If K is a positive integer, what is the remainder when (13^4K+2) +8 is
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GMAT Club Legend Joined: 07 Jun 2014 Posts: 4749 GRE 1: Q167 V156 WE: Business Development (Energy and Utilities) Followers: 93 
In the coordinate plane, points (a, b) and (c, d) are equidi [#permalink]
Expert’s post In the coordinate plane, points (a, b) and (c, d) are equidistant from the origin.
\(a > c\)
A) Quantity A is greater. _________________ Sandy 

Intern Joined: 20 Oct 2018 Posts: 1 Followers: 0 
Re: In the coordinate plane, points (a, b) and (c, d) are equidi [#permalink]
The answer is B 

Intern Joined: 22 Jul 2018 Posts: 41 Followers: 0 
Re: In the coordinate plane, points (a, b) and (c, d) are equidi [#permalink]
1 This post received since both are equidistance from origin both are 5 , so ab (4,3) and cd (3,4) and and B 

Intern Joined: 02 Dec 2018 Posts: 10 Followers: 0 
Re: In the coordinate plane, points (a, b) and (c, d) are equidi [#permalink]
ruposh6240 wrote: since both are equidistance from origin both are 5 , so ab (4,3) and cd (3,4) and and B But d is bigger than b.. I’m sure the answer is wrong. Let’s do it another way. We should have 

Manager Joined: 22 Feb 2018 Posts: 119 Followers: 2 
Re: In the coordinate plane, points (a, b) and (c, d) are equidi [#permalink]
Answer: B
What we have: Distance from (a,b) to origin (0,0) = √((a0)^2 + (b0)^2) = √(a^2 + b^2) As it is said that a > c so we can conclude that a^2 > c^2 [They are both positive, thus when they are in the same power, their comparison result will be just the same as when the power is 1.] Follow your heart 

greprepclubot 
Re: In the coordinate plane, points (a, b) and (c, d) are equidi [#permalink]


Carcass wrote:
A. 0.5
B. 1
C. 1.5
D. 2
E. 2.5
F. 3
G. 3.5
IMPORTANT RULE: If two sides of a triangle have lengths A and B, then: AB < third side < A+B
Now let’s see what measurements we get for each value of x.
A. x = 0.5. So 1, x, and x² become 1, 0.5 and 0.25. Here, it is NOT the case that 0.5 – 0.25 < 1 < 0.5 + 0.25
ELIMINATE A.
B. x = 1. So 1, x, and x² become 1, 1 and 1. Here, it IS the case that 1 – 1 < 1 < 1 + 1
B WORKS
C. x = 1.5. So 1, x, and x² become 1, 1.5 and 2.25. Here, it IS the case that 1.5 – 1 < 2.25 < 1 + 1.5
C WORKS
D. x = 2. So 1, x, and x² become 1, 2 and 4. Here, it is NOT the case that 1 – 2 < 4 < 1 + 2
ELIMINATE D.
E. x = 2.5. So 1, x, and x² become 1, 2.5 and 6.25
Here, it is NOT the case that 1 – 2.5 < 6.25 < 1 + 2.5
ELIMINATE E.
F. x = 3. So 1, x, and x² become 1, 3 and 9. Here, it is NOT the case that 1 – 3 < 9 < 1 + 3
ELIMINATE F.
G. x = 3.5. So 1, x, and x² become 1, 3.5 and 12.25. Here, it is NOT the case that 1 – 3.5 < 12.25 < 1 + 3.5
ELIMINATE G.
Answer: B, C
Cheers,
Brent
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Carcass wrote:
Quantity A 
Quantity B 
p+q 
5 
A. The quantity in Column A is greater
B. The quantity in Column B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Given:\(\frac{pq}{p+q} = \frac{2}{3}\)
Multiply both sides by (p+q) to get: \(pq = (\frac{2}{3})(p+q)\)
Multiply both sides by 3 to get: \(3(pq) = (2)(p+q)\)
Expand both sides to get: \(3p3q = 2p+2q\)
Subtract 2p from both sides to get: \(p3q = 2q\)
Add 3q to both sides to get: \(p = 5q\)
Now take given quantities and replace p with 5q to get:
Quantity A: 5q + q
Quantity B: 5
Simplify:
Quantity A: 6q
Quantity B: 5
Let’s TEST some values of q
CASE i: If q = 1/6, then we get:
Quantity A: 6(1/6) = 1
Quantity B: 5
In this case, Quantity B is greater
CASE ii: If q = 1, then we get:
Quantity A: 6(1) = 6
Quantity B: 5
In this case, Quantity A is greater
Answer: D
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This is the tricky part.
You should infer from the stem. Moreover, this GRE (or GMAT) is all about: a test of reasoning.
Quote:
In 1995 the number of students enrolled in public institutions of higher education was approximately how many times the number of students enrolled in private institutions of higher education?
From the bold part you might infer that you need the %. As such, the second graph.
Regards
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Could you share the source please
The question asks about choosing 2 ppl from 10 people and the probability of the two being male.
There are 4 women and 6 men.
The number of ways we can select 2 men = \(C_2^{6}=\frac{6!}{2! \times 4!}=15\).
The number of ways we can select 2 people = \(C_2^{10}=\frac{10!}{2! \times 8!}=45\).
So the probability of selecting 2 men = \(\frac{15}{45}=0.333\)
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Moderator Joined: 18 Apr 2015 Posts: 5078 Followers: 76 
QOTD #1 Set M consists of all the integers between –2 and 12 [#permalink]
Expert’s post Set M consists of all the integers between –2 and 12, inclusive.
Set N consists of all the integers between 9 and 15, inclusive.
A)The quantity in Column A is greater. _________________ 

