- Staff Selection Commission Mathematics 1999 to 2017 - TIME AND DISTANCE Part 2

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

Ans .

(4) 400 m

Explanation :

Let the length of the train travelling at 48 kmph be x metres.
Let the length of the platform be y metres.
Relative speed of train = (48 + 42) kmph
$$\frac{90*5}{18}$$m/sec
= 25 m./sec. and 48 kmph
$$\frac{48*5}{18}$$=$$\frac{40}{3}$$m./sec.
According to the question,
$$\frac{x+\frac{x}{2}}{25}$$=12
$$\frac{3x}{2x*25}$$=12
3x = 2 × 12 × 25 = 600
x = 200 m.
Also,$$\frac{200+y}{\frac{40}{3}}$$=45
600 + 3y = 40 × 45
3y = 1800 – 600 = 1200
y=$$\frac{1200}{3}$$= 400 m.

Ans .

(2) 12 Noon

Explanation :

Let two trains meet after t hours when the train from town A leaves at 8 AM.
Distance covered in t hours at 70 kmph + Distance covered in (t – 2) hours at 110 kmph = 500km
70t + 110 (t – 2) = 500
70t + 110t – 220 = 500
180 t = 500 + 220 = 720
t=$$\frac{720}{180}$$=4 hours
Hence, the trains will meet at 12 noon.

Ans .

(3) 5 sec

Explanation :

Relative speed
= (68 + 40) kmph = 108 kmph
=$$\frac{108*5}{18}$$ m/s or 30 m/s
Required time
$$\frac{Sum of the lengths of both trains}{Relative speed}$$
$$\frac{70+80}{30}$$second = 5 seconds

Ans .

(3) 12

Explanation :

When a train crosses a telegraph post, it covers its own length.
Speed of first train=$$\frac{120}{10}$$= 12 m/sec.
Speed of second train=$$\frac{120}{15}$$= 8 m/sec.
Relative speed = 12 + 8
= 20 m/sec.
Required time
$$\frac{Total length of trains}{Relative speed}$$
=$$\frac{2*120}{20}$$= 12 seconds

Ans .

(3) 12 sec.

Explanation :

Relative speed = 42 + 48 = 90 kmph
$$\frac{90*5}{18}$$m/s = 25 m/s
Sum of the length of both trains = 137 + 163 = 300 metres
Required time =$$\frac{300}{25}$$= 12 seconds

Ans .

(1) 54 km/hr

Explanation :

Speed of second train = 43.2 kmph
=$$\frac{43.2*5}{18}$$ m/sec
Let the speed of first train be x m per second, then $$\frac{150 + 120}{x + 12}$$=10
27 = x + 12
x = 15 m/s=15*$$\frac{18}{5}$$kmph = 54 kmph

Ans .

(1) 444

Explanation :

Let the trains meet after t hours
Then, 21t – 16t = 60
5t = 60 Þ t = 12 hours
Distance between A and B = (16 + 21) × 12
= 37 × 12 = 444 miles

Ans .

(3) 8 sec

Explanation :

Relative speed = 45 + 54
= 99 kmph
99*$$\frac{5}{18}$$m/sec
Required time =$$\frac{108 + 112}{\frac{55}{2}}$$
=$$\frac{220*2}{55}$$= 8 seconds

Ans .

(3) 3.42 sec

Explanation :

Let the length of each train be x metres
Then, Speed of first train = $$\frac{x}{3}$$m/sec.
Speed of second train =$$\frac{x}{4}$$m/sec.
They are moving in opposite di- rections Relaive speed = $$\frac{x}{3}$$+ $$\frac{x}{4}$$
=$$\frac{4x+3x}{12}$$=$$\frac{7x}{12}$$m/sec
Total length = x + x = 2 x m.
Time taken = $$\frac{2x}{\frac{7x}{12}}$$=$$\frac{24}{7}$$=3.42sec.

Ans .

(2) 85 km/hour

Explanation :

To tal length of both trains = 250 metres
Let speed of second train = x kmph
Relative speed = (65 + x) kmph
=(65 + x )*$$\frac{5}{18}$$m/sec
Time=$$\frac{Sum of length of trains}{Relative speed}$$
6=$$\frac{250}{65 + x * \frac{5}{18}}$$
=6*$$\frac{5}{18}$$*(65 + x ) = 250
65+x=$$\frac{250*3}{5}$$
65 + x = 150
x = 150 – 65 = 85 kmph

Ans .

(3) 100.

Explanation :

Relative speed = (84 + 6)
= 90 kmph
=90*$$\frac{5}{18}$$ m/sec.
= 25 m/sec.
Length of train = Relative speed × Time
= 25 × 4 = 100 metre

Ans .

(3) 54 km/hr

Explanation :

=$$\frac{Speed of X}{Speed of Y}$$
=$$\frac{Time taken by Y}{Time taken by X} ^\frac{1}{2}$$
=$$\frac{45}{y}$$=$$\frac{3 hours 20 min}{4 hours 48 min.} ^\frac{1}{2}$$
=$$\frac{45}{y}$$=$$\frac{200 minutes}{288 minutes.} ^\frac{1}{2}$$
=$$\frac{10}{12}$$
10y = 12 × 45
y=54 kmph.

Ans .

(3) 180

Explanation :

Let P and Q meet after t hours.
Distance = speed × time
According to the question,
30t – 20t = 36
10t = 36
t=3.6 hours.
Distance between P and Q
= 30t + 20t
= 50t = (50 × 3.6) km.
= 180 km

Ans .

