Problem Solving

The cost of sending a package is T cents for the first 1/4 kilogram and T/5 cents for each additional 1/4 kilogram or fraction thereof. What is the cost, in cents, to send a P kilogram package at this rate, where P is an integer greater than 1?


1. PT/20
2. (PT-1)/20
3. (P/4+1)T
4. (6P-1)T/20
5. 4(P+1)T/5
I need help understanding this problem. How can this problem be solved with algebra? Will plugging in number be easier?

I think the answer is C.

nidhis.1408 wrote:The cost of sending a package is T cents for the first 1/4 kilogram and T/5 cents for each additional 1/4 kilogram or fraction thereof. What is the cost, in cents, to send a P kilogram package at this rate, where P is an integer greater than 1?


1. PT/20
2. (PT-1)/20
3. (P/4+1)T
4. (6P-1)T/20
5. 4(P+1)T/5

I need help understanding this problem. How can this problem be solved with algebra? Will plugging in number be easier?

My opinion: Brent´s solution (above) is BJJ (Brazilian Jiu-Jitsu) at its best.
I mean, it´s THE way to study for the exam so that you develop your mathematical maturity AND, as a sub-product, develop your skills to do some STREET FIGHTING when you feel your opponent is "chicken".
Again, in MY opinion, BJJ - not street fighting - will be the difference between life-or-death combats with though opponents.
("The best is to be great at both." Sure. And if you can also be an astrophysicist, date three top-models - simultaneously - and be one of the best cellists in the Royal Concertgebouw is even better, I am certain.)

Now... let´s do some STREET FIGHTING:

Let´s explore the PARTICULAR CASE T = 20 (cents) and P = 2 (kg), because we must choose P an integer greater than 1.

Target value: 48 cents. Reason: 20 cents (per first 1/4 kg) and 4 cents * 7 = 28 cents (per seven additional 1/4 kg´s)

1. (2*20)/20 is not the target... out.
2. (2*20-1)/20 idem.
3. (2/4 + 1)*20 idem.
4. [(6*2-1)*20]/20 idem.
5. 4(2+1)*20/5 = 48, the only survivor. (Hence it must be the right answer.)

Regards,
Fabio.