Manager Joined: 23 Jan 2016 Posts: 137 Followers: 3 
Re: QOTD #1 Set M consists of all the integers between –2 and 12 [#permalink]
1 This post received The common elements of both sets are 9,10,11,12; the smallest is 9. Answer C 

Intern Joined: 06 Dec 2018 Posts: 1 Followers: 0 
Re: QOTD #1 Set M consists of all the integers between –2 and 12 [#permalink]
M = {2,1,0,1,2,3,4,5,6,7,8,9,10,11,12}
N = {9,10,11,12} Only 9 is common from both M, and N set. 

greprepclubot 
Re: QOTD #1 Set M consists of all the integers between –2 and 12 [#permalink]


Let’s see which information we have. Car A has traveled 45 minutes at a speed 8m/h greater than car B speed (x). Car B traveled 15 minutes at speed x m/h.
Since they meet along a road 62 km long, it must be that the distance the two cars traveled sum to 62. Thus, we can proceed by pluggingin the choices we have and check if they do the job. Let’s start from the middle one, C = 10 m.
If car B would have traveled 10 m, its speed would have been 10/(1/4) = 40 m/h (15 minutes is 1/4 of hour). Thus, car A speed would have been 48 m/h and it would have been traveled for 48*(3/4) = 24 m. Then, 10m+24m = 34m < 62 m. Thus, choice C is not our answers. We have to look for something higher.
If we proceed with this same kind of check we get that if car B traveled for 14m, then its speed is 14/(1/4) = 56 m/h so that car B speed is 64 m/h and it traveled for 48m, which summed to 14 gives us 62, the length of the road.
Answer A
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Moderator Joined: 18 Apr 2015 Posts: 5056 Followers: 76 
x is a nonnegative number and the square root of (10 – 3x) [#permalink]
1 This post received Expert’s post x is a nonnegative number and the square root of (10 – 3x) is greater than x.
A) Quantity A is greater. _________________ 

Director Joined: 03 Sep 2017 Posts: 521 Followers: 1 
Re: x is a nonnegative number and the square root of (10 – 3x) [#permalink]
1 This post received We need to solve the equation \(sqrt(103x)>x\). Squaring both sides we get \(103x>x^2\) that can be rewritten as \(x^2+3x10
Since x is non negative, among the solutions we must consider only those between 0 and 2, excluded. Thus, no matter the absolute value, column A is always smaller than 2. Answer B 

