Aptitude - Time and Distance - Discussion
Discussion Forum : Time and Distance - General Questions (Q.No. 11)
11.
In covering a distance of 30 km, Abhay takes 2 hours more than Sameer. If Abhay doubles his speed, then he would take 1 hour less than Sameer. Abhay's speed is:
Answer: Option
Explanation:
Let Abhay's speed be x km/hr.
| Then, | 30 | - | 30 | = 3 |
| x | 2x |
6x = 30
x = 5 km/hr.
Discussion:
209 comments Page 15 of 21.
UDAY said:
8 years ago
Your method is correct and easy to understand @Devi.
Rocky said:
8 years ago
@Mayank.
When Abhay speed was x he reached 2 hours early.
And when it doubled he reached 1 hour early then what it usually took for him.
So he reached 3 hours early when he doubled his speed.
When Abhay speed was x he reached 2 hours early.
And when it doubled he reached 1 hour early then what it usually took for him.
So he reached 3 hours early when he doubled his speed.
Shru said:
8 years ago
Good explanation @Devi.
Neha said:
8 years ago
Why subtraction takes place? Please, tell me the reason.
Wasiq said:
8 years ago
Best and helpful explanation, thanks @Devi.
Tejeshwar Rao said:
8 years ago
Distance=30km.
Let the speeds of abhay be x kmph and sameer be y kmph.
then,
30/x - 30/y =2----i,
When abhay doubles his speed
30/y - 30/2x = 1-----ii,
By i+ii we get,
30/x - 30/2x=3,
6x=30,
x=5km/hr.
Let the speeds of abhay be x kmph and sameer be y kmph.
then,
30/x - 30/y =2----i,
When abhay doubles his speed
30/y - 30/2x = 1-----ii,
By i+ii we get,
30/x - 30/2x=3,
6x=30,
x=5km/hr.
Tejeshwar Rao said:
8 years ago
Distance=30km.
Let the speeds of abhay be x kmph and sameer be y kmph.
Then,
30/x - 30/y =2 ---> I.
When abhay doubles his speed
30/y - 30/2x = 1 ---> II
By I + II we get,
30/x - 30/2x=3
6x=30.
x=5km/hr.
Let the speeds of abhay be x kmph and sameer be y kmph.
Then,
30/x - 30/y =2 ---> I.
When abhay doubles his speed
30/y - 30/2x = 1 ---> II
By I + II we get,
30/x - 30/2x=3
6x=30.
x=5km/hr.
Kishore said:
8 years ago
Speed = distance/time.
Let x be Abhay's speed and y be Sameer's time.
* first condition
speed s = 30/(y+2) ; ------- a.
speed 2s=30/(y-1); ---------b.
Put a ((s=30/(y+2)) in b.
Now you get y=4.
put y=4 in either a or b.
You will get speed as 5kmph.
Let x be Abhay's speed and y be Sameer's time.
* first condition
speed s = 30/(y+2) ; ------- a.
speed 2s=30/(y-1); ---------b.
Put a ((s=30/(y+2)) in b.
Now you get y=4.
put y=4 in either a or b.
You will get speed as 5kmph.
Ram ji said:
8 years ago
You can use this formula for this type of questions d=(s1*s2)/s1-s2*t.
Here d:distance(30) ,t: time difference( 3).
Here d:distance(30) ,t: time difference( 3).
Kazi faruk said:
8 years ago
Thanks for your explanation @Devi.
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