Walking 5/6 of his usual speed, a man is 10 minute too late. The usual time taken by him to cover th

Swarnabha Dey   10 years AGO

he is walking 5/6 th of his speed, i.e
Actual speed : present speed = 6 : 5
as we know speed = distance/Time
so the Time should be = 5 : 6
so , Let actual time taken is 5x
present time taken is 6x
so, 6x-5x =10
=> x=10
=>5x= 50
So the actual ime taken is 50 mins .

Like? Yes (88) | No (25)  

Rajamohan Raju   10 years AGO

S = D/T ------------- (1)
(5/6)S = D/(T+10) -------- (2)
By solving this equations we get T=50min.

Like? Yes (63) | No (26)  

Narendra Jadon   10 years AGO

Let the time taken by him to cover the distance normally is T min.
Let the normal speed to reach in time is S

Distance = S x T
when he is late by 10 min his distance is :
Distance = 5/6 x S x (T+10)

Distances are equal in both cases hence equating them :
S x T = 5 / 6 x S x (T+10)
6T = 5T + 50
T = 50 min.
Option 2.

Like? Yes (24) | No (5)  

Sundar Kumar   10 years AGO

formula p/q=5/6,t=10
usual time= (5*10)/|5-6|=50
Answer: Option 2) 50 minute

Like? Yes (29) | No (20)  

RAKESH   10 years AGO

6t/5 - t = 10
=> t = 50 min

Like? Yes (18) | No (12)  

navya   10 years AGO

s=x/t
(5/6)s=x/(t+10)
on solving these two equations we get t=50min

Like? Yes (13) | No (7)  

Harshad    10 years AGO

New speed=5/6 of the usual speed
New time taken =6/5 of the usual time
So,(6/5 of the usual time) - (usual time)=10 min
=> 1/5 of the usual time =10 min => usual time=50 min

Like? Yes (11) | No (6)  

anushree dey   10 years AGO

let x=speed
t=time taken when speed is x so...
xt=5/6x(t+10)
t=50minutes
option-(2)

Like? Yes (8) | No (3)  

SAYAN CHATTERJEE   10 years AGO

Let, the distance be 100
so, at velocity v km/hr time taken= 100/v
at a velocity, 5v/6 km/hr time taken= (100/(5v/6))=120/v
so, (120/v-100/v)= 1/6
v= 120 km/hr.
so, original time taken= (100/120*60)mins
=50 minutes (option 2)

Like? Yes (3) | No   

Surya Rajitha   10 years AGO

Let his usual speed is v. and usual time is t.

distance he travelled d=vt

at 5/6 th of his speed.. he's late by 10 min.

so (5/6)v*(t+10)=d

vt=(5/6)v*(t+10)

solving for 5 we get

t=(5/6)*(t+10)
6t=5t+50

t=50min

Like? Yes (9) | No (6)  

Saraswathy   10 years AGO

d/s = t
6d/5s = t + 1/6
Solving , we ll get t = 5/6 hrs => 5/6 * 60 = 50 min

Ans : 50 min (Option 2)

Like? Yes (6) | No (4)  

kundan kumar   10 years AGO

s1/s2=t2/t1
time=6/5of usual
so extra time=1/5 of usual=10
so usual time=50 min

Like? Yes (4) | No (2)  

guddu   10 years AGO

as distance is constant so S*T=5/6S*6/5T,so late=6/5T-T=10
t=50

Like? Yes (3) | No (2)  

Ravi Bhalani   10 years AGO

New speed=5/6 of the usual speed
New time taken =6/5 of the usual time
So,(6/5 of the usual time) - (usual time)=10 min
=> 1/5 of the usual time =10 min => usual time=50 min

Like? Yes (5) | No (4)  

HARDIP BHALANI   10 years AGO

New speed=5/6 of the usual speed
New time taken =6/5 of the usual time
So,(6/5 of the usual time) - (usual time)=10 min
=> 1/5 of the usual time =10 min => usual time=50 min

Like? Yes (5) | No (4)  

