- Staff Selection Commission Mathematics 1999 to 2017 - TIME AND DISTANCE Part 3

# Staff Selection Commission Mathematics - TIME AND DISTANCE TYPE-III

Ans .

(3) 12 km.

Explanation :

Let the distance be x km.
$$\frac{x}{10}$$-$$\frac{x}{12}$$=$$\frac{12}{60}$$
x=$$\frac{1}{5}$$*60
= 12 km.

Ans .

(1) 60 km

Explanation :

Let the distance between stations be x km, then speed of train
=$$\frac{x}{\frac{45}{60}}$$=$$\frac{4x}{3}$$
$$\frac{3x}{4x-15}$$=$$\frac{4}{5}$$
16x – 60 = 15x
x = 60 km

Ans .

(1) 13.33 minutes

Explanation :

Speed of train =$$\frac{Distance}{Time}$$=$$\frac{10}{\frac{12}{60}}$$
= 50 kmph
New speed = 45 kmph
Required time =$$\frac{10}{45}$$
$$\frac{2}{9}$$*60 minutes
=$$\frac{40}{3}$$=13.33 minutes.

Ans .

(1) 4 km.

Explanation :

Let the distance of the office
be x km, then $$\frac{x}{5}$$-$$\frac{x}{6}$$=$$\frac{8}{60}$$
x = 2 × 2 = 4 km

Ans .

(2) 4 km.

Explanation :

Let the distance of school be x km,
then $$\frac{x}{3}$$-$$\frac{x}{4}$$=$$\frac{20}{60}$$
$$\frac{x}{12}$$=$$\frac{1}{3}$$
x= 4 km

Ans .

(3) 20 minutes

Explanation :

Distance between stations X and
Y = Speed × Time
= 55 × 4 = 220 km.
New speed = 55 + 5 = 60 kmph
Required time =$$\frac{220}{60}$$=$$\frac{11}{3}$$
= 3 hours 40 minutes.
= 4 hours – 3 hours 40 minutes
= 20 minutes

Ans .

(3) 1 hour

Explanation :

Distance of journey = x km
Difference of time = 12 – 3 = 9 minutes
$$\frac{9}{60}$$ hr=$$\frac{3}{20}$$hr
$$\frac{x}{70}$$-$$\frac{x}{80}$$=$$\frac{3}{20}$$
$$\frac{x}{56}$$=$$\frac{3}{2}$$
x=84 km
Required correct time $$\frac{84}{70}$$hr-12 minutes
=72 – 12 = 60 minutes
= 1 hour

# TYPE-10

Ans .

(4) 45 km/hr

Explanation :

Let the length of train be x me- tres
\ According to question
Speed of the train =$$\frac{x}{10}$$m / sec
Also, the speed of the train $$\frac{x+50}{14}$$m / sec. It passes the platform in 14 seconds]
Both the speeds should be equal, i.e.,
$$\frac{x}{10}$$=$$\frac{x+50}{14}$$
or 14x = 10x + 500
or 14x – 10x = 500
or 4x = 500
\ x = 125 metres
Hence, Speed =$$\frac{125}{10}$$= 12 . 5 m / sec
$$\frac{12.5*18}{5}$$km / hr .
= 45 km/hr.

Ans .

(2) 176

Explanation :

Let length of train be x m
Speed of train $$\frac{x+264}{20}$$
Also, speed of train =$$\frac{x}{8}$$
$$\frac{x}{8}$$=$$\frac{x+264}{20}$$
5x = 2x + 528
5x – 2x = 528
x = 528 ÷ 3 = 176 m

Ans .

(4) 79.2 km/hr

Explanation :

Let the length of train be x me- tres.
Then, speed of train when it passes a telegraph post = $$\frac{x}{8}$$m/sec and speed of train, when it passes the bridge =$$\frac{x+264}{20}$$
Clearly,
$$\frac{x}{8}$$=$$\frac{x+264}{20}$$
5x = 2x + 528
3x = 528
x=176 m
Speed of train
$$\frac{176}{8}$$= 22 m/sec
22*$$\frac{18}{5}$$kmph
= 79.2 kmph

Ans .

