Question

Match Column I with Column II

Column I		Column II
I	Moving $\frac{5}{6}$ th of its usual speed, a train is 10 minutes late. Find its usual time to cover the journey?	P	200m
II	Speed of a train in 20 meters per second. What is the length of train?	Q	1200m
III	A train 50 m long passes a platform 100m long in 10 seconds. The speed of the train in m/sec is	R	50min
IV	A man covers a certain distance at 36 km/h. How many meters does he cover in 2 minutes?	S	15sec

• A. I—Q; II—S; III—P; IV—R
• B. I—R , II— S III—R; IV—Q
• C. I—R ; II—P; III—S; IV—Q
• D. I—S; II—R; III—P; IV—Q

Answer 1

Answer: (C)

(I) Let speed be S and time to cover the journey be ' $\mathrm{T}$ '
$\mathrm{S} \times \mathrm{T}=\mathrm{D}$...(i)
Now, as per the question we have
$\left(\frac{5}{6}\right) s \times(T+10)=D$...(ii)
Equation value of $\mathrm{D}$ from (i) and (ii) we get $\mathrm{T}=50 \mathrm{~min}$
$\mathrm{T}=50 \mathrm{~min}$...(R)
(II) Length of train $=20 \times 10=200$ meters
(III) We have, speed of the train = length of the train + lenght of the platform
Time taken in crossing the platform
$=\frac{50+100}{10}=15 \mathrm{sec}$...(S)
(IV) $ \text { Speed }=\frac{\text { Distance }}{\text { time }} \rightarrow$
$ \text { Distance }=\text { speed } \times \text { time } $
$ \text { Speed }=\frac{36 \mathrm{~km}}{1 \text { hour }}=\frac{30000 \mathrm{~m}}{3600 \mathrm{~s}}=10 \mathrm{~m} / \mathrm{s} $
$\text { Time }=2 \text { minutes }=120 \mathrm{~s} $
$ \text { Distance }=\text { speed } \times \text { time } $
$ =10 \mathrm{~m} / \mathrm{s} \times 120 \mathrm{~s}=1200 \mathrm{~m} $....(Q)
$\mathrm{I}-(\mathrm{R}) ; \mathrm{II}-(\mathrm{P}) ; \mathrm{III}-(\mathrm{S}) ; \mathrm{IV}-(\mathrm{Q})$