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ThatIndianGuy
Joined: 18 Mar 2020
Last visit: 09 Apr 2026
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Location: India
Concentration: Strategy, Technology
WE:General Management (Technology)
Posts: 74
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
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: 25 Aug 2026
Posts: 38
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)
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, \(P=5, Q=3\)
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.
Bunuel
Let there be values P, Q, x, y such that when the dice is thrown P times, the probability of the agate stone appearing all P times is x and when the dice is thrown Q times, the probability of the agate stone appearing all Q times is y. Furthermore, x<y and 1/x+1/y=7,992
probability of agate appearing all P times = (1/6)^p =x
probability of agate appearing all Q times = (1/6)^q =y
if x< y, it means p > q
1/x +1/y = 7992
6^p +6^q=7992
6*6*6*6*6=36*6*6*6=216*6*6=1296*6= 7776
7992-7776= 216= 6*6*6
so p will be 5 and q will be 3
Schools: ESSEC MFin "26 (S)
Posts: 38
Answer: P=5 and Q=3
If X>Y, then P>Q
6^P+6^Q = 7992
6^5(P)= 7776
6^3(Q)=216
If X>Y, then P>Q
6^P+6^Q = 7992
6^5(P)= 7776
6^3(Q)=216
Since we are given the probability of the agate stone when the dice is rolled is \(\frac{1}{6}\)
If the dice is thrown P times, the probablity of the agate stone appearing all times is x
\(1/6 * 1/6 * 1/6 *\) ....... P times = \(x\)
\((1/6)^P = x\)
If the dice is thrown Q times, the probablity of the agate stone appearing all times is y
\(1/6 * 1/6 * 1/6 *\) .......Q times = \(y\)
\((1/6)^Q = y\)
We are also given that x < y
This implies that \((1/6)^P < (1/6)^Q\)
This implies that \(P > Q\)
From the given choices, we can see that P and Q are between 1 to 6.
We are given that \(1/x + 1/y = 7992\)
This is equal to \(6^P + 6^Q = 7992\)
\(6^1 = 6 \)
\(6^2 = 36\)
\(6^3 = 216\)
\(6^4 = 1296\)
\(6^5 = 7776\)
\(6^6 =\) no need to calculate as this is much bigger than 7992.
If we try various sums keeping in mind that P > Q, then we can boil down to P = 5 and Q = 3 (7776 + 216 = 7992)
P = 5 and Q = 3 are the right answer choices.
If the dice is thrown P times, the probablity of the agate stone appearing all times is x
\(1/6 * 1/6 * 1/6 *\) ....... P times = \(x\)
\((1/6)^P = x\)
If the dice is thrown Q times, the probablity of the agate stone appearing all times is y
\(1/6 * 1/6 * 1/6 *\) .......Q times = \(y\)
\((1/6)^Q = y\)
We are also given that x < y
This implies that \((1/6)^P < (1/6)^Q\)
This implies that \(P > Q\)
From the given choices, we can see that P and Q are between 1 to 6.
We are given that \(1/x + 1/y = 7992\)
This is equal to \(6^P + 6^Q = 7992\)
\(6^1 = 6 \)
\(6^2 = 36\)
\(6^3 = 216\)
\(6^4 = 1296\)
\(6^5 = 7776\)
\(6^6 =\) no need to calculate as this is much bigger than 7992.
If we try various sums keeping in mind that P > Q, then we can boil down to P = 5 and Q = 3 (7776 + 216 = 7992)
P = 5 and Q = 3 are the right answer choices.
Bunuel
1/6 ^ p =x
1/6 ^ q =y
1/x = 1/(1/6 ^ p) = 6^p
1/y = 6^ q
x< y
6^5 =7776
6^3 =216
1/x + 1/y =7992
so x = 1/7776 and y =1/216
p = 5 and q=3
said.tojiboev
PhD Forum Moderator
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Schools: Stanford '27
Posts: 65
Bunuel
6^1=6
6^2=36
6^3=216
6^4=1,296
6^5=7,776
6^6=46,656
1/6 - the probability of the agate stone appearing
(1/x)^P - the probability of the agate stone appearing all P times
(1/y)^Q - the probability of the agate stone appearing all Q times
Since 1/x + 1/y = 7,992, P or Q cannot be 6. (1/6)^6 = 1/46,656 which is more than 7,992.