(1) 25.2 km/hour.

Explanation :

Let the length of train be x metres.
When the train crosses the standing man, its speed = $$\frac{x}{9}$$
When the train crosses the plat- form of length 84 m, its speed $$\frac{x+84}{21}$$
Obviously,$$\frac{x+84}{21}$$=$$\frac{x}{9}$$
21x – 9x = 9 × 84
12x = 9 × 84
x=63 m
Required speed =$$\frac{63}{9}$$=$$\frac{63}{9}$$*$$\frac{18}{5}$$= 25.2 kmph

Ans .

(4)45 km/hr.

Explanation :

Suppose length of train be x
According to question
$$\frac{x+50}{14}$$ = $$\frac{x}{10}$$14x = 10x + 500
4x = 500
x=125 m
Therefore, speed $$\frac{125}{10}$$*$$\frac{18}{5}$$= 45 kmph

Ans .

(4) 21.6 kmph.

Explanation :

Let the length of the train be x
According to the question,
Speed of the train $$\frac{x+90}{30}$$=$$\frac{x}{15}$$
x + 90 = 2x
x = 90 m
Speed of train =$$\frac{90}{15}$$
6 m/s =6*$$\frac{18}{5}$$= 21.6 kmph

Ans .

(3) 20 seconds.

Explanation :

Let the length of the train be x metre
Speed of train when it crosses man=$$\frac{x}{10}$$
Speed of train when it crosses platform =$$\frac{x+300}{25}$$
According to the question,
Speed of train=$$\frac{x}{10}$$=$$\frac{x+300}{25}$$
25x = 10x + 3000
15x = 3000
x=200 m
Length of train = 200 metre
Speed of train= $$\frac{x}{10}$$=$$\frac{200}{10}$$= 20 m/sce
Time taken in crossing a 200m long platform = $$\frac{200+200}{20}$$= 20 seconds

Ans .

(4) 330 m

Explanation :

Let the length of the train be x metres.
Speed of train in crossing boy = $$\frac{x}{30}$$
Speed of train in crossing platform $$\frac{x+110}{40}$$
According to the question,
$$\frac{x}{30}$$=$$\frac{x+110}{40}$$
4x = 3x + 330
x = 330 metres

Ans .

(3) 150

Explanation :

Let the length of train be x metre
$$\frac{x}{15}$$=$$\frac{x+100}{25}$$
5x = 3x + 300
2x = 300
x=150 metres

Ans .

(2) 40, 30

Explanation :

Ans .

(2) 52 km/hr, 26 km/hr

Explanation :

Let the speed of trains be x and y metre/sec respectively,
$$\frac{100+95}{x-y}$$=27
x-y=$$\frac{65}{9}$$..(1)
$$\frac{195}{x+y}$$=9
x+y=$$\frac{195}{9}$$...(2)
By equation (i) + (ii)
2x= $$\frac{65}{9}$$+$$\frac{195}{9}$$=$$\frac{260}{9}$$
x = $$\frac{130}{9}$$m/sec.
$$\frac{130}{9}$$*$$\frac{18}{5}$$kmph = 52 kmph
From equation (ii),
y = $$\frac{65}{9}$$m/sec
$$\frac{65}{9}$$*$$\frac{18}{5}$$
= 26 kmph

Ans .

(2) 130.

Explanation :

Let the length of train be x me- tre, then
\ Speed of train
$$\frac{x}{7}$$=$$\frac{x+390}{28}$$
x=$$\frac{390}{3}$$
= 130 metres

Ans .

(3) 15.5 seconds.

Explanation :

Speed of train = 36 kmph
36 * $$\frac{5}{18}$$= 10 m/sec
Length of train = 10 × 10
= 100 metres
Required time= $$\frac{100+55}{10}$$
= 15.5 seconds

Ans .

(2) 300.

Explanation :

Speed of train = 60 kmph
60*$$\frac{5}{18}$$m/sec. =$$\frac{50}{3}$$m/sec.
If the length of platform be
= x metre, then
Speed of train=$$\frac{Length of (train + platform)}{Time taken in crossing}$$
50 × 10 = 200 + x
x = 500 – 200 = 300 metre

Ans .

