Dear all,

please post your doubts of Time and Distance sheet No 2 on the following forum

11 posts
• Page **1** of **2** • **1**, 2

Dear all,

please post your doubts of Time and Distance sheet No 2 on the following forum

please post your doubts of Time and Distance sheet No 2 on the following forum

- sadmin
- Site Admin
**Posts:**228**Joined:**Wed Feb 13, 2013 5:51 am

Sir,

Can u please post the challenge question which you gave at the end of the class ? I think i missed something.

Can u please post the challenge question which you gave at the end of the class ? I think i missed something.

- tpsriram
**Posts:**39**Joined:**Wed Feb 27, 2013 8:43 am

Kindly clear the idea of a billiards game, i have no knowledge about the game, relating to questions 29 to 30(subjective);

Can we co-relate this to some other concept??

Can we co-relate this to some other concept??

- Seema Devi
**Posts:**9**Joined:**Thu Feb 28, 2013 7:13 am

The billiards game questions are based on basic theory of ratios - I shall tackle a couple of questions in the regular classes once we resume them next week - this week is set aside for the back up classes - i n the mean time if you want I can solve a couple of questions on the forum

- sadmin
- Site Admin
**Posts:**228**Joined:**Wed Feb 13, 2013 5:51 am

Question 24 in the practice question, the train A takes 4hrs 48 min and train B takes 3 hrs 20 mins after crossing each other, its given that speed of A is 45km/hr, so speed of B must be more than speed of A, none of the options states that. Am i correct?? If i am wrong kindly clarify my error, i solved the question and got an answer as of Speed of B= 54.

It will be better you discuss it once in class, then i will try them.

It will be better you discuss it once in class, then i will try them.

- Seema Devi
**Posts:**9**Joined:**Thu Feb 28, 2013 7:13 am

The challenge question

Three cars leave A for B in equal time intervals. They reach B simultaneously and then leave for point C , which is 120 km from B. The first car arrives there an hour after the second car, and, the third car, having reached C, immediately reverses the direction and 40 km from C meets the first car. Find the speed of the first car.

Three cars leave A for B in equal time intervals. They reach B simultaneously and then leave for point C , which is 120 km from B. The first car arrives there an hour after the second car, and, the third car, having reached C, immediately reverses the direction and 40 km from C meets the first car. Find the speed of the first car.

- sadmin
- Site Admin
**Posts:**228**Joined:**Wed Feb 13, 2013 5:51 am

Question 24 in the practice questions of the sheet, the train A takes 4hrs 48 min and train B takes 3 hrs 20 mins after crossing each other, its given that speed of A is 45km/hr, so speed of B must be more than speed of A, none of the options states that. Am i correct?? If i am wrong kindly clarify my error, i solved the question and got an answer as of Speed of B= 54.

- Seema Devi
**Posts:**9**Joined:**Thu Feb 28, 2013 7:13 am

q 24 -- A gives B 40m start means, if A covers 1000m , B covesr 960m and

A wins by 19 seconds...means B took 19 second to cover this difference of 40 m... hence speed of B will be... 40m/19sec= 2.1m/s

so B will take how much time to cover 1000m ? = d/s=476 sec

now about A ... if A gives 30 sec to B then B wins by 40 m ,,,means if A starts 30 sec late, B will be 1000 and A will be 960m...

ans as B took 476 sec to travel 1000m , A has taken 476-30 sec= 446 sec , to travel 960 m

so time taken to travel 1000m will be == 446 /960 *1000 sec= 464 sec

is my approach right or totally wrong? i m NOT getting 125 and 150 sec.

q 25-- A , B , C travels in circumference of 12 km at speeds 3,7,13 respectively

1) they meet each other at -- lcm of 12/4,12/6,12/10=1.2 hours

2) they meet each other at start = lcm of 12/3,12/7,12/13=6.2hours

i m not getting 6 and 12 hours... q 26 - sir i checked the maths refernce book of pf, in that the same q was given , but I DONT KNOW , how to calculate GCM? I know only to calculate lcm ,hcf. q-23--of practice questions----please solve sir once ,

A wins by 19 seconds...means B took 19 second to cover this difference of 40 m... hence speed of B will be... 40m/19sec= 2.1m/s

so B will take how much time to cover 1000m ? = d/s=476 sec

now about A ... if A gives 30 sec to B then B wins by 40 m ,,,means if A starts 30 sec late, B will be 1000 and A will be 960m...

ans as B took 476 sec to travel 1000m , A has taken 476-30 sec= 446 sec , to travel 960 m

so time taken to travel 1000m will be == 446 /960 *1000 sec= 464 sec

is my approach right or totally wrong? i m NOT getting 125 and 150 sec.

q 25-- A , B , C travels in circumference of 12 km at speeds 3,7,13 respectively

1) they meet each other at -- lcm of 12/4,12/6,12/10=1.2 hours

2) they meet each other at start = lcm of 12/3,12/7,12/13=6.2hours

i m not getting 6 and 12 hours... q 26 - sir i checked the maths refernce book of pf, in that the same q was given , but I DONT KNOW , how to calculate GCM? I know only to calculate lcm ,hcf. q-23--of practice questions----please solve sir once ,

- gaurav1399
**Posts:**48**Joined:**Wed Feb 27, 2013 7:49 am

To solve the above question - on your copy draw a schematic diagram with X and Y as end points from which the trains start and Z as the meeting point of the two trains such that Z lies between A and B.

Train A takes = 4 hours and 48 minutes to reach Y after meeting train B at Z = 4.8 hours

Train B takes = 3 hours and 20 minutes to reach X after meeting A at Z= 10/3 HOURS

A simple analysis tells us that the time taken by train A to travel from X to Z will be equal to the time taken by train B to travel the distance YZ. ....Point 1

XZ = Sb X 10/3 where Sa IS THE speed of train A ....2

YZ = Sa X 4.8 .......................................................3

XZ/Sa =YZ/Sb

substituting the values of XY and YZ from 2 and 3 above we get

Sb X 10/3 /Sb = Sa X4.8/Sb

on solving the above equation we get the value

Sa/Sb =5/6

45/Sb =5/6 or Sb =54 kmph

Train A takes = 4 hours and 48 minutes to reach Y after meeting train B at Z = 4.8 hours

Train B takes = 3 hours and 20 minutes to reach X after meeting A at Z= 10/3 HOURS

A simple analysis tells us that the time taken by train A to travel from X to Z will be equal to the time taken by train B to travel the distance YZ. ....Point 1

XZ = Sb X 10/3 where Sa IS THE speed of train A ....2

YZ = Sa X 4.8 .......................................................3

XZ/Sa =YZ/Sb

substituting the values of XY and YZ from 2 and 3 above we get

Sb X 10/3 /Sb = Sa X4.8/Sb

on solving the above equation we get the value

Sa/Sb =5/6

45/Sb =5/6 or Sb =54 kmph

- sadmin
- Site Admin
**Posts:**228**Joined:**Wed Feb 13, 2013 5:51 am

In this question your observation is absolutely correct. The question should be read as " The speed of train B =45 kmph"

Thanks for point that out

Thanks for point that out

- sadmin
- Site Admin
**Posts:**228**Joined:**Wed Feb 13, 2013 5:51 am

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