Aptitude - Time and Distance - Discussion
Discussion Forum : Time and Distance - General Questions (Q.No. 3)
                   
                                       
                                3.
If a person walks at 14 km/hr instead of 10 km/hr, he would have walked 20 km more. The actual distance travelled by him is:
 
                                    Answer: Option
                                                    Explanation:
                                                Let the actual distance travelled be x km.
| Then, | x | = | x + 20 | 
| 10 | 14 | 
 14x = 10x + 200
 4x = 200
 x = 50 km.
Discussion:
174 comments Page 12 of 18.
                
                        Raj said: 
                         
                        9 years ago
                
                Here time is constant = t;
No old speed = 10 new speed 10*14/10.
D-14/10d = 20.
D = 20*10/4 = 50 km.
                No old speed = 10 new speed 10*14/10.
D-14/10d = 20.
D = 20*10/4 = 50 km.
                        Feroz said: 
                         
                        9 years ago
                
                14 - 10 = 4,
20/4 = 5,
5 * 10 = 50.
                20/4 = 5,
5 * 10 = 50.
                        Jayant said: 
                         
                        9 years ago
                
                @All.
Thanks, you all expressed clearly.
                Thanks, you all expressed clearly.
                        Krishna Sharma said: 
                         
                        9 years ago
                
                Please solve this.
While covering a distance of 60 km, A takes 4 hours more than B. If A doubles its speed, A took 2 hours less than B. Find the walking speed using a simultaneous equation or quadratic equation.
                While covering a distance of 60 km, A takes 4 hours more than B. If A doubles its speed, A took 2 hours less than B. Find the walking speed using a simultaneous equation or quadratic equation.
                        Satya said: 
                         
                        9 years ago
                
                Given;
14kmph at 20 km MORE i.e extra speed.
So we can write it as 4 kmph -20.,
Each km 20÷4 = 5.
Hence 10 * 5 = 50.
                14kmph at 20 km MORE i.e extra speed.
So we can write it as 4 kmph -20.,
Each km 20÷4 = 5.
Hence 10 * 5 = 50.
                        Vikash said: 
                         
                        9 years ago
                
                Let actual distance x km.
Actual speed =10 kmph T1 = ?
      
Man walk 20 km more,
Distance = x + 20.
Speed = 14 kmph T2 = ?
Now T1 = T2.
D1/S1 = D2/S2.
x/10 = x + 20/14 after solving x (actual distance) = 50 km answer.
                Actual speed =10 kmph T1 = ?
Man walk 20 km more,
Distance = x + 20.
Speed = 14 kmph T2 = ?
Now T1 = T2.
D1/S1 = D2/S2.
x/10 = x + 20/14 after solving x (actual distance) = 50 km answer.
                        Sambath said: 
                         
                        9 years ago
                
                @Gowtham, @Ilesh.
Both are explained very well. Thank you.
                Both are explained very well. Thank you.
                        Sonia choudhary said: 
                         
                        9 years ago
                
                @Kiran.
Yes, I agree with your solution. Thanks for this solution.
                Yes, I agree with your solution. Thanks for this solution.
                        Rajesh H said: 
                         
                        9 years ago
                
                Detailed Explanation:
Cconsider,
Z is actual distance travelled (when going at 10km/hr).
X is time (Unknown).
Case: 1
travelling at normal speed (10Km/hr).
10Km-----------------@1 hr
ZKm-------------------@X hr
after cross multiplying each other we get;
10X=Z*1 means that 10X=Z---------1)
Case 2:
Travelling at 14Km/hr.
|||y
14Km-------------@1hr.
20+ZKm----------@Xhr.
here, 20+Z means 20Km + Z Km travelled by Normal Speed @ same X hour.
by cross multiplying we get,
14X=20+Z-----------------2)
just solve both equation 1) & 2.
sub 1 in 2
Z=10X ,we get 14X=20+10X.
so, 4X=20 , X=5 now, we are not questioned for hours but only we need distance so,sub X directly in equation 2).
We get 70=20+Z so Z = 70-20 = 50Km.
                Cconsider,
Z is actual distance travelled (when going at 10km/hr).
X is time (Unknown).
Case: 1
travelling at normal speed (10Km/hr).
10Km-----------------@1 hr
ZKm-------------------@X hr
after cross multiplying each other we get;
10X=Z*1 means that 10X=Z---------1)
Case 2:
Travelling at 14Km/hr.
|||y
14Km-------------@1hr.
20+ZKm----------@Xhr.
here, 20+Z means 20Km + Z Km travelled by Normal Speed @ same X hour.
by cross multiplying we get,
14X=20+Z-----------------2)
just solve both equation 1) & 2.
sub 1 in 2
Z=10X ,we get 14X=20+10X.
so, 4X=20 , X=5 now, we are not questioned for hours but only we need distance so,sub X directly in equation 2).
We get 70=20+Z so Z = 70-20 = 50Km.
                        White said: 
                         
                        9 years ago
                
                The actual distance 50km(10km) but he travelled 70km by 14kmph, the question is THE ACTUAL DISTANCE TRAVELLED BY HIM? Is it 70 or 50?
                Post your comments here:
 
            
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