Difficult GMAT prep problems

[HarvardDreamin]

For some reason, I seem to find the problems tricky. There are just 3 and would love to know what the solution is as well as the method so I could learn the best way to approach it. Thanks for your help


1) This year Henry will save a certain amount of his income, and he will spend the rest. Next year Henry will have no income, but for each dollar that he saves this year, he will have 1 + r dollars available to spend. In terms of r, what fraction of his income should Henry save this year so that next year the amount he was available to spend will be equal
to half the amount that he spends this year?

A 1 / (r+2)

B 1 / (2r +2)

C 1 / (3r+2)

D 1 / (r+3)

E 1 / (2r+3)


2)
From a group of 3 boys and 3 girls, 4 children are to be randomly selected. What is the probability that equal numbers of boys and girls will be selected?

A. 1/ 10
B. 4/9
C 1/2
D 3/5
E 2/3



3 Q10)

In a certain game, a large bag is filled with blue, green, purple and red chips worth 1, 5, x and 11 points each, respectively. The purple chips are worth more than the green chips, but less than the red chips. A certain number of chips are then selected from the bag. If the product of the point values of the selected chips is 88,000, how many purple chips were selected?

a) 1 b) 2 c)3 d)4 e)5

[ash g]

Solution to [1]

==> This year
Income = I
Savings = s
Spending = I - s

Asked = s / I


==> Next year
"but for each dollar that he saves this year, he will have 1 + r dollars available to spend"

Available to spend next year = s(1+r)


==> Equation
"that next year the amount he was available to spend will be equal to half the amount that he spends this year"

s(1+r) = 0.5 ( I - s)

2 + 2r = (I - s)/s = I/s - 1
I/s = 3 + 2r

s/I = 1 / (3+2r) <Answer> The wordier the problem, the more easier it is normally to solve it.
My approach- Step1: Read it entirely once without feeling that this is difficult and understand it say 60%. Step2: Reread , variablize and convert it into equation. Step3 : Solve.

[tmmyc]

2)

I would try to solve this using combinations.


First, find the total number of possible combinations:

6 total people and we want to select 4.
6 choose 4 = 6!/(4!*2!) = 15


Second, find the total number of combinations with 2 boys and 2 girls (i.e. B B G G).

Boy combinations: 3 choose 2 = 3!/(2!*1!) = 3
Girl combinations: 3 choose 2 = 3!/(2!*1!) = 3
Total combinations of this type: 3 * 3 = 9


Find the probability by dividing the number of combinations you want by the total number of possible combinations.

9/15
-> 3/5

[tmmyc]

3)

First, break 88,000 down to its prime factors

88,000
->8 * 11 * 10 * 10 * 10
-->2*2*2*11*2*5*2*5*2*5
--->(2^6)*11*(5^3)

Since the product of the point values of the selected chips must be equal to the product of those prime factors, we know there is one 11-point red chip and three 5-point green chips.

There is still the issue of how to deal with the 2^6.

Since the question states that purple chips are greater than 5 but less than 11, purple chips can only be 8, which is 2^3. This is the only way we can accommodate the 2^6 above.

2^6 = (2^3)*(2^3) = 8 * 8

Two purple chips were selected.

[ash g]

For [2] , the solution mentioned by tmmyc is very correct.
But if you are amazingly running short of time...like need to solve this 15-20 secs......go through the post below and look at Stuart's reply on "strategic guessing".
It leads you to [D] again.

https://www.beatthegmat.com/i-suck-at-pr ... t8927.html

Follow this thread for Stuart's further replies.

[HarvardDreamin]

Guys, many many thanks. Really appreciate it. all the answers mentioned are correct so will try to learn the method behind it.

For [2] , the solution mentioned by tmmyc is very correct.
But if you are amazingly running short of time...like need to solve this 15-20 secs......go through the post below and look at Stuart's reply on "strategic guessing".
It leads you to [D] again.

https://www.beatthegmat.com/i-suck-at-pr ... t8927.html

Follow this thread for Stuart's further replies.

[cmoney]

Good reply, but I don't think that the math for the Boy Girl problem is correct. There are only 6 combinations of Boys and Girls that is equal.

BBGG
GGBB

BGBG
GBGB

BGGB
GBBG

correct me if I'm wrong.

[cmoney]

Additionally....

I get 6 combinations of Equal girl boy:
BBGG
GGBB
BGBG
GBGB
BGGB
GBBG

And 8 combinations of Unequal Girl Boy....remember only 3 girls and boys, so the combination has a limit:
BBBG
BBGB
BGBB
GBBB

GGGB
GGBG
GBGG
BGGG

My problem is that my answer is 6/14 or 3/7 which isn't even an option, anyone want to help me out here?

[cramya]

Somtimes using combination solution comes in handy for probabaility. Its easy to miss one of the choices in exhausting the various possibilies lik

It will 3c2(selecting 2 out of 3 boys) *3c2(selecting 2 out of 3 girls)/6c4 (selecting 4 children from 6)= 2/5

3C2*3C2/6C4

[cramya]

Correction : The value of 3c2*3c2/6c4 = 3/5 D)