A person is walking with uniform speed and when he has completed half his journey he increased his s

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18 Jul 2013 Problem

A person is walking with uniform speed and when he has completed half his journey he increased his speed 20% and arrives at his destination. Also last half part of his journey be complete 30 minutes earlier than first half part of journey. How long was he walking the first half.

OPtion

1) 1hour
2) 1hour 30 minute
3) 2hour
4) 3hour
5) 4hour
6) 4hour 30 minute
7) 6hour
8) 6hour minute
9) 7hour
10)None of these

Solution

Let v=speed in first half and t=time taken to cover first half
Distance = vt (i)
For the second half;
Speed=v+(20v/100)=6v/5
Time =(t-0.5)hrs
Distance =(6v/5)(t-1/2) (ii)
From (i) and (ii), we get vt = (6v/5)(t-0.5)
t=3 hour
option 4

Correct Option: 4 Best Solution (26)

Pradyumna K S   11 years AGO

1st half:

2nd half:


divide both sides by



The 1st half takes 3 hrs to cover

The 2nd half takes 2 and a half hours to cover
Hence answer is (4)

Like? Yes (13) | No (7)  

ENABOTHULA MOUNIKA   11 years AGO

let time taken for first half distanct be t(min) an d speed be v then speed of second half distance is 6v/5 and time is t-30(min)
as v1/v2=t2/t1 => v/(6v/5)=(t-30)/t =>t=180 min
therfore t=3hours

Like? Yes (5) | No   

Kalidindi Mounika   11 years AGO

let speed to cover first half of distance is 100km/hr
so speed to cover second half of distance will be 120km/hr
Let time taken to cover first half be \'x\' minutes
=> 100*x=120*(x-30)
=> x=180minutes = 3*60=3hrs

Like? Yes (4) | No (1)  

dulal sarkar   11 years AGO

let the distance be 2d.
if x is the speed for first half distance then for next half distance it will be 20% more than first half i.e. 6x/5.
for first half time taken is 30 minutes more than second half. if time taken for second half is t minutesthen for first half it is t+30 minutes.
now, first half distance= second half distance
so x*(t+30)=6*x*t/5;
on solving we get t=150 minutes
so for first half time taken is t+30 minutes=150 +30 minutes=180 minutes=3 hours.

Like? Yes (4) | No (1)  

priyankanshi   11 years AGO

Let total distance = 2D
first half speed = S
first half time = T
So for the 1st part of the journey
D=ST
Similarly for the second half of the journey
D=(1.2S)*(T-30)
ST=(1.2S)*(T-30)
T=36/0.2
180 minutes

Like? Yes (2) | No   

ankur pratap singh   11 years AGO

let speed= x km/min
time taken during half = t min.
1st half
x*t=d;-------1
2nd half
(x+x*20/100)*(t-30)=d;
6x/5*(t-30)=d;
6xt/5-180x/5=d;-----2
from 1 & 2
xt=6xt/5-180x/5;
t=6t/5-36;
6t/5-t=36;
t/5=36;
t=180 min.
3 hours

Like? Yes (2) | No (2)  

Gaurav Sharma   11 years AGO

since time is inversely proportional to the speed.
let time taken to travel first half be t
and that to travel second half would be (t-1/2)hours
and speed for first half is S and that for second half is 1.2S
now we have (S/1.2S)=(t-1/2/t)
t=1.2t-0.6
0.2t=0.6
t=3
thus time taken for first half is 3 hours

Like? Yes (1) | No (1)  

chowthanya   11 years AGO

Distance = Speed x Time
So for the 1st part of the journey
d=s.t
Similarly for the second half of the journey
d=(20% of s).(t-30)
d=(120/100 *s).(t-30)
st=1.2st-36s[Since d=st by 1st half]
t=1.2t-36
t=36/0.2
t=180 mins
so,t=3 hours

Like? Yes (3) | No (4)  

vigneshwari   11 years AGO

Let the total distance be X.
then half of the distance X/2.
Let time taken to cross first half (X/2) be t.
Then the time taken to cross second half (X/2) = t-30 minutes.
Let speed in 1st half be S
Then the speed in 2nd half = 120S/100
Now, distance = speedxtime
X/2 = t x S and X/2 = (t-30)120S/100
t x S = (t-30)120S/100
Solving, t = 180 minutes = 3 hours

Like? Yes (2) | No (4)  

shivendra pratap singh   11 years AGO

1st half
d=s*t .......i)
2nd half
d=1.2*s*(t-30) ........ii)
equate i) and ii)
s*t=1.2*s*(t-30)
we get t=180 i.e he takes 3 hours to complete 1st half
so option 4)3 hour is correct option

