Problem Solving

Problem 1

2+2+2^2+2^3+2^4+2^5+2^6 +2^7+ 2^8 =
Correct answer: 2^9


Problem 2:

5^21 x 4^11= 2 x 10^n =
What is N?

Correct answer: 21

Problem 3

From a bag containing 12 identical bleu ball and y identical yellow balls one ball will be removed at random. If the probability is less than 2/5 that the removed ball will be bleu what is the least number of yellow balls that must be in the bag?

Correct answer: 19


For all positive m m = 3 m when m is odd and m=1/2 when m is even. Which of the following is equal to 9 x 6

Correct answer: 27

I am totally confused with these problems and will appreciate any help.

Thanks

Problem 1: 2+2+2^2+2^3+2^4+2^5+2^6 +2^7+ 2^8 =

Rule: Sum of the series: 2^0+2^1+2^2....2^n = 2^(n+1)-1.

The simplest way to prove this is to just evaluate each term and sum them.

Problem 2: 5^21 x 4^11 = 2 x 10^N. What is N?
Left hand side:5^21 x 4^11 = 5^21 x (2^2)^11 = 5^21 x 2^22

Right hand side: 2 x 10^N = 2 x (2x5)^N = 2^(N+1) x 5^N

5^21 x 2^22 = 2^(N+1) x 5^N ==> N = 21.

Problem 3: From a bag containing 12 identical bleu ball and y identical yellow balls one ball will be removed at random. If the probability is less than 2/5 that the removed ball will be bleu what is the least number of yellow balls that must be in the bag?

# of balls in the bag = 12+y
# of ways to select a blue ball randomly = 12C1 = 12
# of ways to select a ball randomly from 12+y balls = (12+y)C1 = 12+y
P(blue) = 12C1/(12+y)C1 = 12/(12+y) < 2/5

Solve 12/(12+y) < 2/5

60 < 24+2y -->y>18.

So least # of yellow balls = 18+1 = 19.

Problem 4: For all positive m m = 3 m when m is odd and m=1/2m when m is even. Which of the following is equal to 9 x 6

the function m* takes value 3m for any odd # and the value 1/2m for an even #. 9 is odd -> 9* = 3 times 9 = 27, 6* = 6/2 = 3. so 9*.6* = 27*3 = 81.

Now, from the answer choices, 27* = 27*3 = 81. So answer is 27* (not 27).

Gmatcoach, thank you so so much!!