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 6 of 21.
Aditya said:
1 decade ago
@Abhi.
I still don't get it. The "3" from the original solution. He was supposed to be 2 hours behind. How does doubling his speed Which would means he takes 1 hour less than sameer end up being "3".
I still don't get it. The "3" from the original solution. He was supposed to be 2 hours behind. How does doubling his speed Which would means he takes 1 hour less than sameer end up being "3".
Piali said:
1 decade ago
Lets time taken by abhay be x and sameer be y.
Now acc to first line abhay is taking 2hrs more.
Hence , 30/x - 30/y = 2..... eqn 1 // time taken by abhay-sameer = 2.
Acc to second line speed of abhay got increased double.
So now sameer is taking 1 hour more.
30/y - 0/2x = 1.... eqn 2,
Adding eqns 1 and 2.
30/x - 30/2x = 3.
(60-30)/2x = 3 .
6x = 30.
x = 5.
Now acc to first line abhay is taking 2hrs more.
Hence , 30/x - 30/y = 2..... eqn 1 // time taken by abhay-sameer = 2.
Acc to second line speed of abhay got increased double.
So now sameer is taking 1 hour more.
30/y - 0/2x = 1.... eqn 2,
Adding eqns 1 and 2.
30/x - 30/2x = 3.
(60-30)/2x = 3 .
6x = 30.
x = 5.
Ranjani said:
1 decade ago
@Devi, why do you subtract 30/x and 30/2x and t+2 and t-1?
Parvathi said:
1 decade ago
WE CAN OBTAIN SOLUTION EVEN WITHOUT SUBTRACTING:
Let time taken by sameer be t.
Case 1:
30/x = t+2.
On rewriting,
30 = (t+2)x--------equ 1.
Case 2:
30/2x = t-1.
On rewriting,
30 = (t-1)2x-------equ 2.
We can equate equ 1 and 2, since they have same value 30
On equating,
(t+2)x = (t-1)2x.
(t+2) = (t-1)2.
t+2 = 2t-2.
t = 4.
On subs, t=4 in any equ we get,
30/(4+2) = x.
x = 5.
Let time taken by sameer be t.
Case 1:
30/x = t+2.
On rewriting,
30 = (t+2)x--------equ 1.
Case 2:
30/2x = t-1.
On rewriting,
30 = (t-1)2x-------equ 2.
We can equate equ 1 and 2, since they have same value 30
On equating,
(t+2)x = (t-1)2x.
(t+2) = (t-1)2.
t+2 = 2t-2.
t = 4.
On subs, t=4 in any equ we get,
30/(4+2) = x.
x = 5.
Abdul Majid P said:
1 decade ago
Hi friends,
Consider Abhayz speed be x km/hr, and time taken to cover 30km for Sameer is t;
We know, in both the cases, distance are equal.Therefore we can use this equation: v= d/t => d = v*t;
x(t+2) = 2x(t-1),
From the above we get t = 4, time for Sameer, but Abhay is 2hr more
in first case, So t+2 = 4+2 = 6hr.
Therefore x = d/t = 30/6 = 5kmph.
Consider Abhayz speed be x km/hr, and time taken to cover 30km for Sameer is t;
We know, in both the cases, distance are equal.Therefore we can use this equation: v= d/t => d = v*t;
x(t+2) = 2x(t-1),
From the above we get t = 4, time for Sameer, but Abhay is 2hr more
in first case, So t+2 = 4+2 = 6hr.
Therefore x = d/t = 30/6 = 5kmph.
Gagan said:
1 decade ago
Try to solve using options given.
Lets pick up the 1st option i.e. 5kmph.
If Abhay cover the distance of 30Km at 5kmph it means he takes = 6 Hours.
That means Sameer takes = 4 hours (Because Sameer takes 2 hours less than Abhay).
And when Abhay double his speed(10Kmph) then he would take = 3 Hours (which means Abhay take 1 hour less than sameer).
So the answer is 5Kmph.
Lets pick up the 1st option i.e. 5kmph.
If Abhay cover the distance of 30Km at 5kmph it means he takes = 6 Hours.
That means Sameer takes = 4 hours (Because Sameer takes 2 hours less than Abhay).
And when Abhay double his speed(10Kmph) then he would take = 3 Hours (which means Abhay take 1 hour less than sameer).
So the answer is 5Kmph.
Anonymous said:
1 decade ago
30/x = y+2 ->1 [x=speed,y=time].
30/2x = y-1 ->2.
Now 1-2 we get,
(30/x)-(30/2x) = (y+2)-(y-1).
(30/x)-(30/2x) = 3.
30/2x = y-1 ->2.
Now 1-2 we get,
(30/x)-(30/2x) = (y+2)-(y-1).
(30/x)-(30/2x) = 3.
Monpara Ashvin said:
1 decade ago
30/x = y+2 ->1 [x=speed, y=time].
30/2x = y-1 ->2.
Now 1-2 we get,
(30/x)-(30/2x) = (y+2)-(y-1).
(30/x)-(30/2x) = 3.
30/2x = y-1 ->2.
Now 1-2 we get,
(30/x)-(30/2x) = (y+2)-(y-1).
(30/x)-(30/2x) = 3.
Maaz said:
1 decade ago
Don't make it so complicated.
Suppose speed of Abhay is x kmph.
Time taken by sameer is t hr.
From the given statement we will get 2 equation.
30/x = t +2.
30/2x = t -1.
Here we got 2 equations And to unknown Variables. X&t.
Therefore after solving above 2 equations. We'll get the answer which is x=5 kmph.
Thanks.
Suppose speed of Abhay is x kmph.
Time taken by sameer is t hr.
From the given statement we will get 2 equation.
30/x = t +2.
30/2x = t -1.
Here we got 2 equations And to unknown Variables. X&t.
Therefore after solving above 2 equations. We'll get the answer which is x=5 kmph.
Thanks.
Kingstan said:
1 decade ago
Guys all are confusing ! here the simple method.
Ajay = x+2 (2 hours more than sameer?)
= 2(x-1) (doubles and 1 hour less?)
Therefore x=? 4? so sameer is 4 na ajay ? 6?
Then? s-d/t ? 30/6 --- 5 (ANS) !! Simple.
Ajay = x+2 (2 hours more than sameer?)
= 2(x-1) (doubles and 1 hour less?)
Therefore x=? 4? so sameer is 4 na ajay ? 6?
Then? s-d/t ? 30/6 --- 5 (ANS) !! Simple.
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