cost of package

[nidhis.1408]

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?

[Brent@GMATPrepNow]

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?

A. PT/20
B. (PT-1)/20
C. (P/4+1)T
D. (6P-1)T/20
E. 4(P+1)T/5

Since the two different rates are given per 1/4 kilogram, it might be useful to use measurements of 1/4 kg. So, each unit of measurement is 1/4 kg.

Since there are 4 units of measurement per kilogram, and since the package weighs P kilogram, we can say that the package weighs 4P units.

Of these 4P units of weight, the first 1 unit is charged at a rate of T cents. So, the total cost for that 1 unit is T cents.

This leaves 4P-1 units remaining to be charged at a rate of T/5 cents per unit. So, the total cost for those 4P-1 units is (T/5)(4P-1) cents.

So, the total cost is T + (T/5)(4P-1) cents.
When we simplify this, we get [spoiler][4T(1+P)]/5[/spoiler]

Answer = E

Cheers,
Brent

[PhyllisFierro]

I think the answer is C.

[fskilnik@GMATH]

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.

[Scott@TargetTestPrep]

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

Since there are 4P quarter-kilograms in P kilograms, the cost of sending a P-kilogram package is:

T + (T/5)(4P - 1) = 5T/5 + (4PT - T)/5 = (4PT + 4T)/5 = 4T(P + 1)/5

Answer: E/5

[[email protected]]

Hi All,

We're told that 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. We're asked for the cost, in cents, to send a P kilogram package at this rate, where P is an integer greater than 1. This question can be solved in a couple of different ways, including by TESTing VALUES.

IF... T=5 and P=2...
then it costs 5 cents for the FIRST 1/4 kilogram and
5/5 = 1 cent for EACH additional 1/4 kilogram.

A 2 kilogram package is 8/4 kilograms, so the first 1/4 would cost 5 cents and each of the additional seven 1/4s would cost 1 cent each.
The total cost would be 5 + 7 = 12 cents. Thus, we're looking for an answer that equals 12 when T=5 and P=2. There's only one answer that matches...

Final Answer: E

GMAT assassins aren't born, they're made,
Rich