# Quantitative Aptitude: Time, Speed and Distance Set 8

Time Speed Distance Questions for NICL, SBI PO, IBPS PO, NIACL, LIC, Dena Bank PO PGDBF, BOI, Bank of Baroda and other competitive exams.

1. Two trains are running with speed 40kmph and 60kmph in the same direction. A man in the slower train passes by the faster train in 36seconds. Find the length of faster train?
A) 100mtr
B) 150mtr
C) 200mtr
D) 250mtr
E) 300mtr
Option C
Solution:

In the same direction speed … 60 -40 =2okmph
20* 5/18 * 36 =200mtr
2. After travelling two hours a train met with an accident due to this it stops for an hour. After this the train moves at 66(2/3)% speed of its original speed and reaches to destination 3hour late. If the accident would occur at 200km ahead in the same line then the train reaches only 2.5hours late. Then find the distance of journey and the original speed of the train?
A) 2400km,600kmph
B) 1800km,300kmph
C) 2400km,400kmph
D) 1800km,200kmph
E) 2000km,100kmph
Option B
Solution:

Due to 200km it saves 5hrs.
For 3hrs it has to run 200*2*3 =1200km
66(2/3)% = 2/3
.                     after………..normal
speed             2         :       3
time               3         :       2
.                             3-2 =1.
1 =2 ( train stops for 1 hr out 3hrs. so 3-1 =2)
2 =4
1200/4 = 300kmph
so  2hr*300 = 600km
Now total distance = 1200 +600 = 1800km
3. A man travels a distance in three equal parts. He covers first part at 20kmph,second part at 40kmph and third part at 120kmph. Find the distance if he covers total distance in 20hrs.
A) 1400km
B) 1200km
C) 1440km
D) 1600km
E) 1500km
Option C
Solution:

Distance = average speed * time
Average speed will be….72kmph
72*20 =1440km
4. A person who can walk down a hill at the rate of 6kmph and climb up the hill at rate of 4kmph. He ascends and comes down to his starting point in 5hrs. how far did he ascends ?
A) 12km
B) 14km
C) 20km
D) 24km
E) 16km
Option A
Solution:

First find average speed = 2*6*4/(6+4)
Time = 5hrs
distance = 48/10 * 5 =24
one side distance = 24/2 =12km
5. A student walks from his house at a speed of 2(1/2)km per hour and reaches his school 6minutes late. The next day he increases his speed by 1kmph and reaches 6minutes before school time. How far is the school from his home?
A) 5/4km
B) 9/4km
C) 7/4km
D) 11/4km
E) 10/4km
Option C
Solution:

S1*S2/difference of speed *(( late + early)/60)
= (5/2 * 7/2)/1 * 12/60= 7/4km
6. In covering a distance of 60km Abhi takes 2hrs more than Sam. If Abhi triples his speed then he would take 2hrs less than Sam. Abhi speed in kmph is ?
A) 10kmph
B) 12kmph
C) 15kmph
D) 20kmph
E) 14kmph
Option A
Solution:

.              Abhi triples his speed
.                          triple…………….normal
speed                     3……………………1
time                       1…………………..3
.                                       3 -1 =2
2=4hrs
3=6hrs
so Abhi cover60km in 6hrs
60/6 =10kmph
7. Two men start together to walk a certain distance, one at 5kmph and another at 4kmph. The former arrives an hour before the latter. Find the distance?
A) 10km
B) 15km
C) 20km
D) 25km
E) 30km
Option C
Solution:

.              speed          5………….4
.              time            4…………..5
.                                      5 -4 =1
1=1
5=5hrs
distance = 4*5 = 20km
8. A man covers a distance in downstream at 18kmph. If the speed of stream is 2kmph then find his speed in upstream?
A) 12kmph
B) 14kmph
C) 16kmph
D) 18kmph
E) 20kmph
Option B
Solution:

.               18 – 2 – 2= 14kmph
9. The distance between a thief and apolicemen is 300m. the speed of thief is 12m/s and the speed of police is 15m/s. find the distance covered by police to catch the thief?
A) 1000m
B) 1200m
C) 1500m
D) 2000m
E) 1300m
Option C
Solution:

