0 books -> 2
1 book -> 12
2 books -> 10
3 or more -> 6 [ we will call this <3-or-more> ]
Since Arith. mean = 2
(2*0 + 1*12 + 2*10 + <3-or-more>) / 30 = 2
<3-or-more> = 60 - 12 - 20
<3-or-more> = 28
We know that 6 students borrowed >= 3 books. So, the max number
of books taken by a single student can be computed when 5 out of
6 students take only 3 books
5*3 + max = 28
max = 28 - 15 = 13
Is 13 the correct answer ?
BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course
Redeem
Target Test Prep GMAT OnDemand
Scott Woodbury-Stewart’s private virtual classroom — 400 hours of master-class video lessons for the GMAT Focus Edition.
- 715+ score guarantee — highest in the industry (99th percentile)
- 52 chapters · 1,500+ lessons · 4,000+ practice questions
- 400 hours of video · 1,500+ instructor-led HD smartboard lessons
- 300,000+ students accepted to Harvard, Stanford, Wharton, Booth & Sloan & more
- 24/7 live support + weekly Zoom office hours with GMAT instructors
- TTP AI Assist — 24/7 AI-powered virtual tutor for instant help
- 1,200+ flashcards + AI-powered study assistant & daily calendar
- OnDemand, LiveTeach & GMAT Bootcamp formats available
- Also: GRE, SAT Math & Executive Assessment courses
- MBA Admissions Consulting now available
- 🏆 2025 EdTech Breakthrough Award: Test Prep Solution Provider of the Year
- 200,000+ students served
- 5-day free trial — $0 to start, no auto-billing, cancel anytime
★★★★★
5.0
(559 reviews)
130-pt guarantee
$0 to start
then $127/mo
GMAT Prep - Library books
Source: Beat The GMAT — Problem Solving |
To maximise the number of books borrowed by 1 student you have to minimise the books borrowed by the rest.
2 students did not borrow anything; Thus 30-2 = 28 borrowed atleast 1
12 borrowed 1 each
10 borrowed 2 each
Thus 28 - 10 - 12 = 6 borrowed atleast 3 each
Now assume that 5 out of these 6 borrowed the least possible (which is 3 in this case). Let us assume that the last student borrowed y books
Average books borrowed = 2
2 = (12*1 + 10*2 + 5*3 + 1*y)/30
60 = 12 + 20 + 15 + y
Therefore y = 13
2 students did not borrow anything; Thus 30-2 = 28 borrowed atleast 1
12 borrowed 1 each
10 borrowed 2 each
Thus 28 - 10 - 12 = 6 borrowed atleast 3 each
Now assume that 5 out of these 6 borrowed the least possible (which is 3 in this case). Let us assume that the last student borrowed y books
Average books borrowed = 2
2 = (12*1 + 10*2 + 5*3 + 1*y)/30
60 = 12 + 20 + 15 + y
Therefore y = 13
