(3) 9.5 kmph

Explanation :

Speed of train starting from Q = x kmph
Speed of train starting from P = (x + 8) kmph
According to the question, PR + RQ = PQ
(x + 8) × 6 + x × 6 = 162
[Distance = Speed × Time]
6x + 48 + 6x = 162
12x = 162 – 48 = 114
x=$$\frac{114}{12}$$
=9.5 kmph.

Ans .

(1) 875 km.

Explanation :

Let the trains meet after t hours.
Distance = Speed × Time
According to the question,
75t – 50t = 175
25t = 175
t=7hours.
Distance between A and B
= 75t + 50t = 125t
= 125 × 7 = 875 km.

Ans .

(4) 11 seconds

Explanation :

Relative speed = (50 + 58) kmph
=108*$$\frac{5}{18}$$ m/sec
= 30 m/sec
Required time
=$$\frac{Total length of trains}{Relative speed}$$
=$$\frac{150+180}{30}$$ sec
=$$\frac{330}{30}$$=11 sec.

Ans .

(1) 350 km.

Explanation :

Let the trains meet each other after t hours.
Distance = Speed × Time
According to the question,
21t – 14t = 70
7t = 70
t=10
Required distance
= 21t + 14t = 35t
= 35 × 10 = 350 km.

# TYPE-5

Ans .

(3) 8 hours

Explanation :

Since the train runs at $$\frac{7}{11}$$of its own speed, the time it takes is $$\frac{11}{7}$$of its usual speed.Let the usual time taken be t hours.
Then we can write,$$\frac{11}{7}$$t = 22
t=14 hours
Hence, time saved
= 22 – 14 = 8 hours

Ans .

(1) 3.75 hours

Explanation :

$$\frac{3}{5}$$of usual speed will take $$\frac{5}{3}$$of usual time. time & speed are inversely
proportional $$\frac{5}{3}$$ of usual time
= usual time + $$\frac{5}{2}$$
=$$\frac{2}{3}$$ of usual time = $$\frac{5}{2}$$
usual time
$$\frac{5}{2}$$ * $$\frac{3}{2}$$=$$\frac{15}{4}$$=3.75 hours.

Ans .

(1) 35 kmph

Explanation :

1 hr 40 min 48 sec
1 hr 40 + $$\frac{48}{60}$$
1 hr 40 + $$\frac{4}{5}$$
1 hr $$\frac{204}{5}$$
1 + $$\frac{204}{300}$$hr=$$\frac{504}{300}$$hr
Speed = $$\frac{42}{\frac{504}{300}}$$= 25 kmph
Now $$\frac{5}{7}$$*usual speed = 25
Usual speed = $$\frac{25*7}{5}$$= 35 kmph

Ans .

(3) 6 hours

Explanation :

$$\frac{4}{3}$$× usual time – usual time = 2
$$\frac{1}{3}$$usual time = 2
Usual time = 2 × 3 = 6 hours

Ans .

(2) 60 minutes

Explanation :

$$\frac{4}{3}$$of usual time
= Usual time + 20 minutes
$$\frac{1}{3}$$of usual time = 20 minutes
Usual time = 20 × 3
= 60 minutes

Ans .

(2) 4.5 hours

Explanation :

Time and speed are inversely proportional.
$$\frac{4}{3}$$of usual time –usual time
=$$\frac{3}{2}$$
$$\frac{1}{3}$$ * usual time= $$\frac{3}{2}$$
Usual time=$$\frac{3*3}{2}$$=$$\frac{9}{2}$$=4.5 hours

Ans .

(1) 2 hours 30 minutes

Explanation :

Time and speed are inversely proportional.
$$\frac{7}{6}$$* Usual time – Usual time
= 25 minutes
Usual time $$\frac{7}{6}$$-1
= 25 minutes
Usual time × $$\frac{1}{6}$$
= 25 minutes
Usual time = 25 × 6
= 150 minutes
= 2 hours 30 minutes

Ans .

(2) 1 hour 12 minutes

Explanation :

Time and speed are inversely proportional.
Usual time * $$\frac{7}{6}$$– usual time
= 12 minutes
Usual time * $$\frac{1}{6}$$= 12 minutes
Usual time = 72 minutes
= 1 hour 12 minutes

Ans .

(2) 420 km

Explanation :

