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Re: In the coordinate plane, points (a, b) and (c, d) are equidi

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In the coordinate plane, points (a, b) and (c, d) are equidi [#permalink] New post 30 Sep 2018, 15:58

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In the coordinate plane, points (a, b) and (c, d) are equidistant from the origin.
\(|a| > |c|\)

Quantity A

Quantity B

|b|

|d|

A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.

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Re: In the coordinate plane, points (a, b) and (c, d) are equidi [#permalink] New post 28 Oct 2018, 11:48

The answer is B

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Re: In the coordinate plane, points (a, b) and (c, d) are equidi [#permalink] New post 30 Oct 2018, 20:57

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since both are equidistance from origin both are 5 , so ab (4,3) and cd (3,4) and and B

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Re: In the coordinate plane, points (a, b) and (c, d) are equidi [#permalink] New post 10 Dec 2018, 09:50

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
sqrt(b^2 + a^2) = sqrt(c^2 + d^2).
squaring both sides we get
b^2 + a^2 = c^2 + d^2
Since abs(a) > abs(c) we must have abs(b) < abs(d)

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Re: In the coordinate plane, points (a, b) and (c, d) are equidi [#permalink] New post 10 Dec 2018, 17:41
Answer: B
What we have:

Distance from (a,b) to origin (0,0) = √((a-0)^2 + (b-0)^2) = √(a^2 + b^2)
Distance from (c,d) to origin (0,0) = √((c-0)^2 + (d-0)^2) = √(c^2 + d^2)
It is said that points (a, b) and (c, d) are equidistant from the origin. It means their distances from origin are equal, so:
√(a^2 + b^2) = √(c^2 + d^2) , both sides to power2:
a^2 + b^2 = c^2 + d^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.]
Now that we know a^2 > c^2 and a^2 + b^2 = c^2 + d^2, so we can construe that b^2 is less than d^2 and so |b| is less than |d| (b and d don’t necessarily have the same relation) B is bigger than A
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Re: In the coordinate plane, points (a, b) and (c, d) are equidi   [#permalink] 10 Dec 2018, 17:41
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Re: The sides of a triangle are 1, x, and x^2. What are possi

Carcass wrote:

The sides of a triangle are 1, x, and \(x^2\). What are possible values of x? Indicate all possible values.

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: |A-B| < 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

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Re: For positive numbers p and q,

Carcass wrote:

For positive numbers p and q, \(\frac{p-q}{p+q} = \frac{2}{3}\)

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:\(\frac{p-q}{p+q} = \frac{2}{3}\)
Multiply both sides by (p+q) to get: \(p-q = (\frac{2}{3})(p+q)\)
Multiply both sides by 3 to get: \(3(p-q) = (2)(p+q)\)
Expand both sides to get: \(3p-3q = 2p+2q\)
Subtract 2p from both sides to get: \(p-3q = 2q\)
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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Re: In 1985 the number

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.

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Re: Probability with discrete random distribution

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

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QOTD #1 Set M consists of all the integers between –2 and 12 [#permalink] New post 27 Jun 2016, 08:14

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Set M consists of all the integers between –2 and 12, inclusive.

Set N consists of all the integers between 9 and 15, inclusive.

Quantity A

Quantity B

The smallest integer in set M that is also in set N

9

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.

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Re: QOTD #1 Set M consists of all the integers between –2 and 12 [#permalink] New post 29 Jun 2016, 20:12

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The common elements of both sets are 9,10,11,12; the smallest is 9. Answer C

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Re: QOTD #1 Set M consists of all the integers between –2 and 12 [#permalink] New post 08 Dec 2018, 08:41
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.
So, Answer is (C).

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Re: QOTD #1 Set M consists of all the integers between –2 and 12   [#permalink] 08 Dec 2018, 08:41
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Re: Half an hour after car A started traveling from Newtown to

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 plugging-in 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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Re: x is a non-negative number and the square root of (10 – 3x)

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x is a non-negative number and the square root of (10 – 3x) [#permalink] New post 17 Sep 2017, 03:55

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x is a non-negative number and the square root of (10 – 3x) is greater than x.

Quantity A

Quantity B

\(|x|\)

\(2\)

A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.

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Re: x is a non-negative number and the square root of (10 – 3x) [#permalink] New post 21 Sep 2017, 08:35

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We need to solve the equation \(sqrt(10-3x)>x\). Squaring both sides we get \(10-3x>x^2\) that can be rewritten as \(x^2+3x-10

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

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Re: x is a non-negative number and the square root of (10 – 3x) [#permalink] New post 28 Feb 2018, 14:56

IlCreatore wrote:

We need to solve the equation \(sqrt(10-3x)>x\). Squaring both sides we get \(10-3x>x^2\) that can be rewritten as \(x^2+3x-10

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.
Yes I will go with B as well.
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Re: x is a non-negative number and the square root of (10 – 3x) [#permalink] New post 04 Mar 2018, 02:11
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Re: x is a non-negative number and the square root of (10 – 3x) [#permalink] New post 05 Mar 2018, 14:47

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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
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Re: x is a non-negative number and the square root of (10 – 3x) [#permalink] New post 29 Mar 2018, 00:39

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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

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Re: x is a non-negative number and the square root of (10 – 3x) [#permalink] New post 01 Apr 2018, 17:21

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Answer: B
Sure root of 10 – 3x is greater than x and x is a non-negative integer.
First, as x is non-negative, thus |x| equals x in A.
We try numbers. The minimum value for x can be 0.
If x = 0, square root of 10-3x equals 10, which is greater than x.
If x = 1, 10-3 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.

