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805+ Level|   Math Related|         
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Abhijeet24
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GMAT Club Olympics 2024 (Day 3): The king of Trishkinda has specially 10 Jul 2024, 09:26
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Bunuel
The king of Trishkinda has specially designed an 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 a value for Q that are jointly consistent with the given information. Make only two selections, one in each column.

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­The probability of the agate stone appearing all P times is x, i.e., x = (1/6)^P
and when the dice is thrown Q times, the probability of the agate stone appearing all Q times is y,  i.e.,  y = (1/6)^Q

And 
1/x+1/y=7992 is given.
This translates to
6^P + 6^Q = 7992

As x<y, so P>Q.

The possible pair satisfying the equation is 5 and 3.

So, P = 5 and Q = 3.
 
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GMAT Club Olympics 2024 (Day 3): The king of Trishkinda has specially 10 Jul 2024, 09:38
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­P(Agate)=1/6
Given: \((1/6)^P\) = x
\((1/6)^Q\) = y

Given: x < y
--> P > Q  (Because the value decreases when fraction is raised to higher power)

 1/x+1/y = 7,992
(x + y)/xy = 7992

\((1/6)^Q + (1/6)^P\)
________________                 = 7992
\((1/6)^PQ\)­

\(6^Q + 6^P\) = 7992
 \(6^Q + 6^P\) = 6 * 6 * 6 * 37
 \(6^Q + 6^P\) = 6 * 6 * 6 * (36 + 1)
\(6^Q + 6^P\) = \(6^5 + 6^3\)
Q < P
Q=3, P=5­
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GMAT Club Olympics 2024 (Day 3): The king of Trishkinda has specially 10 Jul 2024, 09:40
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Given info. x<y and 1/x+1/y=7,992.
by considering above conditions, we need to find P and Q.
Probability of agate stone appearing in one roll of the dice =1/6,
If the dice is thrown P times, then probability of the agate stone appearing all P times = (1/6)^P = x,
If the dice is thrown P times, then probability of the agate stone appearing all P times = (1/6)^Q =y,
As x = (1/6)^P and y= (1/6)^Q
Convert x and y into reciprocal, 1/x = 6^P & y = 6^Q,
1/x+1/y=7992, 6^P + 6^Q = 7992,
As x<y, P>Q.
By considering options and plugging P (3 & 4) and Q (1 & 2) into 6^P + 6^Q = 7992 would render little value than the required. So, let's start with 5 and 6, then we'll guess further.

If P = 5 & Q= 1, 6^P = 7776 & 6^Q = 6, 7782. Closer to the solution.
If P =5 & Q= 2, 6^P = 7776 & 6^Q = 36, 7812. Closer to the solution.
If P =5 & Q= 3, 6^P = 7776 & 6^Q = 216, 7992. Bingo.

So, P =5 & Q=3.­
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GMAT Club Olympics 2024 (Day 3): The king of Trishkinda has specially 10 Jul 2024, 09:48
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­x = (1/6) ^P or 1/x = 6^p

and 

y = (1/6)^ Q or 1/y = 6^q

We are given 1/x + 1/y = 7992 or 6^p+ 6^q = 7992

By hit and trial,

6^5 + 6^3 = 7992

Since x <y, (1/6)^p < (1/6)^q or p>q

Therefore, P= 5 and Q=3 is our answer.
 
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The question is essentially asking about the number of times a dice has been rolled on two separate equations. The information that we are provided takes the form of probabilities, the relationships between these probabilities and the relationship between the number of rolls. 

The question states that the probability that the agate stone appears when the dice is thrown \(P\) times, is \(x\) and that the probability the same stone appears when the dice is thrown \(Q\) times, is \(y\). We know that there are six sides to the dice and therefore the probability of it landing on agate can be expressed as \(\frac{1}{6}\). The probability that the dice lands on agate twice in two throws is \(\frac{1}{6}\) * \(\frac{1}{6}\) or \(\frac{1}{36}\). The probability then for landing on agate \(N\) times in \(N\) throws then, would be expressed as \((\frac{1}{6})^N\) or simply \(\frac{1}{6^N}\) since we know that \(N\) must be a positive integer. 

Based of this, we can express the probabilities \(x\) and \(y\) as follows. 

\(x = \frac{1}{6^P}\) and \(y = \frac{1}{6^Q}\)

We also know that \(x < y\) and that \(\frac{1}{x} + \frac{1}{y} = 7,992\)

Now all we need to do is plug in the values for \(x\) and \(y\) and we get \(\frac{1}{1/6^P} + \frac{1}{1/6^Q} = 7,992\) or expressed simpler, \(6^P + 6^Q = 7,992\).

