Events & Promotions
|
|
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
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Nov 1811:00 AM PST
-12:00 PM PST
Join us in a live GMAT practice session and solve 30 challenging GMAT questions with other test takers in timed conditions, covering GMAT Quant, Data Sufficiency, Data Insights, Reading Comprehension, and Critical Reasoning questions.
Nov 2211:00 AM IST
-01:00 PM IST
Do RC/MSR passages scare you? e-GMAT is conducting a masterclass to help you learn – Learn effective reading strategies Tackle difficult RC & MSR with confidence Excel in timed test environment- Nov 23
11:00 AM IST
-01:00 PM IST
Attend this free GMAT Algebra Webinar and learn how to master the most challenging Inequalities and Absolute Value problems with ease. 
Nov 2510:00 AM EST
-11:00 AM EST
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors.
ThatIndianGuy
Posts: 80
10 Jul 2024, 23:04
Kudos
Bookmarks
Given:
probability of the agate stone appearing all P times is x
probability of agate stone = 1/6
for all P times, it will be (1/6)^p
Hence, x = (1/6)^p
probability of the agate stone appearing all Q times is y
probability of agate stone = 1/6
for all Q times, it will be (1/6)^q
Hence, y = (1/6)^q
Given, 1/x + 1/y = 7992
6^p + 6^q = 7992 -------(1)
we also know that, x<y
1/x > 1/y
6^p > 6^q
p>q ------(2)
From (1), 7992 can be written as (6^3)*37
= 6^3 * (36+1)
= 6^3 * (6^2 + 1)
= 6^5 + 6^3 ------(3)
Comparing (1) and (2) with constraint as (2)
p = 5, q = 3
probability of the agate stone appearing all P times is x
probability of agate stone = 1/6
for all P times, it will be (1/6)^p
Hence, x = (1/6)^p
probability of the agate stone appearing all Q times is y
probability of agate stone = 1/6
for all Q times, it will be (1/6)^q
Hence, y = (1/6)^q
Given, 1/x + 1/y = 7992
6^p + 6^q = 7992 -------(1)
we also know that, x<y
1/x > 1/y
6^p > 6^q
p>q ------(2)
From (1), 7992 can be written as (6^3)*37
= 6^3 * (36+1)
= 6^3 * (6^2 + 1)
= 6^5 + 6^3 ------(3)
Comparing (1) and (2) with constraint as (2)
p = 5, q = 3
10 Jul 2024, 23:10
Kudos
Bookmarks
Oh yeah.. Day 3 and counting.
We got a really tough challenge with the Orange Team.
But we will make it. We will win, I'm sure.
Let's get started with our explanation for this topic on day three:
First step in Two Parts, before reading the text I understand what is the question and what is it asking for:
Rephrase the Question:
We have P and Q.
P for times we get x
Q for times we get y
We have 1->6 answer choices.
Rephrase - Reading and Understanding the text:
Ok, we got probabilty question.
we can see that is similar to dice with [1][/6] possibility. because each face is different.
we can see that we are looking for 'Agate' face so it is [1][/6] of picking it.
we see an equation given here: [1][/x] + [1][/y] = 7992 (if you don't fimiliar with that kind of equation, it is just implying that x and y are fractions)
and x<y - we need to be focus not being wrong with that x<y because they are fractions!
Let's try getting inside the head of the writer of that question. it is asking for possibilty. we know the possibilities here are [1][/6]. we have an equation.
let's see what happened if we finding the multiples of 6: 6 , 36 , 216 , 1296 , 7776 (for the last one I used a calculator)
Ok, now, can you think of adding two values to get 7992? Yes, we can
try 7776 + 216.
ok we know that 7776 is 6^5 so it is ([1][/6])^5 = [1][/7776], and 216 is 6^3 so it is ([1][/6])^3 = [1][/216]
[1][/7776] < [1][/216] - what is bigger? dont forget x<y
x y P=x Q=y P=5 tosses Q= 3 tosses
We have our answer
THE END
I hope you liked the explanation, I have tried my best here.
Let me know if you have any questions about this question or my explanation.
We got a really tough challenge with the Orange Team.
But we will make it. We will win, I'm sure.
Let's get started with our explanation for this topic on day three:
First step in Two Parts, before reading the text I understand what is the question and what is it asking for:
Rephrase the Question:
We have P and Q.
P for times we get x
Q for times we get y
We have 1->6 answer choices.
Rephrase - Reading and Understanding the text:
Ok, we got probabilty question.
we can see that is similar to dice with [1][/6] possibility. because each face is different.
we can see that we are looking for 'Agate' face so it is [1][/6] of picking it.
we see an equation given here: [1][/x] + [1][/y] = 7992 (if you don't fimiliar with that kind of equation, it is just implying that x and y are fractions)
and x<y - we need to be focus not being wrong with that x<y because they are fractions!
Let's try getting inside the head of the writer of that question. it is asking for possibilty. we know the possibilities here are [1][/6]. we have an equation.
let's see what happened if we finding the multiples of 6: 6 , 36 , 216 , 1296 , 7776 (for the last one I used a calculator)
Ok, now, can you think of adding two values to get 7992? Yes, we can
try 7776 + 216.ok we know that 7776 is 6^5 so it is ([1][/6])^5 = [1][/7776], and 216 is 6^3 so it is ([1][/6])^3 = [1][/216]
[1][/7776] < [1][/216] - what is bigger? dont forget x<y
x y P=x Q=y P=5 tosses Q= 3 tosses
We have our answer
THE END
I hope you liked the explanation, I have tried my best here.
Let me know if you have any questions about this question or my explanation.
a191291r
Joined: 20 Dec 2023
Last visit: 22 Jul 2025
Posts: 40
Given Kudos: 57
Location: India
Concentration: Finance, Leadership
Schools: INSEAD '26 ISB '27 (S) Said '27 Judge '26 LBS '26
GMAT Focus 1: 615 Q82 V81 DI79 
GMAT Focus 2: 665 Q85 V83 DI81 
WE:Operations (Other)
10 Jul 2024, 23:16
Kudos
Bookmarks
Quote:
From the statement, we have
\(x=\frac{1}{6}^P\)
\(y=\frac{1}{6}^Q\)
Given that \(x<y\), which means:
\(\frac{1}{x}>\frac{1}{y}\) or,
\(6^P>6^Q\)
\(P>Q\)
Further, given that:
\(\frac{1}{x} + \frac{1}{y} = 7,992\), which means
\(6^P+6^Q=7992\)
Substituting higher values for P and lower values of Q, we get:
\(6^5+6^3=7992\)
Therefore, \(P=5, Q=3\)
