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Retired Moderator
Joined: 27 Oct 2017
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24 Mar 2020, 03:26
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Hello and welcome to GMAT preparation. You may know me from GMATBusters' Quant Quiz. I hope you are finding the Quiz very useful.

This topic is a new feature on the DS Forum and a way for you to directly interact with me and ask anything about the DS (even PS ), e.g. if you want a certain concept explained or have a particular you question you want me addressed, this is the place to post a link to it or your question.

If you have any doubts/queries about Quant DS (even PS ), I would be happy to address them.
• If the question is general in nature regarding strategy or any concept, you can post it here.
• you can tag me in a post to ask queries related to the particular question.
• Participate in the Quiz, we will discuss the Quiz questions in the webinar here.
• Join the chat group-GMAT Busters, I would be conducting Quant doubt sessions every Saturday 9PM EST.
Here is the Chat group link

Happy Learning!!!
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24 Mar 2020, 23:24
2
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2
Is $$m*k^4$$ > 0?
1) $$m^2-8m+15<0$$
2) $$k^2-6k+8>0$$
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25 Mar 2020, 09:51
2
Myth : k^2n is always Positive
Concept:

k^2n is always Non Negative

St 1) using wavy curve approach:
(m-3)(m-5)<0
hence 3<m<5
m is always positive. also power of K is even.
k can not be negative, but it can be zero or positive.
if k = 0, $$m*k^4$$ = 0
if k is not equal to 0, $$m*k^4$$ >0
NOT SUFFICIENT

St 2: (k-2)(k_4)>0
hence, k >4 or k <2
now, k can also be equal to zero.
Also no info about m, so it is not sufficient.
NOT SUFFICIENT

Even after combining St1 &2, we know m >0 but k can be zero or non zero.
hence NOT SUFFICIENT

GMATBusters wrote:
Is $$m*k^4$$ > 0?
1) $$m^2-8m+15<0$$
2) $$k^2-6k+8>0$$

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30 Mar 2020, 02:10
Hi. Please clarify how you got St 2: (k-2)(k_4)>0
hence, k >4 or k <2.

According to my workings K>4 or K>2. How did you get K<2.Much appreciated

Posted from my mobile device
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30 Mar 2020, 04:37
Use wavy curve method for solving the inequality range problems:

Steps of wavy curve method :

1. All zeros of the function f(x) contained on the left-hand side of the inequality should be marked on the number line with inked (black) circles.

2. All points of discontinuities of the function f(x) contained on the left-hand side of the inequality should be marked on the number line with un-inked (white) circles.

3. From right to left, beginning above the number line (in case of the value of f(x) is positive in step (iii), otherwise, from below the number line), a wavy curve should be drawn to pass through all the marked points so that when it passes through a simple point the curve intersects the number line, and, when passing through a double point the curve remains located on one side of the number line.

4. The appropriate intervals are chosen in accordance with the sign of inequality (the function f(x) is positive whenever the curve is situated above the number line, it is negative if the curve is found below the number line). Their union represents the solution of the inequality.

Remark :

i) A point of discontinuity will never be included in the answer.

ii) If you are asked to the interval where f(x) is non-negative or non-positive then make the interval closed corresponding to the roots of the numerator and let it remain open corresponding to the roots of the denominator.

The point where f(x) vanishes are called zeroes of the function .

The point x=bj are called point of discontinuity of f(x).

If the exponent of the factor is odd then the point is called simple point .

If the exponent of the factor is even the point is called a simple point.

Illustration :

Let f(x) = {(x-3)(x+2)(x+5) }/{(x+1)(x+7)} .

As we can find the critical points as x=3,-2,-5,-1,7.

Now plot the points on the number line as per the rule of the number line . (Left to Right)

Now check for each interval whether the value of f(x) is greater than zero or less than zero.

Note: Just take a value between the interval and calculate f(x). If f(x)>0 then curve lies above the number line or it will lie below the number line.

Follow the procedure for this given problem you will get the curve as shown above.

f(x) >0 for x ∈ [-5,-2] U (-1,3] U (7,∞)

f(x)<0 for x ∈ (-∞,-5] U [-2,-1) U [3,7)

archiemuty wrote:
Hi. Please clarify how you got St 2: (k-2)(k_4)>0
hence, k >4 or k <2.

According to my workings K>4 or K>2. How did you get K<2.Much appreciated

Posted from my mobile device

Attachment:

Capture.JPG [ 14.76 KiB | Viewed 2249 times ]
Attachment:

Capture.JPG [ 18.02 KiB | Viewed 2250 times ]

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30 Mar 2020, 05:10
For the specific doubt on this problem, see the sketch.

archiemuty wrote:
Hi. Please clarify how you got St 2: (k-2)(k_4)>0
hence, k >4 or k <2.