(4) 9:24 am

Explanation :

Let both trains meet after t hours since 7 a.m.
Distance between stations A and B = x Km.
$$\frac{x}{4}$$*t+$$\frac{x}{\frac{7}{2}}$$*(t-1)=x
Speed=$$\frac{Distance}{Time}$$
$$\frac{7t+8t-8}{28}$$=1
15 t – 8 = 28
15 t = 28 + 8 = 36
t=2 hours 24 minutes
Required time = 9 :24 a.m.

Ans .

(2) 12.1 seconds.

Explanation :

Speed of train = 72 kmph.
$$\frac{72*5}{18}$$m/sec
= 20 m./sec.
Required time
$$\frac{Length of train and bridge}{Speed of train}$$
$$\frac{242}{20}$$
= 12.1 seconds

Ans .

(2) 6 seconds.

Explanation :

Relative speed of train
= (60 + 6) kmph.
$$\frac{66*5}{18}$$m/sec
=$$\frac{55}{3}$$m/sec.
Length of train = 110 metre
Required time =$$\frac{110}{\frac{55}{3}}$$
= 6 seconds

# TYPE-11

Ans .

(2) 20 m.

Explanation :

Let the time taken to complete the race by A,B, and C be x min- utes
Speed of A =$$\frac{1000}{x}$$
B =$$\frac{1000-50}{x}$$ =$$\frac{950}{x}$$
C =$$\frac{1000-69}{x}$$= =$$\frac{931}{x}$$
Now, time taken to complete the race by
B=$$\frac{1000}{\frac{950}{x}}$$=$$\frac{1000*x}{950}$$
and distance travelled by C in
$$\frac{1000x}{950}$$min
$$\frac{1000x}{950}$$*$$\frac{931}{x}$$= 980 km.
B can allow C
= 1000 – 980 = 20 m

Ans .

(4) 12 minutes

Explanation :

Ratio of the speed of A, B and
C = 6 : 3 : 1
Ratio of the time taken
=$$\frac{1}{6}$$:$$\frac{1}{3}$$:1=1:2:6
Time taken by A
$$\frac{72}{6}$$= 12 minutes

Ans .

(1) 17.24 seconds.

Explanation :

Let A take x seconds in covering
1000m and b takes y seconds According to the question,
x+20=$$\frac{900}{1000}y$$
x+20=$$\frac{9}{10}y$$...(1)
$$\frac{950}{1000}$$x + 25 = y....(2)
From equation (i),
$$\frac{10x}{9}$$+$$\frac{200}{9}$$=y
$$\frac{10x}{9}$$+$$\frac{200}{9}$$=$$\frac{950}{1000}$$x + 25
$$\frac{200x-171x}{180}$$=$$\frac{225-200}{9}$$
$$\frac{29x}{180}$$=$$\frac{25}{9}$$
x=$$\frac{500}{29}$$=17.24

Ans .

(3) 14.4 kmph

Explanation :

Time taken by Kamal
$$\frac{100}{18*\frac{5}{18}}$$
= 20 seconds
Time taken by Bimal
= 20 + 5 = 25 seconds
Bimal’s speed = $$\frac{100}{25}$$=4 m/sec
=$$\frac{4*18}{5}$$=14.4 kmph.

Ans .

(1) 95 m..

Explanation :

When A runs 1000m, B runs 900m.
\ When A runs 500m, B runs 450 m.
Again, when B runs 400m, C runs 360 m.
\ When B runs 450m, C runs
$$\frac{360}{400}$$*450 = 405 metres
Required distance = 500 – 405
= 95 metres

Ans .

(1) 11.9 metre

Explanation :

According to the question,
\ When A runs 800 metres, B runs 760 metres
\ When A runs 200 metres, B runs= $$\frac{760}{800}$$*200= 190 metres
Again, when B runs 500 metres, C runs 495 metres.
\ When B runs 190 metres, C runs =$$\frac{495}{500}$$*190= 188.1 metres
Hence, A will beat C by 200 – 188.1 = 11.9 metres in a race of 200 metres.