Fixed distance = x km and certain speed = y kmph (let).
Case I,
$$\frac{x}{y+10}$$=$$\frac{x}{y}$$ - 1
=$$\frac{x}{y+10}$$ + 1=$$\frac{x}{y}$$ .....(1)
Case II,
$$\frac{x}{y+20}$$ = $$\frac{x}{y}$$ -1 -$$\frac{3}{4}$$
=$$\frac{x}{y}$$- $$\frac{4+3}{4}$$
$$\frac{x}{y+20}$$+$$\frac{7}{4}$$=$$\frac{x}{y}$$.....(2)
From equations (i) and (ii),
$$\frac{x}{y+10}$$+1=$$\frac{x}{y+20}$$+$$\frac{7}{4}$$
$$\frac{x}{y+10}$$-$$\frac{x}{y+20}$$=$$\frac{7}{4}$$-1
x*( \frac{y + 20 - y - 10}{y + 10 )( y + 20 )} )
$$\frac{7-4}{4}$$=$$\frac{3}{4}$$
$$\frac{x *10}{( y + 10 )( y + 20 )}$$=$$\frac{3}{4}$$
3 (y + 10) (y + 20) = 40 x
$$\frac{3 ( y + 10 )( y + 20 )}{40}$$=x...(3)
From equation (i),
$$\frac{3 ( y + 10 )( y + 20 )}{40(y+10)}$$ + 1
$$\frac{3 ( y + 10 )( y + 20 )}{40y}$$
3 (y +20) + 40
$$\frac{3 ( y + 10 )( y + 20 )}{y}$$
3y 2 + 60y + 40 y = 3(y 2 + 30y + 200)
3y 2 + 100y = 3y 2 + 90y + 600
10y = 600 Þ y = 60
Again from equation (i),
$$\frac{x}{y+10}$$+1=$$\frac{x}{y}$$
$$\frac{x}{60+10}$$+1=$$\frac{x}{60}$$
$$\frac{x}{70}$$+1=$$\frac{x}{60}$$
$$\frac{x+70}{70}$$+1=$$\frac{x}{60}$$
6x + 420 = 7x
7x – 6x = 420
x = 420 km.

Ans .

(2) 20 km/hour

Explanation :

Total distance
= 7 × 4 = 28 km.
Total time
$$\frac{7}{10}$$+$$\frac{7}{20}$$+$$\frac{7}{30}$$+$$\frac{7}{60}$$ hours
$$\frac{42 + 21 + 14 + 7}{60}$$hours
=$$\frac{84}{60}$$hours=$$\frac{7}{5}$$hours
Average speed
=$$\frac{Total distance}{Total time}$$=$$\frac{28}{\frac{7}{5}}$$kmph
=20 kmph

Ans .

(2) 72

Explanation :

1 m/sec = $$\frac{18}{5}$$ kmph
20 m/sec =$$\frac{20*18}{5}$$
= 72 kmph

Ans .

(1) 15 m/sec

Explanation :

1 kmph =$$\frac{5}{18}$$m/sec
54 kmph =$$\frac{5}{18}$$*54
= 15 m/sec.

Ans .

(3) 40 km./hr.

Explanation :

Speed of car = x kmph.
Distance = Speed × Time
= 25x km.
Case II,
Speed of car =$$\frac{4x}{5}$$kmph
Distance covered =$$\frac{4x}{5}$$*25
= 20x km.
According to the question,
25x – 20x = 200
5x = 200
x=40 kmph.

Ans .

(4) 6.6 km. per hour

Explanation :

Speed of car = x kmph.
Relative speed = (x – 4) kmph
Time = 3 minutes =$$\frac{3}{60}$$hour=$$\frac{1}{20}$$hour
Distance = 130 metre
$$\frac{130}{1000}$$km=$$\frac{13}{100}$$km
Relative speed =$$\frac{Distance}{Time}$$
5x – 20 = 13
5x = 20 + 13 = 33
x=6.6 kmph.

# TYPE-6

Ans .

(2) 10.8 km/hr

Explanation :

Total distance = 10 + 12
= 22 km
Total time = $$\frac{10}{12}$$+ $$\frac{12}{10}$$=$$\frac{244}{120}$$hours
Required average speed
$$\frac{Total Distance}{Total Time}$$=$$\frac{22}{\frac{244}{120}}$$=$$\frac{22}{244}$$*120
= 10.8 km/hr.

Ans .

(1) 65.04 km/hr

Explanation :

Total distance = 10 + 12
= 22 km
Total time
$$\frac{600}{80}$$+$$\frac{800}{40}$$ +$$\frac{500}{400}$$ +$$\frac{100}{50}$$
$$\frac{246}{8}$$hr
Average speed
$$\frac{600 + 800 + 500 + 100}{\frac{246}{8}}$$
$$\frac{2000*8}{246}$$
=65.04 km/hr.

Ans .

(2) 36 kmph

Explanation :

Average speed
=$$\frac{Total distance}{Time taken}$$
=$$\frac{30* \frac{12}{60}+45*\frac{8}{60}}{\frac{12}{60}+\frac{8}{60}}$$
= 12 × 3 = 36 kmph

Ans .

(3) 4 km/hr

Explanation :

If the same distance are covered at different speed of x kmph and y kmph, the average speed of the whole journey is given by =$$\frac{2xy}{x+y}$$kmph
Required average speed =$$\frac{36}{9}$$=4 kmph

Ans .

(3) 6

Explanation :

If two equal distances are cov- ered at two unequal speed of x
kmph and y kmph, then average =$$\frac{2xy}{x+y}$$kmph
=$$\frac{96}{16}$$= 6 kmph

Ans .

(1) 3 km/hour more

Explanation :

Remaining distance
= (3584 – 1440 – 1608) km
= 536 km.
This distance is covered at the rate of $$\frac{536}{8}$$= 67 kmph.
Average speed of whole journey =$$\frac{3584}{56}$$=64 kmph
Required difference in speed = (67 – 64) kmph i.e. = 3 kmph more

Ans .

(1) 8

Explanation :

Total distance
= 24 + 24 + 24 = 72 km.
Total time
=$$\frac{24}{6}$$+$$\frac{24}{8}$$+$$\frac{24}{12}$$
= (4 + 3 + 2) hours = 9 hours
\ Required average speed
=$$\frac{Total distance}{Total time}$$=8 kmph

Ans .

