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In a room filled with 7 people, 4 people have exactly 1 sibling
In a room filled with 7 people, 4 people have exactly 1 sibling in the room and 3 people have exactly 2 siblings in the room. If two individuals are selected from the room at random, what is the probability that those two individuals are NOT siblings?(A) 5/21
(B) 3/7
(C) 4/7
(D) 5/7
(E) 16/214 people have exactly 1 sibling in the room
Translate: #1 is siblings with #2. #3 is siblings with #4. Or you could mix the numbers around, but however way you arrange them, there are TWO pairs of siblings in this group.3 people have exactly 2 siblings in the room
Translate: #1 is siblings with #2 and #3 – they’re all siblings with each other. So this is ONE triplet of siblings.So in total, you have one pair, one pair, and then one triplet.
You see the word “NOT” in the question – so you can consider doing the popular (1-prob that they ARE siblings) method.
prob (2 picked people are siblings) = ?
Well, 2 people picked can only siblings if they belong in the same pair or same triplet. We already divided up the people in the the pair, pair, and triplet.
So prob (2 picked people are siblings) = (2 choose 2) + (2 choose 2) + (3 choose 2)
Total # of possibilities = (7 choose 2)In each case you are choosing 2 people. But your broke down the problem from a large set of 7 people to smaller groups of size 4 and size 3.
So 2nCr2 + 2nCr2 + 3nCr2 = 1 + 1 + 3 = 5 ways to pick siblings
Total # of possibilities = 7nCr2 =
7*6* (5!)
———- = 42/2 = 21 possibilities
2 * (5!)So # of ways to pick siblings is 5 out of a total of 21.
We want the opposite of this so we do 1 – (5/21) =16/21
Having gone through this, note that of all the answer choices – (D) is the last one I would have picked. Usually, with a question like this, if (5/21) is the correct answer, then whatever 1 – (5/21) is will surely also be an answer choice to trick people who forgot to do the 1 – operation.
Looking at the answer choices, I see that (b) 3/7 and (c) 4/7 add up to 1. So those two can potentially be correct.
(A) 5/21 and (e) 16/21 add up to 1, so they can potentially be correct.
Then (d) just comes out of nowhere as 5/7 by itself.
If I were to guess, I’d probably guess between A, B, C, and E and would eliminate D from guessing.
Just our 2 cents. Hope that helps!
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Here’s how to solve OG Data Sufficiency #112, Rate Problem in < 1 minute
112. Machines X and Y produced identical bottles at different constant rates. Machine X, operating alone for 4 hours, filled part of a production lot; then Machine Y, operating alone for 3 hours, filled the rest of this lot. How many hours would it have taken Machine X operating alone to fill the entire production lot?(1) Machine X produced 30 bottles per minute.
(2) Machine X produced twice as many bottles in 4 hours as Machine Y produced in 3 hours.Think of this as D=RT
Only, here we have distance = the entire production
R is split into the rate for X, and the rate for Y.
T is split into 4 hours for X, and 3 hours for Y.So the entire production = 4 hours @ rate X + 3 hours @ rate Y
= 4X + 3YThe question asks us how many hours of X is required for the entire production?
Well, we already know the entire production is 4X + 3Y, so we set that = (? hours) @ rate X
4X + 3Y = (?) X
So we have 3 pieces: 2 variables (X, Y) and 1 missing number (?)
(1) tells us what X is. Do you really think you can solve 1 equation with 2 variables? NO!! You need more information.
So (1) is no good. Don’t even need to think of anything else.
Let’s look at (2): X produces twice as many bottles in 4 hours as Y in 3 hours. Oh, so they’re telling you what X is in relation to Y.
Well, that would help you rewrite the equation:
4X + 3Y = (?)Xin terms of just (X’s and the ?) or (Y’s and the ?)
I don’t know the calculations, but it would like something like:
4X + somethingX = (?) X
combine and you get(somethingX) = (?)X
Oh wait, there’s X on both sides. If you cancel out the X’s, you’ll end up with some number.
I don’t know what that number is, but at this point, I can confidently say that (2) gives me enough information.(2) gave me X in terms of Y and vice versa. From there, I can get rid of one variable, and be left 1 variable. So I have 1 equation and 1 unknown – I just need that missing number (?). We know 1 equation/1 unknown is solvable, and then from there you can get the actual number.
Don’t waste time thinking about what that number is. Just know that you have enough info to solve the problem with (2).
When (1) is no good and (2) is good, then choose (B) as your final answer.
Hope that helps.
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GMAT Official Guide 12: PS #148
OG12: PS: #148If x, y, and k are positive numbers such that
(x/x+y)(10)+(y/x+y)(20)=k and if x is less than y, which of the following could be the value of k?A. 10
B. 12
C. 15
D. 18 (Correct answer)
E. 30Zeke– is there any possible way of going about this problem without having to simplify and then find the boundaries of x and y? The explanation in the OG12 is very confusing. I wanted to know if there was a more efficient way to solve this problem.
Thanks!
KatieHi Katie,
The question implies that there are certain values that k cannot be. Indeed, the OG explanation is confusing. Most people would not be able to simplify expressions to that level of depth.
But because you know that k cannot be certain values, you can always try the various values of x, y, and k. Put together a chart and do some quick math in your head. Make sure you write it out and that you only test values of x
You can simplify it up to the expression 10y/(x+y)..then put the following table together
x y expression (10y/(x+y)
———————
1 2 6.7ish
1 3 7.5
2 3 6
2 8 8
2 20 ~10So it seems like we found values of k in the range of 6-10ish is doable. You need to add 10 to it since the expression you simplified it to was
10 + 10y/(x+y)
So the range of k values is roughly 16-20. Of the answer choices, only 18 fits in (D). And you can assume that the reason why you haven’t been able to generate values outside of that range is because those are invalid values of k. So you can safely assume all the other answer choices are no good.
Hope that helps!
GMAT Pill Support Team
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