BITU KUMAR NIRALA   10 years AGO

1/(5/6)=(t+10)/t
hence t=50 min

Answer :- 50 min (option-2)

Like? Yes (3) | No (2)  

Kushagra Tiwari   10 years AGO

let x=speed
t=time taken when speed is x so...
xt=5/6x(t+10)
t=50 minutes
Hence proof

So option is 2

Like? Yes (4) | No (3)  

Saurabh Kumar   10 years AGO

{d/(5/6x)}- d/x=10min

where d/x is the normal speed

Like? Yes (3) | No (2)  

sompon temtheerakij   10 years AGO

let X = The usual time
Distance = rate * Time
5/6*(X+10) = 1*X
5X+50 = 6X
X = 50

Like? Yes (2) | No (1)  

Ramesh Sriram   10 years AGO

distance is constant
let distance=d, time=t, speed=x
d=x*t=5/6*x*(t+10)
5/6*x*(t+10)=x*t
t=50 minutes

Like? Yes (10) | No (9)  

Sumit Ranjan   10 years AGO

original speed=s ,new speed=5s/6.
original time=t,new time=t+10.
now,distance is constant.....
so..s/(5s/6)=(t+10)/t
t=50 min.....

Like? Yes (3) | No (2)  

vishnuvardhan   10 years AGO

let t=time taken
(6/5*t)-t=10min
t/5=10
t=50min

Like? Yes (3) | No (2)  

RIPON DEB   10 years AGO

let usual velocity is x km/hr. so present velocity is (5x)/6 km/hr.
so we have {(6*60)/(5*x)}-(60/x)=10
x=6/5
so 6/5 km goes in 60 min
so 1 km goes in50 min.
ans is option 2.

Like? Yes (3) | No (2)  

dheeraj   10 years AGO

let original speed be x and time be t
x*t=5/6 x(t+10)

t=50

Like? Yes (4) | No (3)  

ABHINAV THAKUR   10 years AGO

Let the normal speed of man be 's' and time taken be 't'. So, distance covered will be s*t. Now, the man walks with 5/6 of his usual speed, so present speed = 5/6x and time taken = t+10. So distance covered = 5/6x(t+10). Since distt is same, so xt=5/6x(t+10), solving this we get t= 50 mins.

Like? Yes (3) | No (3)  

Tark Patel   10 years AGO

Let usual speed is s and usual time is t,
so distance is same
( (5 / 6) s ) ( t + 10 ) = ( s ) ( t )
-> 5 t + 50 = 6 t
-> t = 50minutes

Like? Yes (2) | No (2)  

Devendra Marghade   10 years AGO

If distance=d & speed=x
Usual time taken =d/x
Given, d/x=[d/(5x/6)] - (10/60)
On solving, d/x =5/6 Hrs= 50 Minutes

Like? Yes (2) | No (2)  

BHANU KUMAR   10 years AGO

let the usual speed be x m/s.
and it takes t second to cover the distance.

as speed is inversely proportional to time so
x:5/6x=t+10:t
hence t=50 minutes (OPTION 2)

Like? Yes (3) | No (3)  

sanghavi    10 years AGO

let usual time be t and seed s
then
(5/6)*s * (t+10)=s * t
st-(5/6)*st=50s/6
s/6(6t-5t)=50s/6
1t=50

Like? Yes (2) | No (2)  

DIPAK TIWARI   10 years AGO

Let normal speed be Xkm/hr
And total distance be Ykm/hr
Hence, (6X/5Y)-(10/60)=X/Y
=>(6X-5X)/5Y=1/6
=>X/Y=5/6 hour=50 minutes

Like? Yes (2) | No (2)  

raj ranjan verma   10 years AGO

let distance be d & x be the original speed
then,
d/(5x/6)-d/x=10 by solving this eqn we will get....d/x=50 i.e his normal time taken to cover the same distance

Like? Yes (2) | No (2)  

KRISHNA CHANDRA UPADHYAY   10 years AGO

The ratio of his usual speed & present speed is 6:5
So the ratio of usual time & present time taken by
him is 5:6
We can say that in this ratio in present, he is 1 min late. But in actual question, he is 10 min late .
So simply, multiply the gap with 10 to get 10 from 1.
So the usual time & present time taken by him is 50 min & 60 min.
Hence usual time taken by him = 50 minute.
hence, option (2) is correct.

