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.
10 Jul 2024, 07:01
Kudos
Bookmarks
P: 5
Q: 3
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
47% (03:04) correct
53%
(03:22)
wrong
based on 505
sessions
History
Date
Time
Result
Not Attempted Yet
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.
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.
|
| P | Q | |
| 1 | ||
| 2 | ||
| 3 | ||
| 4 | ||
| 5 | ||
| 6 |
ShowHide Answer
Official Answer
P: 5
Q: 3
10 Jul 2024, 07:14
Kudos
Bookmarks
Bunuel
Experts' Global Explanation:
Probability of agate stone appearing when the dice is rolled 1 time = (1/6)
Probability of agate stone appearing when the dice is rolled 2 times = (1/6) x (1/6) = (1/6)2
Probability of agate stone appearing when the dice is rolled P times = (1/6)^P = x
Probability of agate stone appearing when the dice is rolled Q times = (1/6)^Q = y
1/x + 1/y = 7992
1/(1/6)^P + 1/(1/6)^Q = 7992
6^P + 6Q = 7992
6^P + 6^Q = 7992
Taking 6Q as the common term:
6^Q(6^(P – Q) + 1) = 7992 (Equation I)
6^Q(1 + 6^(P – Q)) = 7992
6^Q(1 + 6^(P – Q)) = 6 x 6 x 6 x 37
6^Q(1 + 6^(P – Q)) = 6 x 6 x 6 x (1 + 36)
6^Q(1 + 6^(P – Q)) = 6^3 x (1 + 6^2)
Comparing both sides:
Q = 3
P – Q = 2
P – 3 = 2
P = 5
Note that if we had taken 6^P as the common term in Equation I, we would have received P = 3 and Q =5 as the answer. Both (P = 5, Q = 3) and (P = 3 and Q =5) satisfy Equation I, however since the question specifies that x < y, it follows that (1/6)^P < (1/6)^Q, and thus P =5 and Q =3 is the only valid solution.
Hence, for “P” column, “5”, and for “Q” column, “3” is the correct combination of the answer choices.
10 Jul 2024, 07:38
Kudos
Bookmarks
7992 = (6^3) x 37
37=36+1
36=6^2
So, we can write 7992 = (6^3) x (6^2 + 1) = 6^5 + 6^3 equation (i)
x=1/(6^P) and y=1/(6^Q)
We are given that x<y, so we can conclude that Q<P.
Using equation (i), we can assign the values.
P=5 and Q=3
37=36+1
36=6^2
So, we can write 7992 = (6^3) x (6^2 + 1) = 6^5 + 6^3 equation (i)
x=1/(6^P) and y=1/(6^Q)
We are given that x<y, so we can conclude that Q<P.
Using equation (i), we can assign the values.
P=5 and Q=3
