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AG2907
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gmatophobia
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gmatophobia
DS Question 1 - May 31 Is p > q ? (1) p^2 > q (2) p^3 < q Source: Expert’s Global | Difficulty: Hard

gmatophobia
PS Question 1 - May 31 Let x be a two-digit number. If the sum of the digits of x is 9, then the sum of the digits of the number (x + 10) is (A) 1 (B) 8 (C) 10 (D) either 8 or 10 (E) either 1 or 10 Source: Others | Difficulty: Hard

DS Question 1 - June 6

For each landscaping job that takes more than 4 hours, a certain contractor charges a total of r dollars for the first 4 hours plus 0.2*r dollars for each additional hour or fraction of an hour, where r > 100, Did a particular landscaping job take more than 10 hours.

(1) The contractor charges a total of $288 for the job.
(2) The contractor charges a total of 2.4r dollars for the job.

Source: Official Guide | Difficulty: Hard

PS Question 1 - June 6

9 people, including 3 couples, are to be seated in a row of 9 chairs. What is the probability that only one couple is sitting together

A. 1/63
B. 1/6
C. 8/21
D. 61/126
E. 7/9

Source: Others | Difficulty: Hard
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gmatophobia
DS Question 1 - June 6 For each landscaping job that takes more than 4 hours, a certain contractor charges a total of r dollars for the first 4 hours plus 0.2*r dollars for each additional hour or fraction of an hour, where r > 100, Did a particular landscaping job take more than 10 hours. (1) The contractor charges a total of $288 for the job. (2) The contractor charges a total of 2.4r dollars for the job. Source: Official Guide | Difficulty: Hard
B

A is not sufficient because value of r is unknown
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gmatophobia
PS Question 1 - June 6 9 people, including 3 couples, are to be seated in a row of 9 chairs. What is the probability that only one couple is sitting together A. 1/63 B. 1/6 C. 8/21 D. 61/126 E. 7/9 Source: Others | Difficulty: Hard
Hi gmatphobia
I checked twice but I’m getting 11/63. Could you point out the error in below.
P(Only 1C) = P(T)-[p(0C)+P(2+ C)]
{9!-[(9!-8!*2!) + 6!*4!]} / 9! = 11/63
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Got it. It’s C - 8/21

P(only 1C) = 3{P(1+ C)-[2*P(only 2C)+P(only 3C)]}
P(only 3C) = 6!*2!*2!*2!/9!
P(only 2C) = P(2+ C)-P(only 3C)= (7!*2!*2!-6!*2!*2!*2!)/9! (for only 1 pair of 2couple, *3 for all )
P(1+ C)=2*8!/9! (for only 1 couple, *3 for all )
Substitute in first eqn to get 8/21
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What is the value of x?

(1) x−2x=1


2

=
1

(2) xx=1

What is the value of x? (1) x^(-2x) = 1 (2) x^x = 1

can we take zero to the power zero as 1?
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c

for 1 it could be 1 as well as-1

Anki111
can we take zero to the power zero as 1?
if this is the case then it should be E right ?
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the answer is B

BUT if we consider zero to the power zero ..then it should be E
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Anki111
can we take zero to the power zero as 1?
No, the gmat does not use zero to the power of zero
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B looks surficient for this

x^x=1

x can only be 1, as if it is -1 it does not change signs
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Yea i got B as well, just clarifying the doubt
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and btw, 0^0 is undefined
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No, it is undefined and can be 1 depends on what you are asking
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because if you consider: 0^(1-1) which implies that 0^0=0/0

so that is why gmat does not use it
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yes then the answer stands B
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To practice quant questions, which test prep has the closest gmat like questions, anyone ?
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Is x/y > 0? (1) x - y > 0 (2) xy > 0
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wickedvikram
P(only 1C) = 3{P(1+ C)-[2*P(only 2C)+P(only 3C)]} P(only 3C) = 6!*2!*2!*2!/9! P(only 2C) = P(2+ C)-P(only 3C)= (7!*2!*2!-6!*2!*2!*2!)/9! (for only 1 pair of 2couple, *3 for all ) P(1+ C)=2*8!/9! (for only 1 couple, *3 for all ) Substitute in first eqn to get 8/21
Can you explain this in words?

Anki111
What is the value of x? (1) x^(-2x) = 1 (2) x^x = 1
D
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