GMAT Questions Chat Group

C. three

(3+1)(5+1) = 24 factors but z^4 is a perfect square so it will have odd factors
since 24 is not odd
then x and y are the same prime number
Hence, x^8=z^4
x^2=z
(2+1)=3 factors.

If |y| > |x|, is x – y < 0?

(1) |x| + |y| > |x - y|
(2) x + y < 0\

can someone explain this

I think the answer should be 2 ..
Because x will be a factor, y will be a factor and 1 and z itself is a factor.

Answer should be B
For QS to be true "y" must be +ve, else for false it must be -ve
stmt1 - it can be achieved if both x & y have same sign. Thus, y can be -ve or =ve;
Insufficient
stmt2- by this statement y must be -ve; |y|>|x|
Ans: False ; Sufficient

Is it clear?

A; because x=1; stmt 2 is not needed

why y must be positive

c

How do u get x=1?

Y +ve can also result in tru tho, if x<y and x>0

HL emp =x; Total HL emp count =x
each supervises ML emp = x^2 ; Total ML emp count =x^3; (note each x has x^2)
each supervises LL emp =x^3 ; Total LL emp count = x^6; (by above logic )
stmt 1 says x^6<60; for x=2, it will be 64; Thus, x can only be 1 (assuming office is not employing fractional employees )

What is your question?
I chose B because, A gives both +ve and -ve for Y. This is contradiction
B only gives -ve value for Y

Please read the full sentence. Also, please read =ve as +ve

Ah make sense. Thanks!

PS Question 1 - Apr 28

A Manager needs 4 different teams (named P, Q, R, S) for doing projects A, B, C, D respectively. In how many different ways can a group of 8 people be divided into 4 teams of 2 people each for carrying out the Projects?
A. 90
B. 105
c. 168
D. 420
E. 2520

Source: Others | Difficulty: Hard

DS Question 1 - Apr 28

When A and B are positive integers, is AB a multiple of 4?

1) The greatest common divisor of A and B is 6
2) The least common multiple of A and B is 30

Source: Math Revolution | Difficulty: Hard

Ps question explanation pls

In the problem solving question, we have to find the number of ways in which the manager can divide the 8 people into 4 groups of two people each and then assign the each group one project (from the four available ones).

GCD of A & B is 6 means , 6 is a factor of A and B, we can write
A= 6*a
B=6*b
AB = 36 ab= 4 *(9ab) 
Multiple of 4

for statement B:
A=6, B=10 YES
A=3, B=10 NO

A

A or B, unsure between the 2

Actually I agree

A should be the answer

Can you help with the calc please, i think its a mix of permutation n combination formula but unable to solve

Here you go!

The solution to this question involves two steps -

Step 01: Create four groups of two people in each group
Step 02: Divide four tasks among the four groups

1) Create four groups of two people in each group

We can do this in

8! / [(2)^4 * 4!]

If you’re wondering how did I arrive at this value, follow along, or else skip to step 2.

Let’s assume we have four boxes and 8 distinct balls. We have to put two balls in each box.

_ _ _ _

The first box can be filled in 8C2 ways
The second box can be filled in 6C2 ways (We have already selected 2 balls from the available 8, hence for the second box can be filled with two out of six remaining balls)
The third box can be filled in 4C2 ways
The fourth box can be filled in 2C2 ways

Now as we have considered a specific arrangement, divide by 4! to de-arrange the positions.

So, total way of forming the groups =

8C2 * 6C2 * 4C2 * 2C2 * (1/4!) = 8! / [(2)^4 * 4!]

2) Once you’ve created four groups of two people each, we can divide four tasks among the group the task in 4! ways.

So the total number of ways of performing both steps =

8! / [(2)^4 * 4!] * 4! = 8! / 2^4 = 2520

Option E.