Intern Joined: 11 Jan 2018 Posts: 44 Followers: 0 
Re: x is a nonnegative number and the square root of (10 – 3x) [#permalink]
IlCreatore wrote: We need to solve the equation \(sqrt(103x)>x\). Squaring both sides we get \(103x>x^2\) that can be rewritten as \(x^2+3x10
Since x is non negative, among the solutions we must consider only those between 0 and 2, excluded. Thus, no matter the absolute value, column A is always smaller than 2. Answer B Good explanation. Persistence >>>>>>> Success Don’t say thanks, just give KUDOS. 

Manager Joined: 26 Jun 2017 Posts: 104 Followers: 0 
Re: x is a nonnegative number and the square root of (10 – 3x) [#permalink]
so
Attachments _________________ What you think, you become. 

Moderator Joined: 18 Apr 2015 Posts: 5056 Followers: 76 
Re: x is a nonnegative number and the square root of (10 – 3x) [#permalink]
Expert’s post Please boxing,
your replies are very delightful. However, please as a test and not as a screenshot. Thank you so much for your collaboration. The board is cleaner that way. regards 

Intern Joined: 28 Nov 2017 Posts: 44 Followers: 1 
Re: x is a nonnegative number and the square root of (10 – 3x) [#permalink]
1 This post received Short & Quick way to get the Answer is :
3x has to be less than 10 as we can’t find out the Sqaure root of a a negative number so x can only take values of 0,1,2 & exclude non negative solutions so 2 is rejected. Answer b 

Manager Joined: 22 Feb 2018 Posts: 118 Followers: 2 
Re: x is a nonnegative number and the square root of (10 – 3x) [#permalink]
1 This post received Answer: B
Sure root of 10 – 3x is greater than x and x is a nonnegative integer. First, as x is nonnegative, thus x equals x in A. We try numbers. The minimum value for x can be 0. If x = 0, square root of 103x equals 10, which is greater than x. If x = 1, 103 is greater than 1 If x = 2, 10 6 is not greater than 2. So, x can be either 0 or 1 and always less than 2. The answer is B. _________________ Follow your heart 

Intern Joined: 20 Mar 2018 Posts: 34 GRE 1: Q164 V150 Followers: 1 
Re: x is a nonnegative number and the square root of (10 – 3x) [#permalink]
Yiddo_Bhushan wrote: Short & Quick way to get the Answer is :
3x has to be less than 10 as we can’t find out the Sqaure root of a a negative number so x can only take values of 0,1,2 & exclude non negative solutions so 2 is rejected. Answer b However, it is not mentioned that x has to be an integer. x can be just less than 10/3 or 3 for that matter. 

Intern Joined: 20 Mar 2018 Posts: 34 GRE 1: Q164 V150 Followers: 1 
Re: x is a nonnegative number and the square root of (10 – 3x) [#permalink]
1 This post received Given: x>0
and \sqrt{(103x)} > x gives 5 Ignoring negative solutions. 0<x<2 and x would be less than 2 

Intern Joined: 02 Dec 2018 Posts: 4 Followers: 0 
Re: x is a nonnegative number and the square root of (10 – 3x) [#permalink]
Hey guys, I understand your solution, but could you point out the fallacy or mistake in my solution.
sqrt(103x) > x So D because the absolute value of x can be greater than 2 or smaller than 2. 

greprepclubot 
Re: x is a nonnegative number and the square root of (10 – 3x) [#permalink]