(4) 88.89 km/hr

Explanation :

If same distance are covered at two different speed of x and y kmph, the average speed of journey =$$\frac{2xy}{x+y}$$
=$$\frac{2*100*80}{100+80}$$
= 88.89 kmph

Ans .

(2) $$\frac{2xy}{x+y}$$

Explanation :

Required average speed $$\frac{2xy}{x+y}$$
Since, can be given as corollary If the distance between A and B be z units, then
Average speed =$$\frac{Total speed}{Time taken}$$
$$\frac{z+z}{\frac{z}{x}+\frac{z}{y}}$$
=$$\frac{2xy}{x+y}$$

Ans .

(1) 48 km/hr

Explanation :

Average speed
$$\frac{2xy}{x+y}$$
$$\frac{2*40*60}{40+60}$$
= 48 kmph

Ans .

(1) 14*$$\frac{2}{5}$$km/hr

Explanation :

Average speed
$$\frac{2xy}{x+y}$$
$$\frac{2*12*18}{12+18}$$
=14*$$\frac{2}{5}$$

Ans .

(2) 33*$$\frac{1}{3}$$ km/hr

Explanation :

Let the total distance be x km
Total time =$$\frac{\frac{x}{3}}{25}$$+$$\frac{\frac{x}{4}}{30}$$+$$\frac{\frac{5x}{12}}{50}$$
=$$\frac{4x+5x}{300}$$
=$$\frac{3x}{100}$$
Average speed=$$\frac{Distance}{Time}$$
=$$\frac{x}{\frac{3x}{100}}$$
=33*$$\frac{1}{3}$$ km/hr

Ans .

(1) 7 km/hr

Explanation :

Time taken to cover 30km at 6 kmph=$$\frac{30}{6}$$= 5 hour
Time taken to cover 40 km = 5 hours
\ Average speed=$$\frac{Total Distance}{Time}$$
$$\frac{30+40}{10}$$
=7 km/hr

Ans .

(1) 40 km/hr

Explanation :

Here same distances are covered at different speeds.
\ Average speed
$$\frac{2xy}{x+y}$$
=$$\frac{2*36*45}{36+45}$$
=40 kmph

Ans .

(1) 120 kmph

Explanation :

Here, the distances are equal.
\ Average speed=$$\frac{2*100*150}{100+150}$$
=120 kmph

Ans .

(2) 5*$$\frac{1}{3}$$

Explanation :

Total distance = 5 × 6 + 3 × 6
= 30 + 18 = 48 km
Total time = 9 hours
\ Average speed
$$\frac{48}{9}$$
=5*$$\frac{1}{3}$$

Ans .

(3) 70 km

Explanation :

Let the length of journey be x km, then $$\frac{x}{35}$$-$$\frac{x}{40}$$=$$\frac{15}{60}$$=$$\frac{1}{4}$$ x= 70 km

Ans .

(3) 20 km/hr

Explanation :

Average speed =$$\frac{Distance}{Time}$$ =$$\frac{12}{\frac{3}{10}+\frac{3}{20}+\frac{3}{30}+\frac{3}{60}}$$ $$\frac{12*60}{3*12}$$ =20 km/hr

Ans .

(1) 30 km/hr

Explanation :

Distance covered 35*$$\frac{10}{60}$$+20*$$\frac{5}{60}$$
=$$\frac{45}{6}$$km
Total time = 15 minutes=$$\frac{1}{4}$$hr
Required average speed =$$\frac{Distance}{Time}$$
=30 kmph

Ans .

(2) 40

Explanation :

Total distance = 100 km.
Total time $$\frac{50}{50}$$+$$\frac{40}{40}$$+$$\frac{10}{20}$$
=$$\frac{5}{2}$$hr
Average speed =$$\frac{100*2}{5}$$
= 40

Ans .

(4) 24 km/hr

Explanation :

Required average speed $$\frac{2*30*20}{30+20}$$
= 24 km/hr

Ans .

(3) 9.00 a.m.

Explanation :

If A and B meet after t hours, then
4 t + 6 t = 20
10 t = 20
t = 2 hr
Hence, both will meet at 9 a.m.

Ans .

(3) 24

Explanation :

Average speed =$$\frac{2*20*30}{20+30}$$
= 24

Ans .

(1) 37.5

Explanation :

Average speed of whole journey $$\frac{2*50*30}{50+30}$$
= 37.5 kmph

Ans .

(2) 4 km

Explanation :

Required distance of office from house = x km. (let)
Time =$$\frac{Distance}{Speed}$$
According to the question,
$$\frac{x}{5}$$-$$\frac{x}{6}$$=$$\frac{2}{15}$$
$$\frac{x}{30}$$=$$\frac{2}{15}$$
x= 4 km

Ans .

(4) 14 hrs

Explanation :

Time =$$\frac{Distance}{Speed}$$=$$\frac{1050}{75}$$
= 14 hrs

Ans .

(2) 45 km/hr

Explanation :

Total distance covered by train in 5 minutes
= (500 + 625 + 750 + 875 + 1000)
metre = 3750 metre
= 3.75 km.
Time = 5 minutes =$$\frac{5}{60}$$=$$\frac{1}{12}$$hr
Speed of train=$$\frac{3.75}{\frac{1}{12}}$$
= (3.75 × 12) kmph
= 45 kmph

Ans .

