Forum Home > GMAT > Quantitative > Problem Solving (PS)
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.
Dec 1511:00 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.
Dec 1608:00 AM PST
-11:59 PM PST
GMAT Club 12 Days of Christmas is a 4th Annual GMAT Club Winter Competition based on solving questions. This is the Winter GMAT competition on GMAT Club with an amazing opportunity to win over $40,000 worth of prizes!- Dec 17
09:00 AM PST
-10:00 AM PST
Join Manhattan Prep instructor Whitney Garner for a fun—and thorough—review of logic-based (non-math) problems, with a particular emphasis on Data Sufficiency and Two-Parts. - Dec 18
10:00 AM PST
-11:00 AM PST
Here is the essential guide to securing scholarships as an MBA student! In this video, we explore the various types of scholarships available, including need-based and merit-based options. - Dec 21
11:00 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
10 Apr 2020, 04:16
Kudos
Bookmarks
E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
59% (02:46) correct
41%
(03:00)
wrong
based on 70
sessions
History
Date
Time
Result
Not Attempted Yet
In 2010, a library contained a total of 11500 books in two categories - fiction and nonfiction. In 2015, the library contained a total of 12760 books in these two categories. During this period, there was 10% increase in the fiction category while there was 12% increase in the non-fiction category. How many fiction books were in the library in 2015?
A. 5500
B. 6000
C. 6160
D. 6500
E. 6600
Are You Up For the Challenge: 700 Level Questions
A. 5500
B. 6000
C. 6160
D. 6500
E. 6600
Are You Up For the Challenge: 700 Level Questions
Updated on: 18 Jun 2024, 08:26
Originally posted by GMATWhizTeam on 10 Apr 2020, 04:42.
Last edited by Bunuel on 18 Jun 2024, 08:26, edited 2 times in total.
Last edited by Bunuel on 18 Jun 2024, 08:26, edited 2 times in total.
Kudos
Bookmarks
Bunuel
In 2010, a library contained a total of 11500 books in two categories - fiction and nonfiction. In 2015, the library contained a total of 12760 books in these two categories. During this period, there was 10% increase in the fiction category while there was 12% increase in the non-fiction category. How many fiction books were in the library in 2015?
A. 5500
B. 6000
C. 6160
D. 6500
E. 6600
Are You Up For the Challenge: 700 Level Questions
A. 5500
B. 6000
C. 6160
D. 6500
E. 6600
Are You Up For the Challenge: 700 Level Questions
Solution:
Let f and n be the number of fiction and nonfiction book in library in 2010, respectively
- • f + n = 11500…(i)
- • Number of fiction books = \(f+\frac{10}{100}*f = \frac{110}{100}*f\)
• Number of nonfiction book = \(n + \frac{12}{100}*n = \frac{112}{100}*n\)
- o Total books = \(\frac{110}{100}*f + \frac{112}{100}*n = 12760\)
o \(110*f + 112*n = 12760*100\)…(ii)
- • n = 11500 – f
• On substituting this value in equation (ii), we get
• 110*f + 112(11500-f) = 12760*100
• 110*f – 112*f +115*112*100 = 12760*100
- o 2*f = (12880-12760)*100
o \(f = \frac{120*100}{2} = 6000\)
Number of fiction books in 2015 = 6000*1.1 = 6600
Hence, the correct answer is Option E.
10 Apr 2020, 05:24
Kudos
Bookmarks
Let \(f\) represent fiction books and \(n\) represent non-fiction books.
In 2010, \(f+n=11500\) \(----(1)\)
In 2015 \(1.1f+1.12n=12760\) \(----(2)\)
We are to find \(1.1f\)
\((1)*1.12\)===> \(1.12f+1.12n=12880 -----(3)\)
\((3)-(2)\)===> \(0.02f=120\)
\(f=6000\)
\(1.1f=6600\)
The answer is E.
In 2010, \(f+n=11500\) \(----(1)\)
In 2015 \(1.1f+1.12n=12760\) \(----(2)\)
We are to find \(1.1f\)
\((1)*1.12\)===> \(1.12f+1.12n=12880 -----(3)\)
\((3)-(2)\)===> \(0.02f=120\)
\(f=6000\)
\(1.1f=6600\)
The answer is E.
