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Portada del sitio > Recursos GMAT > GMAT QUANTITATIVE QUESTION > Problem Solving Word Problems > GMAT Distance / Rate Problem, 6 EXERCISES AND ANDWERS SOLVED

# GMAT Distance / Rate Problem, 6 EXERCISES AND ANDWERS SOLVED

Jueves 12 de julio de 2018,

Distance / Rate Problem N5, N6, N7
1.-Anne traveled from City A to City B in 4 hours, and her speed was between 25 miles per hour and 45 miles per hour. John traveled from City A to City B along the same route in 2 hours, and his speed was between 45 miles per hour and 60 miles per hour. Which of the following could be the distance, in miles, from City A to City B?
A. 95
B. 115
C. 125
D. 160
E. 180
2.Aaron will jog from home at x miles per hour and then walk back home on the same route at y miles per hour. How many miles from home can Aaron jog so that he spends a total of t hours jogging and walking?
A. xt/y
B. (x+t)/xy
C. xyt/(x+y)
D. (x+y+t)/xy
E. (y+t)/x-t/y
3.- A driver completed the first 20 miles of a 40-mile trip at an average speed of 50 miles per hour. At what average speed must the driver complete the remaining 20 miles to achieve an average speed of 60 miles per hour for the entire 40-mile trip? (Assume that the driver did not make any stops during the 40-mile trip.)
(A) 65 mph
(B) 68 mph
(C) 70 mph
(D) 75 mph
(E) 80 mph
4.- Car A and Car B are each traveling along the same stretch of highway in the same direction. Car A leaves at 10:00am traveling at a constant rate of 40 miles per hour and Car B leaves at 10:30am traveling at a constant rate of 50 miles per hour. At what time will Car B be 40 miles ahead of Car A?
A. 3:30pm
B. 4:00pm
C. 4:30pm
D. 5:00pm
E. 5:30pm

5.- A certain vehicle is undergoing performance adjustments during a set of three trials on an 8-mile track. The first time it travels around the track at a constant rate of p miles per minute, the second time at a constant rate of p^2 miles per minute, and the third time at a constant rate of p^3 miles per minute. If the vehicle takes the same time to travel the first 18 miles of these trials as it does to travel the last 8 miles, how many minutes does it take to complete all three trials (not counting any time between trials)?
(A) 18
(B) 36
(C) 60
(D) 96
(E) 112

6.-Recently, fuel price has seen a hike of 20%. Mr X is planning to buy a new car with better mileage as compared to his current car. By what % should the new mileage be more than the previous mileage to ensure that Mr X’s total fuel cost stays the same for the month? (assuming the distance traveled every month stays the same)[/b]
(A) 10%
(B) 17%
(C) 20%
(D) 21%
(E) 25%

Estas invitado a enviar tus respuestas, con un pequeño desarrollo al correo clasesgmatchile@gmail.com, para publicarlas y comentarlas.

Saludos

claudio hurtado

Resp: 1/b, 2/c, 3/d, 4/c, 5/e, 6/b

SOLVED

1.-Distance range per Anne’s speed : 25*4 - 4*45 ---> 100 - 180 ...(1)
Distance range per John’s speed : 45*2 - 60*2 ----> 90-120 ....(2)
(2) eliminates C,D,E and (1) eliminates A.
B is the correct answer.

2.- Say the distance Aaron jogs is d miles, notice that the distance Aaron walks back will also be d miles (since he walks back home on the same route).
Next, total time t would be equal to the time he spends on jogging plus the time he spends on walking:d/x+d/y=t-→ d(1/x+1/y)=t-→ d=xyt/(x+y).

Answer: C.
3.-average speed=(total distance)/(total time) =40/(total time) =60. This iimplies that for the average time to be 60 miles per hour, the total time must be 40/60 = 2/3 hours.
Now, the first 20 miles were covered in (time) = (distance)/(speed) = 20/50 = 2/5 hours.
Thus, the remaining 20 miles should be covered in 2/3 - 2/5 = 4/15 hours, which means that the remaining 20 miles should be covered at an average speed (distance)/(time) = 20/(4/15) = 75 miles per hour.

Answer: D.

4.- Relative speed = 10 mph
At 10.30 car A would have traveled 20 miles
If car B has to be 40 miles ahead of car A, then it has cover the above 20 miles and be ahead by 40 miles. i.e. Net additional distance car B has to cover = 60 miles
Time taken to complete 60 miles = 60/10 = 6 hours
Car B will be ahead of car A by 40 miles at 10.30 am + 6hrs = 4.30 pm

Answer: C
5.- 1. Time taken for first 18 miles = 8/p+8/p^2+2/p^3
2. For the last 8 miles =8/p^3
As per the problem,
8/p+8/p^2+2/p^3=8/p^3
Solving for p,
p=1/2
Therefore, time taken for all the 3 trials =8/(1/2)+8/(1/4)+8/(1/8)= 16 +32 + 64 = 112
Ans: E

6.- Answer: B
Lets say Price and Mileage before hike are P1 and M1 respectively. After Hike, they are P2 and M2 respectively.
If price per mile is to stay the same, we need to have this equation.

P1/M1=P2/M2P1/M1=P2/M2

M2=P2∗M1/P1M2=P2∗M1/P1

but P2 is 1.2 times P1.

P1/M1=1.2P1/M2∗P1P1/M1=1.2P1/M2∗P1

M2=1.2M1M2=1.2M1

Difference is 1.2 or 20%.

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