(1) 12$$\frac{1}{2}$$ km/hr

Explanation :

Distance covered in first 2 hours = 2 × 20 = 40 km.
Remaining distance = 100 – 40 = 60 km.
Time taken in covering 60 km at
10 kmph $$\frac{60}{10}$$=6 hr
Required average speed=$$\frac{Distance}{Time}$$
$$\frac{100}{2+6}$$
=12$$\frac{1}{2}$$ km/hr

Ans .

(1) 68 kmph

Explanation :

Difference of time = 5 + 3 = 8 minutes
$$\frac{8}{60}$$=$$\frac{2}{15}$$hr
If the speed of motorbike be x kmph, then
$$\frac{25}{50}$$-$$\frac{25}{x}$$=$$\frac{2}{15}$$
11x = 25 × 30
x=68.18 kmph
x= 68 kmph

Ans .

(4) 4

Explanation :

Let the speed of cyclist while returning be x kmph.
\ Average speed $$\frac{2*16*x}{16+x}$$
6.4 × 16 + 6.4x = 32x
32x – 6.4x = 6.4 × 16
25.6x = 6.4 × 16
x= 4 kmph

Ans .

(3) 40 km/hr

Explanation :

Total distance covered = 400 km. Total time =$$\frac{25}{2 }$$hr $$\frac{3}{4}$$of total journey $$\frac{3}{4}$$ * 400 = 300 km. Time=$$\frac{Distance}{Speed}$$ $$\frac{300}{30}$$=10 Remaining time =$$\frac{25}{2}$$-10 =$$\frac{5}{2}$$ Remaining distance = 100 km. \ Required speed of car $$\frac{100}{\frac{5}{2}}$$ =40 km/hr

Ans .

(3) 160 minutes

Explanation :

Durga’s average speed
$$\frac{2*5*15}{5+15}$$
=$$\frac{15}{2}$$ kmph
Distance of School = 5 km.
Smriti’s speed =$$\frac{15}{4}$$
Required time =2*$$\frac{5}{\frac{15}{4}}$$
=$$\frac{8}{3}$$hr
$$\frac{8}{3}$$*60 =160 minutes

Ans .

(4) 35.55 kmph

Explanation :

Here, distances are equal.
\ Average speed
$$\frac{2*32*40}{32+40}$$
$$\frac{320}{9}$$
= 35.55 kmph

Ans .

(1) 48 km/h

Explanation :

Here, distance is same.
Average speed=$$\frac{2xy}{x+y}$$
=$$\frac{2*40*60}{40+60}$$
=48 km/h

Ans .

(1) 54 km/hr

Explanation :

Total distance covered by the bus = 150 km. + 2 × 60 km.
= (150 + 120) km.
= 270 km.
\ Average speed=$$\frac{Distance}{Time}$$
$$\frac{270}{5}$$
= 54 km/hr

Ans .

(3) 10.9 kmph

Explanation :

Here distances are same
=$$\frac{2*12*10}{12+10}$$
=$$\frac{240}{22}$$
= 10.9 kmph

Ans .

(1) 18 kmph.

Explanation :

Total distance covered
= (50 + 40 + 90) km
= 180 km
Time = $$\frac{Distance}{Speed}$$
Total time taken
$$\frac{50}{25}$$+$$\frac{40}{20}$$ +$$\frac{90}{15}$$hours
= (2 + 2 + 6) hours
= 10 hours
Average speed
=$$\frac{Total distance}{Total time taken}$$
=$$\frac{180}{10}$$
=18 kmph

Ans .

(3) 560 m.

Explanation :

Distance = Speed × Time
= (80 × 7) km.
= 560 km.

Ans .

(4) 18.75 metre/second

Explanation :

Required speed of car=$$\frac{Distance}{Time}$$
$$\frac{216}{3.2}$$kmph
$$\frac{216}{3.2}$$*$$\frac{5}{18}$$m/sec
= 18.75 m./sec.

# TYPE-7

Ans .

(1) 2 hours

Explanation :

Let the distance of destination be D km
Let the speed of A = 3x km/hr
then speed of B = 4x km/hr
\ According to question,
$$\frac{D}{3x}$$-$$\frac{D}{4x}$$=30 min
=$$\frac{1}{2}$$hour
$$\frac{D}{12x}$$=$$\frac{1}{2}$$
$$\frac{D}{3x}$$=$$\frac{4}{2}$$= 2 hours
Hence, time taken by A to reach destination = 2hrs.

Ans .

(1) 1.33 hour..

Explanation :

Ratio of speed = 3 : 4
Ratio of time taken = 4 : 3
Let the time taken by A and B be 4x hours and 3 x hours respec- tively.
Then, 4x–3x =$$\frac{20}{60}$$
x=$$\frac{1}{3}$$
Time taken by A = 4x hours
4*$$\frac{1}{3}$$
=1.33 hour

Ans .

(3) 25 : 18.

Explanation :

Required ratio
$$\frac{5}{6}$$:$$\frac{3}{5}$$
$$\frac{5*30}{6}$$:$$\frac{30*3}{5}$$
= 25 : 18

Ans .

(2) 3 : 2.

Explanation :

Required ratio of the speed of two trains
= $$\frac{√9}{√4}$$
3 : 2

Ans .

(3) 78 km/hr

Explanation :

Speed of second train
$$\frac{364}{4}$$
= 91 kmph
7x = 91
6x=$$\frac{91}{7x}$$*6x
=78 kmph

Ans .

(3) 3 : 4.

