Aptitude - Time and Distance - Discussion
Discussion Forum : Time and Distance - General Questions (Q.No. 15)
15.
A man covered a certain distance at some speed. Had he moved 3 kmph faster, he would have taken 40 minutes less. If he had moved 2 kmph slower, he would have taken 40 minutes more. The distance (in km) is:
Answer: Option
Explanation:
Let distance = x km and usual rate = y kmph.
| Then, | x | - | x | = | 40 |  2y(y + 3) = 9x ....(i) |
| y | y + 3 | 60 |
| And, | x | - | x | = | 40 | y(y - 2) = 3x ....(ii) |
| y -2 | y | 60 |
On dividing (i) by (ii), we get: x = 40.
Discussion:
118 comments Page 9 of 12.
Jeevitha said:
1 decade ago
2y(y+3) = 9x.
(y cancel each other and 9x/3x=3).
y(y-2) = 3x.
After that we get:
2(y+3) = 3(y-2).
2y+6 = 3(y-2).
2y+6 = 3y-6.
6+6 = 3y-2y.
12 = y.
Therefore y = 12.
(y cancel each other and 9x/3x=3).
y(y-2) = 3x.
After that we get:
2(y+3) = 3(y-2).
2y+6 = 3(y-2).
2y+6 = 3y-6.
6+6 = 3y-2y.
12 = y.
Therefore y = 12.
Sruthi said:
1 decade ago
How to get y=12?
Once explain it.
Once explain it.
Lakshmi said:
1 decade ago
I solved the above two equations but I'm not getting the answer. So, please explain clearly.
Aayush said:
1 decade ago
3X-2X = X.
40(common)40 = X.
No logic, just this.
40(common)40 = X.
No logic, just this.
Jay said:
1 decade ago
@Snega,
Why you take t = d/x.
Just consider 2 and 4 equation.
Solve the equation. But we can't get the answer. What is the problem in that?
Just explain me please.
Why you take t = d/x.
Just consider 2 and 4 equation.
Solve the equation. But we can't get the answer. What is the problem in that?
Just explain me please.
Kunal said:
1 decade ago
How to divide this two equation to find the value of y?
Sneha said:
1 decade ago
Let the actual speed be xkmph and time be t hours.
Then distance (d) = x*t.
And so t = d/x.......(i).
Then from the question, d=(x+3)(t-40/60).......(ii).
Putting (i) in (ii),
d=(x+3)(d/x-40/60).
Then 40/60 =d/x-d/(x+3).....(iii).
Then taking the second condition from the given question,
d=(x-2)(t+40/60).......(iv).
Putting (i) in (iv),
d=(x-2)(d/x+40/60).
Then on solving 40/60 =d/(x-2)-d/x......(v).
Now on dividing (iv) and (v) , and on solving v get,
x=12kmph.
And putting the value of 'x' in (iii) and solving we get 'd=40km'.
Then distance (d) = x*t.
And so t = d/x.......(i).
Then from the question, d=(x+3)(t-40/60).......(ii).
Putting (i) in (ii),
d=(x+3)(d/x-40/60).
Then 40/60 =d/x-d/(x+3).....(iii).
Then taking the second condition from the given question,
d=(x-2)(t+40/60).......(iv).
Putting (i) in (iv),
d=(x-2)(d/x+40/60).
Then on solving 40/60 =d/(x-2)-d/x......(v).
Now on dividing (iv) and (v) , and on solving v get,
x=12kmph.
And putting the value of 'x' in (iii) and solving we get 'd=40km'.
Dewashish said:
1 decade ago
Solve the minimum value of sin^2x+cos^2x.
Kirti said:
1 decade ago
All I have is v+3=s/ (t-2/3).
And v-2=s/ (t+2/3). With these two eqns. How do we solve it?
And v-2=s/ (t+2/3). With these two eqns. How do we solve it?
Nive said:
1 decade ago
Still am not able to understand, its confusing !
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