Ans .

(3) 29 metres..

Explanation :

According to the question,
Q When B runs 200 m metres, A runs 190 metres
\ When B runs 180 metres, A runs=$$\frac{190}{200}$$*180= 171 metres When C runs 200m, B runs 180 metres.
Hence, C will give a start to A by = 200 – 171 = 29 metres

Ans .

(1) 31.25 metre

Explanation :

According to the question,
When A covers 1000m, B covers = 1000 – 40 = 960 m
and C covers =1000 – 70 = 930 m
When B covers 960m, C covers 930 m.
\ When B covers 1000m, C covers=$$\frac{930}{960}$$*1000 = 968.75 metre
Hence, B gives C a start of = 1000 – 968.75 = 31.25 metre

Ans .

(2) 20 min.

Explanation :

Relative speed
= 95 – 75 = 15 kmph
Required Time=$$\frac{Distance}{Relative speed}$$
$$\frac{5}{15}$$*60
=20 minutes

Ans .

(1) 15 minutes

Explanation :

Time taken by C = t hours
Time taken by B =$$\frac{t}{3}$$hours
Time taken by A =$$\frac{t}{6}$$hours
Here,t=$$\frac{3}{2}$$hours
Required time taken by A
$$\frac{3}{\frac{2}{6}}$$=$$\frac{1}{4}$$
$$\frac{1}{4}$$*60= 15 minutes

Ans .

(2) 40 min

Explanation :

2 hours 45 minutes
2 + $$\frac{45}{60}$$hours=$$\frac{11}{4}$$hours
Distance = Speed × Time
4 * $$\frac{11}{4}$$= 11 km.
Time taken in covering 11 km at 16.5 kmph
=$$\frac{11}{16.5}$$
=40 minutes.

Ans .

(2) 35 km.

Explanation :

Let the total distance be x km.
Time =$$\frac{Distance}{Speed}$$
According to the question,
$$\frac{10}{6}$$ +$$\frac{20}{16}$$ +$$\frac{x-30}{3}$$=4 $$\frac{35}{60}$$
=4$$\frac{7}{12}$$
$$\frac{5}{3}$$ +$$\frac{5}{4}$$ +$$\frac{x}{3}$$-10 =$$\frac{55}{12}$$
x=$$\frac{140}{12}$$*3=35 km

Ans .

(1) 1 hour.

Explanation :

Usual time = x minutes
New time =$$\frac{4x}{3}$$
Speed ∝ $$\frac{1}{Time}$$
According to the question,
$$\frac{4x}{3}$$– x = 20
x= 1hour

Ans .

(2) 4 km/hr.

Explanation :

Let, A’s speed = x kmph.
\ B’s speed = (7 – x) kmph
Time=$$\frac{Distance}{Speed}$$
According to the question,
$$\frac{24}{x}$$+$$\frac{24}{7-x}$$=14
$$\frac{24*7}{x(7-x)}$$=14
x (7 – x) = 12 = 4 × 3 or 3 × 4
Þ x (7 – x) = 4 (7 – 4) or 3 (7 – 3)
Þ x = 4 or 3
\ A’s speed = 4 kmph.

Ans .

(3) 2 hours

Explanation :

Relative speed
= 12 + 10 = 22 kmph Distance covered
= 55 – 11 = 44 km
\ Required time
$$\frac{44}{22}$$
= 2 hours

Ans .

(2) 200.

Explanation :

Required time = LCM of 40
and 50 seconds
= 200 seconds

Ans .

(1) 2 km.

Explanation :

Distance between starting
point and multiplex = x metre
Time =$$\frac{Distance}{Speed}$$
According to the question,
$$\frac{x}{3}$$ -$$\frac{x}{4}$$= $$\frac{5+5}{60}$$
$$\frac{x}{12}$$ =$$\frac{1}{6}$$
x=2 km

# TYPE-12

Ans .

