DS Problem from Mprep CAT 1

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DS Problem from Mprep CAT 1

by nsuen » Thu Apr 28, 2016 3:46 am
Hi all,

I have two DS problems that I couldnt understand ( I wont type the answers as we all know what A, B, C,D, E, mean) Thank you in advance.

1) Is d negative?

(1) e + d = -12
(2) e - d < -12

My problem: Is there an easier way than below to understand? I tested no. but couldn't be sure whether it is C or E and I ran out of time.

Explanation: (1) INSUFFICIENT: When two numbers sum to a negative value, we have two possibilities:
Possibility A: Both values are negative (e.g., e = -4 and d = -8)
Possibility B: One value is negative and the other is positive.(e.g., e = -15 and d = 3).
(2) INSUFFICIENT: When the difference of two numbers produces a negative value, we have three possibilities:
Possibility A: Both values are negative (e.g., e = -20 and d = -3)
Possibility B: One value is negative and the other is positive (e.g., e = -20 and d = 3).
Possibility C: Both values are positive (e.g., e = 20 and d = 30)
(1) AND (2) SUFFICIENT: When d is ADDED to e, the result (-12) is greater than when d is SUBTRACTED from e. This is only possible if d is a positive value. If d were a negative value than adding d to a number would produce a smaller value than subtracting d from that number (since a double negative produces a positive). You can test numbers to see that d must be positive and so we can definitively answer the question using both statements.


2) A certain panel is to be composed of exactly three women and exactly two men, chosen from x women and y men. How many different panels can be formed with these constraints?

(1) If two more women were available for selection, exactly 56 different groups of three women could be selected.
(2) x = y + 1
Source: — Data Sufficiency |

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by GMATGuruNY » Thu Apr 28, 2016 4:03 am
Is d negative?

(1) e + d = -12

(2) e - d < -12
Clearly, neither statement is sufficient on its own.
Substituting e = -d-12 into e-d < -12, we get:
(-d-12) - d < -12
-2d - 12 < -12
-2d < 0
d > 0.
SUFFICIENT.

The correct answer is C.
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by GMATGuruNY » Thu Apr 28, 2016 4:05 am
A certain panel is to be composed of exactly three women and exactly two men, chosen from x women and y men. How many different panels can be formed with these constraints?

(1) If two more women were available for selection, exactly 56 different groups of three women could be selected.

(2) x = y + 1
Statement 1:
No information about the number of men.
INSUFFICIENT.

Statement 2:
Since x and y can be any two consecutive nonnegative integers, INSUFFICIENT.

Statements combined:
Statement 1:
Since 56 = 7*8, at least 7 women are required to yield 56 different groups of 3.
TEST CASES:
From 7 women, the number of groups of 3 that can be formed = 7C3 = (7*6*5)/(3*2*1) = 35.
From 8 women, the number of groups of 3 that can be formed = 8C3 = (8*7*6)/(3*2*1) = 56.
From 9 women, the number of groups of 3 that can be formed = 9C3 = (9*8*7)/(3*2*1) = 84.
As indicated by the red case, 8 women are required to form 56 different groups of 3.
Since 8 is equal to two more than the ACTUAL number of women, x = 8-2 = 6.

Statement 2:
Since x=6 and x = y+1, y=5.

Since the values of x and y are known, it is possible to determine the number of 3-women, 2-men panels that can be formed.
SUFFICIENT.

The correct answer is C.
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by MartyMurray » Thu Apr 28, 2016 4:25 am
nsuen wrote:Hi all,

I have two DS problems that I couldnt understand ( I wont type the answers as we all know what A, B, C,D, E, mean) Thank you in advance.

1) Is d negative?

(1) e + d = -12
(2) e - d < -12

My problem: Is there an easier way than below to understand? I tested no. but couldn't be sure whether it is C or E and I ran out of time.

Statement 1: e + d = -12

Since e could have any value, d could have any value.

e = 0 d = -12

e = -12 d = 0

Insufficient.

Statement 2: e - d < -12

Once again, since e could have any value, d could have any value.

e = -1000 d = 0

e = 500 d = 1000

You could test big, simple numbers like those to see that b could be positive, negative or zero.

