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poojaarora1818 wrote:
Maybe option E

Can anyone pls confirm what would be right answer ?
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mr.safin wrote:
4 odd numbers ?

I went with a too first but apparently when we take m as 9.5 and n as 0.5 we get more than 4 odd integers
Now statement 2 tells us m is a positive integer(which eliminates the possibility of taking 9.5 and 0.5 or decimals in general)
Then we get 4 odd integers in every case thus we need both answer is C
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poojaarora1818 wrote:
Can anyone pls confirm what would be right answer ?

The answer is B(120) for the PS one
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Nikhil418 wrote:
The answer is B(120) for the PS one

How??
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C is the right answer for DS
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Yes, it is confimed C is the correct answer for DS.
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Tanks
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Nikhil418 wrote:
I went with a too first but apparently when we take m as 9.5 and n as 0.5 we get more than 4 odd integers Now statement 2 tells us m is a positive integer(which eliminates the possibility of taking 9.5 and 0.5 or decimals in general) Then we get 4 odd integers in every case thus we need both answer is C

M can still be 9 and N 0

Then there would be 3 odd integers
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GmatKing8 wrote:
M can still be 9 and N 0 Then there would be 3 odd integers

1.3,5,7(4 odd integers) between 9 and 0
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1,3,5,7*
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Thanks
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Answer for ps is E
Every thousand is divisible by 8 and all the options which are divisible by 8 except 12 and 210
The greatest among them is 320. Let m be the number
M= n + 8k
n would be multiple of by 1000
8k would be the remainder when divided by 1000
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Dhanxtomybeat wrote:
Answer for ps is E Every thousand is divisible by 8 and all the options which are divisible by 8 except 12 and 210 The greatest among them is 320. Let m be the number M= n + 8k n would be multiple of by 1000 8k would be the remainder when divided by 1000

Yeah, I have used the same logic and getting E as the correct answer.
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Answer for ps is E
Every thousand is divisible by 8 and all the options are divisible by 8 except 12 and 210
The greatest among them is 320. Let m be the number
M= n + 8k
n would be multiple of 1000
8k would be the remainder when divided by 1000
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Heres the explanation for the answer being B: https://gmatclub.com/forum/n-is-the-gre ... 2#p2696601
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Thank you so much.
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gmatophobia wrote:
DS Question 1 - September 27 How many odd numbers are there between m and n? (1) m – n = 9 (2) m is a positive integer. Source: Others | Difficulty: Hard

C
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It’s my humble to all that pls stop giving kudos on my wrong answers. So, whosoever doing this pls stop this. Thanks!
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An automobile rental agency purchases new cars from manufacturer in lots of 5 and releases the oldest cars in lots of 3. Last year the number of cars owned by the agency increased by 14. If a purchase of a lot and release of a lot each count as one transaction, which of the following is least number of transactions that could have occurred last year?

a) Three
b) Four
c) Five
d) Six
e) Seven
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Option E
IMO
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Nope
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It is option d
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5x-3y = 14 (x=4, y=2)
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gmatophobia wrote:
DS Question 1 - September 27 How many odd numbers are there between m and n? (1) m – n = 9 (2) m is a positive integer. Source: Others | Difficulty: Hard

C
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anyone from chd?
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are these tests reliable?
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chd here
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whats up vaibhv
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sourabhmulik wrote:
5x-3y = 14 (x=4, y=2)

When purchase and release each count as 1 transaction then X=Y
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Can someone please go through the above question and help me with the answer?
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gmatophobia wrote:
DS Question 1 - September 27 How many odd numbers are there between m and n? (1) m – n = 9 (2) m is a positive integer. Source: Others | Difficulty: Hard

4 odds number
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If PQ > 0 and |P|^Q = 1, then what is the value of P?

(1) |P| = P
(2) |Q| = Q
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D
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AsiaticLion wrote:
If PQ > 0 and |P|^Q = 1, then what is the value of P? (1) |P| = P (2) |Q| = Q

D
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Kaisa D?
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I mean, isn’t it A?
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Got it now. Thank you.
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👍
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can anyone explain this
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what is the rightmost non zero digit in the integer value of 12!?
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SATYAM7777 wrote:
what is the rightmost non zero digit in the integer value of 12!?

It is a unit digit question, remove 2, 5, and 10 because this gives unit digit as 0. Then remove group of numbers which gives unit digit as 6. then 11 will be left. when you multiply 11 with 6 you get unit digit as 6.
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SATYAM7777 wrote:
what is the rightmost non zero digit in the integer value of 12!?