Like? Yes (3) | No (3)  

BISHVAMBHAR KUMAR   10 years AGO

1-1/6=5/6
1/6 mai 10min
1 mai 10*6=60
5/6mai=60*5/6=50

Like? Yes (2) | No (2)  

Abhishek Vishwakarma   10 years AGO

let n=speed
t=time taken when speed is n so...
nt=5/6n(t+10)
t=50 minute

So option is 2

Like? Yes (2) | No (2)  

AMAN KUMAR   10 years AGO

let the distance traveled is z km
usual speed x km/hr
so, z/(5x/6)-z/x= 1/6 hr.
hence x=6z/5 km/hr
so time taken to cover distance is =z/(6z/5)hr
time taken =5/6 hr=50 min

Like? Yes (2) | No (2)  

Anshu sharma   10 years AGO

Speed=5/6
time=6/5
6/5x-x=10
x/5=10
x=50min

Like? Yes (2) | No (2)  

perni.raviteja   10 years AGO

usual time =late time/{1/(5/6-1)

=10/(6/5-1)
=10/(1/5)
=50min

Like? Yes (2) | No (2)  

Saranya   10 years AGO

new speed=5/6 of usual speed
new time taken=6/5 of usual time
1/5 usual time=10 min
usual time=50 min

Like? Yes (2) | No (2)  

Divya   10 years AGO

speed=dist/time
s1/s2=t2/t1
s1=5/6 of s
s2=s; t1=t+10;t2=t
so substituing in above formula.u wil get ansr=50 min

Like? Yes (3) | No (3)  

RANDHIR KUMAR   10 years AGO

distance=x*t,present spd=5x/6,nd tym=t+10,so x*t=5x/6*(t+10)=50 minute..

Like? Yes (2) | No (2)  

naina   10 years AGO

5/6 th speed,extra time =1/5th of time =10 min.
usual time = 50 min

Like? Yes (2) | No (2)  

venkatesh   10 years AGO

let x=speed
t=time taken when speed is x so...
xt=5/6x(t+10)
t=50 minutes

Like? Yes (2) | No (2)  

Sneha Goel   10 years AGO

6/5T-T=10
T=50 mins

Like? Yes (2) | No (2)  

natuva aditya   10 years AGO

5/6=x/x-10
cross multiplying the above equation,
we get 5x-50=6x,
x=10 [neglecting the x -ve value]

Like? Yes (2) | No (2)  

Deepanwita   10 years AGO

Let the speed be x.
Distance is D.
Then time taken is =D/x

now speed=(5/6)*x
time taken=D/x +10

so we can write:
D=(5x/6)*((D/x)+10)
D=50x

Therefore:
time=D/x=50x/x=50 minutes

hence option (2)

Like? Yes (2) | No (2)  

Varunraj Reddy   10 years AGO

Ratio of speeds between usual & present = 6:5
So ratio of time taken= 5:6
Difference in ratio 1 part
According to problem 1 part of time=10 min
So 5 parts of usual time taken=5*10=50 min

Option 2)

Like? Yes (2) | No (2)  

SWATHI   10 years AGO

Assume Usual speed=x
So Speed(new)=5/6(x)
If the speed(new) is a:b the time(new) will be b:a.
time(new)=6/5(x);
So ,(6/5x)-10 min=(x)
.;50min.

