Time And Distance Problem

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A man travels 35 km partly at 4 kmph and partly at 5 kmph. If he covers the former distance at 5kmph and the later distance at 4 kmph, he could cover 2 km more in the same time. The time taken to cover the whole distance at the original rate (in hours) is ??

(a) 4.5

(b) 7

(c) 8

(d) 9

[talaangoshtari]

Hi parthdpatel,

We know that rt=d:

Suppose the time taken for the first part of the trip is t1, and the time taken for the second part of the trip is t2,

4t1+5t2=35
5t1+4t2=37

So, t1=5 and t2=3, and the sum is equal to 8.

[Asmathumairy]

Hi talaangoshtari
i could not understand how you find t1 and t2 !
may you re-answer it in different way please?

[talaangoshtari]

Hi parthdpatel,

Multiply the first equation by 5 and the second equation by 4,

(5)[4t1 + 5t2 = 35] => 20t1 + 25t2 = 175

(4)[5t1 + 4t2 = 37] => 20t1 + 16t2 = 148

Now we can subtract the two equations:

25t2 - 16t2 = 175 - 148 = 27 => 9t2 = 27,

So t2 = 3, and

20t1 + 25(3) = 175 => t1 = 5

[GMATGuruNY]

Hi parthdpatel,

Multiply the first equation by 5 and the second equation by 4,

(5)[4t1 + 5t2 = 35] => 20t1 + 25t2 = 175

(4)[5t1 + 4t2 = 37] => 20t1 + 16t2 = 148

Now we can subtract the two equations:

25t2 - 16t2 = 175 - 148 = 27 => 9t2 = 27,

So t2 = 3, and

20t1 + 25(3) = 175 => t1 = 5

Since the question stem asks for the value of t�+t₂, we can save time by simply adding together 4t�+5t₂=35 and 5t�+4t₂=37:

(4t�+5t₂) + (5t�+4t₂) = 35+37
9t�+9t₂ = 72
t�+t₂ = 8.

The correct answer is C.

[DavidG@VeritasPrep]

A man travels 35 km partly at 4 kmph and partly at 5 kmph. If he covers the former distance at 5kmph and the later distance at 4 kmph, he could cover 2 km more in the same time. The time taken to cover the whole distance at the original rate (in hours) is ??

(a) 4.5

(b) 7

(c) 8

(d) 9

We can also use a little logic. If he walked at 4 km/h for the whole 35 km, it would take him 35/4 = 8 3/4 hours to do the trip. If he walked at 5 km/h for the full 35 km, it would take him 35/5 = 7 hours. So if he's doing part of the trip at 4km/h and part at 5km/h, he'll have to do the trip in something in between 7 and 8 3/4 hours. The only answer choice in that range is 8.

[ahmedshafea]

could you pls let me know what steps i should follow for solved it

[Matt@VeritasPrep]

If you want to use a longer but more predictable approach:

Say his first distance = r and his second distance = s.
Say his time traveling the first distance = x and his time traveling the second distance = y.

We know:

r + s = 35
r = 4x
s = 5y

by substitution:

4x + 5y = 35

We also know that if he went 5x then 4y, we'd have 5x + 4y = 37.

From there, we have a simple two equation system:

4x + 5y = 35
5x + 4y = 37

Adding those together, 9x + 9y = 72, and (x + y) = 8. x + y is the sum of the times, so this is our answer and we're done.

[Matt@VeritasPrep]

Another idea that's a little quicker (but harder to think up):

Our guy travels somewhere 35 km at a rate between 4 and 5 km/r.

His minimum time = distance/faster time = 35/5 = 7

His maximum time = distance/slower time = 35/4 = 8.75

He can't travel 4 OR 5 km for the whole trip, since he spends some time at each rate, meaning

7 < his time < 8.75

and the only answer that fits is 8.

[Scott@TargetTestPrep]

A man travels 35 km partly at 4 kmph and partly at 5 kmph. If he covers the former distance at 5kmph and the later distance at 4 kmph, he could cover 2 km more in the same time. The time taken to cover the whole distance at the original rate (in hours) is ??

(a) 4.5

(b) 7

(c) 8

(d) 9

Let s = the time the man travels at 4 kmph and t = the time he travels at 5 kmph. We can the equations:

4s + 5t = 35

and

5s + 4t = 37

Adding the two equations together, we obtain:

9s + 9t = 72

9(s + t) = 72

s + t = 8

Therefore, the total travel time for the 35 km is s + t = 8 hours.

Answer: C