.               300/(15-12) =100sec
15*100 = 1500m
10. Two trains of same length passes each other in 36sec. if the speed of trains are 40kmph and 20kmph respectively, then find the length of train?
A) 200 m
B) 400 m
C) 600 m
D) 300 m
E) 500 m
Option D
Solution:

2x/(40+20) * 18/5 = 36
2x= 600
x = 300m

## 55 Thoughts to “Quantitative Aptitude: Time, Speed and Distance Set 8”

1. Prasanjit Nayak

Can u plz explain how average speed is 72 km in question no 3

2. Sneha

explain the third qus

3. AVI™ (Sinchen loveR)

In Q 2 option C also follow all conditions but only .1 difference comes so can we accept it as a ans or its wrong ??

4. _/_

q5 ?

1. AVI™ (Sinchen loveR)

u can easily solve with the help of options so go for it
its also easy
u can understand Q 3?

1. _/_

by option how?

1. AVI™ (Sinchen loveR)

use these given value and solve

5. _/_

qs 8 ?

1. AVI™ (Sinchen loveR)

its very simple
downstream speed =18
stream speed 2 but when we swim against stream speed it will be double so now stream speed 4
so 18-4=14ans

1. _/_

ty

2. \$~\$PrAdeEP\$K\$~\$~\$

downstream =x+y
upstream =x-y
now (x-y)^2 = (x+y)^2 -4xy
=(18)^2 -4*14*4 =14 km/hr

6. _/_

Q4. There are 3 containers A, B and C which contain water, milk and acid respectively in equal quantities. 10% of the content of A is taken out and poured into B. Then, the same 10% from B is transferred to C, from which again the same 10% is transferred to A. What is the proportion of milk in container A at the end of the process ?
(a) 9/10
(b) 1/11
(c) 1/121
(d) 10/1011
(e) None of these

1. AVI™ (Sinchen loveR)

process kitni baar ki h ye to bathoo 1,2,3,4???

1. _/_

etna hi hai qs me

1. AVI™ (Sinchen loveR)

na phir to nhi hoga ese kese hoga process to btani pdegi ki kitni baar kiya h
2 baar kiya hoga to alag aayga 3 baar kiya hoga to alag % aayega

2. Jellyfish

quantity A 100x (water) Ataken 10% water poured into B
means A remaining 90X
now,B=100Xmilk710X water
10%poured into c means remaining 10x,x
C contained 100x acid 10x milk x water
10% poured into A
(90x+x/10) water x milk 10x acid
X=1011/10=101.1=10/1011

7. AVI™ (Sinchen loveR)

done
plz explain Q 3 in details
>>>TY,,,TY<<<

8. _/_

qs 3 ?
mere kuch or aa rha

9. Oliver Queen

Que-3 ?

1. Oliver Queen

ye kaun sa method h?

1. \$~\$PrAdeEP\$K\$~\$~\$

given spd
s(1) =20km/hr
s(2) =40km/hr
s(3) =120km/hr
take lcm =120
let distance =120
d(1) =d(2) =d(3) =40
now t1 +t2+t3 =20
(40/20) +(40/40)+(40/120) =20
=>10/3 =20 =>1 =6
so 120 =720 => d =720 km

10. Jellyfish

Que no 9 easier way
X/12=X+300/15
X=1200
1200+300=1500 thief covered dist &police covered dist 1200km

11. Jellyfish

Mam qye 3 explain
Average of 3quantity is
3*a*b*c/(ab)+(BC)+(ca) when dist is same
According to this formula
3*30*40*120/8000=36*20=720 ans ???

1. aayat

same

1. Jellyfish

i think mam ka galat ho gaya hoga
aayat math ke liye konsa book follow karti ho ??