Since we know that P and Q must be positive integers between 1 and 6, it is not very difficult to work out this equation. We can quickly calculate the first powers of 6:

\(6^1=6 ; 6^2=36 ; 6^3=216 ; 6^4=1296, 6^5=7776\). No need to calculate further as 7,776 is quite close to 7,992 and there is no posibility that it will be \(6^6\). Since there are only two integers we are adding we know that one value must be larger than half of 7,992, which in this case is 7776 or \(6^5\). The second value we can calculate by subtracting 7,776 from 7,992 which is 216 or \(6^3\). Therefore \(6^5+6^3=7992\)

Now we just need to assign \(P\) and \(Q\) to the right values. To help with that, we have one piece of key information in that  \(x < y\). The larger \(P\) is the smaller \(x\) will be and the smaller \(Q\) is the larger \(y\) will be therefore \(P>Q\) must be true.

Of the two values, 5 is larger than 3 and so \(P=5\) and \(Q=3\)
Bunuel
The king of Trishkinda has specially designed an 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 a value for Q that are jointly consistent with the given information. Make only two selections, one in each column.

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GMAT Club Olympics 2024 (Day 3): The king of Trishkinda has specially 10 Jul 2024, 10:35
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Since we know P (agate)= 1/6
When the dice is thrown 1 time, the probability of the agate stone appearing is = 1/6
When the dice is thrown 2 times, the probability of the agate stone appearing 2 times is = (1/6)^2
When the dice is thrown P times, the probability of the agate stone appearing P times is = (1/6)^P = x
When the dice is thrown Q times, the probability of the agate stone appearing Q times is = (1/6)^Q = y
.
Its given that x<y , and since (1/6) is a decimal, when raised to any positive power its value decreases. 
hence x<y implies that P>Q
.
It is also given that
\(\frac{1}{x}+\frac{1}{y}=7992\)

i.e
\(\frac{1}{(1/6)^P} + \frac{1}{(1/6)^Q}=7992\)

\(6^P+6^Q = 7992\)

Looking at the answer choices, possible values are \(6^4 +  6^3\) only.
Since we established P>Q
P=4 and Q=3
 ­
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GMAT Club Olympics 2024 (Day 3): The king of Trishkinda has specially 10 Jul 2024, 10:37
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Bunuel
The king of Trishkinda has specially designed an 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 a value for Q that are jointly consistent with the given information. Make only two selections, one in each column.

­
 


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­Total outcomes of rolling dice for 1 time = 6
Total outcomes of rolling dice for 2 times = 36 = 6^2
Total outcomes of rolling dice for P times = 6^P
Total outcomes of rolling dice for Q times = 6^Q

Now same stone will appear only once for all P times and all Q times. 

x = 1/6^P, y = 1/6^Q

As we know x < y => 6^P > 6^Q => P > Q.

1/x + 1/y = 7992
= > 6^P + 6^Q = 7992

Considering P > Q, we can get P = 5(1st column), Q = 3(2nd column)
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GMAT Club Olympics 2024 (Day 3): The king of Trishkinda has specially 10 Jul 2024, 11:01
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Bunuel
The king of Trishkinda has specially designed an 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 a value for Q that are jointly consistent with the given information. Make only two selections, one in each column.

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­All the given content can be converted into a single statement as below:

\(x = (1/6)^p &  y = (1/6)^q\)
We can also break down  7,992 as => \(6^5 + 6^3\)

So from the given equation p, q {5,3}

​​​​​​​Hence  p is 5 and q is 3 so that x < y; 
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­The king of Trishkinda has specially designed an 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 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.
when the dice is thrown P times, the probability of the agate stone appearing all P times is x.
Therefore x = \((1/6)^P\)­, Means 1/x = \(6^P\)­
Similarly,
when the dice is thrown Q times, the probability of the agate stone appearing all Q times is y.
y = \((1/6)^Q\)­, Means 1/y = \(6^Q\)­
Furthermore, x<y
\(\frac{1}{y}\)<\(\frac{1}{x}\)
\(6^Q\)­ < \(6^P\)­
and \(\frac{1}{x}\)+\(\frac{1}{y}\)=7992
Means,  \(6^P\)­ + \(6^Q\)­  = 7992­  ---Eq1
Therefore, for \(6^P\)­ lets start with value more than half of 7992 i.e. more than 3996
and for P=5(Because \(6^4\)­ = 1296), \(6^P\)­ = \(6^5\)­ = 7776
Now substitute in eq1
Get the value for \(6^Q\)­ = 7992 - 7776 = 216 = \(6^3\)­
Therefore, Q=3­­
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GMAT Club Olympics 2024 (Day 3): The king of Trishkinda has specially 10 Jul 2024, 11:17
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­This is quite a straight question but tricks you in a way when you dont notice the equation 1/x + 1/y = 7992.

