CLEAR
Toprankers-Info

+91 7676564400

Mobile App

Get it on Google Play
Discussions
Select Date
Tags:

Articles

Discussions
Submit Discussion
Shashank Shah

· started a discussion

· 1 Months ago

PLEASE HELP ANYBODY (Question: 269915.0 )

Lets Distance = x
Then, ATQ, (x )/20-(x )/30=1/2 (Difference of times)
Then x=150 kms
Now , new time limit =6:15
Therefore new time gap= 6:30-6:15=15 min=1/4 hrs
Let new required speed be =S
New equation, (150 )/30-(150 )/S=1/4
Then S=31.4 approx

Please HELP

Question:

Ashraf went to a mall for shopping. He spends 1 hour on shopping. If he walks at speed of 20 km an hour, he returns to home at 7:00 PM. If he walks at 30 km an hour, he returns to home at 6:30 PM. How fast must he walk in order to return at 6:15 PM?

Options:
A)

20 kmph

B)

30 kmph

C)

40 kmph

D)

50 kmph

Solution:

Ans: (c)

  

Varun

· commented

· 1 Months ago

x=30 aayega

Yogesh

· commented

· 1 Months ago

consider initial time limit 45 min
-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