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Moderator Joined: 18 Apr 2015 Posts: 5051 Followers: 76 
The points A( 0, 0), B( 0, 4a – 5), and C( 2a + 1, 2a + 6) f [#permalink]
Expert’s post The points A( 0, 0), B( 0, 4a – 5), and C( 2a + 1, 2a + 6) form a triangle. If ∠ ABC = 90 °, what is the area of triangle ABC?
A. 102 B. 120 C. 132 D. 144 E. 156 _________________ 
Intern Joined: 05 Dec 2018 Posts: 8 Followers: 0 
Re: The points A( 0, 0), B( 0, 4a – 5), and C( 2a + 1, 2a + 6) f [#permalink]
As angle ABC is 90. This is a right triangle with base AB and height BC. So area of Triangle is 1/2 (AB). BC 
Supreme Moderator Joined: 01 Nov 2017 Posts: 197 Followers: 3 
Re: The points A( 0, 0), B( 0, 4a – 5), and C( 2a + 1, 2a + 6) f [#permalink]
Expert’s post Carcass wrote: The points A( 0, 0), B( 0, 4a – 5), and C( 2a + 1, 2a + 6) form a triangle. If ∠ ABC = 90 °, what is the area of triangle ABC?
A. 102 B. 120 C. 132 D. 144 E. 156 90° is at B so area will be (AB*BC)/2 Area = \(\frac{12*17}{2}=6*17=102\) A Some useful Theory. 
greprepclubot 
Re: The points A( 0, 0), B( 0, 4a – 5), and C( 2a + 1, 2a + 6) f [#permalink]

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Intern Joined: 06 Oct 2018 Posts: 6 Followers: 0 
Set A {300,400,500,700,900} Set B {100,200,300,500,900} [#permalink]
1 This post received Set A {300,400,500,700,900}
Set B {100,200,300,500,900}
A) Quantity A is greater. 

Supreme Moderator Joined: 01 Nov 2017 Posts: 197 Followers: 3 
Re: Set A {300,400,500,700,900} Set B {100,200,300,500,900} [#permalink]
Expert’s post Arun1992 wrote: Set A {300,400,500,700,900}
Set B {100,200,300,500,900}
A) Quantity A is greater. The standard deviation is dependent on the spread of the elements. Set A {300,400,500,700,900}… Mean = \(\frac{2800}{5}=580\) B Some useful Theory. 

GRE Instructor Joined: 10 Apr 2015 Posts: 1225 Followers: 45 
Re: Set A {300,400,500,700,900} Set B {100,200,300,500,900} [#permalink]
Expert’s post Arun1992 wrote: Set A {300,400,500,700,900}
Set B {100,200,300,500,900}
A) Quantity A is greater. For many standard deviation questions on the GRE, we can often just “eyeball” the numbers in each set to get a feel for which set has the greatest dispersion (aka deviation). If we do so, we can see that the numbers in Set B are more dispersed, which means Set B has the greater standard deviation Answer: B ——————————— Brent Hanneson – Creator of greenlighttestprep.com 

greprepclubot 
Re: Set A {300,400,500,700,900} Set B {100,200,300,500,900} [#permalink]


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nadeem790 wrote:
a) 5
b) 10
c) 20
d) \(33 \frac{1}{3}\)
e) 50
Let’s keep track of the acid
A container holds 10 liters of a solution which is 20% acid.
20% of 10 liters = 2 liters
So, there are 2 liters to start
6 liters of pure acid are added to the container
2 + 6 = 8
So, there are now 8 liters of acid in the RESULTING solution.
What percent of the resulting mixture is acid?
We added 6 liters to 10 liters, to get a RESULTING solution with volume = 16 liters
8/16 = 1/2 = 50%
Answer: E
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sandy wrote:
Quantity A 
Quantity B 
m 
3 
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
GIVEN: \(\frac{(10^3)(0.027)}{(900)(10^{2})}=(3)(10^m)\)
Rewrite some values: \(\frac{(10^3)(27)(10^{3})}{(9)(100)(10^{2})}=(3)(10^m)\)
Rewrite more values: \(\frac{(10^3)(3^3)(10^{3})}{(3^2)(10^2)(10^{2})}=(3)(10^m)\)
Simplify: \(\frac{(10^0)(3^3)}{(3^2)(10^0)}=(3)(10^m)\)
Simplify: \(\frac{3^3}{3^2}=(3)(10^m)\)
Simplify: \(3=(3)(10^m)\)
This tells us that \(10^m = 1\), which means m = 0
We get:
QUANTITY A: 0
QUANTITY B: 3
Answer: B
Cheers,
Bret
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