(2) 19 minutes

Explanation :

Two ways walking time = 55 min...(i)
One way walking + One way riding time = 37 min.....(ii)
By 2 × (ii) – (i),
2 ways riding time = 2×37–55 = 19 minutes.

Ans .

(3) 50 km.

Explanation :

Let the distance be x km
Time taken by A =$$\frac{x}{40}$$hrs
Time taken by B =$$\frac{x}{50}$$hrs
$$\frac{x}{40}$$-$$\frac{x}{50}$$=$$\frac{15}{60}$$
$$\frac{5x-4x}{200}$$=$$\frac{15}{60}$$
x=50 km

Ans .

(1) 5 km/hr.

Explanation :

Let the speed of man be x kmph
30 x – 30{x-$$\frac{x}{15}$$} = 10
30{x-x+$$\frac{x}{15}$$}=10
$$\frac{x}{15}$$=$$\frac{10}{30}$$
x= 5kmph

Ans .

(1) 46 minutes 12 seconds

Explanation :

Required time = LCM of 252,
308 and 198 seconds.
Now, 252 = 2 × 2 × 3 × 3 × 7
308 = 2 × 2 × 7 × 11
198 = 2 × 3 × 3 × 11
\ LCM = 2 × 2 × 3 × 3 × 7 × 11
= 36 × 77 seconds
$$\frac{36*77}{60}$$minutes
= 46 minutes 12 seconds

Ans .

(4) 6 hours

Explanation :

Suppose, time taken while walking be x hours And, time taken on riding be y hours
\ According to question
x+y=4$$\frac{1}{2}$$hr
Then, 2y = 3 hours
y=1$$\frac{1}{2}$$hr
x=4$$\frac{1}{2}$$ - 1$$\frac{1}{2}$$=3 hr
Time required to walk both ways = 6 hours

Ans .

(4) 9 km

Explanation :

Let the required distance be x km
$$\frac{x}{\frac{9}{2}}$$+$$\frac{x}{3}$$=5
x{$$\frac{2+3}{9}$$=5
x=9 km

Ans .

(4) 26.7 km.

Explanation :

Distance covered by A in 4
hours = 4 × 4 = 16 km
Relative speed of B with respect
to A = 10 – 4 = 6 km/hr
Time taken to catch A
$$\frac{16}{6}$$=$$\frac{8}{3}$$hr
Required distance
=$$\frac{8}{3}*10$$
= 26.67 km.= 26.7 km

Ans .

(2) 480 km.

Explanation :

Suppose distance be x km
$$\frac{x}{2*40}$$+$$\frac{x}{2*60}$$=10
$$\frac{3x+2x}{240}$$=10
x=480 km

Ans .

(1) 2 min 25 sec.

Explanation :

If A covers the distance of 1 km in x seconds, B covers the distance of 1 km in (x + 25) sec- onds. If A covers the distance of 1 km, then in the same time C covers only 725 metres.
If B covers 1 km in (x + 25) sec- onds, then C covers 1 km in (x + 55) seconds.
Thus in x seconds, C covers the distance of 725 m.
$$\frac{x}{725}$$*1000=x+55
x = 145
A co vers the di stance of 1 km in 2 minutes 25 seconds.

Ans .

(4) 250 m

Explanation :

Ans .

(2) 205 seconds

Explanation :

A beats B by 30 seconds and B beats C by 15 seconds. Clearly, A beats C by 45 seconds. Also, A beats C by 180 metres. Hence, C covers 180 metres in 45 seconds
Speed of C =$$\frac{180}{45}$$= 4 m/sec
Time taken by C to cover 1000 m =$$\frac{1000}{4}$$=250 sec
Time taken by A to cover 1000 m = 250–45 = 205 sec

Ans .

(2) 27.

Explanation :

Difference of time
= 6 min. – 5 min. 52 sec.
= 8 seconds
Distance covered by man in 5 min. 52 seconds
= Distance covered by sound in 8 seconds
= 330 × 8 = 2640 m.
\ Speed of man
$$\frac{2640 m}{5 min. 52 sec.}$$=$$\frac{2640}{352}$$
$$\frac{2640}{352}$$*$$\frac{18}{5}$$kmph
= 27 kmph

Ans .

