MISCELLANEOUS PROBLEM –
Example.1) A shopkeepers purchase a certain number of books for $960, if the cost per book was $8 less, the number of books that could be purchased for the same amount would be 4 more. Write an equation, taking the original cost of each book to be $x and solve it to find the original cost of the book.
Ans.) Let the original cost of each book be $x.
960
Number of books purchased for $960 = --------
x
New cost of each book = $ (x – 8)
960
Number of books purchased for $960 at the new rate = --------
(x – 8)
As per the given condition, we have -
960 960
So, -------- - ------- = 4
(x – 8) x
1 1 4
=> --------- - ------- = -------
(x – 8) x 960
x – (x – 8) 1
=> -------------- = --------
x (x – 8) 240
(x – x + 8) 1
=> ------------- = ------
(x² - 8x) 240
=> (8 X 240) = x² - 8x
=> x²- 8x – 1920 = 0
=> x²- (48 – 40)x – 1920 = 0
=> x²- 48x + 40x – 1920 = 0
=> x (x – 48) + 40 (x – 48) = 0
=> (x – 48) (x + 40) = 0
=> (x – 48) = 0 or (x + 40) = 0
=> x = 48 or x = - 40
=> x = 48 [x ≠ - 40, because cost of book cannot be negative]
Hence the cost of original book is $48 (Ans.)
Example.2) $480 is divided equally among x children. If the number of children were 20 more, then each would have got $12 less. Find the value of x.
Ans.) Let the total number of children is ‘x’.
480
Then, share of each = $ -------
x
Now, as per the given condition, new number of children is (x + 20)
480
Share of each now = $ ---------
(x + 20)
480 480
So, -------- - --------- = 12
x (x + 20)
1 1 12
=> ------ - --------- = --------
x (x + 20) 480
(x + 20) – x 1
=> -------------- = --------
x (x + 20) 40
20 1
=> ---------- = -------
(x² + 20x) 40
=> 800 = x² + 20x
=> x² + 20x – 800 = 0
=> x² + (40 – 20)x – 800 = 0
=> x² + 40x – 20x – 800 = 0
=> x (x + 40) – 20 (x + 40) = 0
=> (x + 40) (x – 20) = 0
=> (x + 40) = 0 or (x – 20) = 0
=> x = - 40 or x = 20
=> x = 20 [x ≠ - 40, as number of children cannot be negative]
Hence, the value of x is 20 (Ans.)
Example.3) In a cricket ground, seats were arranged in rows and columns. The number of rows was equal to number of seats in each row. When the number of rows was doubled and the number of seats in each row was reduced by 10, the total number of seats increased by 300.
Find – (i) The number of rows in the original arrangement,
(ii) The number of seats in the auditorium after rearrangement.
Ans.) Originally, let the number of rows be ‘x’
Then, number of seats in each row = x
So, the total number of seats = x²
Again the number rows = 2x
Number of seats in each row = (x – 10)
So, total number of seats = 2x (x – 10)
As per the given condition, we have –
2x (x – 10) - x² = 300
=> 2x²- 20x - x² = 300
=> x²- 20x – 300 = 0
=> x²- (30 – 10) x – 300 = 0
=> x²- 30x + 10x – 300 = 0
=> x (x – 30) + 10 (x – 30) = 0
=> (x – 30) (x + 10) = 0
=> (x – 30) = 0 or (x + 10) = 0
=> x = 30 or x = - 10
=> x = 30 [x ≠ - 10, as the number of rows cannot be negative]
Thus, we have –
(i) Number of rows in original arrangement = 30
(ii) Number of seats after rearrangement = (30 X 30) + 300
= 1200 (Ans.)
Example.4) The hotel bill for a number of people for overnight stay is $14400. If there were 4 more people, the bill each person had to pay would have reduced by $600. Find the number of people staying overnight.
Ans.) Let the number of people staying overnight be x
Total expenditure = $14400
14400
Expenditure per head = $ ---------
x
New number of people to stay = (x + 4)
14400
New, expenditure per head = $ ----------
(x + 4)
As per the given condition –
14400 14400
---------- - ---------- = 600
x (x + 4)
1 1 600
=> ------ - -------- = ---------
x (x + 4) 14400
(x + 4) – x 1
=> ------------- = -------
x (x + 4) 24
4 1
=> ---------- = ------
(x² + 4x) 24
=> x²+ 4x = 96
=> x²+ 4x – 96 = 0
=> x²+ (12 – 8)x – 96 = 0
=> x²+ 12x – 8x – 96 = 0
=> x (x + 12) – 8 (x + 12) = 0
=> (x + 12) (x – 8) = 0
=> (x + 12) = 0 or (x – 8) = 0
=> x = - 12 or x = 8
=> x = 8 [x ≠ - 12, since number of people cannot be negative]
Hence, the number of people staying overnight is 8 (Ans.)
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