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Sovon

· started a discussion

· 1 Months ago

What if I want to use the formula? (Question: 106819.0 )

can't we solve it by the formula??

Distance = (speed1*speed2)*(Time Difference)/(Speed Difference)

Question:
I have to be at a certain place at a certain time and I find that I shall be 15 minutes too late, if I walks at 4 km an hour; and 10 minutes too soon, if I walks at 6 km an hour. How far have I to walk?
Options:
A) 3km
B) 5km
C) 6km
D) 8km
Solution:

According to question

Distance covered is same

4 km./hr. × \((t+\frac{1}{4})\) hr. = 6 km./hr × \((t-\frac{1}{6})\) hr.

(15 min. = \(\frac{1}{4}\) hr. & 10 min.  =  \(\frac{1}{6}\) hr.) or 4t + 1hr. = 6t - 1hr.

T = 1 hr.

So, the required distance

= 4 km./hr. × \((1+\frac{1}{4})\) hr. = 5 kms.

Vaishali

· commented

· 1 Months ago

x/4 - x/6 = 25/60 (x=distance)

JEEVA

· commented

· 1 Months ago

(6*4)/(6-4)*25/60

Sovon

· commented

· 1 Months ago

Yet to be approved!

Knowledge Expert

· commented

· 1 Months ago

its applicable in the question which you have solved right now.....but remember if any typical question appear then please go through step by step....:)...

Sovon

· commented

· 1 Months ago

please mention where to use and where not to & please give a solution to the mentioned problem

Knowledge Expert

· commented

· 1 Months ago

the formula you have mentioned can not be used everywhere....

Sovon

· commented

· 1 Months ago

please help me in this problem, don't leave me unanswered

Sovon

· commented

· 1 Months ago

Please solve this question

If a boy goes to work at 10 km/hr and reaches 2 hour late and if he goes with 20 km/hr reaches 3 hour early. find the distance.
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