Explanation :

Speed of truck
= 550m/minute
Speed of bus =$$\frac{33000}{45}$$
Required ratio = 550 :$$\frac{2200}{3}$$
=3:4

Ans .

(2) 1 : 3 : 9

Explanation :

Required ratio =$$\frac{1}{3}$$:$$\frac{2}{2}$$:$$\frac{3}{1}$$
=$$\frac{1}{3}$$:1:3
=$$\frac{1}{3}$$*3:1*3:3*3
= 1 : 3 : 9

Ans .

(3) 3

Explanation :

The winner will pass the other, one time in covering 1600m. Hence, the winner will pass the other 3 times in completing 5km race

Ans .

(3) 3 : 4

Explanation :

Distance covered on the first day
$$\frac{4}{5}$$*70= 56 km
Required ratio = 42 : 56
= 3 : 4

Ans .

(1) 1 : 4

Explanation :

Let speed of cyclist = x kmph
& Time = t hours
Distance= $$\frac{xt}{2}$$while time = 2t
Required ratio =$$\frac{xt}{2*2t}$$:x
= 1 : 4

Ans .

(3) 3 : 4

Explanation :

Speed of train = x kmph
Speed of car = y kmph
Case 1:
$$\frac{120}{x}$$+$$\frac{600-120}{y}$$=8
$$\frac{15}{x}$$+$$\frac{60}{y}$$=1...(1)
Case 2
$$\frac{200}{x}$$+$$\frac{400}{y}$$= 8 hours 20 min
$$\frac{24}{x}$$+$$\frac{48}{y}$$=1...(2)
$$\frac{15}{x}$$+$$\frac{60}{y}$$=$$\frac{24}{x}$$+$$\frac{48}{y}$$
$$\frac{9}{x}$$=$$\frac{12}{y}$$
$$\frac{x}{y}$$=$$\frac{9}{12}$$
=3:4

Ans .

(2) 3 : 4

Explanation :

Let the speed of train be x kmph. and the speed of car be y kmph
Time=$$\frac{Distance}{Speed}$$
$$\frac{120}{x}$$+$$\frac{480}{y}$$=8
$$\frac{15}{x}$$+$$\frac{60}{y}$$=1.....(1)
$$\frac{200}{x}$$+$$\frac{400}{y}$$=$$\frac{25}{3}$$
$$\frac{24}{x}$$+$$\frac{48}{y}$$=1....(2)
From equations (i) and (ii),
$$\frac{15}{x}$$+$$\frac{60}{y}$$=$$\frac{24}{x}$$+$$\frac{48}{y}$$
$$\frac{x}{y}$$=$$\frac{9}{12}$$
=3:4

Ans .

(3) 3 : 4.

Explanation :

Speed of truck =$$\frac{550 metre}{60 second}$$
$$\frac{55}{6}$$m/sec
Speed of bus =$$\frac{33 * 1000 metre}{\frac{3}{4}*60*60sec}$$
=$$\frac{440}{36}$$
Required ratio =$$\frac{55}{6}$$:$$\frac{440}{36}$$
= 55 × 6 : 440
= 3 : 4

Ans .

(1) 2 : 3

Explanation :

Speed =$$\frac{Distance}{Time}$$
Speed of car : Speed of train
=$$\frac{80}{2}$$:$$\frac{180}{3}$$
= 40 : 60 = 2 : 3

Ans .

(3) 15 : 5 : 3

Explanation :

Speed ∝ $$\frac{1}{Time}$$
Required ratio of time
1:$$\frac{1}{3}$$:$$\frac{1}{5}$$
=15:$$\frac{1}{3}$$*15:$$\frac{1}{5}$$*15
= 15 : 5 : 3

# TYPE-8

Ans .

(1) 100 m.

Explanation :

Relative speed of police
= 11 – 10 = 1 kmph
=$$\frac{5}{18}$$
Distance decreased in 6 min- =$$\frac{5}{18}$$ × 6×60 = 100 m
Distance remained between them = 200–100 = 100 m

Ans .

(1) 85 km/hr

Explanation :

Suppose the speed of first train be x kmph
Speed of second train
= 30 kmph
$$\frac{30*1000}{60}$$= 500 m per min.
According to question
$$\frac{Total distance}{Relative speed}$$
$$\frac{(66 + 88 )}{x-500}$$=0.168
0.168x – 84 = 154
0.168x = 238
x=$$\frac{238}{0.168}$$
$$\frac{238*1000}{168}$$*$$\frac{3}{50}$$
= 85 kmph

Ans .

(1) 19 minutes.

Explanation :

The gap of 114 metre will be filled at relative speed. Required time
$$\frac{114}{21-15}$$
=19 minutes

Ans .

(4) 25 seconds.

Explanation :

Both trains are moving in the same direction.
\ Their relative speed
= (68 – 50) kmph = 18 kmph
=18*$$\frac{5}{8}$$= 5 m/sec
Total length = 50 + 75 = 125 m
\ Required time =$$\frac{Total length}{Relative speed}$$=$$\frac{125}{5}$$
=25 seconds.

Ans .

(2) 12 minutes

Explanation :

The constable and thief are running in the same direction
\ Their relative speed
= 8 – 7 = 1km.
1*$$\frac{5}{18}$$
Required time =$$\frac{200}{\frac{5}{18}}$$
=720 sec
=$$\frac{720}{60}$$
=12 minutes

Ans .