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Re: x is a non-negative number and the square root of (10 – 3x) [#permalink] New post 04 Apr 2018, 07:38

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.

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Re: x is a non-negative number and the square root of (10 – 3x) [#permalink] New post 04 Apr 2018, 07:41

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Given: x>0
and
\sqrt{(10-3x)} > x gives -5

Ignoring negative solutions. 0<x<2 and |x| would be less than 2

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Re: x is a non-negative number and the square root of (10 – 3x) [#permalink] New post 07 Dec 2018, 15:23
Hey guys, I understand your solution, but could you point out the fallacy or mistake in my solution.

sqrt(10-3x) > x
x>=0 so sqrt(10-3x)>0
then 10-3x >0
Thus x < (10/3)

So D because the absolute value of x can be greater than 2 or smaller than 2.

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Re: x is a non-negative number and the square root of (10 – 3x)   [#permalink] 07 Dec 2018, 15:23
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Re: The points A( 0, 0), B( 0, 4a – 5), and C( 2a + 1, 2a + 6) f

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The points A( 0, 0), B( 0, 4a – 5), and C( 2a + 1, 2a + 6) f [#permalink] New post 30 Nov 2018, 18:11

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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

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Re: The points A( 0, 0), B( 0, 4a – 5), and C( 2a + 1, 2a + 6) f [#permalink] New post 06 Dec 2018, 01:37

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
BC is perpendicular to AB. And A is at origin (0,0).
The Y-Coordinates of B & C should be same as they are on the same line i.e BC.
4a-5 = 2a+6
a=11/2
So
A= (0,0)
B=(0,4a-5)=(0,17)
C=(2a+1,2a+6)=(12,17)
Find distances AB & BC
AB = 17
BC = 12
Area of triangle = 1/2 (17)(12) = 102
It’s A.

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Re: The points A( 0, 0), B( 0, 4a – 5), and C( 2a + 1, 2a + 6) f [#permalink] New post 06 Dec 2018, 18:32

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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
Since coordinates of x in A and B are the same that is 0, meaning A and B lie on y-axis, and 90° is at B, the y-coordinates of B and C should be the same.
So \(4a-5=2a+6…….2a=11…..a=\frac{11}{2}\)..
Now AB is nothing but y-coordinates of B and BC is nothing but x-coordinates of C
1)So X coordinates of C are \(2a+1=2*\frac{11}{2}+1=11+1=12\)
2) y coordinates of B = \(2a+6=2*\frac{11}{2}+6=11+6=17\)

Area = \(\frac{12*17}{2}=6*17=102\)

A
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2. Effect of Arithmetic Operations on fraction : https://greprepclub.com/forum/effects-of-arithmetic-operations-on-fractions-11573.html?sid=d570445335a783891cd4d48a17db9825
3. Remainders : https://greprepclub.com/forum/remainders-what-you-should-know-11524.html
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5. Absolute Modulus and Inequalities : https://greprepclub.com/forum/absolute-modulus-a-better-understanding-11281.html

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Re: The points A( 0, 0), B( 0, 4a – 5), and C( 2a + 1, 2a + 6) f   [#permalink] 06 Dec 2018, 18:32
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Re: Set A {300,400,500,700,900} Set B {100,200,300,500,900}

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Set A {300,400,500,700,900} Set B {100,200,300,500,900} [#permalink] New post 05 Dec 2018, 09:32

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Set A {300,400,500,700,900}
Set B {100,200,300,500,900}

Quantity A

Quantity B

Standard deviation of Set A

Standard deviation of Set B

A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given

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Re: Set A {300,400,500,700,900} Set B {100,200,300,500,900} [#permalink] New post 05 Dec 2018, 10:22

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Arun1992 wrote:

Set A {300,400,500,700,900}
Set B {100,200,300,500,900}

Quantity A

Quantity B

Standard deviation of Set A

Standard deviation of Set B

A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given

The standard deviation is dependent on the spread of the elements.
Although it is highly unlikely that we will be required to calculate a standard deviation, but we should be able to relate the spread of elements in sets.

Set A {300,400,500,700,900}… Mean = \(\frac{2800}{5}=580\)
Set B {100,200,300,500,900}….Mean = \(\frac{2000}{5}=400\)
we can see that set B has more spread around its mean, thus B>A

B
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Re: Set A {300,400,500,700,900} Set B {100,200,300,500,900} [#permalink] New post 06 Dec 2018, 08:12

Expert’s post

Arun1992 wrote:

Set A {300,400,500,700,900}
Set B {100,200,300,500,900}

Quantity A

Quantity B

Standard deviation of Set A

Standard deviation of Set B

A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given

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

———————————
Alternatively, we can apply the informal definition of standard deviation (covered in the video below) to conclude that Set B has the greater standard deviation.

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Re: Set A {300,400,500,700,900} Set B {100,200,300,500,900}   [#permalink] 06 Dec 2018, 08:12
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Re: A container holds 10 liters of a solution which is 20% acid.

nadeem790 wrote:

A container holds 10 liters of a solution which is 20% acid. If 6 liters of pure acid are added to the container, what percent of the resulting mixture is acid?

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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Re: calculate the value of m

sandy wrote:

\(\frac{(10^3)(0.027)}{(900)(10^{-2})}=(3)(10^m)\)

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
_________________

Brent Hanneson – Creator of greenlighttestprep.com
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