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ThatIndianGuy
Posts: 74
10 Jul 2024, 23:04
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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
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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: 26 Aug 2026
Posts: 38
Given Kudos: 56
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
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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, \(P=5, Q=3\)
10 Jul 2024, 23:28
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Bookmarks
Point to note: If a dice is rolled once, the probablity of any one (among 6) to come in 1/6. This porbabilty decreases if we roll more dice together and wish to see the same face on all dices. i.e., for two dice rolled the probabilty to see same face on both roll will be 1/36, for three it will be 1/216 and so on.
Now, we know that when the dice is rolled once the probability of agate stone apperaing is 1/6.
If dice is thrown P times, the probabilty is x.
If dice is thrown Q times, the probabilty is y.
Also, x<y.
With the first point we are certain x and y to be among (1/36 or 1/216 or 1/6^4 or 1/6^5 or 1/(6^6)).
Also it is given than 1/x + 1/y = 7992.
We can factorize 7992 as 6*6*6*37.
We can also write 6*6*6*37 as 6*6*6*(36+1) or 6*6*6*36 + 6*6*6 or 6^5 + 6^3.
So, 1/x & 1/y could be 6^5 or 6^3: Implying x or y as 1/6^5 or 1/6^3.
Given x<y, x shall definetly be 1/6^5, Thus P=5 and y as 1/6^3, so Q as 3.
Now, we know that when the dice is rolled once the probability of agate stone apperaing is 1/6.
If dice is thrown P times, the probabilty is x.
If dice is thrown Q times, the probabilty is y.
Also, x<y.
With the first point we are certain x and y to be among (1/36 or 1/216 or 1/6^4 or 1/6^5 or 1/(6^6)).
Also it is given than 1/x + 1/y = 7992.
We can factorize 7992 as 6*6*6*37.
We can also write 6*6*6*37 as 6*6*6*(36+1) or 6*6*6*36 + 6*6*6 or 6^5 + 6^3.
So, 1/x & 1/y could be 6^5 or 6^3: Implying x or y as 1/6^5 or 1/6^3.
Given x<y, x shall definetly be 1/6^5, Thus P=5 and y as 1/6^3, so Q as 3.