(4) 140

Explanation :

Relative speed
= (58 – 30) km/hr
28*$$\frac{5}{18}$$
$$\frac{70}{9}$$m/sec.
Length of train =$$\frac{70}{9}$$*18
= 140 metres

Ans .

(3) 75.

Explanation :

Relative speed
= 56 – 29 = 27 kmph
27*$$\frac{5}{18}$$
$$\frac{15}{2}$$
Distance covered in 10 sec- onds $$\frac{15}{2}$$*10
= 75 m

Ans .

(1) 27 km/hr

Explanation :

Let the speed of the truck be x kmph
Relative speed of the bus
= 45 - x kmph
Time=$$\frac{Distance}{Relative speed}$$
$$\frac{30}{60*60}$$=$$\frac{\frac{150}{1000}}{45-x}$$
(45 – x ) = 18
x=27 kmph

Ans .

(2) 50 m.

Explanation :

Let the length of each train be x metre.
Relative speed
= 46 – 36 = 10 kmph
=$$\frac{25}{9}$$
=$$\frac{2x}{\frac{25}{9}}$$=36
x = 50 metre

Ans .

(3) 3 km 750 m

Explanation :

Relative speed
= 45– 40 = 5 kmph
Required distance
5*$$\frac{45}{60}$$
$$\frac{15}{4}$$km
= 3 km 750

Ans .

(3) 18.6

Explanation :

Let the speed of Scooter be x
Distance covered by cycling in 3$$\frac{1}{2}$$hours = Distance covered
by scooter in 2$$\frac{1}{4}$$ hours
12*$$\frac{7}{2}$$=x*$$\frac{9}{4}$$
x=$$\frac{56}{3}$$
= 18.6 kmph

Ans .

(2) 400 m

Explanation :

Relative speed
$$\frac{1000}{8}$$-$$\frac{1000}{10}$$=$$\frac{1000}{40}$$
Required time = 4 m/minute
Distance covered by the thief =$$\frac{1000}{10}$$*4
= 400 metres

Ans .

(1) 27.7 m

Explanation :

Relative speed = 40 – 20
= 20 km/hour
=$$\frac{20*5}{18}$$
Length of the faster train =$$\frac{250}{9}$$= 27.7 metres

Ans .

(4) 90 km/h

Explanation :

Distance = Speed × Time
= 80 × 4.5 = 360 km
Required speed = $$\frac{360}{4}$$
= 90 kmph.

Ans .

(2) 9

Explanation :

Required time =$$\frac{Sum of the lengths of trains}{Relative speed}$$
Relative speed = 65 + 55
= 120 kmph
$$\frac{120*5}{18}$$
Required time = $$\frac{180+120}{\frac{120*5}{18}}$$
= 9 seconds

Ans .

(1) 125

Explanation :

When two trains cross each other, they cover distance equal to the sum of their length with relative speed.
Let length of each train = x metre
Relative speed = 90 – 60
= 30 kmph
$$\frac{30*5}{18}$$
=$$\frac{25}{3}$$m/sec
$$\frac{2x}{\frac{25}{3}}$$=30
2x = 250
x = 125 metres

Ans .

(4) 72.

Explanation :

Relative speed = 35 – 25
= 10 kmph
=$$\frac{10*5}{18}$$m/sec
Total length = 80 + 120
= 200 metres
Required time =$$\frac{Sum of the length of trains}{Relative speed}$$
=$$\frac{200*18}{10*5}$$
= 72 seconds

Ans .

(1) 24

Explanation :

Distance covered by the first goods train in 8 hours = Distance covered by the second goods train in 6 hours.
18 × 8 = 6 * x
x=$$\frac{18*8}{6}$$
= 24 kmph

Ans .

(3) 12.

Explanation :

Relative speed
= (33 + 39) kmph
= 72 kmph
$$\frac{72*5}{18}$$m/sec
= 20 m/sec.
\ Time taken in crossing
=$$\frac{Length of both trains}{Relative speed}$$
=$$\frac{240}{20}$$
=12 seconds

Ans .

(2) 4 p.m..

Explanation :

Distance covered by the thief in half an hour =$$\frac{1}{2}$$*40 =20 km
Relative speed of car owner
= 50 – 40 = 10 km
\ Required time
=$$\frac{Difference of distance}{Relative speed}$$
$$\frac{20}{10}$$
= 2 hours
i.e. at 4 p.m.

Ans .

(1) 50 m

Explanation :

Length of each train = x metre

Relative speed = 46 – 36
= 10 kmph
=$$\frac{10*5}{18}$$
=$$\frac{25}{9}$$m/sec
Time taken in crossing
$$\frac{Length of both trains}{Relative speed}$$
36=$$\frac{2x}{\frac{25}{9}}$$
x = 50 metre

Ans .

(3) 1320 km

Explanation :

Let both trains meet after t hours.
\ Distance = speed × time
60t – 50t = 120
10t = 120 t = 12 hours
Required distance
= 60t + 50t
= 110t = 110 × 12
= 1320 km

Ans .

(3) 6.

Explanation :

Let both cars meet at C after t hours.
Distance covered by car A = AC = 35t km
Distance covered by car B
= BC = 25t km
AC – BC = AB = 60 km.
35t – 25t = 60
10t = 60
t = 6 hours

Ans .

(2) 88.

Explanation :

Let the speed of train C be x kmph.
Relative speed of B
= (100 – x ) kmph.
Time taken in crossing
$$\frac{Length of both trains}{Relative speed}$$
$$\frac{2}{60}$$=$$\frac{\frac{150+250}{1000}}{100-x}$$
100 – x = 12
x = 100 – 12 = 88 kmph.