(1) 70 km

Explanation :

Let the required distance be x km.
Difference of time
= 15 + 5 = 20 minutes
=$$\frac{1}{3}$$hr
According to the question,
$$\frac{x}{35}$$-$$\frac{x}{42}$$=$$\frac{1}{3}$$
$$\frac{x}{210}$$=$$\frac{1}{3}$$
x=70 km

Ans .

(3) 7 hours 30 minutes

Explanation :

1-$$\frac{5}{6}$$ of time taken by B
=1 hour 15 minutes
\ Time taken by B
= 1 hour 15 minutes × 6
=7 hours 30 minutes

Ans .

(1) 5

Explanation :

Abhay’s speed = x kmph
Sameer’s speed = y kmph
$$\frac{30}{x}$$-$$\frac{30}{y}$$=12
$$\frac{30}{y}$$-$$\frac{30}{2x}$$=1
$$\frac{30}{y}$$-$$\frac{30}{2x}$$=3
$$\frac{30}{2x}$$=3
x = 5 kmph

Ans .

(3) 4 hours 45 minutes

Explanation :

Time taken in walking both ways = 7 hours 45 minutes ....(i)
Time taken in walking one way and riding back = 6 hours 15 minutes ....(ii)
By equation (ii) × 2 – (i), we have
Time taken by the man to ride both ways
= 12 hours 30 minutes – 7 hours
45 minutes
= 4 hours 45 minutes

Ans .

(1) 25 km/hr.

Explanation :

Let the total distance be 100 km.
Average speed
$$\frac{Total distance covered}{Time taken}$$
$$\frac{100}{ \frac{30}{20}+ \frac{60}{40}+ \frac{10}{10}}$$
$$\frac{100*2}{8}$$=25 kmph

Ans .

(2) 6 km/hr

Explanation :

Let the speed of A = x kmph and that of B = y kmph
According to the question,
x × 6 + y × 6 = 60
x + y = 10
and $$\frac{2}{3}$$x* 5 + 2 y * 5 = 60
10x + 30y = 180
x + 3y = 18 ...(ii)
From equations (i) × (3) – (ii)
3x + 3y – x – 3y = 30 – 18
12x = 12
x = 6 kmph.

Ans .

(2) 81 km

Explanation :

Let the trains meet after t hours, then
24t – 18t = 27
6t = 27
t=$$\frac{9}{2}$$hours
QR = 18t = 18 * $$\frac{9}{2}$$= 81 km

Ans .

(3) 8 km/hr

Explanation :

Let the speed of Ravi be x kmph
then, Ajay’s speed = (x + 4) kmph
Distance covered by Ajay
= 60 + 12 = 72 km
Distance covered by Ravi
= 60 – 12 = 48 km.
According to the question
$$\frac{72}{x+4}$$=$$\frac{48}{8}$$
$$\frac{3}{x+4}$$=$$\frac{2}{x}$$
3x = 2x + 8
x = 8 kmph

Ans .

(2) 16 km

Explanation :

Let man walked for t hours.
then, t × 4 + (9 – t) × 9 = 61
4t + 81 – 9t = 61
81 – 5t = 61
5t = 20
t = 4
Distance travelled on foot
= 4 × 4 = 16 km.

Ans .

(1) 6

Explanation :

Let the required distance be x km, then
$$\frac{x}{5}$$-$$\frac{x}{6}$$=$$\frac{1}{5}$$
$$\frac{x}{30}$$=$$\frac{1}{5}$$ x=6 km

Ans .

(4) 6 km.

Explanation :

Let the required distance be x km.
$$\frac{x}{3}$$-$$\frac{x}{4}$$=$$\frac{30}{60}$$
$$\frac{x}{12}$$=$$\frac{1}{2}$$
x= 6km

Ans .

