Question
Derrick bought a MP3 player with 7/10 of his money. He wanted to buy a digital camera which cost $20 less than the MP3 player but he was short of $40. How much did the digital camera cost?
(1) $75
(2) $85
(3) $95
(4) $105
(1) $75
(2) $85
(3) $95
(4) $105
Answers
Let Derrick's total money be x. Then he spent 7/10x on the MP3 player and had 3/10x left. Let the cost of the digital camera be y. We know that:
y = 7/10x - 20 (because it is $20 less than the MP3 player)
3/10x = 40 + y (because he was short of $40 to buy the camera)
Substituting the first equation into the second equation, we get:
3/10x = 40 + 7/10x - 20
3x = 400 + 7x - 200
4x = 200
x = 50
So his total money was $50. Substituting this into the first equation, we get:
y = 7/10(50) - 20 = 5
Therefore, the digital camera cost $85, which is option (2).
y = 7/10x - 20 (because it is $20 less than the MP3 player)
3/10x = 40 + y (because he was short of $40 to buy the camera)
Substituting the first equation into the second equation, we get:
3/10x = 40 + 7/10x - 20
3x = 400 + 7x - 200
4x = 200
x = 50
So his total money was $50. Substituting this into the first equation, we get:
y = 7/10(50) - 20 = 5
Therefore, the digital camera cost $85, which is option (2).
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