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Bunuel
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Practice Questions Chat 13 Apr 2024, 01:49
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pranavkohli16
How to solve this?
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Discussed in detail here: how-many-positive-integers-less-than-500-have-a-421279.html
Searching the PS forum would have saved you time...
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Practice Questions Chat 13 Apr 2024, 02:18
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Am ia beginner i need help
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Practice Questions Chat 13 Apr 2024, 02:50
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can someone help with shortest approach to solve this question
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Practice Questions Chat 13 Apr 2024, 02:57
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Bhagwan212
can someone help with shortest approach to solve this question
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in-the-figure-above-x-and-y-represent-locations-in-a-district-of-a-ce-305848.html
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Badsha5796
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Practice Questions Chat 13 Apr 2024, 03:37
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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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Practice Questions Chat 13 Apr 2024, 03:40
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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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Practice Questions Chat 13 Apr 2024, 03:53
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pranavkohli16
How do I approach a question like this?
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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

Badsha5796
Why did you do 6*3 ? Can you please elaborate?
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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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Practice Questions Chat 13 Apr 2024, 04:28
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can anyone please tell me the answer asap
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Bunuel
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Practice Questions Chat 13 Apr 2024, 04:30
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aayushimehta12
can anyone please tell me the answer asap
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Check here: 

https://gmatclub.com/forum/if-x-3-2-whi ... 19682.html
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Practice Questions Chat 13 Apr 2024, 05:16
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Edoua
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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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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Practice Questions Chat 13 Apr 2024, 05:40
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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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Practice Questions Chat 13 Apr 2024, 05:43
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but -1 (2x - 3) = x -5 will be for value of X = 3/8
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Practice Questions Chat 13 Apr 2024, 05:44
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Number of odd integers from 1 to 79 is 40
Number of odd integers from 1 to 23 is 12
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Practice Questions Chat 13 Apr 2024, 05:45
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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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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
but -1 (2x - 3) = x -5 will be for value of X = 3/8
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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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Practice Questions Chat 13 Apr 2024, 05:50
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Thank you all

but can anyone explain question 2

further 2
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Practice Questions Chat 13 Apr 2024, 05:55
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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
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Practice Questions Chat 13 Apr 2024, 05:59
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anyone cal solve the compound interest math
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aayushimehta12
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Practice Questions Chat 13 Apr 2024, 06:01
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Shwarma
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
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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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Practice Questions Chat 13 Apr 2024, 06:02
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Nidhi1715
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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Because we are filling up three numbers in 6 cases
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