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 3 of 21.
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.
Sakshi said:
1 decade ago
Its a little bit long bt simple way to solve it.
Let suppose Sameer takes x hrs to cover distance
so, Abhay will take x+2 hrs
speed of Abhay = dis/time = 30/x+2
when Abhay doubles his speed = 2*30/x+2 = 60/x+2
now, time taking by Abhay = dis/speed = 30/60/x+2 = x+2/2..0k
Abhay wiil take 1 hr less than Sameer
Therrfore, x+2/2= x-1; x=4
Abhay's time= x+2= 6
now Abhay's speed= 30/6= 5km/hr.
Let suppose Sameer takes x hrs to cover distance
so, Abhay will take x+2 hrs
speed of Abhay = dis/time = 30/x+2
when Abhay doubles his speed = 2*30/x+2 = 60/x+2
now, time taking by Abhay = dis/speed = 30/60/x+2 = x+2/2..0k
Abhay wiil take 1 hr less than Sameer
Therrfore, x+2/2= x-1; x=4
Abhay's time= x+2= 6
now Abhay's speed= 30/6= 5km/hr.
Vikas said:
9 years ago
We have to find the speed of Abhay.
So,
let speed of Abhay = x km/hr.
Time taken by Sameer = t hrs.
Now, first condition,
Speed = distance/time.
So, Abhay takes 2 hrs more than Sameer that comes.
(30/t+2)= x ----> eq(1).
Now condition second,
If abhay doubles his speed that is x km/hrs to 2xkm/hr. Time t reduces to t-1,
{30/(t+2)}=2x ----> eq(2).
By solving equation1 and 2.
x = 5km/hr.
So,
let speed of Abhay = x km/hr.
Time taken by Sameer = t hrs.
Now, first condition,
Speed = distance/time.
So, Abhay takes 2 hrs more than Sameer that comes.
(30/t+2)= x ----> eq(1).
Now condition second,
If abhay doubles his speed that is x km/hrs to 2xkm/hr. Time t reduces to t-1,
{30/(t+2)}=2x ----> eq(2).
By solving equation1 and 2.
x = 5km/hr.
Shivakar said:
9 years ago
Let Sameer speed is X km/h and time taken by Sameer is T.
Let Abhay speed is Y km/h.
Then, According to question.
30/X = T........(i)
30/Y = T+2....(ii)
Now,
Abhay speed is Double means 2Y
so, time taken by Abhay
30/2Y=T-1.....(iii)
Substitute the Eqn..(iii) in Eqn...(ii).
30/x-30/2Y = T+2-(T-1),
30/X-30/2Y = T+2-T+1,
(60-30)/2Y = 2+1,
30/2Y = 3,
30 = 6Y,
Hence,
Y = 5 kmph.
Let Abhay speed is Y km/h.
Then, According to question.
30/X = T........(i)
30/Y = T+2....(ii)
Now,
Abhay speed is Double means 2Y
so, time taken by Abhay
30/2Y=T-1.....(iii)
Substitute the Eqn..(iii) in Eqn...(ii).
30/x-30/2Y = T+2-(T-1),
30/X-30/2Y = T+2-T+1,
(60-30)/2Y = 2+1,
30/2Y = 3,
30 = 6Y,
Hence,
Y = 5 kmph.
Anki said:
1 decade ago
Let us look at ths prb this way:
Let abhays's speed be x and sameers time be y
Case 1: Abhays speed=x
Abhays time =t+2
Case 2: abhays speed=2x
Abhays time =t-1
We knw : Speed is inversely proportional to time
So, S2/S1=t1/t2
2x/x=t+2/t-1
2=t+2/t-1
solving we get t=4
hence we can calculate
Speed of abhay=Distance/time
=30/(4+2) ..Since t=4
30/6=5
Let abhays's speed be x and sameers time be y
Case 1: Abhays speed=x
Abhays time =t+2
Case 2: abhays speed=2x
Abhays time =t-1
We knw : Speed is inversely proportional to time
So, S2/S1=t1/t2
2x/x=t+2/t-1
2=t+2/t-1
solving we get t=4
hence we can calculate
Speed of abhay=Distance/time
=30/(4+2) ..Since t=4
30/6=5
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.
Harsh said:
5 years ago
Let's make it simple.
Let Abhay speed be x and Sameer be y
Equation 1be like this:
30/x - 30/y = 2.
Equation 2 be like this;
30/2x - 30/y = - 1(its because the double speed of Abhay than Sameer make a difference of 1 hour, i.e, Abhay take less time than Sameer that's why I put - 1).
On Solving both the equation you get something like this;
-30y= - 6xy.
6x=30 or x=5.
Let Abhay speed be x and Sameer be y
Equation 1be like this:
30/x - 30/y = 2.
Equation 2 be like this;
30/2x - 30/y = - 1(its because the double speed of Abhay than Sameer make a difference of 1 hour, i.e, Abhay take less time than Sameer that's why I put - 1).
On Solving both the equation you get something like this;
-30y= - 6xy.
6x=30 or x=5.
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.
Anudeep said:
6 years ago
Distance=30,
Time of sameer=s,
Time taken by abhay = 2hrs+s.
The Speed of Abhay= 30/(2hrs+s),
Time is taken by Abhay after doubling speed = s-1hr.
Speed of abhay= 60/(s-1hr),
We know that both are the speeds of Abhay, So both the speeds are equal.
30/(2hrs+s) = 60/(s-1hr),
30(s-1hr) = 60(2hr+s).
150hr = 30s,
s=5.
So, the final answer is 5.
Time of sameer=s,
Time taken by abhay = 2hrs+s.
The Speed of Abhay= 30/(2hrs+s),
Time is taken by Abhay after doubling speed = s-1hr.
Speed of abhay= 60/(s-1hr),
We know that both are the speeds of Abhay, So both the speeds are equal.
30/(2hrs+s) = 60/(s-1hr),
30(s-1hr) = 60(2hr+s).
150hr = 30s,
s=5.
So, the final answer is 5.
Hasan said:
4 years ago
Distance = 30km/hr.
The time taken by Sameer is t.
Abhay traveling with a Speed X.
Case 1:
Time taken by Abhay to cover 30km is 30/X = t+2.
Case 2(doubles speed):
Time taken by Abhay to cover 30km is 30/2X = t-1.
(30/X)- (30/2X) = (t+2)-(t-1),
(30/X) - (30/2X) = t+2-t+1,
(30 * 2X) - (30 * X) = 3 (2X^2) //LCM.
60X - 30X = 6 X^2,
30X = 6 X^2,
30 = 6 X,
X = 5.
The time taken by Sameer is t.
Abhay traveling with a Speed X.
Case 1:
Time taken by Abhay to cover 30km is 30/X = t+2.
Case 2(doubles speed):
Time taken by Abhay to cover 30km is 30/2X = t-1.
(30/X)- (30/2X) = (t+2)-(t-1),
(30/X) - (30/2X) = t+2-t+1,
(30 * 2X) - (30 * X) = 3 (2X^2) //LCM.
60X - 30X = 6 X^2,
30X = 6 X^2,
30 = 6 X,
X = 5.
(2)
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