It is seen that the sum of 3rd and 5th power of 6 is 7,992. So the answers will be 3 and 5.
Since x<y, 3 is for P and 5 is for Q.
Quote:
Probability of agate coming all P times
\(x = (\frac{1}{6})^P\)
Probability of agate coming all Q times
\(y = (\frac{1}{6})^Q\)
Therefore, substituting in given equation,
\(6^P + 6^Q\) = 7992
Since x < y, P > Q
\(6^Q (6 ^(^P^-^Q^) + 1) = 7992 = 6^3 (37)\)
Q = 3
P = 5
d_patel
Joined: 16 May 2024
Last visit: 24 Nov 2024
Posts: 56
Schools: ESSEC MFin "26 (A)
GMAT Focus 1: 685 Q89 V84 DI79
We are given that
\(P(agate) = \frac{1}{6}\)
When dice is thrown \(P\) times, the probability of the agate stone appearing all P times is x.
This can be written as,
P(agate appearing P times) = P(agate) X P(agate) X P(agate) X .... P times
because each dice role is independent from the previous one.
\(x = [P(agate)]^P\)
\(x = (\frac{1}{6})^P\)
When dice is thrown \(Q\) times, the probability of the agate stone appearing all Q times is y.
This can be written as,
P(agate appearing Q times) = P(agate) X P(agate) X P(agate) X .... Q times
\(y = [P(agate)]^Q\)
\(y = (\frac{1}{6})^Q\)
Now, \(x < y\)
And as we are working with number between 0 and 1, lower power means larger number.
Ex. \((\frac{1}{2})^2\) = \(\frac{1}{4}\) and \((\frac{1}{2})^3\) = \(\frac{1}{8}\) and \(\frac{1}{4} > \frac{1}{8}\)
So,
if \(x < y\) then \(P > Q\)
Also,
\(\frac{1}{x} + \frac{1}{y} = 7992\)
Using values of x and y,
\(1/(\frac{1}{6})^P + 1/(\frac{1}{6})^Q = 7992\)
\(6^P + 6^Q = 7992\)
Now looking at the options, we can be sure that value of P and Q will be between 1 to 6.
And we also know P > Q.
Looking at all the \(6^n\) for n between 1 to 6 we get,
\(6^1 = 6\), \(6^2 = 36\), \(6^3=216\), \(6^4= 1296\), \(6^5=7776\), \(6^6\) will be greater than \(10000\) so we don't need to check.
Looking at \(6^5\)=7776 we can see it is closest to \(7992\) and substracting it from \(7992\) gives us \(216\).
So,
final answer
\(P = 5\) and \(Q = 3 \)
\(P(agate) = \frac{1}{6}\)
When dice is thrown \(P\) times, the probability of the agate stone appearing all P times is x.
This can be written as,
P(agate appearing P times) = P(agate) X P(agate) X P(agate) X .... P times
because each dice role is independent from the previous one.
\(x = [P(agate)]^P\)
\(x = (\frac{1}{6})^P\)
When dice is thrown \(Q\) times, the probability of the agate stone appearing all Q times is y.
This can be written as,
P(agate appearing Q times) = P(agate) X P(agate) X P(agate) X .... Q times
\(y = [P(agate)]^Q\)
\(y = (\frac{1}{6})^Q\)
Now, \(x < y\)
And as we are working with number between 0 and 1, lower power means larger number.
Ex. \((\frac{1}{2})^2\) = \(\frac{1}{4}\) and \((\frac{1}{2})^3\) = \(\frac{1}{8}\) and \(\frac{1}{4} > \frac{1}{8}\)
So,
if \(x < y\) then \(P > Q\)
Also,
\(\frac{1}{x} + \frac{1}{y} = 7992\)
Using values of x and y,
\(1/(\frac{1}{6})^P + 1/(\frac{1}{6})^Q = 7992\)
\(6^P + 6^Q = 7992\)
Now looking at the options, we can be sure that value of P and Q will be between 1 to 6.
And we also know P > Q.
Looking at all the \(6^n\) for n between 1 to 6 we get,
\(6^1 = 6\), \(6^2 = 36\), \(6^3=216\), \(6^4= 1296\), \(6^5=7776\), \(6^6\) will be greater than \(10000\) so we don't need to check.