(2) 60 km/hr

Explanation :

Let the speed of train be x kmph and that of car be y kmph, then
$$\frac{60}{x}$$+$$\frac{240}{y}$$=4
and ,$$\frac{100}{x}$$+$$\frac{200}{y}$$=$$\frac{25}{6}$$
$$\frac{4}{x}$$+$$\frac{8}{y}$$=$$\frac{1}{6}$$
By equation (i ) – equation (ii) × 30
$$\frac{60}{x}$$+$$\frac{240}{y}$$-$$\frac{120}{x}$$-$$\frac{240}{y}$$=4-5
-$$\frac{60}{x}$$=-1
x=60 kmph

Ans .

(2) 38 minutes.

Explanation :

Ratio of the speed of A and B
= A : B = 2 : 1 = 6 : 3
B : C = 3 : 1
A : B : C = 6 : 3 : 1
Ratio of their time taken
$$\frac{1}{6}$$:$$\frac{1}{3}$$:1=1:2:6
Time taken by B
$$\frac{2}{6}$$*114 minutes
= 38 minutes

Ans .

(3) 54 km/hr

Explanation :

Ans .

(2) 800

Explanation :

Total distance of trip
=$$\frac{1200*5}{2}$$= 3000 km
Part of journey covered by train
1-$$\frac{2}{5}$$-$$\frac{1}{3}$$=$$\frac{4}{15}$$
Distance covered by train
3000*$$\frac{4}{15}$$= 800 km

Ans .

(1) 1.6 minutes

Explanation :

A’s speed =$$\frac{1000}{5}$$ = 200 m/minute
B’s speed =$$\frac{1000}{8}$$ = 125 m/minute
C’s speed =$$\frac{1000}{10}$$ = 100 m/minute
Distance covered by C in 2 min- utes = 200 metre
Distance covered by B in 1 minute = 125 metre
Relative speed of A with respect to C = 100 metre
Time=$$\frac{200}{100}$$= 2 minutes
Relative speed of A with respect
ot B = 75 metre
Time=$$\frac{125}{75}$$= $$\frac{5}{3}$$ minutes =1.6minutes

Ans .

(2) v 1 : v 2 = 4 : 5.

Explanation :

Ans .

(1) 1.20 am..

Explanation :

Time taken in covering 999km
$$\frac{999}{55.5}$$= 18 hours
Required time = 18 hours + 1
hour 20 minutes
= 19 hours 20 minutes
i.e. 1 : 20 am

Ans .

(1) 12.5 metre/seccond.

Explanation :

Speed = 45 kmph
$$\frac{45*1000}{60*60}$$metre/second
$$\frac{45*5}{18}$$metre/second
= 12.5 metre/second

Ans .

(1) 610 m..

Explanation :

Distance covered in 2nd
minute = 90 – 50 = 40 metre
Distance covered in 3rd minute
= 130 – 90 = 40 metre
\ Required distance
= 50 + 40 × 14
= 50 + 560 = 610 metre

Ans .

(3) 7: 56 AM

Explanation :

Here distance is constant.
Speed ∝ $$\frac{1}{Time}$$
Ratio of the speeds of A and B
$$\frac{\frac{7}{2}}{4}$$=7:8
A’s speed = 7x kmph (let)
B’s speed = 8x kmph
AB = 7x × 4 = 28x km.
Let both trains cross each other after t hours from 7 a.m. According to the question,
7x (t + 2) + 8x × t = 28x
7t + 14 + 8t = 28
15t = 28 – 14 = 14
t=$$\frac{14}{15}$$hours
$$\frac{14}{15}$$*60
=56 minutes.
Required time = 7 : 56 A.M.

Ans .

(4) 9 hours

Explanation :

Speed of plane = $$\frac{Distance}{Time}$$
=$$\frac{6000}{8}$$= 750 kmph
New speed = (750 + 250) kmph
= 1000 kmph
Required time =$$\frac{9000}{1000}$$
= 9 hours

Ans .

