Sainik school Entrance coaching Academy Tamil nadu
Defence Academy Coimbatore A leading coaching academy for Sainik, Military and RIMC Schools, NDA, IMA, AFCAT SSB, FSB Interview http://www.ssbcoaching.com call 094437 20076
We offer Residential Coaching for RIMC- RMS- Sainik School- Entrance Exam Preparation from- Oct 2nd till the day of exam. |
All India Sainik - RMS- RIMC School Entrance Examination Coaching Academy
|
We offer:-
Duration:-
Three months for Sainik, 2.5 Molnths for RMS and 2 moinths for RIMC
Maths:
Problems on Trains
For sainik- RMS-RIMC Entrance Coaching and Self-Preparation Guides contact:-Defence Academy |
Cell: 094437 20076 or land line: 04224043076 |
We offer Residential Coaching for RIMC- RMS- Sainik School- Entrance Exam Preparation from- Oct 2nd till the day of exam. |
FOR SSB- NDA- CDSE- AFCAT Contact 094437 20076
⇒ Click to Order Online
⇒ Problems on Train Questions
Q1. A train moves with the speed of 180 km/hr. Find the
speed in m/sec ?
(a) 20 m/sec
(b) 30 m/sec
(c) 50 m/sec
(d) 70 m/sec
Answer:-
(c) 50 m/sec>
Solution:-
180 km/hr = 180 X (5/18) = 50 m/sec
Q2. A 150 metres long train moving at the speed of 126
km/hr. What time it will take to cross an electric pole ?
(a) 3.5 seconds
(b) 4.2 seconds
(c) 5 seconds
(d) 6 seconds
Answer: (b) 4.2 seconds
Solution:-
126 km/hr = 126 X (5/18) = 35 m/sec
So the time taken by the train = (150/35) = 4.2 sec
Q3. A train of length 250 meter long running at the speed
of 10 m/sec. How much time it will take to cross a platform
of 650 meter length ?
(a) 60 seconds
(b) 72 seconds
(c) 85 seconds
(d) 90 seconds
Answer: (d) 90 seconds
Solution:-
Total length to cross the platform = 250 + 650 = 900 meter
So the time taken by the train = (900/10) = 90 sec
Q4. A train with a uniform speed crosses a pole in 2
seconds and a bridge of length 250m in 7 seconds. The length
of the train is ?
(a) 85 metre
(b) 100 metre
(c) 105 metre
(d) 125 metre
Answer: (b) 100 metre
Solution:-
Let the length of the train is x metre
Speed of the train is,
x/2 = (x + 250)/7
7x = 2x + 500
x = 100 metre
Q5. A train of 260 m long passed a pole in 26 secconds.
How long will it take to pass a platform of 740 m long ?
(a) 78 seconds
(b) 96 seconds
(c) 100 seconds
(d) 115 seconds
Answer: (c) 100 seconds
Solution:-
Speed of the train = 260/26 = 10 m /sec
So required time to cross the platform is,
=(260 + 740 )/10
=100 sec
Q6. A train travels first 160 Km at 64 Km/hr and the next 160 Km at 80 Km/hr. The average speed for the first 320 Km of the tour is
(a) 57.24 km/hr
(b) 71.11 km/hr
(c) 78.32 km/hr
(d) 91 km/hr
Answer: (b) 71.11 km/hr
Solution:
Total time taken =( (160/64) + (160/80)) hrs= 9/2 hrs
Average Speed
=(320 X (2/9)) km/hr
=71.11 km/hr
For sainik- RMS-RIMC Entrance Coaching and Self-Preparation Guides contact:-Defence Academy |
Cell: 094437 20076 or land line: 04224043076 |
We offer Residential Coaching for RIMC- RMS- Sainik School- Entrance Exam Preparation from- Oct 2nd till the day of exam. |
FOR SSB- NDA- CDSE- AFCAT Contact 094437 20076
⇒ Click to Order Online
Paper - I - Mathematics and Language Ability- Class VI
Time: 2 Hours Maximum Marks 200
PART A MATHEMATICS- SECTION - 1
(Each question carries two marks)
(Each question carries two marks)
Q.1. Find the lowest Common multiple of 2, 4 and 5
Solution:-
Step 1:
Find the factors of 2, 4 and 5
Step 2:
The factors of 2 = 1 and 2
The factors of 4 = 1, 2 and 4 Or (1 *2 *2)
The factors of 5 = 1 and 5
Step 3:
Taking the common factors of these we get: 1 * 2 * 2* 5 Find the factors of 2, 4 and 5
Step 4:
To find the LCM of 2, 4 and 5 we will mulitply the common factors i.e.1* 2 *2 * 5 = 20
Step 5:
Therefore the LCM of 2, 4 and 5 is 20 (Answer)
Q.2. What least number should be added to 322516 so that the result is exactly divisible by 543?
Solution:-
Step 1:
First find: 322516 Divided by 543 = ?
Step 2: Method:
Divide and subtract the remainder from 543. the answer is to be added to 322516
On dividing 322516 by 543 we get a remainder of 517
To make the given number exactly divisible by 543 we have to add 26 to the Main number
Step 3:
This we get by subtractring the remainder from the number 543i.e 543-516 - 26
Step 4:
By adding we getg 322516 + 26 =322542, this is exactly divisible by 543
Step 5:
Therefore the Leat num ber to be added is 26 (Answer)
Find the factors of 2, 4 and 5
The factors of 2 = 1 and 2
Taking the common factors of these we get: 1 * 2 * 2* 5 Find the factors of 2, 4 and 5
To find the LCM of 2, 4 and 5 we will mulitply the common factors i.e.1* 2 *2 * 5 = 20
Therefore the LCM of 2, 4 and 5 is 20 (Answer)
Q.2. What least number should be added to 322516 so that the result is exactly divisible by 543?
Solution:-
First find: 322516 Divided by 543 = ?
Divide and subtract the remainder from 543. the answer is to be added to 322516
This we get by subtractring the remainder from the number 543i.e 543-516 - 26
By adding we getg 322516 + 26 =322542, this is exactly divisible by 543
Therefore the Leat num ber to be added is 26 (Answer)
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