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# Quant Question of the Day Chat

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Re: Quant Question of the Day Chat [#permalink]
Bhagwan212 wrote:
can someone help with shortest approach to solve this question

in-the-figure-above-x-and-y-represent-locations-in-a-district-of-a-ce-305848.html
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Re: Quant Question of the Day Chat [#permalink]
Let’s assume 20 students name start with différents letters from A to T

We selected 3 out of 20 then reduce the different cases of 3 consecutives letters

C²⁰₃ =20!/3 !×17!=1140

Then let’s reduce how many case 3 names can be consecutive

6*3 =18
Hence 1140-18=1122

Why did you do 6*3 ? Can you please elaborate?
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Re: Quant Question of the Day Chat [#permalink]
Please check detailed discussion on this question here: in-the-figure-above-x-and-y-represent-locations-in-a-district-of-a-ce-305848.html
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Re: Quant Question of the Day Chat [#permalink]
pranavkohli16 wrote:
How do I approach a question like this?

8⁴=4⁶
8¹⁶=4¹⁶2¹⁶=4¹⁶4⁸
16⁸=4⁸4⁸

we have (4⁶+4¹⁶4⁸ )/ (4⁸+4⁸4⁸)

Divide by 4⁶
—->> (1+4²4¹⁶ ) /( 4²+4²4⁸)
1+4¹⁸/4²+4¹⁰

1+4¹⁸≈4¹⁸
4²+4¹⁰ =4²(1+4⁸ )≈4¹⁰ because 1+4⁸≈4⁸

4¹⁸/4¹⁰≈4⁸ >50 000

Why did you do 6*3 ? Can you please elaborate?

I count the favorable case
a b c d e … t

abc def ghi Jkl mno pqr —st— (6cases)

—a—. bcd efg hij klm nop qrs —t —(6cases)

—a b— cde fgh ijk lmn opq rst (6 cases)

6*3
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Re: Quant Question of the Day Chat [#permalink]
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Re: Quant Question of the Day Chat [#permalink]
aayushimehta12 wrote:

Check here:

https://gmatclub.com/forum/if-x-3-2-whi ... 19682.html
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Re: Quant Question of the Day Chat [#permalink]
Edoua wrote:
I count the favorable case a b c d e … t abc def ghi Jkl mno pqr —st— (6cases) —a—. bcd efg hij klm nop qrs —t —(6cases) —a b— cde fgh ijk lmn opq rst (6 cases) 6*3

Hey while counting the favourable cases here "consecutive names on the list"
Why will we not multiply 6*3 by 3!? 3! assuming that say abc (three consecutive people) can be selected in 6 ways?
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Re: Quant Question of the Day Chat [#permalink]
For the first question, on solving the equation, whatever answers you’ll get for X, if you’ll put them again in the equation to cross check, you’ll see that RHS will be negative. And absolute value of any equation can never be negative. Hence ans will be 0
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Re: Quant Question of the Day Chat [#permalink]
but -1 (2x - 3) = x -5 will be for value of X = 3/8
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Re: Quant Question of the Day Chat [#permalink]
Number of odd integers from 1 to 79 is 40
Number of odd integers from 1 to 23 is 12
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Re: Quant Question of the Day Chat [#permalink]
The first question
Solve when |2x-3| is positive = x value is -2 so, RHS becomes -7 which is not possible as LHS has to be positive
When |2x-3| is negative, x=8/3, even then RHS will be -7/3 which is again negative, so 0 solution is the answer
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Re: Quant Question of the Day Chat [#permalink]
the theory behind it is that, as the left side is an absolute value which must be equal or greater than zero, the left side must be as well, this means that x-5 > 0 or x>5. As you can see none of the possibilities are greater than 5 which means they’re not valid therefore the answer is 0.

aayushimehta12 wrote:
but -1 (2x - 3) = x -5 will be for value of X = 3/8

for future refernce, if you cant seem to figure out a question, just type the question in the search bar above and you’ll likely find a detailed explenation of that particular question on the forum!
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Re: Quant Question of the Day Chat [#permalink]
Thank you all

but can anyone explain question 2

further 2
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Re: Quant Question of the Day Chat [#permalink]
question 2:
formula given -- first 5 odd numbers consecutive - aka 1,3,5,7-

so you need to find the formula for n^2 upto 25 --> n =13 (13 odd numbers)-> but since 26 is inclusive it will be 12
79--> N=40
so you find the difference between them
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Re: Quant Question of the Day Chat [#permalink]
anyone cal solve the compound interest math
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Re: Quant Question of the Day Chat [#permalink]
Shwarma wrote:
question 2: formula given -- first 5 odd numbers consecutive - aka 1,3,5,7- so you need to find the formula for n^2 upto 25 --> n =13 (13 odd numbers)-> but since 26 is inclusive it will be 12 79--> N=40 so you find the difference between them 40^2-12^2 to get the answer

I understand how to get 40, but I thought it should be 40 and 13 not 40 and 12, because using An= a + (n-1) 2, formula on both we get 40 and 13 as answers.
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Re: Quant Question of the Day Chat [#permalink]
Nidhi1715 wrote:
Hey while counting the favourable cases here "consecutive names on the list" Why will we not multiply 6*3 by 3!? 3! assuming that say abc (three consecutive people) can be selected in 6 ways?

Because we are filling up three numbers in 6 cases
Re: Quant Question of the Day Chat [#permalink]
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