Like? Yes (4) | No (6)  

mohan lal garg   11 years AGO

let speed x
after half distance speed=1.2x
total distance=y
y/2x-y/2.4x=1/2
y/x=6
in half distance time 3 hour

Like? Yes (2) | No (4)  

sameer singh   11 years AGO

10/12=t/t+30
t=150 min
t+30=180 min =3 hrs

Like? Yes (2) | No (4)  

Bagmita Chanda   11 years AGO

Let he covers first half way in x mins.
with 100% speed he covers half way in x min
with 1% speed he covers in x*100 min
with 120% speed he covers in (x*100)/120=5x/6min

Now with 20% increase in speed ie, with 120% speed he cover the second half way in less than 30 min than the first half way.
so we can say-
5x/6=x-30
5x=6x-180
x=180
now 180 min=3 hours

Like? Yes (2) | No (4)  

Roopam Jain   11 years AGO

first half time : t
first half speed : v
seconf half time : t-30
second half speed : v+ 20v/100
Equating the distance:
v*t = (6v/5)*(t-30)
On solving t = 180 mins = 3 hours

Like? Yes (2) | No (4)  

debabrata sarkar   11 years AGO

let the distance be 2d.
if x is the speed for first half distance then for next half distance it will be 20% more than first half i.e. 6x/5.
for first half time taken is 30 minutes more than second half. if time taken for second half is t minutesthen for first half it is t+30 minutes.
now, first half distance= second half distance
so x*(t+30)=6*x*t/5;
on solving we get t=150 minutes
so for first half time taken is t+30 minutes=150 +30 minutes=180 minutes=3 hours.

Like? Yes (2) | No (4)  

Balakrishnan M   11 years AGO

First half speed S.
Total distance D.
First half time t.
In first half,
S=(D/2)/t
==>(D/2) = st -----(1)
Second half,
Increased speed = s+s(20/100)=6s/5
time taken=(t-30)min
==> (D/2)=(6s/5)(t-30)-----(2)
From (1)&(2),
st = (6s/5)(t-30)
5st = 6st-180s
180s=6st-5st
st=180s
t=180 min = 3 hrs

Like? Yes (3) | No (5)  

Devendra Marghade   11 years AGO

If the total distance=2d, Speed =s
Time taken to cover first half=d/s
Time taken to cover last half=d/1.2s
Given d/s - d/(1.2s)=1/2 ,0.4d=1.2s or d/s=3 hour

Like? Yes (3) | No (5)  

H.BALA SUBRAHMANYAM   11 years AGO

let the uniform speed be s
let the time taken be t
then for first half the distance is given by s*t
for the second half the distance is given by (6s/5)*(t-30)
hence s*t = (6s/5)*(t-30)
hence t = 180 min
= 3 hours

Like? Yes (2) | No (4)  

shantanu sharma   11 years AGO

spped = d/t
t=1.2t+36

Like? Yes (2) | No (4)  

Md Shahnawaz Shahab   11 years AGO

1.2s*(t-0.5) = s*t
from above equation we can find out t(time).
NB: s=speed in the first half of the journey.

Like? Yes (2) | No (4)  

MUKESH KUMAR   11 years AGO

1st half:
d=r*t
2nd half:
d=(r.2r)*(t-30)
r*t=1.2*r*t-36r

divide both sides by r
t=1.2t-36
.2t=36
t=180

The 1st half takes 3 hrs to cover

Like? Yes (3) | No (5)  

jitendra singh   11 years AGO

3 hrs

Like? Yes (2) | No (4)  

ravi singh   11 years AGO

let speed =s and time in minute = t

since distance are same hance

s*t = (120/100)*s*(t-30)
by solving this t= 180 minute hence ans is 3hours

Like? Yes (2) | No (4)  

tushar singh   11 years AGO

let speed=s,
distance=2d, and time t min
in the 1st half
d/s=t

after increasing speed by 20%
speed=s+s20%=s+s*20/100
=6s/5
in 2nd half
d/6s/5=t-30

s*t*5/6s=t-30 (d=s*t)
t=180 minute
=3 hour

Like? Yes (2) | No (4)  

santhosh reddy vermareddy   11 years AGO

let speed be x
given x*(t+30)=1.2*x*t
t=150min
t+30=180min=3hrs

Like? Yes (2) | No (4)  

raksha agrawal   11 years AGO

let the speed for half distance is x, the for next half distance the speed is 1.2x
let the time to complete first half is t then time to complete second half is t-30

because distance travelled is same,
thus
x*t= 1.2x*(t-30)
Solving this
t=180 min
t=3 hr

Like? Yes (2) | No (6)