The probabilty will always be 1/6^n where n = the number of times dice is thrown.

calculate power 6's and try inserting in the above equation. As we know x<y, so 1/x= (6^3 = 216) and 1/y = (6^5 = 7776)
the number of times thrown is 3 and 5
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GMAT Club Olympics 2024 (Day 3): The king of Trishkinda has specially 10 Jul 2024, 11:18
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Understand the probability conditions:

The probability of the agate stone appearing once is 1/6
The probability of the agate stone appearing P times when the dice is thrown P times is x.
The probability of the agate stone appearing Q times when the dice is thrown Q times is y.
It's given that x<y and 1/x + 1/y = 7.992.

2. Set up the probability formulas:

The probability of getting the agate stone P times in P throws: x= (1/6)^p.
The probability of getting the agate stone Q times in Q throws: y= (1/6)^q.

3.Use the equation given:

Substitute x and y into the equation 1/x + 1/y = 7.992.

4.Solve for P and Q.

Let's start by calculating the values for x and y for different P and Q values and then check the consistency with 1/x + 1/y = 7.992.

We'll compute x and y and then find 1/x and 1/y for P and Q from 1 to 6.

Let's perform these calculations systematically.

Calculations:

1. for P = 1

x = (1/6)^1 = 1/6

1/x = 6

2. for p = 2

x = (1/6)^2 = 1/36

1/x = 36

3. for p = 3

x = (1/6)^3 = 1/216

1/x = 216

4. for p = 4

x = (1/6)^4 = 1/1296

1/x = 1296

5. for p = 5

x = (1/6)^5 = 1/7776

1/x = 7776

6. for p = 6

x = (1/6)^6 = 1/46656

1/x = 46656


We'll do the same for Q values.

Now, we need to find the combination where 1/x + 1/y = 7.992.

try to find matching pairs systematically. I'll continue these calculations.

First, let's look at the possibilities for 1/y when 1/x = 6 and check if their sum equals 7.992.

1/x = 6

7.992−6=1.992

Similarly for others:

By checking each possible pair, it becomes evident that the only plausible pairs for P and
Q can be calculated from the options given in the problem. For instance, If P=1 and Q=3 then:

1. x = (1/6)^1 = 1/6, and 1/x = 6
2. y = (1/6)^3 = 1/216, and 1/y = 216

thus,

1/x + 1/y =6+216=222


Which doesn't fit. So continue this:

Finally, After checking P=2 and Q=3:

Where x = (1/6)^2 = 1/36 , y = (1/6)^5 = 1/7776

So, P=2 and Q=6 correctly fits into provided sum of 7.992.­
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GMAT Club Olympics 2024 (Day 3): The king of Trishkinda has specially 10 Jul 2024, 11:19
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­P= 1/6
Probabilty if thrown P times = (1/6)^P= X
Probabilty if thrown Q times = (1/6)^Q=Y

1/x+1/y= 7992

6^p+ 6^q= 7992
So P =3 , q=5
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Bunuel
The king of Trishkinda has specially designed an 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 a value for Q that are jointly consistent with the given information. Make only two selections, one in each column.

­
 


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­x= (1/6)^P
y=(1/6)^Q
x<y, therefore P>Q
1/x +  1/y = 7992
6^P+6^Q= 7992

6^1=6
6^2=36
6^3=216
6^4=1296
6^5= 7776
6^5+6^3 = 7992

Ans: P=5, Q=3
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GMAT Club Olympics 2024 (Day 3): The king of Trishkinda has specially 10 Jul 2024, 11:48
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­P = 5
Q = 3

Given a dice with six exotic stones named against each face instead of a number (1 - 6)
Probability of rolling "agate" stone = 1/6

P : a number
Q : a number
x : when the dice is thrown P times, the probability of the "agate" stone appearing all P times is x
y : when the dice is thrown Q times, the probability of the "agate" stone appearing all P times is y

condition: x < y (probability of appearing P times is more than the probability of appearing Q times) ----- (1)

1/x + 1/y = 7992

x = 1/(6)^P
y = 1/(6)^Q

1/x + 1/y = (6)^P + (6)^Q = 7992 (prime factorise this number)

Given x < y => 1/x > 1/y =>  (6)^P > (6)^Q ------- (2)

\((6)^P + (6)^Q = (2^3)(3^3)(37)\)
\((6)^P + (6)^Q = (2^3)(3^3)(36 + 1)\)
\((6)^P + (6)^Q = (6^3)(6^2 + 1)\)
\((6)^P + (6)^Q = (6)^5 + (6)^3\)

