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Help on average & f(n) type of problem

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Help on average & f(n) type of problem

by crackgmat007 » Wed May 06, 2009 9:35 am
At a certain food stand, the price of each apple is ¢ 40 and the price of each orange is ¢ 60. Mary
selects a total of 10 apples and oranges from the food stand, and the average (arithmetic mean)
price of the 10 pieces of fruit is ¢ 56. How many oranges must Mary put back so that the average
price of the pieces of fruit that she keeps is ¢ 52?
A. 1
B. 2
C. 3
D. 4
E. 5

I tried plug in, is there a much easier way to do this problem?

OA – E

The function f is defined for each positive three-digit integer n by f(n) = 2^x*3^y*5^z, where x, y and z are the hundreds, tens, and units digits of n, respectively. If m and v are three-digit positive
integers such that f(m) = 9f(v), then m-v = ?
A. 8
B. 9
C. 18
D. 20
E. 80

OA-D

How do I go about this problem?
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by DanaJ » Wed May 06, 2009 9:59 am
Q1. Make some notations:
A = number of apples
O = number of oranges
You also know that A + O = 10.

You get that (40A + 60O)/(A + O) = 56 - this means that 40A + 60O = 56(A + O).

Now, she puts some oranges back, say x oranges. The new mean will be:
[40A + 60(O - x)]/(A + O - x) = 52. Again, this means that 40A + 60(O - x) = 52(A + O - x) or that 40A + 60O - 60x = 52(A + O) - 52x.

Now, let's replace 40A + 60O by using the first mean:

56(A + O) - 60x = 52(A + O) - 52x
4(A + O) = 8x
x = (A + O)/2 = 10/2 = 5.

Q2. This is actually quite easy once you get the meaning of it.
So you have m = abc, where a, b and c are digits
v = def, where d, e and f are digits as well.

You know that:
f(m) = (2^a)*(3^b)*(5^c)
f(v) = (2^d)*(3^e)*(5^f).

You also know that f(m) = 9f(v) = (3^2)f(v). This translates into:

(2^a)*(3^b)*(5^c) = (3^2)*(2^d)*(3^e)*(5^f)

Because the difference between the two numbers is a power of 3, a = d and c = f and the only difference between m and v is actually in the tens digit (since this digit is used as the exponent for the power of 3). This is why m will be smth like abc, while v will be smth like a(b-2)c. Some examples of numbers:
m = 123
v = 103

m = 679
v = 659.

Since the difference between the two numbers lies solely in the tens digit, the answer will be 20.
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by crackgmat007 » Wed May 06, 2009 11:02 am
Thanks much for the explanation. Can you clarify how you were able to determine that the diff is in tens digit and not units or hudreths?

"Because the difference between the two numbers is a power of 3, a = d and c = f and the only difference between m and v is actually in the tens digit (since this digit is used as the exponent for the power of 3). This is why m will be smth like abc, while v will be smth like a(b-2)c."
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by DanaJ » Wed May 06, 2009 11:10 am
The difference is in the tens digit because 3 is raised to the tens digit in the function f(n). You can't get 3s from anywhere else in the structure of f(n), since you've only got 2s and 5s. This is why it's the tens digit that makes the cut.
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