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:
211 comments Page 9 of 22.
Meenakshi said:
5 years ago
Actually use simple logic.
Initially, he took 2 hrs more than Sameer, then he took one hr less.
which means the difference in time that has occurred in 3hrs.
So, Intial time - final time = 3hrs.
Initially, he took 2 hrs more than Sameer, then he took one hr less.
which means the difference in time that has occurred in 3hrs.
So, Intial time - final time = 3hrs.
Rocky said:
9 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.
Karim Mirazul said:
1 decade ago
Hey just stick with anyone solution then you will definitely sure about the solution, don't go for each and every solution. Because it will confuse you. And yes nice explanation by @Devi :).
Ankan Banerjee said:
8 years ago
Let time taken by Sameer = T hrs.
Before doubling speed, time taken by Abhay = T+2 hrs --> (1).
After doubling the speed time taken = T-1 hrs --> (2).
(1) - (2) = (T+2) - (T-1) = 3 hrs.
Before doubling speed, time taken by Abhay = T+2 hrs --> (1).
After doubling the speed time taken = T-1 hrs --> (2).
(1) - (2) = (T+2) - (T-1) = 3 hrs.
VIMAL said:
1 decade ago
Lets abhays speed x km/hr and sameer takes T hour to complete 30 km journey
X=30/T+2
2X= 30/T-1
THEN 2(30/T+2)=30/T-1
2T-2=T+2
T=4 hour
X= 30/T+2
30/4+2= 5 HOURS
X=30/T+2
2X= 30/T-1
THEN 2(30/T+2)=30/T-1
2T-2=T+2
T=4 hour
X= 30/T+2
30/4+2= 5 HOURS
Kanisetty Ashok said:
7 years ago
Abhay's difference in covering the distance after doubling the speed is 2+1 = 3 hrs.
So, Abhay's speed after doubling the speed is 30/3 = 10,
So, the current speed of Abhay is 10/2 = 5.
So, Abhay's speed after doubling the speed is 30/3 = 10,
So, the current speed of Abhay is 10/2 = 5.
Ashish Jalit said:
1 decade ago
Let the abhay speed be x then,
30=x*(t+2).........I
30 2x(t-1)..........II
t-1=30/2x
t=30/2x+1 ....Put this value in equation I , we get
30=x*(15/x+3)
30=15+3x
x=15/3.
x=5 km/hrs.
30=x*(t+2).........I
30 2x(t-1)..........II
t-1=30/2x
t=30/2x+1 ....Put this value in equation I , we get
30=x*(15/x+3)
30=15+3x
x=15/3.
x=5 km/hrs.
Jafar said:
7 years ago
Given that Abhay+2 hours= Sameer.
And 2Abhy-1=Sameer.
On solving these two equations;
We will get Abhy = 3.
And Sameer = 5.
We need to find the Abhay speed that's why we take 3.
And 2Abhy-1=Sameer.
On solving these two equations;
We will get Abhy = 3.
And Sameer = 5.
We need to find the Abhay speed that's why we take 3.
Amit said:
1 decade ago
Guys it could be like this also....
Abhay speed =x
sameer speed=y
(30/x)- (30/y)= 2 ....eq-1
-(30/2x)+(30/y)= 1 ...eq-2
Solving 1 and 2 we get => 6x=30 so x=5 answer.
Abhay speed =x
sameer speed=y
(30/x)- (30/y)= 2 ....eq-1
-(30/2x)+(30/y)= 1 ...eq-2
Solving 1 and 2 we get => 6x=30 so x=5 answer.
Niraj katwal said:
3 years ago
Let the speed of Sameer be x.
Then, Ajay's speed = x+2.
When Ajay doubles his speed,
2(x+2) = x -1
or,2x+4 = x-1
Since speed can never be negative.
x=5.
Then, Ajay's speed = x+2.
When Ajay doubles his speed,
2(x+2) = x -1
or,2x+4 = x-1
Since speed can never be negative.
x=5.
(105)
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