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 11 of 12.
Md Parvez Islam said:
6 years ago
Thanks @Sachin Bhagat.
Nisha said:
6 years ago
Thanks @Sachin Bhagat.
(1)
Sandjive said:
6 years ago
@Pauli.
For Jake sum.
S*t=90.....1.
(S+15)(t-30/60)=90.......2.
Solving 2 eqn, answer: 45km/hr.
For Jake sum.
S*t=90.....1.
(S+15)(t-30/60)=90.......2.
Solving 2 eqn, answer: 45km/hr.
Shweta said:
6 years ago
Thank you @Sachin bhagat.
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)
Rajasekhar said:
6 years ago
d=(S1S2/S1-S2)*t.
(S(S+3) / 3)*(40/60)=(S(S-2)/2)*(40/60).
2S+6= 3S-6.
S=12Kmph.
d=( (12*15)/3)*40/60.
d=40km.
(S(S+3) / 3)*(40/60)=(S(S-2)/2)*(40/60).
2S+6= 3S-6.
S=12Kmph.
d=( (12*15)/3)*40/60.
d=40km.
(4)
Dinesh said:
5 years ago
@All.
In the above question, only time and speed are varying so distance remains constant.
So, First convert time 40 min=40/60 hr.
Now let's begin.
we know v=d/t. therefore d=vt--> (1)
Now by the first condition.
v+3=d/(t-40/60) .
Therefore d=(v+3)(t-40/60).
By multiplying the brackets.
d=(vt)-(2v/3)+(3t)-2 --> (2).
Now we know distance is constant
Therefore equating equations (1) and (2).
vt = vt-2v/3+3t-2.
Solving this we get -2v/3+3t=2.
now multiply it by 3.
We get -2v+9t=6 ---> (3).
Now By the second condition.
v-2=d/(t+40/60),
therefore d=(v-2)(t+40/60),
therefore d=(vt) +(2v/3)-(2t)-(4/3)--> (4).
Now equating equations (1) and (4)
vt=vt+2v/3-2t-4/3
By solving we get 2v/3-2t=4/3
Now multiply by 3 on both sides
2v-6t=4 ---> (5).
Now adding equations (3) & (5)
-2v+9t=6
+2v-6t=4
We get 3t=10 therfore t=10/3 hr
Now put t=10/3 in equation (5).
2v-6(10/3)=4.
And 2v=4+20.
Therfore
v=12 kmph.
Now finally from equation (1)
d=vt=12*(10/3)
d=40 km..
Hope it helps.
In the above question, only time and speed are varying so distance remains constant.
So, First convert time 40 min=40/60 hr.
Now let's begin.
we know v=d/t. therefore d=vt--> (1)
Now by the first condition.
v+3=d/(t-40/60) .
Therefore d=(v+3)(t-40/60).
By multiplying the brackets.
d=(vt)-(2v/3)+(3t)-2 --> (2).
Now we know distance is constant
Therefore equating equations (1) and (2).
vt = vt-2v/3+3t-2.
Solving this we get -2v/3+3t=2.
now multiply it by 3.
We get -2v+9t=6 ---> (3).
Now By the second condition.
v-2=d/(t+40/60),
therefore d=(v-2)(t+40/60),
therefore d=(vt) +(2v/3)-(2t)-(4/3)--> (4).
Now equating equations (1) and (4)
vt=vt+2v/3-2t-4/3
By solving we get 2v/3-2t=4/3
Now multiply by 3 on both sides
2v-6t=4 ---> (5).
Now adding equations (3) & (5)
-2v+9t=6
+2v-6t=4
We get 3t=10 therfore t=10/3 hr
Now put t=10/3 in equation (5).
2v-6(10/3)=4.
And 2v=4+20.
Therfore
v=12 kmph.
Now finally from equation (1)
d=vt=12*(10/3)
d=40 km..
Hope it helps.
(12)
Kajal kumari said:
5 years ago
First we find the speed using this concept.
2s1*s2/s1-s2.
2*3*2/3-2=12km/hr.
Now, if we increase speed by 3km/hr, it will become 12+3=15km/hr and 40min less
same as,if we decrease the speed by 2km/hr then it become 12-2=10km/hr and 40min more
we have s1=15kmhr and s2=10kmhr and the time difference is 80/60.
Let distance be x.
x/10 - x/15 =80/60,
3x-2x/30=4/3,
x=4/3*30=40km/hr.
2s1*s2/s1-s2.
2*3*2/3-2=12km/hr.
Now, if we increase speed by 3km/hr, it will become 12+3=15km/hr and 40min less
same as,if we decrease the speed by 2km/hr then it become 12-2=10km/hr and 40min more
we have s1=15kmhr and s2=10kmhr and the time difference is 80/60.
Let distance be x.
x/10 - x/15 =80/60,
3x-2x/30=4/3,
x=4/3*30=40km/hr.
(10)
Feroze Ahmad kumhar said:
5 years ago
Distance is same.
Therefore d = d.
3(x-40) = 2(x+40),
3x+120 = 2x+80,
3x-2x = 120-80,
X = 40km.
Therefore d = d.
3(x-40) = 2(x+40),
3x+120 = 2x+80,
3x-2x = 120-80,
X = 40km.
(134)
Samir Mendhe said:
5 years ago
3times more -> Dist= 3 * 40= 120.
2tines less -> Dist= -2* 40= -80.
So, 120 - 80 = 40kmph.
2tines less -> Dist= -2* 40= -80.
So, 120 - 80 = 40kmph.
(248)
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