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 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.
siddharthkapoor
GMAT Club Reviews PM Intern
Joined: 10 Apr 2018
Last visit: 12 Sep 2025
Posts: 527
Given Kudos: 522
Location: India
Schools: ISB'22 (D)
GMAT 1: 680 Q48 V34
GPA: 3.3
Products:
12 Oct 2019, 06:42
Kudos
Bookmarks
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
67% (02:47) correct
33%
(02:29)
wrong
based on 700
sessions
History
Date
Time
Result
Not Attempted Yet
Every day, Tom walks from his home to his office, via the same route, covering s feet at a speed of x feet per minute. Today he took a different route and ended up walking 10% more than he usually does, at a speed of x meters per minute. What is the percentage change in the time he took today compared to the time he takes on a usual day? (1 feet = 0.3 meter)
A. 10% decrease
B. 10% increase
C. 67% decrease
D. 67% increase
E. 200% decrease
A. 10% decrease
B. 10% increase
C. 67% decrease
D. 67% increase
E. 200% decrease
12 Oct 2019, 19:29
Kudos
Bookmarks
iamsiddharthkapoor
(I) Substitute values for variables..
Let s = 100 and x=10..
Time taken while going 100 at 10 feet per minute= \(\frac{100}{10}=10\) minutes
Time taken on return 1.1*100 at 10 meter per minute or @\(\frac{10}{0.3}\) feet per minute= \(110/\frac{10}{0.3}=\frac{33}{10}=3.3\) minutes
Change or decrease = \(\frac{10-3.3}{10}*100=67\)%
(II) Algebra..
Time taken while going s feet at x feet per minute= \(\frac{s}{x}\) minutes
Time taken on return 1.1*s at x meter per minute or @\(\frac{x}{0.3}\) feet per minute= \(1.1s/\frac{x}{0.3}=\frac{0.33s}{x}\) minutes
Change or decrease = \((\frac{s}{x}-\frac{0.33s}{x})/\frac{s}{x}*100=(1-0.33)*100=67\)%
C
General Discussion
Archit3110
Major Poster
13 Oct 2019, 05:29
Kudos
Bookmarks
let distance s feet = 10feet = 3 mtr
speed = 10 feet per minute ; total time ; 3/10 ; .3 min
new distance ; 3*1.1 ; 3.3 mtrs
speed ; 10/.3 feet per min; time taken ; 3.3/33.33 = .099 min
% ∆ in time = .099-.3/.3 ; (67%)
IMO C
speed = 10 feet per minute ; total time ; 3/10 ; .3 min
new distance ; 3*1.1 ; 3.3 mtrs
speed ; 10/.3 feet per min; time taken ; 3.3/33.33 = .099 min
% ∆ in time = .099-.3/.3 ; (67%)
IMO C
iamsiddharthkapoor