Insufficient.

Statements Combined:

One way to handle this is to subtract Statement 1 from Statement 2. You could do that as the two sides of Statement 1 are equal. So you would be subtracting equal quantities from both sides of Statement 2.

..e - d < -12
- e + d = -12

..-2d < 0

So -d < 0 and 0 < d Therefore d must be positive.

Alternatively you could just notice that e - d < -12 = e + d.

If e - d < e + d, then d must be positive.

Sufficient.

The correct answer is C.

2) A certain panel is to be composed of exactly three women and exactly two men, chosen from x women and y men. How many different panels can be formed with these constraints?

(1) If two more women were available for selection, exactly 56 different groups of three women could be selected.
(2) x = y + 1
Statement 1:

From this statement we can determine the number of women, because there is only one number of women, 8, from which we can choose 3 women, to get 56 unique groups. 8C3 = 56.

So the number of women is 8 - 2 = 6.

However, this statement does not provide information for determining how many men there are.

Insufficient.

Statement 2:

This statement provides information on the relationship between the number of women and the number of men, but does not provide information on the exact number of either women or men.

Insufficient.

Statements Combined:

Statement 1 provides information leading to the number of women, and then that information can be plugged into statement 2 to get the number of men.

Sufficient.

The correct answer is C.
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by Brent@GMATPrepNow » Thu Apr 28, 2016 6:07 am
A certain panel is to be composed of exactly three women and exactly two men, chosen from x women and y men. How many different panels can be formed with these constraints?

(1) If two more women were available for selection, exactly 56 different groups of three women could be selected.

(2) x = y + 1
Target question: How many different panels can be formed with these constraints?

First recognize that, since the order of the selected people does not matter, we can use combinations to solve this. We can select 3 women from x women in xC3 ways, and we can select 2 men from y men in yC2 ways. So, the total number of possible panels = (xC3)(yC2)

As you can see,the answer to the target question will depend solely on the individual values of x and y.

Statement 1: If two more women were available for selection, exactly 56 different groups of three women could be selected.
Since there's no information about the number of men, statement 1 is NOT SUFFICIENT

Statement 2: x = y + 1
There are several pairs that meet this condition. Here are two:
Case a: x = 3 and y = 2, in which case there's only 1 possible panel (since we'd have no choice but to select all 5 people)
Case b: x = 201 and y = 200, in which case there are TONS of possible panels
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Statement 1 basically tells us that (x+2)C3 = 56 ["x+2 CHOOSE 3 equals 56"]
Do we need to solve for x? NO. We need only recognize that we COULD solve for x.
Let's start checking a few possible values.
3C3 = 1
4C3 = 4
5C3 = 10
6C3 = 20

Aside: if anyone is interested, we have a free video on calculating combinations (like 6C3) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789

We can see that the numbers keep increasing, so there will be ONLY ONE value of x such that (x+2)C3 = 56 [incidentally, 8C3 = 56. So, (x+2) = 8, which means x = 6. Of course, that doesn't really matter since we need only recognize that we COULD determine the value of x]
So, from statement 1, we COULD determine the value of x
Once we know the value of x, we can use statement 2 to determine the value of y.
At this point, we can answer the target question with certainty, so the combined statements are SUFFICIENT

Answer = C

Cheers,
Brent
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by Matt@VeritasPrep » Fri Apr 29, 2016 2:21 pm
The first one is MUCH easier than the explanation suggests. Just telescope the two inequalities:

e + d = -12 > e - d

so

e + d > e - d

so

d > -d

so

2d > 0

so

d > 0

and you're set!

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by Matt@VeritasPrep » Fri Apr 29, 2016 2:24 pm
Now for the second one. The prompt tells us we're picking 3 women (out of x) and 2 men (out of y), so we have a total of (x choose 3) * (y choose 2) panels.

S1::

((x + 2) choose 3) = 56

(x+2)! / (3! * (x - 1)!) = 56

x! = 3! * 56

That gives us the number of women, but NOT the number of men.

S2::

This won't give us any actual numbers.

S1 + S2::

S1 gives us x, and S2 tells us that y = x - 1, so now we have both.