6
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AsiaticLion wrote:
If PQ > 0 and |P|^Q = 1, then what is the value of P? (1) |P| = P (2) |Q| = Q

D
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If a and b are positive integers, what is the units digit of b ?

(1) b is 25% greater than a.

(2) If b is decreased by 50%, the result is not an integer.
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can someone explain this plz?
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From 1: b is a multiple of 5 (units digit can be 0/5) - not sufficient

From 2: b is odd (not sufficient)

1&2: units digit is 5, so both together are sufficient
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Isn’t the answer E ?

Because from 1: b =5a/4 => a is even => b is even

From 2: b is odd

So there in no single solution

Please feel free to correct me if I’m wrong here
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My bad, there’s an error on my end
Please ignore
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answer is D
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I got B
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sorry C
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C is the answer but i got B
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I know B isnt correct so its either E or C
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but I cant see C being right
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It’s C

From 2: we know b is odd

From 1: we get a = 4b/5

So for a to be an integer > 0, be has to be an integer divisible by 5

So the unit’s digit is either 0 or 5
But since we know from 2 that b is odd, the unit’s digit can only be 5

Hope this helps
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AsiaticLion wrote:
If PQ > 0 and |P|^Q = 1, then what is the value of P? (1) |P| = P (2) |Q| = Q

(1) Since |P| = P, P^Q = 1. Ln both sides, so Q ln P = 0. Q or ln P = 0. Since PQ>0, Q cannot be 0. Thus P = 1. Sufficient

(2) Since |Q| = Q, P^Q =1. Same as before, apply ln both sides and reject Q=0. So again, ln P = 0 which means P =1. Sufficient

D
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If the probability is 0.7 that event X will occur and 0.85 that event Y will occur, what is the least possible value of the probability that both events will occur?

(A) 0.15
(B) 0.55
(C) 0.595
(D) 0.7
(E) 0.85
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Can someone help with this one
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aryannn_1712 wrote:
If the probability is 0.7 that event X will occur and 0.85 that event Y will occur, what is the least possible value of the probability that both events will occur? (A) 0.15 (B) 0.55 (C) 0.595 (D) 0.7 (E) 0.85

C?
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aryannn_1712 wrote:
If the probability is 0.7 that event X will occur and 0.85 that event Y will occur, what is the least possible value of the probability that both events will occur? (A) 0.15 (B) 0.55 (C) 0.595 (D) 0.7 (E) 0.85

A?
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sourabhmulik wrote:
An automobile rental agency purchases new cars from manufacturer in lots of 5 and releases the oldest cars in lots of 3. Last year the number of cars owned by the agency increased by 14. If a purchase of a lot and release of a lot each count as one transaction, which of the following is least number of transactions that could have occurred last year? a) Three b) Four c) Five d) Six e) Seven

D =20-6= added 4 lots and sold 2 lots
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AsiaticLion wrote:
If PQ > 0 and |P|^Q = 1, then what is the value of P? (1) |P| = P (2) |Q| = Q

D
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SATYAM7777 wrote:
what is the rightmost non zero digit in the integer value of 12!?

12!= 12*11*10****2*1 split into Prime factors = 2^10 * 3^5 * 5^2 * 7^1 * 11^1 now exclude 0’s i.e. 5*2 for non 0 digit = 2^8* 3^5 *7^1 * 11^1 = 4*3*7*1 = 4 (Ans)
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aryannn_1712 wrote:
If the probability is 0.7 that event X will occur and 0.85 that event Y will occur, what is the least possible value of the probability that both events will occur? (A) 0.15 (B) 0.55 (C) 0.595 (D) 0.7 (E) 0.85

Should be C. As P(X will occur) * P(Y will occur) = 0.70 * 0.85 = 0.595
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aryannn_1712 wrote:
If the probability is 0.7 that event X will occur and 0.85 that event Y will occur, what is the least possible value of the probability that both events will occur? (A) 0.15 (B) 0.55 (C) 0.595 (D) 0.7 (E) 0.85

B: 0.55.

P(X) = 0.7
B(Y) = 0.85
P(X and Y) = P(X) + P(Y) - P(X or Y)
P(X and Y) = 0.7 + 0.85 - P(X or Y)
P(X and Y) = 1.55 - P(X or Y)

Now, we know that probability can not be less than 0, and can not be more than 1. Thus, to minimize P(X and Y) we need to maximize P(X or Y). P(X or Y) can be set as much as 1. And thus,

P(X and Y) = 1.55 - 1 = 0.55.