Like? Yes (2) | No (2)  

prem kumar   10 years AGO

present speed=5/6*usual speed
present time=6/5*usual time
extra time taken=6/5(usual time)-usual time
1/5*usual time
1/5 of usual time=10 minutes
hence usual time=50 minutes
so option 2

Like? Yes (2) | No (2)  

T.VIGNESWARA REDDY   10 years AGO

s*t=5/6(s*(t+10)) (distance is same) d=s*t so s canceled on both sides 6t=5t+50 t= 50

Like? Yes (2) | No (2)  

pavan teja   10 years AGO

for constant distance
s1*t1=s2*t2
so let the actual distance and time be x and y respectively. then
x*y=(5/6)*x*(y+10)

Like? Yes (2) | No (2)  

MRITUNJAY KUMAR   10 years AGO

let d an x be the distance & usual speed of the man resp.
6d/5x-d/x=10 => d/x(1/5)=10
=> d/x=t=50 minutes.

Like? Yes (2) | No (2)  

Shivangi Rawat   10 years AGO

equating both distances initial one as s*t and later one as (5/6)s(t+10) we get 50 minutes.

Like? Yes (2) | No (2)  

Ritika   10 years AGO

Let distance=x and usual time is t
Then usual speed=x/t
Now,time=t+10 so,speed=x/(t+10)
A/Q,

x/(t+10)=5/6(x/t)
6t=5t+50
t=50min

Like? Yes (2) | No (2)  

AJAY KUMAR   10 years AGO

(6/5)x-(1/x)=600
1/x=3000/60=50 min

Like? Yes (2) | No (2)  

Vikram Saini   10 years AGO

its 50 minutes to cover the distance

Like? Yes (2) | No (2)  

Dharmendra   10 years AGO

50

Like? Yes (2) | No (3)  

Yogesh   10 years AGO

Given Ratio of speed=5/6
Gap in the ratio is 1 minute.
He is 10 minutes late.
Multiply this by given ratio i.e.
50:60
His Usual taken is 50 Minute

Like? Yes (2) | No (3)  

Arijit Acharyya   10 years AGO

d=s*t--------(1)
d=(5/6)*s*(t+10)-----(2)
by solving....
t=50 minute
option 2)

Like? Yes (2) | No (3)  

K S SREEKRISHNAN   10 years AGO

New speed = 5/6 of usual speed
As distance is same
New time = 6/5 of usual time
So,
New time-usual time=10 minutes
6/5 usual time-usual time=10 minutes
1/5 usual time=10 minutes
Therefore
usual time = 50 minutes (option 2)

Like? Yes (2) | No (3)  

Urvashi Jain   10 years AGO

let speed be s and time taken to cover that distance is t
we know that speed *time =distance
s*t=d...................(1)
according to the question
(5/6)*s*(t+10)=d.......................(2)
put eqn2 in 1
s*t=(5/6)*s*(t+10)
st-(5/6)st=50/6s
(1/6)st=(50/6)s
t=50 min
option 2

Like? Yes (2) | No (3)  

Tania Mondal   10 years AGO

let initial speed= x unit/min
new speed=(5/6)*x unit/min
(5/6)*x=x/(6/5)
the distance covered in 1min takes now 6/5 min
so late by (6/5)-1= 1/5min
1/5min late for 1min
then10min late for 10/(1/5)=50min
so ans 2 is correct

Like? Yes (1) | No (3)  

RAJIB BANERJEE   10 years AGO

Let,man's usual speed 6 m/min so,5/6 of his usual speed means 5 m/min,
Now,usual time taken 50 mins,so distance covered by the man is 300 meters.
300 meters covered in 1 hour (10 mins greater than usual) when man's speed is 5/6 of his speed... so the usual time taken is 50 mins.

Like? Yes (1) | No (3)  

Eswari   10 years AGO

speed is inversly proportional to time.
s1/s2=t2/t1
s/(5/6s)=(t+10)/t
6t=5t+50
t=50

Like? Yes (1) | No (3)  

Logesh   10 years AGO

Simply by using logical calculations

Like? Yes (1) | No (3)  

prabhat kumar   10 years AGO

according to formula p/q=5/6 and t=10minut
usual time=10*5/|5-6|=50minut

Like? Yes (3) | No (5)