By solving we'll get
=> (x-16) (x+20) = 0
=> x= 16 or x= -20.
But number of books can never be -ve.
So, number of books he bought = 16
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1
Answer:
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Step-by-step explanation:
Let he bought x books for rs 80
So the cost of each book was 80/x
If the cost of each books were re 1 less then he can buy 4 more books
So, he can buy x+4 books
So the cost of each of the book would be
80/(x+4)
According to the question,
80/x-80/(x+4)=1
(80x+320-80x)/(x^2 +4x)=1
x^2+4x=320
x^2+4x-320=0
x^2+4x-320=0x^2+20x-16x-320=0
x^2+4x-320=0x^2+20x-16x-320=0x(x+20)-16(x+20)=0
x^2+4x-320=0x^2+20x-16x-320=0x(x+20)-16(x+20)=0(x+20)(x-16)=0
x^2+4x-320=0x^2+20x-16x-320=0x(x+20)-16(x+20)=0(x+20)(x-16)=0Therefore, x= -20, 16
As x wouldnt be negative because quantity of book cant be negative so x=16
So he bought 16 books
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