Karen travelled x miles from her home to office, at a speed of y miles

A simple way would be to take values for x and y

Let x=40 miles and y=5.... time taken while goons =40/5=8
Now while coming back
10 miles at 5 mph means 10/5 or 2 hrs. One hour break, adding up to 3 hrs
So now he has to cover 40-10=30 miles in 5 hrs. Speed =30/5=6 mph

Change in speed (6-5)/5*100=100/5=20%

Check choices
B. 100y/(x-y-10)=100*5/(40-5-10)=500/25=20%

B
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Dear chetan2u

By this approach, it will be very time consuming during exam.
As Firstly we have to suppose the values & then find solutions
Then putting values in options for same result.
It will take around 3-4mins to solve by this method,& during GMAT Exam, it will be difficult to solve in time.

Is there any other approach?

The other way is to solve with variables - straight forward one. But it is not something that one can avert mistakes.
Ans yes it time consuming as well...
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IN questions like these on the GMAT, the Algebraic approach would only do one thing – put you at a disadvantage in terms of time taken to solve the question. The best approach to solve this question within 2 minutes (or at least close to) is to plug in simple values for the unknowns.

Let’s take x as 100 miles and y as 10 miles per hour. This means that she takes 10 hours to reach her office from her home.

On her way home, she travels for 10 miles at the same speed. This means she’s spent one hour travelling this distance. She then stops for one hour. This means that she will have to travel the remaining 90 miles in the remaining 8 hours since the total time has to be the same in both the onward and return journeys. This works out to an average speed of 11.25 miles per hour for this segment.
Compared to her original speed of 10 miles per hour, this is a percentage increase of 12.5%. Now, the only thing left is to find the option that gives us this percentage change.

Observing the options, we see that the denominator is either (x-y-10) or (y-x-10). Remember that y is going to be smaller than x and hence (y-x-10) will turn out to be negative. Since there has to be a percentage increase in the speed (remember, the average speed has to increase to maintain the time), we can’t have negative values.

Therefore, we eliminate all options that have (y-x-10) as the denominator. Options C, D and E can be eliminated. The possible answer options are A or B.

Answer option A has a substantially large numerator for us to consider that as a logical percentage increase. Substituting the values of x and y in answer option B, we see that
\(\frac{{100y} }{ {(x-y-10)}}\) = \(\frac{{100 * 10} }{ {100-10-10}}\) = 1000/80 = 12.5.

This is the number we were looking for. We can conclusively rule out option A now. The correct answer option is B.

A point to note is that you will have to practice the plugging in and elimination strategies regularly when you are preparing for the GMAT. Only then will you have the confidence to use it on the actual test. Also remember that GMAT is not a test of your math skills alone, your reasoning and logical skills are tested too. Considering this, if you always take recourse to traditional methods of problem solving in Quant, you’d be a one-dimensional problem solver. That’s not what any of us want to be on the GMAT.

Hope that helps!

Given: Karen travelled x miles from her home to office, at a speed of y miles per hour and on her way back, she takes the same route, and travels for 10 miles at a speed of y miles per hour before stopping for an hour.

Asked: Now, by what percentage should she increase her speed so that the overall time taken to reach back home from the office is the same as that taken to reach the office from home?

From home to office :
Distance = x miles
Speed = y miles/hr
Time taken = x/y hrs

From office to home:
Distance 1= 10 miles
Speed 1= y miles/hr
Time taken 1 = 10/y hrs

Idle time = 1 hr

Distance 2 = (x -10) miles
Time taken 2 = x/y - 10/y - 1= (x-y-10)/y hrs
Speed 2 = (x-10)y/(x-y-10) miles/hr

Percentage increase in speed = {(x-10)y/(x-y-10) / y - 1}*100% = {(x-10)/(x-y-10) -1}*100 = 100(x-10-x+y+10)/(x-y-10)% = 100y/(x-y-10) %

IMO B
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IMO, this problem can be solved algebraically within the time limit.

Let Karen's speed on the last leg of her return trip be 'p' times her original speed 'y'. Then the percentage increase in her speed is (p-1)*100.
Total time for return = (10/y) + 1 + (x-10)/p*y = x/y (Karen's home-to-office time). Multiplying both sides of this equation by 'y':

10 + y + (x-10)/p = x ....> p = (x-10)/(x-y-10)
% increase in speed = (p-1)*100 = {(x-10)/(x-y-10) - 1}*100 = 100y/(x-y-10)
ANS: B

x/y -1 = 10/y + (x-10)/k

(x-y-10)/y = (x-10)/k

k/y = (x-10)/(x-y-10)

(k/y-1)*100 = 100y/(x-y-10)

B

Her usual speed is y. Her usual time taken for the entire one side journey = x/y.
She needs to increase her speed on (x - 10) mile stretch so that she makes up the 1 hr lost from her usual time taken on this stretch which is (x/y - 10/y)

\(Speed = \frac{Distance}{Time} = \frac{(x - 10)}{(x/y - 10/y - 1)} = \frac{y(x - 10)}{(x - 10 - y)}\)

Percentage Increase in speed = \(\frac{\frac{y(x - 10)}{(x - 10 - y)} - y}{y} * 100 = \frac{100y}{x - y - 10}\)

Answer (B)
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