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.
Apr 2512:30 AM EDT
-01:30 AM EDT
At one point, she believed GMAT wasn’t for her. After scoring 595, self-doubt crept in and she questioned her potential. But instead of quitting, she made the right strategic changes. The result? A remarkable comeback to 695. Check out how Saakshi did it.
Apr 2601:30 AM EDT
-02:30 AM EDT
Learn how Keshav, a Chartered Accountant, scored an impressive 705 on GMAT in just 30 days with GMATWhiz's expert guidance. In this video, he shares preparation tips and strategies that worked for him, including the mock, time management, and more.
Apr 2610:00 AM EDT
-11:00 AM EDT
The Target Test Prep course represents a quantum leap forward in GMAT preparation, a radical reinterpretation of the way that students should study. Try before you buy with a 5-day, full-access trial of the course for FREE!
Apr 2711:00 AM EDT
-12:00 PM EDT
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
Apr 2808:00 AM PDT
-11:00 AM PDT
Whether you are just beginning your MBA journey or fine-tuning your path to a 700+ score, GMAT Day is designed to give you a competitive edge.
May 0111:00 AM PDT
-12:00 PM PDT
Free- full-length test + 15 concept videos + 150+ short videos + study plan!
08 Jul 2015, 02:55
Kudos
Bookmarks
B
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:
44% (03:14) correct
56%
(03:13)
wrong
based on 257
sessions
History
Date
Time
Result
Not Attempted Yet
The students at Natural High School sell coupon books to raise money for after-school programs. At the end of the coupon sale, the school selects six students to win prizes as follows:
From the homeroom with the highest total coupon-book sales, the students with the first-, second- and third-highest sales receive $50, $30, and $20, respectively; from the homeroom with the second-highest total coupon-book sales, the three highest-selling students receive $10 each. If Natural High School has ten different homerooms with eight students each, in how many different ways could the six prizes be awarded? (Assume that there are no ties, either among students or among homerooms.)
A. \((2^7)(3^2)(5)(7^2)\)
B. \((2^8)(3^3)(5)(7^2)\)
C. \((2^9)(3)(5^2)(7^2)\)
D. \((2^9)(3^4)(5)(7^2)\)
E. \((2^8)(3^5)(5)(7^2)\)
Kudos for a correct solution.
From the homeroom with the highest total coupon-book sales, the students with the first-, second- and third-highest sales receive $50, $30, and $20, respectively; from the homeroom with the second-highest total coupon-book sales, the three highest-selling students receive $10 each. If Natural High School has ten different homerooms with eight students each, in how many different ways could the six prizes be awarded? (Assume that there are no ties, either among students or among homerooms.)
A. \((2^7)(3^2)(5)(7^2)\)
B. \((2^8)(3^3)(5)(7^2)\)
C. \((2^9)(3)(5^2)(7^2)\)
D. \((2^9)(3^4)(5)(7^2)\)
E. \((2^8)(3^5)(5)(7^2)\)
Kudos for a correct solution.
08 Jul 2015, 03:25
Kudos
Bookmarks
Bunuel
METHOD-1
Since there are 10 homeroom and we need to award to only two of the 10 homeroom so the homerooms to be awarded can be selected in 10C2 ways = 45 ways
Now the two homerooms have to be placed at position First and Position second which can happen in = 2! ways
From the first homeroom three awardee out of 8 Students can be selected in 8C3 ways = 56 ways
The three different awards of $50, $30 and $20 can be given to the selected 3 awardee students in = 3! = 6 ways
From the Second homeroom three awardee out of 8 Students can be selected in 8C3 ways = 56 ways
But there is no need to arrange the awards among them as all the awards are same $10 = 1 way to arrange
Total Ways to make this even of award giving happen = (10C2 *2!) * (8C3 *3!) * (8C3)*1)
Total Ways to make this even of award giving happen = (45 *2* 56*6) * (56*1) = (9*5 *2* 8*7*3*2) * (8*7) = \(2^8*3^3*7^2*5\)
Answer: Option B
METHOD-2
The two homerooms out of 10 for position 1 and 2 can be identified in = 10*9 ways
Three Awardees out of 8 of First positioned homeroom can be placed in = 8*7*6 ways [All awards are different]
Three Awardees out of 8 of Second positioned homeroom can be placed in = (8*7*6)/3! ways [Divided by 3! because All awards are same $10 so no arrangement required]
Total ways to give awards = 10*9*8*7*6*[8*7*6/3!] = \(2^8*3^3*7^2*5\)
Answer: Option B
General Discussion
ENGRTOMBA2018
Joined: 20 Mar 2014
Last visit: 01 Dec 2021
Posts: 2,319
Own Kudos:
Given Kudos: 816
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE:Engineering (Aerospace and Defense)
Products:
08 Jul 2015, 04:11
Kudos
Bookmarks
Bunuel
Total number of selections from 10 homerooms = 10C2 and they can be further selected in 2C1 ways to have either 3 $10 prizes or 1 each of 50,30 and 20$ prizes . Thus total ways (1) = 10C2*2C1
Now, after selecting the rooms above, number of ways of awarding 3 different (50,30,20$ prizes) out of 8 students (2) = 8C3 * 3! (factor of 3! to account for 3 different prizes, this could have been 8P3 as well)
Finally, selecting 3 students for 3 10$ prizes out of 8 students (3) = 8C3
Thus, total arrangements possible = 1 X 2 X 3 = 10C2*2C1*8C3*3!*8C3 = \((2^8 )(3^3 )(5)(7^2 )\).
B is the correct answer
