CLEAR
Toprankers-Info

+91 7676564400

Mobile App

Get it on Google Play
Discussions
Select Date
Tags:

Articles

Discussions
Submit Discussion
Muskan Tanwar

· started a discussion

· 1 Months ago

b (Question: 217528.0 )

sollution

Question:
On walking at 1 \(\cfrac{1}{2}\) times of his original speed, a person reaches his office 40 minutes early. If his original speed is 10 km/hour, then what is the distance between his home and office ?
Options:
A) 10 km
B) 15 km
C) 20 km
D) 25 km
Solution:
Ans: (c) Original speed = 10 km/hr

Increased speed = 10 × \(\cfrac{3}{2}\) = 15 km/hr


1 hours = 60 minutes 

If he reaches 60 minutes earlier then distance = 30 km

If he reaches 40 minutes earlier then distance = \(\cfrac{30}{60}\) × 40 = 20 km

-logo
We at aim to provide most comprehensive content & test for practice which is carefully divided into various chapters and topics to help you focus on your weak areas.Our Practice mock test gives you unmatched analytics to help you realize your strong areas,time management and improvement areas.Our Short & Crisp notes on each topic will help you to quickly revise the topic along with the tips & tricks of the exams.
More About
-Info
-Info

Mobile App

Get it on Google Play
About Us Campus Ambassador Careers Contact Us Refund Policy

All Rights Reserved Top Rankers

User Policy
Disclaimer
Terms & Conditions