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 $\mathrm{T}=50\min$
$\mathrm{T}=50\min$...(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
$=\frac{50+100}{10}=15\sec$...(S)
(IV) $\text{ Speed }=\frac{\text{ Distance }}{\text{ time }}\to$
$\text{ Distance }=\text{ speed }\times \text{ time }$
$\text{ Speed }=\frac{36\mathrm{km}}{1\text{ hour }}=\frac{30000\mathrm{m}}{3600\mathrm{s}}=10\mathrm{m}/\mathrm{s}$
$\text{ Time }=2\text{ minutes }=120\mathrm{s}$
$\text{ Distance }=\text{ speed }\times \text{ time }$
$=10\mathrm{m}/\mathrm{s}\times 120\mathrm{s}=1200\mathrm{m}$....(Q)
$\mathrm{I}-(\mathrm{R});\mathrm{II}-(\mathrm{P});\mathrm{III}-(\mathrm{S});\mathrm{IV}-(\mathrm{Q})$