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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]  17 Sep 2017, 03:55
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36% (00:56) correct 63% (01:14) wrong based on 90 sessions

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.
[Reveal] Spoiler: OA

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Re: x is a non-negative number and the square root of (10 – 3x) [#permalink]  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 $$x^2+3x-10<0$$. The left hand side can be rewritten as $$(x-5)(x+2)$$, thus the solution of the equation is $$-5<x<2$$.

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.

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Re: x is a non-negative number and the square root of (10 – 3x) [#permalink]  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 $$x^2+3x-10<0$$. The left hand side can be rewritten as $$(x-5)(x+2)$$, thus the solution of the equation is $$-5<x<2$$.

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.

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]  04 Mar 2018, 02:11
so
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Re: x is a non-negative number and the square root of (10 – 3x) [#permalink]  05 Mar 2018, 14:47
Expert's post

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]  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]  01 Apr 2018, 17:21
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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.

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Re: x is a non-negative number and the square root of (10 – 3x) [#permalink]  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]  04 Apr 2018, 07:41
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Given: x>0
and
\sqrt{(10-3x)} > x gives -5<x<2

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]  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]  12 Dec 2018, 07:52
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 $$x^2+3x-10<0$$. The left hand side can be rewritten as $$(x-5)(x+2)$$, thus the solution of the equation is $$-5<x<2$$.

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.

Perhaps it is a simple typo, but what you should have (x+5)(x-2) as the factor, then it would mean that -5>x>2.
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Re: x is a non-negative number and the square root of (10 – 3x) [#permalink]  13 Dec 2018, 00:16
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QuantumWonder wrote:
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.

You are wrong in replacing X with 0.
$$\sqrt{10-3x}>x$$..
So the LHS and RHS are interconnected, as increase in X will increase both sides ..

You cannot simply replace X with 0...
That would ok if you were simply finding out if some term is negative or positive
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Some useful Theory.
1. Arithmetic and Geometric progressions : https://greprepclub.com/forum/progressions-arithmetic-geometric-and-harmonic-11574.html#p27048
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
4. Number properties : https://greprepclub.com/forum/number-property-all-you-require-11518.html
5. Absolute Modulus and Inequalities : https://greprepclub.com/forum/absolute-modulus-a-better-understanding-11281.html

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Re: x is a non-negative number and the square root of (10 – 3x) [#permalink]  14 Dec 2018, 13:15
Okay, here is a solution that is purely algebraic ( boxing has a mistake in inequality direction).

sqrt (10-3x) > x
10-3x > x^2
0 > x^2 + 3x - 10
x^2 + 3x - 10 < 0
(x+5)(x-2) < 0

There are two options, x+5 < 0 and x- 2 > 0 || or || x+5 > 0 & x-2< 0
Why? Because each option, when multiplied, would still give you a negative.
First option: x< -5 and x > 2, x is nonnegative so you can throw out first inequality out. For the second inequality, plug this in the criteria 10-3x>x^2 and you will see it doesn't work. Thus, throw it out as well.
Second option: x>-5 and x<2, again, here we throw out the negative solution x>-5.

So x < 2.

Thus B.
Re: x is a non-negative number and the square root of (10 – 3x)   [#permalink] 14 Dec 2018, 13:15
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