A computer company prices its models using the formula P = {1.25^s}a,

Question

A computer company prices its models using the formula \(P = 1.25^sa\), where P is the price per unit, a is a constant, and s is the speed rating, from 1 to 20, of the product. The company also grants a discount of 10% on orders of 100 to 249 units and a discount of 25% on orders of 250 or more units. Lotte orders 300 computers, all with the same speed rating. How many computers could she have purchased for the same price if she had chosen models with a speed rating 2 higher than the ones she chose?

A) 160
B) 192
C) 230
D) 240
E) 675

Kudos for a correct solution.

A) 160
B) 192
C) 230
D) 240
E) 675

GeneraleCanaglia · Answer

Let be Pt1=300*1,25^s*a*(1-25%) the total price for 300 slower computer.
The same price for faster computer will be Pt2=x*1,25^(s+2)*a*k(x), where x is the amount of computer that you can buy at that price and k(x) the discount depending on x itself.

So has to be Pt1= Pt2 → 300*1,25^s*a*(1-25%)= x*1,25^(s+2)*a*k(x)
300*1,25^s*a*(1-25%)= x*1,25^s*1,25^2*a*k(x)
x*k(x)=144
So, k(x) is something like (1-d) with d either 0%, 10%, or 25%, than the max value for k(x) is 1 and the minimum is 0,75, so 108x<=k(x)*x<=144. You are in the range of 10% discount → x=144/(1-10%)=144/0,9=160

Answer A

Khuranasup · Answer

total cost of all 300 computers= 300*(1.25^s)*a*(1-.25)
=225*(1.25^s)*a

let us assume the second time he can buy x no. of computers from the same amount.
so, 225*(1.25^s)*a=x*(1.25^{2+s})*a*(1-p), where p is the discount prcentage/100
on solving we get
x*(1-p)=144
now we take 2 cases

CASE 1
p=0.25 i.e. discount is 25%
we get x=172
so, they cannot avail a 25% discount since the ordered quantity should have been above 250

CASE2
p=0.1 i.e. discount is 10%
we get x=160
so, this is the correct value of p as the quantity ordered falls in the corresponding eligible discount group.

O.A. A

Give a kudo if you like the explanation

Bunuel · Answer

MANHATTAN GMAT OFFICIAL SOLUTION:

The total amount Lotte spent is:

Total cost = (original # of units purchased)(original price per unit)
Total cost = (300)(original price per unit)

Remember that because Lotte is buying 300 units, she is receiving a discount of 25% of the full price. (You don’t need to do anything with this information right now.)

If Lotte opts for a speed rating 2 higher, she will see an increase in unit price from 1.25^s to 1.25^{s+2}. Find the increase in P as follows:

1.25^{s+2}=(1.25^s)(1.25^2)a=(1.25^s)(5/4)^2a=(1.25^s)(25/16)a

The price of the faster models is 25/16 that of the price of the slower models. The new purchase would be:
Total cost = (new # of units purchased)(original price per unit)(25/16)

To keep the total cost the same, then, the new number of units purchased must be 16/25 the original number:

Total cost = (original # of units purchased)(16/25)(original price per unit)(25/16)

The new number of units purchased, then is (300)(16/25) = 192 computers.

But wait! You’re not quite done yet. At this quantity her discount drops from 25% to 10%. Because her discount is not as great, she can’t actually buy 192 computers; she has to buy fewer. Glance at the answers; only answer (A) is smaller than 192, so it must be the correct answer.

Here’s how to calculate that (but you don’t need to!). First, Lotte is losing the 25% discount, so divide the original price per unit by 75%, or 0.75: the new price per unit = original price per unit / 0.75. This was the true original price without any discount applied.

Next, Lotte is gaining a 10% discount, so multiply the price by 90%, or 0.9. The new price per unit = the original price per unit × 0.9 / 0.75.

This is now the new price per unit with the 10% discount applied. (Note: you cannot simply subtract and say that 25% – 10% = 15%. The discounts are taken on different starting values!)

Simplify:
New price per unit = 0.9 / 0.75 (original price per unit)
New price per unit = 90 / 75 (original price per unit)
New price per unit = 6 / 5 (original price per unit)

If the new price per unit is 6/5 of the original price per unit, then Lotte has to purchase 5/6 as many computers in order to keep the same total cost:

Total cost = (original # of units purchased)(5/6)(original price per unit)(6/5)

The new number of computers is 192 (5/6) = 160

The correct answer is (A).

mvictor · Answer

oh damn..took me some time to solve this bad boy...
suppose s=1. and the better computers with s=3.

so...
p=5^1*a*300/4 *3/4 = 5a*225/4 - price paid for 300 computers.
since s would be 3, it is logically to consider that for the same price, the company would buy fewer computers. we can eliminate E right away.

now..
new price:
(5^3)*a/4^3.

the price can be discounted by 10%, by 25%, or not discounted at all..
suppose it's discounted by 25%

(5^3)*a/4^3 * 3/4 = (5^3)*a*3/4^4.

now, how many pc's?
5a*225/4 * 4^4 / (5^3)*a*3

cancel what's possible:
9*5^3 * 4^3 / 5^3*3 = 3*4^3 = 3*64 = 192. but wait..192 - this means that we would not get a 25% discount...so it should be 10% discount instead.

the new price must thus be:
(5^3)*a*9/10*4^3

again # of pc's we can find if we divide the cost of the 300 computers paid by the new price:
5a*225/4 * 10*4^3 / (5^3)*a*9

cancel what's possible:
(5^3)*9*10*4^2 / (5^3)*9 => 10*4^2 = 10*16 = 160.

A

mejia401 · Answer

Same here.

Pick numbers strategy.
The constant  a  is irrelevant and yet I choose 64 to make the division easier

P  (first speed) =  \frac{5a}{4} = \frac{{5*64}}{4} = 80

P  (third speed) =  \frac{125a}{64} = \frac{{125*64}}{64} = 125

Calculating the total amount for three hundred first speed computers
 80 * 300 * \frac{3}{4} = 18000

\frac{18000}{125} = 144

However, since 144 computer grants a discount, the calculation was redone but with  (1 - \frac{1}{10})  Thus,

125*\frac{9}{10} = 18000  or  \frac{{2^4*3^2*5^3}}{{3^2*5^3}}*10 = 160

IMO A

peachfuzz · Answer

125∗9/10=18000 is incorrect. 125∗9/10=112.5

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A computer company prices its models using the formula P = {1.25^s}a,

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