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# Got tricks?

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Got tricks? [#permalink]  19 Jul 2006, 16:29
Hi everyone,

Below are four questions I encountered on a PowerPrep exam that threw me off my time schedule. Could someone please explain efficent ways to solve these problems? I know this test is all about tricks and minimizing the amount of actual calculation you have to do...these just stumped me!

1. On July 1 of last year, the total # of employees at Company E was decreased by 10%. Without any change in salaries of the remaining employees, the average salary was 10% more after the decrease in the # of employees than before the decrease. The total (sum) of combined salaries after July last year was what percent of that before July 1 last year?

a) 90%
b) 99%
c) 100%
d) 101%
e) 110%

2. Three grades of milk are 1%, 2% and 3% by volume. If x gallons of the 1%, y gallons of the 2% and z gallons of the 3% are mixed to give x+y+z gallons of a 1.5% grade, what is x in terms of y?

a) y + 3z
b) (y+4)/4
c) 2y + 3z
d) 3y + z
e) 3y + 4.5z

3. For which of the following functions is f(a+b) = f(a) + f(b) for all positive numbers a+b?

a) f(x) = x^2
b) f(x) = x + 1
d) f(x) = 2/x
e) f(x) = -3x

4. For every integer k from 1-10 inclusive, the Kth term of a certain sequence is given by (-1)^k+1 x (1/(2^k)). If T is the sum of the first 10 terms in the sequence, then T is:

a) greater than 2
b) between 1 and 2
c) between 1/2 and 1
d) between 1/4 and 1/2
e) less than 1/4

Muchas gracias!!!
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Intern
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Q 2)

X Gallons -> 0.01 X (Contribution of fat)
Y Gallons -> 0.02 Y
Z Gallongs -> 0.03 Z

ie ... 0.01 * X + 0.02 * Y + 0.03 * Y = 0.015 (X + Y + Z)

Solve to get X = Y + 3Z (Option A)

3) Straightforward E, by substituting the values

4) Assuming the equation is

(-1)^K + 1 / ((2) ^K )

(-1)^K = 1 for even even number and 1 for every odd number. so they cancel themselves out

So the eqn is 1 / ((2) ^K ) for each increasing value of K the number is less than half the old number 0.5 + 0.25 + 0.125 etc etc..

So the value is lesser than 1 but obviously greater than 0.5.

I may be wrong.. Do correct me, this is my first soln post :D
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Wait....#3 [#permalink]  19 Jul 2006, 19:19
Thanks for responding!

Actually, #3 wasn't so straightforward for me. Usually I don't have a problem with "symbol" or "function" questions, but this one isn't clicking.

I don't understand what the relationship between the answers and the stem are. Can you please explain?

Thanks!

(Also, the answer to #4 is actually supposed to be between 1/4 and 1/2)
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Re: Got tricks? [#permalink]  19 Jul 2006, 19:20
1. B
Let # of emp. = E, Avg Salary = S
before decrease, Total = E*S
after decrease Total = 0.9E*1.1*S = 0.99ES
Thus 99%
B

2. A
0.01x+0.02y+0.03z = 0.015(x+y+z)
5x = 5y+15z
x = y+3z

3. E
e) f(x) = -3x
f(a) = -3a, f(b) = -3b, f(a)+f(b) = -3(a+b), f(a+b) = -3(a+b)

4. D
1st : 1/2
2nd : -1/2^2
3rd : 1/2^3
4th : -1/2^4
..
9th : 1/2^9
10th : -1/2^10
1st+2nd 1/2-1/4 = 1/4
3rd+4th 1/8-1/16 = 1/16
thus, T = 1/4 + 1/16 + 1/64 + 1/256 + 1/1024
We know T is between 1/4 and 1/2.
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Re: Got tricks? [#permalink]  19 Jul 2006, 23:13
peachtree wrote:
1. On July 1 of last year, the total # of employees at Company E was decreased by 10%. Without any change in salaries of the remaining employees, the average salary was 10% more after the decrease in the # of employees than before the decrease. The total (sum) of combined salaries after July last year was what percent of that before July 1 last year?

a) 90%
b) 99%
c) 100%
d) 101%
e) 110%

2. Three grades of milk are 1%, 2% and 3% by volume. If x gallons of the 1%, y gallons of the 2% and z gallons of the 3% are mixed to give x+y+z gallons of a 1.5% grade, what is x in terms of y?

a) y + 3z
b) (y+4)/4
c) 2y + 3z
d) 3y + z
e) 3y + 4.5z

1) Let Number of employees be E.
Let average Salary be A
After July 1st
Number of employees = 9E/10
Average Salary = 11A/10

Total salary before July 1st = EA
Total salary after July 1st = 99/100 * EA

Hence B

2) Amount of milk(by volume) in x gallons = 1/100 * X
Amount of milk(by volume) in y gallons = 2/100 * y
Amount of milk(by volume) in z gallons = 3/100 * z

Hence
1/100 * X + 2/100 * y + 3/100 * z = 1.5/100 * (x+y+z)

-5x/100 = -5y/1000 - 15z/1000
x = y +3z

Hence A
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Q#3 clarification [#permalink]  26 Aug 2006, 20:13
As a newbie I am shocked, surprised and dismayed by the expertise of the responders and the ease of the elegant solutions! Perhaps one day in the long distant future

I had the same confusion as Rajiv... freetheking seems to have understood something neither of us have:

(-1)^k+1 x (1/(2^k))

If we use the BODMAS rule:

This seems to me to follow the pattern (-1)^1 + 1 x (1/2^1)
= -1 + 1/2

The 2nd term would be therefore 1+1/4 etc.

As Rajiv pointed out, in this case all the -1s and 1s would cancel out and we would be left with a value between 1/2 and 1.

Now if we interpret it this way:
((-1)^1+1) * 1/(2^1)_ then it would be 0, 1/4, 0, 1/16 etc.. leading to a value between 1/4 and 1/2.

Is that what is going on? Is the question wrong or am I missing something?

MG
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1. B
2. A
3 . E
4. C The 1's cancel out + -ve so that leaves 1/2 + 1/4 + 1/8 + 1/16..

so the answer is between 1/2 and 1 - c
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(-1)^k+1 x (1/(2^k))

the reason why I got D is that..

(-1)^(k+1) x (1/(2^k))

look at the original version don't u guys think the red part is weird??
(-1)^k + 1 x (1/(2^k))
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Senior Manager
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I agree , it is redundant.. but hey maybe they want to confuse us
Senior Manager
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gk3.14 wrote:
I agree , it is redundant.. but hey maybe they want to confuse us

No it's not, as far as I know there's no trick like this one..

K+1 is just a superscript in a word file, if you copy it.. it just truned into a normal word font. That's why I assumed ^(k+1).
Moreover, OA is D.. thus ^(k+1) makes perfect sense.
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Senior Manager
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freethinking: I see what you mean.. It makes sense now..
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