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A man purchased a set of identical pictures for $100. Five of the pictures were later damaged. The remainder were sold at a gain of $2 on the original cost of each. As a result $20 was gained on the whole transaction. Calculate original number of pictures.

What is known:

A set of identical pictures were purchased for $100.
Five pictures were later damaged.
Remaining pictures were sold at a gain of $2 on original cost of each.
$20 was gained on complete transaction.

Want to know:

Original number of pictures.

Let P = original number of pictures.

$100 / P= cost of each picture

P - 5 = remainder of pictures

How do I proceed from here?

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@Randy Dandy,

([100/p] +2)(p-5)=120
Does this make sense?

Solve for p
p=25
p=-10

1 Reply

@neologist,

I understand the equation, but I am not clear on how to solve it.

1 Reply

@Randy Dandy,

Can you perform the multiplication on the left side of the equation?

1 Reply

@neologist,

([100/p] +2)(p-5)=120

I am not sure, but I think it is FOIL.

100P + 500 + P^2 + 2P = 120
102P + P^2 + 500 - 500= 120 - 500
102P + P^2 = -380

I am stuck.

1 Reply

@Randy Dandy,

Randy Dandy wrote:

([100/p] +2)(p-5)=120

I am not sure, but I think it is FOIL.

100P + 500 + P^2 + 2P = 120
102P + P^2 + 500 - 500= 120 - 500
102P + P^2 = -380

I am stuck.

([100/p] +2)(p-5)=120
Multiply by distribution:
([100/p]+2) (p) =100+2p, and ([100/p]+2) (-5)= [-500/p] -10 , so
100 + 2p -500/p - 10 = 120
2p -500/p = 30
multiply by p
2p^2 - 500 = 30 p
divide by 2 and transpose
p^2 -15p -250 = 0
(p-25)(p+10) = 0
P=25, p = -10

1 Reply

@neologist,

Thanks.

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