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.
Sankaran jack said:
1 decade ago
@All.
Certain distance = x.
Some speed = y.
If 3 kmph faster and 40 min less,
(x/y+3)-40/60..........1.
if 2 kmph slower and 40 min more means,
(x/y-2)+40/60..........2.
Now equate these 2 choices.
(x/y+3)-40/60 = (x/y-2)+40/60.
y = 12.
So
x = 40.
I hope you understood this equation.
Certain distance = x.
Some speed = y.
If 3 kmph faster and 40 min less,
(x/y+3)-40/60..........1.
if 2 kmph slower and 40 min more means,
(x/y-2)+40/60..........2.
Now equate these 2 choices.
(x/y+3)-40/60 = (x/y-2)+40/60.
y = 12.
So
x = 40.
I hope you understood this equation.
Hariom saini said:
7 years ago
You can also solve it like;
If T1=T2=40 than
V=2V1V2/V1-V2
=2*3*2/3-2
=12.
Than X=V*T1(1+V/V1)
= 12*40/60(1+12/3)
= 8*5
=40KM.
T1=40/60hr V=speed V1=12km/hr.
If T1=T2=40 than
V=2V1V2/V1-V2
=2*3*2/3-2
=12.
Than X=V*T1(1+V/V1)
= 12*40/60(1+12/3)
= 8*5
=40KM.
T1=40/60hr V=speed V1=12km/hr.
Pooja Gupta said:
8 years ago
Man goes to his office at a certain speed. If he travels at a speed 4kmph more than his usual speed then he will reach his office early by 12 minutes. If he travels at a speed 8 kmph less than his usual speed he will be late by 36 minutes. Find his original speed.
How would we solve it?
How would we solve it?
Teja said:
1 decade ago
2y(y+3) = 9x --- equation 1.
y(y-2) = 3x --- equation 2.
Equation 1 we can also write that in this form.
2y(y+3) = 3*3x --- equation 3.
Then substitute equation 2 in equation 3.
Then 2y(y+3) = 3*(y(y-2)).
Cancel y on both sides then 2y+6 = 3y-6.
y = 12.
y(y-2) = 3x --- equation 2.
Equation 1 we can also write that in this form.
2y(y+3) = 3*3x --- equation 3.
Then substitute equation 2 in equation 3.
Then 2y(y+3) = 3*(y(y-2)).
Cancel y on both sides then 2y+6 = 3y-6.
y = 12.
Mani said:
1 decade ago
Can any one explain this plz
X/y-2 + 40/60 = X/y+3 -40/60 .... (1) (40 minute is converted in hrs so 40/60)
3x+2y-4/3y-6 = 3x-2y-6/3y+9
After that we got
Y=12, Now put the value of Y in Eqsn .... (1)
I didnt get this.....
X/y-2 + 40/60 = X/y+3 -40/60 .... (1) (40 minute is converted in hrs so 40/60)
3x+2y-4/3y-6 = 3x-2y-6/3y+9
After that we got
Y=12, Now put the value of Y in Eqsn .... (1)
I didnt get this.....
Jayasri said:
6 years ago
d = st--->1.
d = (s+3)(t-2/3)------>2.
d = (s-2)(t+2/3)------>3.
Solving 1and 2 we get 2s/3 = 3t-2----->4.
2s/3 = 2t-4/3--------->5.
By solving 4and 5 we get t = 10/3 and s = 12 so d = st = 40km.
d = (s+3)(t-2/3)------>2.
d = (s-2)(t+2/3)------>3.
Solving 1and 2 we get 2s/3 = 3t-2----->4.
2s/3 = 2t-4/3--------->5.
By solving 4and 5 we get t = 10/3 and s = 12 so d = st = 40km.
(1)
Arjun Singh said:
10 years ago
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.
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.
Jyo said:
8 years ago
Formulae:speed=distance/time.
Case 1:
3=d/t-40
3t-120=d ---> eq(1)
Case 2:
2=d/t+40
2t+80=d ---> eq(2).
After solving eq(1) and eq(2), we get.
t=49.
Case 1:
3=d/t-40
3t-120=d ---> eq(1)
Case 2:
2=d/t+40
2t+80=d ---> eq(2).
After solving eq(1) and eq(2), we get.
t=49.
Sumit said:
8 years ago
How will you get Y=12?
my calculation is:
3x+2y-4/3y-6 = 3x-2y-6/3y+9,
(3y+9)(3x+2y-4) =( 3y-6)(3x-2y-6),
9xy+6y^2+6y+27x-36 = 9xy-6y^2-6y-18x+36,
12y^2+12y+45x-72 = 0.
please explain me.
my calculation is:
3x+2y-4/3y-6 = 3x-2y-6/3y+9,
(3y+9)(3x+2y-4) =( 3y-6)(3x-2y-6),
9xy+6y^2+6y+27x-36 = 9xy-6y^2-6y-18x+36,
12y^2+12y+45x-72 = 0.
please explain me.
P RAJENDRAKUMAR said:
6 years ago
The solution is;
2y(y + 3) = 9x --> (i)
y(y - 2) = 3x --> (ii)
So, 2y(y + 3) = 3*3x = 2y(y + 3) = 3* y(y - 2).
After Solving y=12 put y value any above 1 and 2 equation, and get x=40.
2y(y + 3) = 9x --> (i)
y(y - 2) = 3x --> (ii)
So, 2y(y + 3) = 3*3x = 2y(y + 3) = 3* y(y - 2).
After Solving y=12 put y value any above 1 and 2 equation, and get x=40.
(1)
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