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Ashutosh Singh trainer in Mumbai

Ashutosh Singh

Maths Teacher

Nalasopara East, Mumbai, India - 401208.

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Overview

I have been teaching mathematics since 2014. I teach mathematics to classes 6, 7, 8, 9, and 10.

Languages Spoken

English

Hindi

Education

Banara Hindu University 2004

Bachelor of Arts (B.A.)

IITTM Gwalior 2009

Master of Business Administration (M.B.A.)

Address

Nalasopara East, Mumbai, India - 401208

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Answers by Ashutosh Singh (1)

Answered on 09/01/2015 Learn Tuition/Class VI-VIII Tuition

There are two methods to solve this question. First:- Suppose Train A and train B meet after traveling the distance of X k.m. As we know Time =total distance/ speed So time taken by train A is X/60 And time taken by train B is X/75 But we know train B is faster and start after 2 hours of train... ...more
There are two methods to solve this question. First:- Suppose Train A and train B meet after traveling the distance of X k.m. As we know Time =total distance/ speed So time taken by train A is X/60 And time taken by train B is X/75 But we know train B is faster and start after 2 hours of train A. So train B take 2 hour less then train A So equation would be (X/60) –( X/75) = 2 (5X-4X)/300 = 2 X= 600 k.m. Second:- Suppose both trains meet after X hours. So distance travel by train A is 60X And distance travel by train B is 75X Because we know time*speed= total distance But we know that train B take 2 hour less in compare to train A So distance travel by train B= 75(X-2) So the equation would be 60X = 75(X-2) 60X= 75X-150 15X = 150 X = 10 Train A would take 600 k.m. distances to meet train B.
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Answers by Ashutosh Singh (1)

Answered on 09/01/2015 Learn Tuition/Class VI-VIII Tuition

There are two methods to solve this question. First:- Suppose Train A and train B meet after traveling the distance of X k.m. As we know Time =total distance/ speed So time taken by train A is X/60 And time taken by train B is X/75 But we know train B is faster and start after 2 hours of train... ...more
There are two methods to solve this question. First:- Suppose Train A and train B meet after traveling the distance of X k.m. As we know Time =total distance/ speed So time taken by train A is X/60 And time taken by train B is X/75 But we know train B is faster and start after 2 hours of train A. So train B take 2 hour less then train A So equation would be (X/60) –( X/75) = 2 (5X-4X)/300 = 2 X= 600 k.m. Second:- Suppose both trains meet after X hours. So distance travel by train A is 60X And distance travel by train B is 75X Because we know time*speed= total distance But we know that train B take 2 hour less in compare to train A So distance travel by train B= 75(X-2) So the equation would be 60X = 75(X-2) 60X= 75X-150 15X = 150 X = 10 Train A would take 600 k.m. distances to meet train B.
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Ashutosh Singh describes himself as Maths Teacher. He conducts classes in Class IX-X Tuition and Class VI-VIII Tuition. Ashutosh is located in Nalasopara East, Mumbai. Ashutosh takes at students Home and Online Classes- via online medium. He has 10 years of teaching experience . Ashutosh has completed Bachelor of Arts (B.A.) from Banara Hindu University in 2004 and Master of Business Administration (M.B.A.) from IITTM Gwalior in 2009. He is well versed in English and Hindi.

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