Ans .

(1) 32 kmph

Explanation :

Let the speed of goods train be x kmph.
Distance covered by goods train in 10 hour = distance cov- ered by passenger train in 4 hours
10x = 80 × 4
x = 32 kmph.

Ans .

(4) 3.75 km..

Explanation :

Relative speed = 45 – 40
= 5 kmph.
Gap between trains after 45 minutes = 5*$$\frac{45}{60}$$
= 3.75 km.

Ans .

(3) 500 metre

Explanation :

Distance between thief and policeman = 400 metre
Relative speed of policeman with respect to thief
= (9 – 5) kmph
= 4 kmph
4*$$\frac{5}{18}$$
$$\frac{10}{9}$$m/sec
Time taken in overtaking the thief
$$\frac{400}{\frac{10}{9}}$$
= 360 second
Distance covered by thief
= Speed × Time
=5*$$\frac{5}{18}$$*360
= 500 metre

Ans .

(4) 50 m

Explanation :

Let the length of each train be x metre.
Relative speed = (46 – 36) kmph
= 10 kmph
10*$$\frac{5}{18}$$
=$$\frac{25}{9}$$m/sec
$$\frac{2x}{\frac{25}{9}}$$=36
x=50 metre

# TYPE-9

Ans .

(4) 50 minutes

Explanation :

Time taken to cover 20 km at the speed of 5km/hr = 4 hours.
\ Fixed time = 4 hours – 40 min- utes
= 3 hour 20 minutes
Time taken to cover 20 km at the speed of 8 km/hr =$$\frac{20}{8}$$=2 hours 30 minutes
Required time = 3 hours 20 minutes – 2 hours 30 minutes = 50 minutes

Ans .

(1) 2.

Explanation :

Since man walks at $$\frac{2}{3}$$of usual speed, time taken wil be $$\frac{3}{2}$$ usual time.
=usual time + 1 hour.
$$\frac{3}{2}$$-1 of usual time = 1
usual time = 2 hours.

Ans .

(3) 5 km

Explanation :

Let x km. be the required dis- tance.
Difference in time
= 2.5 + 5 = 7.5 minutes
=$$\frac{7.5}{60}$$=$$\frac{1}{8}$$hr
$$\frac{x}{8}$$-$$\frac{x}{10}$$=$$\frac{1}{8}$$
x=$$\frac{40}{8}$$= 5 km.

Ans .

(4) 40

Explanation :

Let the distance be x km and initial speed be y kmph. According to question,
$$\frac{x}{y}$$-$$\frac{x}{y+3}$$=$$\frac{40}{60}$$.....(1)and
$$\frac{x}{y-2}$$-$$\frac{x}{y}$$=$$\frac{40}{60}$$......(2)
From equations (i) and (ii),
$$\frac{x}{y}$$-$$\frac{x}{y+3}$$=$$\frac{x}{y-2}$$-$$\frac{x}{y}$$
3 (y – 2) = 2 (y + 3)
Þ 3y – 6 = 2y + 6
Þ y = 12
From equation (i),$$\frac{x}{12}$$-$$\frac{x}{15}$$=$$\frac{40}{60}$$
x=40
Distance = 40 km.

Ans .

(3) 19 .

Explanation :

If the distance be x km, then
$$\frac{x}{40}$$-$$\frac{x}{50}$$=$$\frac{6}{60}$$
x = 20 km.
Required time
$$\frac{20}{40}$$hr-11 mimnutes
= 19 minutes

Ans .

(1) 1.75 km

Explanation :

Let the required distance be x km.
Difference of time
= 6 + 6 = 12 minutes = $$\frac{1}{5}$$ hr
According to the question,
$$\frac{x}{\frac{5}{2}}$$-$$\frac{x}{\frac{7}{2}}$$=$$\frac{1}{5}$$
$$\frac{14x-10x}{35}$$=$$\frac{1}{5}$$
x=$$\frac{7}{4}$$=1.75 km

Ans .

(4) 6 km

Explanation :

Let the required distance be x km.
According to the question,
$$\frac{x}{4}$$-$$\frac{x}{5}$$=$$\frac{18}{60}$$
x=$$\frac{3}{10}$$* 20= 6 km

Ans .

(2) 40 km/hour.

Explanation :

Let the initial speed of the car be x kmph and the distance be y km.
Then,y=$$\frac{9}{2}$$x
and, y = 4 (x + 5)
9x = 8x + 40
x = 40 kmph

Ans .

(3) 22 km

Explanation :

Let the distance of office be x km
$$\frac{x}{24}$$-$$\frac{x}{30}$$=$$\frac{11}{60}$$
$$\frac{x}{120}$$=$$\frac{11}{60}$$
x=22 km

Ans .

(3) 3 km.

Explanation :

Let the required distance be x km.
$$\frac{x}{3}$$-$$\frac{x}{5}$$=$$\frac{24}{60}$$
$$\frac{2x}{3}$$=2
2x = 2 × 3
x = 3 km

Ans .

(2) 4

Explanation :

Let the required distance be x
km $$\frac{x}{\frac{5}{2}}$$-$$\frac{x}{3}$$=$$\frac{16}{60}$$
$$\frac{6x-5x}{15}$$=$$\frac{4}{15}$$
x = 4 km.

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.