Looking at \(6^5\)=7776 we can see it is closest to \(7992\) and substracting it from \(7992\) gives us \(216\).
So,
final answer
\(P = 5\) and \(Q = 3 \)
x = \(\frac{1}{6^P} => \frac{1}{x} = 6^P\)
y = \(\frac{1}{6^Q} => \frac{1}{y} = 6^Q\)
\(x < y => \frac{1}{x} > \frac{1}{y} => 6^P > 6^Q => P > Q\)
\(\frac{1}{x} + \frac{1}{y} = 6^P + 6^Q = 7992\)
\(7992 = 6 * 1332 = 6^2 * 222 = 6^3 * 37 = 6^3 (1+36) = 6^3 + 6^5\)
=> P = 5, Q = 3
y = \(\frac{1}{6^Q} => \frac{1}{y} = 6^Q\)
\(x < y => \frac{1}{x} > \frac{1}{y} => 6^P > 6^Q => P > Q\)
\(\frac{1}{x} + \frac{1}{y} = 6^P + 6^Q = 7992\)
\(7992 = 6 * 1332 = 6^2 * 222 = 6^3 * 37 = 6^3 (1+36) = 6^3 + 6^5\)
=> P = 5, Q = 3
Bunuel
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_________________
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Express x and y in terms of P and Q:
x=(1/6)^p. y=(1/6)^q = >
6^p + 6^q= 7992
So if P=5, and q=3 => 7776+216=79992
The correct values are P=5, Q=3
x=(1/6)^p. y=(1/6)^q = >
6^p + 6^q= 7992
So if P=5, and q=3 => 7776+216=79992
The correct values are P=5, Q=3
Bunuel
Given the probability of rolling the dice and getting the agate stone is 1/6 we need to find values for P and Q such that:
1. The probability of rolling the agate stone P times in a row is x=(1/6) power P .
2. The probability of rolling the agate stone \( Q \) times in a row is y=(1/6) power Q.
3. x<y and 1/x+1/y=7,992
We start by expressing x and y in terms of P and Q :
- x=(1/6) power P implies 1/x =6 power p.
- y = (1/6) power Q implies 1/x =6 power Q.
Given x < y , it follows that P > Q . We are also given:
1/x+1/y=7,992
6P+6Q=7,992
Now, we need to find values for P and Q that satisfy this equation. Let's check the powers of 6 that sum to 7,992:
So, we test 6^5 + 6^3
7,776 + 216 = 7,992
Therefore:
P = 5
Q = 3
The probability of the agate stone appearing =\(\frac{1}{6}\)
When the dice is thrown P times
X=\(\frac{1}{(6)^{P}}\)
When the dice is thrown Q times
Y=\(\frac{1}{(6)^{Q}}\)
\(\frac{1}{x}+\frac{1}{y}=7,992\)
\(6^p+6^q =7,992\)
Factorizing 7992= \(37*2^3*3^3 = 37*6^3\)
\(6^p+6^q = 37*6^3\)
x<y which means p> q
\(6^5+6^3= 37*6^3\)
p=5 and q=3
_________________
When the dice is thrown P times
X=\(\frac{1}{(6)^{P}}\)
When the dice is thrown Q times
Y=\(\frac{1}{(6)^{Q}}\)
\(\frac{1}{x}+\frac{1}{y}=7,992\)
\(6^p+6^q =7,992\)
Factorizing 7992= \(37*2^3*3^3 = 37*6^3\)
\(6^p+6^q = 37*6^3\)
x<y which means p> q
\(6^5+6^3= 37*6^3\)
p=5 and q=3
_________________
"Arise ! , Awake ! and Stop Not untill the Goal is reached. ~ Swami Vivekananda
DG1989
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The king of Trishkinda has specially designed ivory dice. Six exotic stones - agate, jasper, chalcedony, sapphire, topaz, garnet - are embedded on the six sides of the dice, such that one side has exactly one type of stone and no two sides have the same type of stone. When the dice is rolled, the probability of the agate stone appearing is \(\frac{1}{6}\). Let there be values P, Q, x, y such that when the dice is thrown P times, the probability of the agate stone appearing all P times is x and when the dice is thrown Q times, the probability of the agate stone appearing all Q times is y. Furthermore, x < y and \(\frac{1}{x}\) + \(\frac{1}{y}\) = 7,992.
In the table, select a value for P and Q that are jointly consistent with the given information. Make only two selections, one in each column.
