Collectively, the five children in the Kramer family have 30 trophies. : Problem Solving (PS)

My Profile Logout

Test's Subscription Expires:

My Follow Feed Kudos My Error Log

Messages and Replies

Settings Mark All Read

See All

Applicant Notifications

Settings Mark All Read

See All

Global notifications

Settings Mark All Read

See All

Collectively, the five children in the Kramer family have 30 trophies.

Sort by Date

Sort by Kudos

Author

Message

TAGS:

Hide Tags

Math Expert

Joined: 02 Sep 2009

Posts: 50585

Collectively, the five children in the Kramer family have 30 trophies. [#permalink]

Show Tags

09 Apr 2018, 00:52

00:00

Difficulty:

25% (medium)

Question Stats:

71% (01:29) correct

29% (01:44) wrong

based on 137 sessions

HideShow timer Statistics

Collectively, the five children in the Kramer family have 30 trophies. If each child has at least one trophy and no two children have the same number of trophies, what is the greatest number of trophies that the child with the second-highest number of trophies could have?

A. 10
B. 11
C. 12
D. 13
E. 14

Spoiler: OA

_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.

What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Manager

Joined: 30 May 2017

Posts: 122

Location: United States

Schools: HBS '21

GMAT 1: 690 Q50 V32

GRE 1: Q168 V164

GPA: 3.57

Collectively, the five children in the Kramer family have 30 trophies. [#permalink]

Show Tags

Updated on: 09 Apr 2018, 02:22

Bunuel wrote:

Let the 5 children be named M, N, P, Q and R. assign minimum possible values starting from M up to P. Q and N will then have the highest possible values.

M = 1 trophy
N = 2 trophies
P = 3 trophies

At this point M+N+P = 1+2+3 = 6. Therefore Q +R will be 30-6 = 24. Since Q and R cannot have the same number of trophies: Q will be 11 trophies and R 13 trophies.

Q = 11 trophies (second-highest number of trophies )
R = 13 trophies

Hence option B (11 trophies) is the answer.

Give me +1 Kudos if you like the explanation. Thanks.
_________________

Kindly press the +1Kudos if you like the explanation. Thanks a lot!!!

Originally posted by houston1980 on 09 Apr 2018, 02:17.
Last edited by houston1980 on 09 Apr 2018, 02:22, edited 1 time in total.

Senior Manager

Joined: 14 Feb 2018

Posts: 379

Re: Collectively, the five children in the Kramer family have 30 trophies. [#permalink]

Show Tags

09 Apr 2018, 02:18

Let , the number of trophies be in the order -1, 2, 3, x, y and we have to find x.
1 + 2 + 3 = 6
Therefore, x + y = 24

Now, the possible values of x and y could be -

x = 5, 6, 7, 8, 9, 10, 11
Correspondingly, y = 19, 18, 17, 16, 15, 14, 13

Thus, the greatest possible value of x = 11

Hence, option B.

Please correct me if I am wrong and if I am right, please give some Kudos !!

Sent from my Lenovo K53a48 using GMAT Club Forum mobile app

Target Test Prep Representative

Status: Head GMAT Instructor

Affiliations: Target Test Prep

Joined: 04 Mar 2011

Posts: 2830

Re: Collectively, the five children in the Kramer family have 30 trophies. [#permalink]

Show Tags

20 Aug 2018, 10:38

Bunuel wrote:

We can let the three children who have the least number of trophies have 1, 2, and 3 trophies, respectively, since no child can have the same number of trophies as any other sibling. Thus, these three children have a total of 1 + 2 + 3 = 6 trophies.

So we are left with a total of 30 - 6 = 24 trophies for the two children with the greatest number of trophies. Since we want the second-most number of trophies to be a large as possible, we will ensure that the greatest number of trophies is as small as possible while still retaining its role as the greatest number. We see that a 12-12 split almost works, but the two numbers are not allowed to be the same value.

Thus, we can let the child with the greatest number of trophies have 13 trophies, and hence, the child with the second most trophies has 11.

Answer: B
_________________

Jeffery Miller
Head of GMAT Instruction