Bob bikes to school every day at a steady rate of x miles

Here's a different algebraic solution:

Let d = the TOTAL distance to school.

Bob had a flat tire exactly halfway to school
So, d/2 = distance spent biking
and d/2 = distance spent walking

We can write: (time spent biking) + (time spent walking) = t
time = distance/speed
We get: (d/2)/x + (d/2)/y = t
Simplify: d/2x + d/2y = t
Find a common denominator of 2yx to get: dy/2yx + dx/2yx = t
Combine terms: (dy + dx)/2yx = t
Multiply both sides by 2yx to get: dy + dx = 2xyt
Factor: d(y + x) = 2xyt
Divide both sides by (x + y) to get: d = 2xyt/(x+y)

Answer: C

Ans C.

D = (Ave Speed) * t

D = (D/t1+t2) * t

D = (D / ((D/2)/X) + (D/2)/Y)) * t

D = ( 4xy/ (2x+2y)) * t

D = 2xyt / (x+Y)

Hi Guys,

Doubt :How can the below Q be solved by picking nos rather than conventional approach

Bob bikes to school every day at a steady rate of x miles per hour. On a particular day, Bob had a flat tire exactly halfway to school. He immediately started walking to school at a steady pace of y miles per hour. He arrived at school exactly t hours after leaving his home. How many miles is it from the school to Bob's home?

A) (x + y) / t

B) 2(x + t) / xy

C) 2xyt / (x + y)

D) 2(x + y + t) / xy

E) x(y + t) + y(x + t)


Thanks,

harikris

Distance you can obtain as Rate x Time. Also, you can add/subtract only quantities that have the same units, i.e. you cannot add speed to time. Thus, you can immediately eliminate B, D and E.

A cannot be the correct answer, as it has units of speed over time, not units of distance.
You are left with C as the correct answer.

I understood the problem as one of the average speed problems. One quick way to solve is with the direct formula of average speed. Since the speeds for different halves are mentioned, we can use the formula Avg speed = 2ab/(a+b), where a is constant speed for 1st half of journey and b is constant speed for 2nd half of journey.
Per the Question Avg speed for total trip = 2xy/(x+y). Therefore the total distance is 2xyt/(x+y).

Hence Answer is C.

The definition of average speed is \(\frac{Total \, distance}{Total \, time}\). For a particular case when on a fraction \(F\) of the distance \(D\) the speed was \(x\) and on the remaining fraction \((1-F)\) of \(D\) the speed was \(y\) , we can calculate the average speed as:

\(\frac{D}{\frac{FD}{x}+\frac{(1-F)D}{y}}=\frac{xy}{(1-F)x+Fy}\)

In our case, \(F=\frac{1}{2}\), and the average speed is indeed \(\frac{2xy}{x+y}\).

I wouldn't say this formula is a must to remember. The definition of average speed, YES and then, depending on the data of the specific question, the average speed can be easily worked out.

Your approach is correct and it follows the path of "let's work out a solution". Although the elimination method is not always possible, I would like to suggest that there is a major takeaway lesson from this question: when variables are used (in our case \(x\) and \(y\) as speeds, \(t\) as time) which have specific units attached, pay attention to the basic rule saying that addition and subtraction has a meaning only when all the terms have the same units. In our case, you cannot add speed to time, so answers involving expressions like \(x+t\) or \(y+t\) can be automatically eliminated.

Can you please explain the specifics about how you went from D = (D / ((D/2)/X) + (D/2)/Y)) * t to D = ( 4xy/ (2x+2y)) * t ?

20 seconds if you just read the question and go for the units...I say always checkk the units first for all the options in such type of questions

When Distance is same:
Average Speed = \(\frac{2 (Rate 1 * Rate 2)}{(Rate 1 + Rate 2)}\)
=\(\frac{2 xy}{(x + y)}\)

So total distance in t hours = \(\frac{2 xyt}{(x + y)}\)
Hence, Ans is C

Let total time=t, total distance=w
Total Time = Time for biking + Time for walking
t = w/(2x) + w/(2y)
4xyt = w (2y + 2x)

w = 2xyt/(y+x)

Therefore, Ans = C.

I got this right but found it to be very irritating to solve. Ok so if you're like me, the key is to keep it simple. I didn't but eventually learned to keep it simple.

s1 = x; s2 = y
d/2 and d/2
t 1 + t 2 = t

now, we know that d = s * t

we can derive that t1 = d/2x and t2 = d2y

d/2x + d/2y = t

take d common --> d (1/2x + 1/2y) = t

solve for d which will give us 2xyt/x+y

C

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