Last visit was: 26 Jul 2024, 17:46 It is currently 26 Jul 2024, 17:46
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Close
Request Expert Reply
Confirm Cancel
SORT BY:
Date
Tags:
Show Tags
Hide Tags
Math Expert
Joined: 02 Sep 2009
Posts: 94619
Own Kudos [?]: 644161 [0]
Given Kudos: 86770
Send PM
User avatar
Intern
Intern
Joined: 10 Apr 2024
Posts: 3
Own Kudos [?]: 0 [0]
Given Kudos: 0
Send PM
User avatar
Intern
Intern
Joined: 17 Apr 2023
Posts: 9
Own Kudos [?]: 2 [0]
Given Kudos: 1
Location: India
Send PM
Senior Manager
Senior Manager
Joined: 06 Oct 2023
Posts: 325
Own Kudos [?]: 112 [0]
Given Kudos: 54
Send PM
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
Intern
Intern
Joined: 10 Apr 2024
Posts: 11
Own Kudos [?]: 3 [0]
Given Kudos: 0
Send PM
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?
Math Expert
Joined: 02 Sep 2009
Posts: 94619
Own Kudos [?]: 644161 [0]
Given Kudos: 86770
Send PM
Re: Quant Question of the Day Chat [#permalink]
Expert Reply
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
Manager
Manager
Joined: 18 Feb 2024
Posts: 67
Own Kudos [?]: 35 [0]
Given Kudos: 278
Send PM
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

Badsha5796 wrote:
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
Manager
Manager
Joined: 23 Jan 2024
Posts: 115
Own Kudos [?]: 26 [0]
Given Kudos: 21
Send PM
Re: Quant Question of the Day Chat [#permalink]
can anyone please tell me the answer asap
Math Expert
Joined: 02 Sep 2009
Posts: 94619
Own Kudos [?]: 644161 [0]
Given Kudos: 86770
Send PM
Re: Quant Question of the Day Chat [#permalink]
Expert Reply
aayushimehta12 wrote:
can anyone please tell me the answer asap

Check here: 

https://gmatclub.com/forum/if-x-3-2-whi ... 19682.html
Intern
Intern
Joined: 20 Dec 2023
Posts: 18
Own Kudos [?]: 3 [0]
Given Kudos: 15
Send PM
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?
Intern
Intern
Joined: 20 Dec 2023
Posts: 18
Own Kudos [?]: 3 [0]
Given Kudos: 15
Send PM
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
Manager
Manager
Joined: 23 Jan 2024
Posts: 115
Own Kudos [?]: 26 [0]
Given Kudos: 21
Send PM
Re: Quant Question of the Day Chat [#permalink]
but -1 (2x - 3) = x -5 will be for value of X = 3/8
Senior Manager
Senior Manager
Joined: 12 Jul 2023
Posts: 422
Own Kudos [?]: 198 [0]
Given Kudos: 107
Location: India
Concentration: Entrepreneurship, Finance
GMAT Focus 1:
535 Q77 V78 DI74
GMAT Focus 2:
575 Q83 V74 DI78
GMAT 1: 420 Q31 V17
GPA: 3.2
WE:Business Development (Finance)
Send PM
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
Manager
Manager
Joined: 10 Sep 2023
Posts: 181
Own Kudos [?]: 153 [0]
Given Kudos: 58
Send PM
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
Intern
Intern
Joined: 05 Nov 2023
Posts: 9
Own Kudos [?]: 10 [0]
Given Kudos: 12
Send PM
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!
Manager
Manager
Joined: 23 Jan 2024
Posts: 115
Own Kudos [?]: 26 [0]
Given Kudos: 21
Send PM
Re: Quant Question of the Day Chat [#permalink]
Thank you all

but can anyone explain question 2

further 2
Manager
Manager
Joined: 10 Sep 2023
Posts: 181
Own Kudos [?]: 153 [0]
Given Kudos: 58
Send PM
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
40^2-12^2 to get the answer
Manager
Manager
Joined: 09 Jan 2024
Posts: 93
Own Kudos [?]: 22 [0]
Given Kudos: 29
Location: Bangladesh
GMAT 1: 760 Q56 V59
Send PM
Re: Quant Question of the Day Chat [#permalink]
anyone cal solve the compound interest math
Manager
Manager
Joined: 23 Jan 2024
Posts: 115
Own Kudos [?]: 26 [0]
Given Kudos: 21
Send PM
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.
Intern
Intern
Joined: 10 Apr 2024
Posts: 11
Own Kudos [?]: 3 [0]
Given Kudos: 0
Send PM
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
GMAT Club Bot
Re: Quant Question of the Day Chat [#permalink]
   1  ...  212   213   214   215   216  ...  292   
Moderator:
Math Expert
94619 posts