According to my workings K>4 or K>2. How did you get K<2.Much appreciated

Posted from my mobile device
Attachment:

Capture.JPG [ 92.83 KiB | Viewed 2245 times ]

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Concentration: Sustainability, Marketing
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30 Mar 2020, 23:25
GMATBusters wrote:
Is $$m*k^4$$ > 0?
1) $$m^2-8m+15<0$$
2) $$k^2-6k+8>0$$

#1
$$m^2-8m+15<0$$

can be written as ; ( m-5)*(m-3)<0
which is possible only when m=4
value of k not know so insufficient
#2
k^2-6k+8>0

can be written as

(k-4)*(k-2)>0

value of k will fall ; 1<k>4
i.e k can be 0 or -ve or >4
value of m not know ; insufficient
from 1*2
$$m*k^4$$ > 0
since k can be 0 or integer so answer to above is both yes and no
hence insufficient
IMO E
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30 Mar 2020, 23:57
You are right...

Archit3110 wrote:
GMATBusters wrote:
Is $$m*k^4$$ > 0?
1) $$m^2-8m+15<0$$
2) $$k^2-6k+8>0$$

#1
$$m^2-8m+15<0$$

can be written as ; ( m-5)*(m-3)<0
which is possible only when m=4
value of k not know so insufficient
#2
k^2-6k+8>0

can be written as

(k-4)*(k-2)>0

value of k will fall ; 1<k>4
i.e k can be 0 or -ve or >4
value of m not know ; insufficient
from 1*2
$$m*k^4$$ > 0
since k can be 0 or integer so answer to above is both yes and no
hence insufficient
IMO E

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07 Apr 2020, 19:56
Hi Guys!!!

Let us schedule the doubt chat session at 10 PM IST at this group - Group link

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08 Apr 2020, 09:51
In rectangle ABCD, two diagonals cut at point O, if some ratio of length and breadth is given,can we find out angle AOB without using any trigonometry?

Posted from my mobile device
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08 Apr 2020, 09:58
The GMAT would ask angle only if the angle is evident owing to an isosceles triangle, or 30-60-90 or 45-45-90 triangle.

gurmukh wrote:
In rectangle ABCD, two diagonals cut at point O, if some ratio of length and breadth is given,can we find out angle AOB without using any trigonometry?

Posted from my mobile device

Attachment:

30-60-90-example-diagram.png [ 5.54 KiB | Viewed 1852 times ]
Attachment:

454590 triangle.jpg [ 8.89 KiB | Viewed 1850 times ]

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08 Apr 2020, 10:04
Ok. That means there is no ratio.
I was also thinking same

Posted from my mobile device
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08 Apr 2020, 10:08
Yes, the trigonometric ratio is not in GMAT syllabus

gurmukh wrote:
Ok. That means there is no ratio.
I was also thinking same

Posted from my mobile device

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08 Apr 2020, 10:31
GMATBusters wrote:
Yes, the trigonometric ratio is not in GMAT syllabus

gurmukh wrote:
Ok. That means there is no ratio.
I was also thinking same

Posted from my mobile device

I think in rectangle ABCD in which two diagonals cut at point O. Angle AOB can be calculated when ratio of length and breadth are given Sq of 3: 1.
One angle at diagonal will come out 60 and one angle will come out 120.
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08 Apr 2020, 18:30
The angle in any triangle depends on the ratio of sides and vice versa using Sine rule. But that is not tested.

gurmukh wrote:
GMATBusters wrote:
Yes, the trigonometric ratio is not in GMAT syllabus

gurmukh wrote:
Ok. That means there is no ratio.
I was also thinking same

Posted from my mobile device

I think in rectangle ABCD in which two diagonals cut at point O. Angle AOB can be calculated when ratio of length and breadth are given Sq of 3: 1.
One angle at diagonal will come out 60 and one angle will come out 120.

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08 Apr 2020, 18:41
When the ratio of sides is given sqt of 3 : 1 then we take out angle using 90:30:60 triangle, this identity of trigonometry we use in GMAT and through this we can take out angles. Otherwise trigonometry is not required.

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08 Apr 2020, 18:45
yes, you are right... only specific cases are tested.

gurmukh wrote:
When the ratio of sides is given sqt of 3 : 1 then we take out angle using 90:30:60 triangle, this identity of trigonometry we use in GMAT and through this we can take out angles. Otherwise trigonometry is not required.

Posted from my mobile device

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16 Apr 2020, 09:49
(k-2)(k_4)>0
hence, k >4 or k <2
now, k can also be equal to zero.
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16 Apr 2020, 10:24
yes you are right.

for (k-2)(k-4)>0
hence, k >4 or k <2
now, k can be anything less than 2 even 0 or -ve

(
Izzy77 wrote:
(k-2)(k_4)>0
hence, k >4 or k <2
now, k can also be equal to zero.

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18 Apr 2020, 06:48
Hello, I have a question regarding "3L of milk are drawn from a container containing 30L of milk. It is replaced by water and the process is repeated 2 times. What is the ratio of milk to water at the end?"

Solution to this is concentration after 3 rounds :- (1 - 3/30)^3 = 721/1000 (Because in each round the concentration for the previous round changes and eventually turns out to be this in the end)

However, I wanted to understand how can I extend this concept to much harder or rather complicated questions for eg:-

3L of solution is drawn from a container containing 30L of solution that is 20% concentrated. It is replaced by another solution that is 10% concentrated and the process is repeated 2 times. What is the concentration of solution after 3 rounds?

Shameek

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