Solution: Given that
The probability of rolling agate (one specific stone) P(agate) = \(\frac{1}{6}\)
If the dice is rolled P times, the probability that agate appears all P times is x = \((\frac{1}{6})^P\)
Similarly, for Q times, Probability is y = \((\frac{1}{6})^Q\)
Since x < y, it implies that
\((\frac{1}{6})^P\) < \((\frac{1}{6})^Q\)
which means P > Q
Also, \(\frac{1}{x}\) + \(\frac{1}{y}\) = 7,992
or \(6^P\) + \(6^Q\) = 7992
As, P > Q
the only possible combination of P and Q that satisfies the above condition is (5, 3)
Thus P = 5 and Q = 3
GMAT-Club-Forum-e66t8fe4.png [ 23.97 KiB | Viewed 529 times ]
In the table, select a value for P and Q that are jointly consistent with the given information. Make only two selections, one in each column.
Solution: Given that
The probability of rolling agate (one specific stone) P(agate) = \(\frac{1}{6}\)
If the dice is rolled P times, the probability that agate appears all P times is x = \((\frac{1}{6})^P\)
Similarly, for Q times, Probability is y = \((\frac{1}{6})^Q\)
Since x < y, it implies that
\((\frac{1}{6})^P\) < \((\frac{1}{6})^Q\)
which means P > Q
Also, \(\frac{1}{x}\) + \(\frac{1}{y}\) = 7,992
or \(6^P\) + \(6^Q\) = 7992
As, P > Q
the only possible combination of P and Q that satisfies the above condition is (5, 3)
Thus P = 5 and Q = 3
Attachment:
GMAT-Club-Forum-e66t8fe4.png [ 23.97 KiB | Viewed 529 times ]
Given : Dice 6 faces, probablility of A appearing = 1/6
To find : P and Q
Soln : X = 1/(6^p) ; as die is thrown P times and probablilty of appearing is x ;
for eg; if a die is thrown twice probability is : 1/6X1/6 = 1/36
Similarly : Y = 1/(6^q)
Substitue in given eqn : 6^p+6^q = 7992
X<Y hence P>Q
6^p+6^q = 6^4(6^2+1)
Hence comparing P=6 and Q =4
To find : P and Q
Soln : X = 1/(6^p) ; as die is thrown P times and probablilty of appearing is x ;
for eg; if a die is thrown twice probability is : 1/6X1/6 = 1/36
Similarly : Y = 1/(6^q)
Substitue in given eqn : 6^p+6^q = 7992
X<Y hence P>Q
6^p+6^q = 6^4(6^2+1)
Hence comparing P=6 and Q =4
P = 5, Q = 3
given Prob(a) = 1/6
and (1/6)^p =x, (1/6)^q = y,
given x<y, hence p>q
now,
1/x + 1/y = 7992
x+y/xy = 7992
substituting values and solving we get
6^p + 6^q = 7992
6^5 + 6^3 = 7992
hence p = 5 and q = 3
given Prob(a) = 1/6
and (1/6)^p =x, (1/6)^q = y,
given x<y, hence p>q
now,
1/x + 1/y = 7992
x+y/xy = 7992
substituting values and solving we get
6^p + 6^q = 7992
6^5 + 6^3 = 7992
hence p = 5 and q = 3
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Schools: NUS BU (A$) Simon '27 (A$$) Tepper (A$)
GPA: 8
Schools: NUS BU (A$) Simon '27 (A$$) Tepper (A$)
Posts: 125
The probability of the agate stone appearing when the dice is thrown once is 1/6.
When the dice is thrown P times, the probability of the agate stone appearing every single time will be (1/6)^P
When the dice is thrown Q times, the probability of the agate stone appearing every single time will be (1/6)^Q
Given, 1/x + 1/y = 7992 (where y>x)
we get, (6)^P + (6)^Q = 7992
where y>x that means P>Q
Hence we get, P = 5, Q = 3
When the dice is thrown P times, the probability of the agate stone appearing every single time will be (1/6)^P
When the dice is thrown Q times, the probability of the agate stone appearing every single time will be (1/6)^Q
Given, 1/x + 1/y = 7992 (where y>x)
we get, (6)^P + (6)^Q = 7992
where y>x that means P>Q
Hence we get, P = 5, Q = 3
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