A shopkeeper buys a no. of bookd for Rs.80 if he had bought 4 more books for the same amount each book would have cost Rs 1 less. how many books did he buy
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AT FIRST, LET NO. OF BOOKS=a AND PRICE OF 1 BOOK=b
I.E. ab=80 I.E. a=80/b................1
IN SECOND CASE,
NO. OF BOOKS=a+4 AND PRICE OF 1 BOOK=b-1
I.E. (a+4)(b-1)=80.....................2
BY PUTTING THE VALUE OF b FROM 1 IN 2, WE GET,
(80/b+4)(b-1)=80
I.E (80+4b)(b-1)/b=80
I.E. (80+4b)(b-1)=80b
I.E. 80b-80+4b^2-4b=80b
I.E. 4b^2-4b-80=0
I.E. b^2-b-20=0
I.E. EITHER b=5 OR b=-4
BECAUSE b CAN'T BE NEGATIVE, THEREFORE b=5,
BY PUTTING IN 1, WE GET
a=80/5=16
I.E. THE REQUIRED NO. OF BOOKS=16
I.E. ab=80 I.E. a=80/b................1
IN SECOND CASE,
NO. OF BOOKS=a+4 AND PRICE OF 1 BOOK=b-1
I.E. (a+4)(b-1)=80.....................2
BY PUTTING THE VALUE OF b FROM 1 IN 2, WE GET,
(80/b+4)(b-1)=80
I.E (80+4b)(b-1)/b=80
I.E. (80+4b)(b-1)=80b
I.E. 80b-80+4b^2-4b=80b
I.E. 4b^2-4b-80=0
I.E. b^2-b-20=0
I.E. EITHER b=5 OR b=-4
BECAUSE b CAN'T BE NEGATIVE, THEREFORE b=5,
BY PUTTING IN 1, WE GET
a=80/5=16
I.E. THE REQUIRED NO. OF BOOKS=16
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