Hence, (B) 0.55 is the answer.
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IshanSaini wrote:
Should be C. As P(X will occur) * P(Y will occur) = 0.70 * 0.85 = 0.595

Agree B is right, I would solve using sets but yes same concept. The problem here and this is something to always be mindful of in 2 event problems as ppl tend to skip through is - The 2 events are not mentioned as independent [in which case we can take P (A&B)= P(A) * P(B)]. Hope this helps :)
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Is n divisible by 11 with no remainder?

(1) n is divisible by m with integer result
(2) m is divisible by 5.5 with integer result
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isn’t it E? E because if n is 5.5, it can be divisible by 5.5 but not divisible by 11. But the answer sheet seems to disagree with me.
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aryannn_1712 wrote:
If the probability is 0.7 that event X will occur and 0.85 that event Y will occur, what is the least possible value of the probability that both events will occur? (A) 0.15 (B) 0.55 (C) 0.595 (D) 0.7 (E) 0.85

I actually used Venn diagram for this which I believe works too. Probability of both events occurring will be the overlapping region. And since it needs to be minimised, prob of X + prob of Y - prob of X and Y occurring = 1.
P(X) + P(Y) - overlapped region = 1
Overlapped region = 1.55 - 1
So B
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sldisek783 wrote:
isn’t it E? E because if n is 5.5, it can be divisible by 5.5 but not divisible by 11. But the answer sheet seems to disagree with me.

I also thought it’s E. Was there an explanation given? I saw the responses to the post and I don’t quite agree with them either.
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If x^2 < |x| and x≠0, which of the following are true?
I. x<0
II. x^2 < 1
III. x^1/2 = integer
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Only II?
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Rohinivish wrote:
If x^2 < |x| and x≠0, which of the following are true? I. x<0 II. x^2 < 1 III. x^1/2 = integer

1 and 2 only?
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1
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1,2
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Rohinivish wrote:
If x^2 < |x| and x≠0, which of the following are true? I. x<0 II. x^2 < 1 III. x^1/2 = integer

I) Is false, consider the case of x=1/2

Ii) Is true. For any integer value of x, x^2 must be larger. So x must be some fraction (neg or pos)

III) root(x)=int. Not necessIrly true.
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sldisek783 wrote:
Is n divisible by 11 with no remainder? (1) n is divisible by m with integer result (2) m is divisible by 5.5 with integer result

1) Clearly ins
2) 5.5=11/2
If m/(11/2)=int then 2m/11=int
M is some multiple of 11, since 2 is jot a factor of 11. Sufficient

B
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hazeljj wrote:
I also thought it’s E. Was there an explanation given? I saw the responses to the post and I don’t quite agree with them either.

The answer is C in fact. The question asks if n is divisible by 11 with no remainder, not if n is divisible by 11 with an integer result and with no remainder.
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sldisek783 wrote:
The answer is C in fact. The question asks if n is divisible by 11 with no remainder, not if n is divisible by 11 with an integer result and with no remainder.

Good catch! It is not given that m is an int, making 2 ins.
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mysterymanrog wrote:
1) Clearly ins 2) 5.5=11/2 If m/(11/2)=int then 2m/11=int M is some multiple of 11, since 2 is jot a factor of 11. Sufficient B

Say n = m = 5.5- yeah so? So S1 n/m =1 is true. S2 m/5.5 = 1 is also true but now your answer to the Qn stem is: False. In case n= m = 11x ( x is an integer) your answer is true. Based on these 2 points. Ans should be E. What is the source of this question??
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IS coordinate Geometry is a part of Gmat focus?
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shivam7 wrote:
IS coordinate Geometry is a part of Gmat focus?

Yes
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danthanh wrote:
Only II?

Only I, bro. It has to be less than 0, only then the square root of it will satisfy the modulus comparison
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Yes
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Akhilyadav2122 wrote:
Only I, bro. It has to be less than 0, only then the square root of it will satisfy the modulus comparison

No, it can be a positive or negative fraction
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shivam7 wrote:
IS coordinate Geometry is a part of Gmat focus?

aye, counts as algebra ( cuz it is )
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Ok
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Does anyone have any GMAT preparation resources for reference? Much appreciated, thanks.
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mysterymanrog wrote:
No, it can be a positive or negative fraction

In case of a positive fraction, it must always be less than 0, else how the modulus condition will be satisfied?
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Ok
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A cube is painted red on two opposite faces and green on the remaining four faces. It has now been cut into 125 identical small cubes.

How many small cubes have none of their face painted ?
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Anyone
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27?
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No sorry 75
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lol sorry 27 was correct
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125 pieces means 5x5x5 which is 5 on each edge, so eliminating one cube on each end gives a 3x3x3 cube inside the 5x5x5 cube which will have no face painted

= 27
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27
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Yes
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