From (1) and (2), we can say that P = 5, and Q = 3
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GMAT Club Olympics 2024 (Day 3): The king of Trishkinda has specially 10 Jul 2024, 12:12
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X is 3 and y is 5

As we know 6^5=7776
6^3=216
Sum of both number is equivalent to 7992
It is given that x<y
So y will be 5 and x is 3

Posted from my mobile device
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GMAT Club Olympics 2024 (Day 3): The king of Trishkinda has specially 10 Jul 2024, 12:15
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­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 = 7992

 

x = (1/6)^P

= 1/ (6^P)

 

y = (1/6)^Q

= 1/ (6^Q)

 

1/x and 1/y are both squares or cubes or some power to the number 6, which add up to 7992

 

6^2 = 36

6^3 = 216

6^4 = 1296

6^5 = 7776

 

7992-7776 = 216

Or 

7776 + 216 = 7992

6^5 + 6^3 = 7992

 

y > x

1/y < 1/x

P = 5

Q = 3
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GMAT Club Olympics 2024 (Day 3): The king of Trishkinda has specially 10 Jul 2024, 12:25
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To solve the problem, let's break it down step-by-step.

Understanding the Probability:

There are 6 sides on the dice, each with a different stone.
The probability of rolling any specific stone (including the agate) is 1/6.
Probability of Agate Appearing All Times:

When the dice is rolled P times, the probability of agate appearing all P times is (1/6)^P.
When the dice is rolled Q times, the probability of agate appearing all Q times is (1/6)^Q.
Given Equations and Relationships:

We are given x = (1/6)^P and y = (1/6)^Q.
It is given that x < y and 1/x + 1/y = 7,992.
Expressing the Equations:

Since x < y, it implies P > Q (since as the exponent increases, the probability decreases).
Therefore, 1/x = 6^P and 1/y = 6^Q.
We can rewrite the given equation as: 6^P + 6^Q = 7,992.
Solving for P and Q:

We need to find integers P and Q such that 6^P + 6^Q = 7,992.
Testing Possible Values:

7,992 is close to 8,000. Let’s see if 7,992 can be expressed as the sum of powers of 6.
6^5 = 7,776.
If P = 5, then 6^P = 7,776.
7,992 - 7,776 = 216.
6^3 = 216.
Therefore, if Q = 3, then 6^Q = 216.
Verification:

We check if P = 5 and Q = 3 satisfy the equation:
6^5 + 6^3 = 7,776 + 216 = 7,992.
This satisfies the equation, and thus the values of P and Q are 5 and 3 respectively.

Therefore, the values of P and Q are:
5 and 3­
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sg787
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GMAT Club Olympics 2024 (Day 3): The king of Trishkinda has specially 10 Jul 2024, 12:42
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­1/6 is the probability that the agate face will show up once. 

If we want to see the agate face all times out of n times, then the probability of getting agate face is 
­1/6*­1/6*­1/6.....n times. 
The probability of getting agate face P times is denoted by x. Therefore, x=­1/6*­1/6*­1/6....multiplied P times.
Similarly, the probability of getting agate face Q times is denoted by y. Therefore, y=­1/6*­1/6*­1/6....multiplied Q times.

Now, note that x will become smaller the more number of times that we keep multiplying ­1/6. And it is already provided that x < y, therefore P is a greater number than Q (because greater the value of P, the greater the number of times ­1/6 is multiplied - making x smaller)

Now the question can be solved by taking each possible selection from the answer and adding it. First, calculate 6^6  = 46656. It is too big for our given choice of 7992. Add \(6^5+6^3\) by trial and error to arrive at a total of 7992, leading to 5 and 3. As we have concluded previously that P>Q, therefore P will be 5 and Q will be 3. 
The presence of a calculator in the DI section helps in such situations. ­
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pranjalshah
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GMAT Club Olympics 2024 (Day 3): The king of Trishkinda has specially 10 Jul 2024, 12:46
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­Answer: P = 5, Q = 3

Given, P(A) = 1/6

x = (1/6)^P

and y = (1/6)^Q

also given that 1/x + 1/y = 7992

Writing the powers of 6, we get 6, 36, 216, 1296, 7776.

We can see that 7776+216=7992

Therefore 1/x and 1/y must be equal to these two. 

And since x<y => 1/x>1/y

so, x must be 1/7776 and y must be 1/216

which can be re-written as x=(1/6)^5 and y=(1/6)^3

Therefore P = 5 and Q = 3.
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GMAT Club Olympics 2024 (Day 3): The king of Trishkinda has specially 10 Jul 2024, 12:47
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x = (1/6)^p
y=(1/6)^q

since y>x, so, p>q

6^p + 6^q = 7992
matches for 6^5 = 7776 +  6^3 = 216
so p=5, q=3­
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