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secondary 4 | E Maths
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Please help thanku
Date Posted: 5 years ago
Views: 209
Original cost of each bulb
= $ 258/x
Number of bulbs sold = x - 3
Price each bulb was sold at
= $(258/x + 3)
= $(258 + 3x)/x
Amount of money collected from sale of the bulbs
= $(258+3x)/x × (x - 3)
= $(258 + 3x)(x - 3)/x
= $(258x - 774 + 3x² - 9x)/x
= $(3x² + 249x - 774)/x
Since amount collected also = $360,
(3x² + 249x - 774)/x = 360
3x² + 249x - 774 = 360x
3x² - 111x - 774 = 0
x² - 37x - 258 = 0
(x + 6)(x - 43) = 0
x = -6 or x = 43
(x = -6 is rejected as number of bulbs can't be negative)
So number of bulbs bought = 43
Selling price = $(258/x + 3)
= $(258/43 + 3)
= $(6 + 3)
= $9
= $ 258/x
Number of bulbs sold = x - 3
Price each bulb was sold at
= $(258/x + 3)
= $(258 + 3x)/x
Amount of money collected from sale of the bulbs
= $(258+3x)/x × (x - 3)
= $(258 + 3x)(x - 3)/x
= $(258x - 774 + 3x² - 9x)/x
= $(3x² + 249x - 774)/x
Since amount collected also = $360,
(3x² + 249x - 774)/x = 360
3x² + 249x - 774 = 360x
3x² - 111x - 774 = 0
x² - 37x - 258 = 0
(x + 6)(x - 43) = 0
x = -6 or x = 43
(x = -6 is rejected as number of bulbs can't be negative)
So number of bulbs bought = 43
Selling price = $(258/x + 3)
= $(258/43 + 3)
= $(6 + 3)
= $9
See 1 Answer
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Date Posted:
5 years ago
Thanku! :)
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