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 3 of 12.
Durgesh Singh said:
6 years ago
S + s1/s1*t1=S+s2/s2*t2=T.
So S+3/3*40=S-2/2*40,
S= 12.
Then T= 12+3/3*40/60 = 10/3.
ST = D.
So 10/3*12 = 40.
So S+3/3*40=S-2/2*40,
S= 12.
Then T= 12+3/3*40/60 = 10/3.
ST = D.
So 10/3*12 = 40.
Manzoor said:
6 years ago
Diff = 1speed.80/60time.
Add =5
Multiple = 6.
80/60 * 6 * 5/1 = 40.
Add =5
Multiple = 6.
80/60 * 6 * 5/1 = 40.
Sachin Bhagat said:
6 years ago
For simplifying method, follow easy steps which are lengthy but clear understanding.
(x/y)-(x/y+3)=40/60 ---> (1)
(x/y-2)-(x/y)=40/60 ---> (2)
Let solve eq. (1)
(x/y)-(x/y+3) = 40/60,
(x(y+3)-y(x)) / y(y+3) = 2/3,
((xy+3x-xy) / (y^2+3y)) = 2/3,
3(3x)=2(y^2+3y),
9x=2y(y+3) ---------> (A).
Now let solve eqn. (2)
(x/y-2)-(x/y) = 40/60,
((xy-x(y-2)) /(y(y-2))) = 2/3,
((xy-xy+2x) / (y^2-2y)) = 2/3,
3(2x) = 2(y^2-2y)
6x = 2y(y-2)
now divide by 2 we get
3x = y(y-2) --------------> (B)
now we get two equations as per given in answer are
9x = 2y(y+3) -------------->(A).
3x = y(y-2) -------------->(B).
Now we divide eqn (A) by (B) we get;
(2y(y+3)) / y(y-2) = 9x/3x,
2(y+3) / (y-2) = 3,
y+3 / y-2 = 3/2,
2(y+3) = 3(y-2),
y = 12.
After putting y=12 in equation (A) we get x = 40.
(x/y)-(x/y+3)=40/60 ---> (1)
(x/y-2)-(x/y)=40/60 ---> (2)
Let solve eq. (1)
(x/y)-(x/y+3) = 40/60,
(x(y+3)-y(x)) / y(y+3) = 2/3,
((xy+3x-xy) / (y^2+3y)) = 2/3,
3(3x)=2(y^2+3y),
9x=2y(y+3) ---------> (A).
Now let solve eqn. (2)
(x/y-2)-(x/y) = 40/60,
((xy-x(y-2)) /(y(y-2))) = 2/3,
((xy-xy+2x) / (y^2-2y)) = 2/3,
3(2x) = 2(y^2-2y)
6x = 2y(y-2)
now divide by 2 we get
3x = y(y-2) --------------> (B)
now we get two equations as per given in answer are
9x = 2y(y+3) -------------->(A).
3x = y(y-2) -------------->(B).
Now we divide eqn (A) by (B) we get;
(2y(y+3)) / y(y-2) = 9x/3x,
2(y+3) / (y-2) = 3,
y+3 / y-2 = 3/2,
2(y+3) = 3(y-2),
y = 12.
After putting y=12 in equation (A) we get x = 40.
(5)
Madhavi said:
7 years ago
Thanks @Amit Kr. Soni.
(1)
Magnus said:
7 years ago
@ Ruchir,
For Converting Minutes to an hour you divide by 60.
For Converting Minutes to an hour you divide by 60.
Ruchir said:
7 years ago
How come 40/60? Please explain in detail.
Kamu said:
7 years ago
Why subtracting the equation?
Shreeshail karajol said:
7 years ago
How come 40/60? Please explain.
Varun said:
7 years ago
d=s*t,
t=40 mins.
s=3 km/hr-2km/hr =1 km/h,
d=1*40=40km/hr.
t=40 mins.
s=3 km/hr-2km/hr =1 km/h,
d=1*40=40km/hr.
Damu said:
7 years ago
Just from simple trick, it is.
3 * 40 - 2 * 40 = 40.
3 * 40 - 2 * 40 = 40.
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