(1) 45

Explanation :

Let speed of train be x kmph.
Speed of car = y kmph.
Case I,
Time = $$\frac{Distance}{Speed}$$
$$\frac{240}{x}$$ + $$\frac{210}{y}$$=8$$\frac{2}{3}$$
$$\frac{240}{x}$$ + $$\frac{210}{y}$$=8$$\frac{26}{3}$$...(1)
Case II,
$$\frac{180}{x}$$+$$\frac{270}{y}$$=9....(2)
By equation (i) × 3 – (ii) × 4,
$$\frac{720}{x}$$ +$$\frac{630}{y}$$ -$$\frac{720}{x}$$ -$$\frac{1080}{y}$$
= 26 – 36
=$$\frac{-450}{y}$$= –10
= –10
y = 45 kmph

Ans .

(3) 81 km/h.

Explanation :

Difference of time = 11 min- utes 45 seconds – 11 minutes = 45 seconds
Distance covered by sound in 45 seconds = Distance covered by train in 11 minutes
330 × 45 = 11 × 60 × Speed of train
Speed of train
$$\frac{330*45}{11*60}$$m/sec.
$$\frac{45}{2}$$ $$\frac{18}{5}$$
=81 kmph

Ans .

(2) 45

Explanation :

Distance covered in 3 hours
36 minutes i.e. 3$$\frac{36}{60}$$hours
=5$$\frac{18}{5}$$= 18 km.
Time taken at 24 kmph.
$$\frac{18}{24}$$hours
$$\frac{18}{24}$$*minutes
=45 minutes

Ans .

(3) 2 hours

Explanation :

Let the origi nal speed of aeroplane be x kmph.
According to the question,
$$\frac{1200}{x-300}$$-$$\frac{1200}{x}$$=2
1200{$$\frac{x-x+300}{x(x-300)}$$}=2
x (x – 300) =$$\frac{1200*300}{2}$$
x (x – 300) = 600 × 300
x (x – 300) = 600 (600 – 300)
x = 600 kmph.
Scheduled duration of flight =$$\frac{1200}{600}$$= 2 hours

Ans .

(4) 240.

Explanation :

Consumption of petrol in cov- ering 540 km=$$\frac{540}{45}$$= 12 litres
Required expenses
= Rs. (12 × 20)
= Rs. 240

Ans .

(2) 6 cm

Explanation :

18 km o 1.5 cm
1 km= $$\frac{1.5}{18}$$cm
72=$$\frac{1.5*72}{18}$$cm = 6 cm

Ans .

(2) 16 km.

Explanation :

Length of journey on foot = x km. (let).
\ Length of journey on cycle = (61 – x ) km.
According to the question,
Time = $$\frac{Distance}{Speed}$$
$$\frac{x}{4}$$+$$\frac{61-x}{9}$$=9
$$\frac{9x+244-4x}{36}$$=9
5x + 244 = 36 × 9 = 324
5x = 324 – 244 = 80
x=16 km

Ans .

(1) 16 km.

Explanation :

Let the distance covered on foot be x km.
Distance covered on cycle = (61 – x) km.
Time=$$\frac{Distance}{Speed}$$
$$\frac{x}{4}$$+$$\frac{61-x}{9}$$=9
$$\frac{x}{4}$$-$$\frac{x}{9}$$=9-$$\frac{61}{9}$$
$$\frac{5x}{36}$$=$$\frac{20}{9}$$
x=16 km

Ans .

(4) 3300 metre

Explanation :

Distance = Speed × Time
= 330 × 10 = 3300 metre

Ans .

(2) 800 km.

Explanation :

Let total distance covered be 2x km.
Total time = 14 hours 40 min- utes
=14$$\frac{40}{60}$$hours=$$\frac{44}{3}$$hours
Time = $$\frac{Distance}{Speed}$$
According to the question,
$$\frac{x}{60}$$\)+$$\frac{x}{50}$$=$$\frac{44}{3}$$
$$\frac{11x}{300}$$=$$\frac{44}{3}$$
x=400
Total distance = 2x = 2 × 400 = 800 km

Ans .

(2) 80

Explanation :

Distance between both don- keys = 400 metre.
Relative speed = (3 + 2) m./sec.
= 5 m./sec.
Required time
=$$\frac{Distance}{Relative Speed}$$
=$$\frac{400}{5}$$= 80 seconds