2. AB

After travelling two hours a train met with an accident due to this it stops for an hour. After this the train moves at 66(2/3)% speed of its original speed and reaches to destination 3hour late. If the accident would occur at 200km ahead in the same line then the train reaches only 2.5hours late. Then find the distance of journey and the original speed of the train?
A) 2400km,600kmph
B) 1800km,300kmph
C) 2400km,400kmph
D) 1800km,200kmph
E) 2000km,100kmph

appka soln chahiye mam

1. aayat

Ye mujhe bhi nai aya try karti hu

2. aayat

1. AB

ty maam :))

2. aayat

2nd question smjh aya ??

1. Jellyfish

2nd muze problem ho rahi hai

12. Gusto

Qn2 —
diff of time= 3 – 2.5 = 1/2 hr
original spd = V
66 2/3% of its original spd = 2/3*V
1/2 = 200/(2V/3) – 200/V
1/2 = 200*/2V
V = 200 km/hr
let normal time = T
extra time = 3T/2 – T = T/2
T/2 = 3
T = 6 hrs
total distance = 200 + 200*6 = 1400 km

1. aayat

Sir ratio wale metd se kaise karenge??

1. Gusto

ratio wale se mujhe nhi ata sorry

1. aayat

Ok ty

1. Gusto

🙂

2. \$~\$PrAdeEP\$K\$~\$~\$

spd ratio =2:3
=> time ratio =3:2
now 3/2 -1 =30 mins
=>1 =60mins
since in time 1 it travelled 200km
=>200 =60min=>200km/hr =spd
now again 3/2 -1 =3
=>1= 6 hrs
=>distance in 6 hrs =6*200 =1200
=> totl dist =1200+200 =1400

1. Gusto

vgud

2. \$~\$PrAdeEP\$K\$~\$~\$

..:)))

2. Jellyfish

dear
x/y-x/2/3*y=2…1
x/y-x+200/4y=1/2..2
solve 1&2 ye method galat hai kya?

1. Gusto

isse to X & Y ki value hi nhi ayega

1. Jellyfish

But dist &speed unknown hai to Lena hoga ??

2. Jellyfish

thx got it..

1. Gusto

ok 🙂

13. Gusto

Qn3 —
let distance of each equal part = D
then total distance = 3D
T1 +T2 +T3 = 20
D/20 +D/40 +D/120 = 20
10D/120 = 20
D = 240
total distance = 3D = 720

1. _/_

ys same here

1. Gusto

_/_

14. Bubli

ques no 3 ka ans to optin se differ aa ra 😛

1. \$~\$PrAdeEP\$K\$~\$~\$

yes ans is 200/1400

1. Bubli

ok sir 😛 apna solution do

1. \$~\$PrAdeEP\$K\$~\$~\$

spd ratio =2:3
=> time ratio =3:2
now 3/2 -1 =30 mins
=>1 =60mins
since in time 1 it travelled 200km
=>200 =60min=>200km/hr =spd
now again 3/2 -1 =3
=>1= 6 hrs
=>distance in 6 hrs =6*200 =1200
=> totl dist =1200+200 =1400

1. Bubli

thanKU :))

2. AMAZON

@@disqus_0OrBOABGUE:disqus Please dear tell me in this question , i didn’t get after 2=4 and 3=6 fir time 6 kese aaya hain
kindly explain?

6 : In covering a distance of 60km Abhi takes 2hrs more than Sam. If Abhi triples his speed
then he would take 2hrs less than Sam. Abhi speed in kmph is ?

. Abhi triples his speed
. triple…………….normal
speed 3……………………1
time 1…………………..3
. 3 -1 =2
2=4hrs
3=6hrs
so Abhi cover60km in 6hrs
60/6 =10kmph

1. \$~\$PrAdeEP\$K\$~\$~\$

see here…:)
let initial spd s(i) of abhi =1 now trippling his speed mns 1*3 =3
now ini.to increased spd ratio
s(i)————-s(c)
1—————-3
=> initial to increased time ratio
T(i)———–T(c)
3—————-1
now 2 ratio =(+2-(-2))=4 => 3 ratio = inital time = 6hrs
now ini.spd =totl dist/ini.time =60/6 =10km/h

2. \$~\$PrAdeEP\$K\$~\$~\$

6hrs is the initial time of abhi with which he covers 60 kms