Quant - Rate Flashcards
Reiko drove from point A to point B at a constant speed, and then returned to A along the same route at a different constant speed. Did Reiko travel from A to B at a speed greater than 40 miles per hour?
(1) Reiko’s average speed for the entire round trip, excluding the time spent at point B, was 80 miles per hour.
(2) It took Reiko 20 more minutes to drive from A to B than to make the return trip.
Ans = A
Hence, in any one leg, your speed cannot be less than half the average.
What you missed to note is that avg of 30 mph and 130 mph is 80 mph when he travels at the two speeds for the same TIME.
If someone travels at two speeds for the same amount of time, say an hr at one speed and an hr at another speed, then the average speed is (Speed1 + Speed 2)/2
Here, the case is different. The two speeds are for the same distance. He travels at Speed1 from A to B and at Speed2 from B to A (distance in the two cases are same, time taken would be different). So here, the avg is not (Speed1 + Speed 2)/2.
If D is the distance from A to B,
Avg Speed = 2D/(D/Speed1 + D/Speed2) = Total distance/Total time
Avg Speed = 2Speed1Speed2/(Speed1 + Speed2) (this is the avg when one travels for the same distance)
Now, what happens if one of the speeds is half the avg speed?
Say avg speed was 80 mph. Say distance from A to B is 80 miles. Since avg speed for to and fro journey is 80 mph, we need 2 hrs to cover the journey. Now, what happens when the speed while going from A to B is 40 mph? He takes 2 hrs to go from A to B i.e. the entire time allotted for the return trip has gotten used up on the first leg of the journey itself. Of course, it doesn’t matter what your speed is in the second leg, you will take more than 2 hrs for the entire journey and hence your avg speed will be less than 80 mph.
Hence, in any one leg, your speed cannot be less than half the average.
If you want to see it algebraically,
From above, A = 2S1S2/(S1 + S2)
If S1 = A/2,
A = 2*(A/2)*S2/(A/2 + S2) S2 = A/2 + S2
A man cycling along the road noticed that every 12 minutes a bus overtakes him and every 4 minutes he meets an oncoming bus. If all buses and the cyclist move at a constant speed, what is the time interval between consecutive buses? (Assume that the busses all travel at the same constant speed and are ‘equally spaced out’ - and that the cyclist travels at a different constant speed)
A. 5 minutes B. 6 minutes C. 8 minutes D. 9 minutes E. 10 minutes
There is only one thing you need to understand in this question - When buses are approaching him from both the sides at a constant speed, it doesn’t matter whether the man is standing still or cycling, the number of buses that he will meet will be the same. Convince yourself by imagining the case where the man is standing still. He will meet a bus from each side after every few mins. When he starts cycling in a direction, he is cycling away from buses of one side but towards buses of the other side. Since in 12 mins he meets total 4 buses (1 + 3), in 6 mins he meets 2 buses, one from each side, if he were standing still. So buses ply at a frequency of 6 mins each.
Twist: Same scenario. If a man is sitting inside one bus, at what frequency will a bus from opposite side cross him?
Also try the same question by changing the time taken by buses to meet the man to 10 min and 8 min respectively (instead of 12 mins and 4 mins)
2nd qtn:
same question by changing the time taken by buses to meet the man to 10 min and 8 min respectively (instead of 12 mins and 4 mins)
i take LCM of 10 and 8 that is 40 now, in 40 mins, i meet 9 buses (4+5) ==> 9buses --> 40 mins ==> 1 bus --> 40/9 mins ==> 2 buses --> 80/9 mins, that is the time interval b/w 2 buses
Relative speed
Case 1: A is standing still and B is moving at a speed of 5 mph.
What is B’s speed relative to A?
Case 2: A is walking due east at a speed of 5 mph and B is walking due east at a speed of 2 mph.
What is A’s speed relative to B?
Case 3: A is walking due east at a speed of 5 mph and B is walking due west, away from A, at a speed of 2 mph.
What is A’s speed relative to B?
Case 1: A is standing still and B is moving at a speed of 5 mph.
What is B’s speed relative to A? B is moving away from A at a rate of 5 miles every hour. So B’s speed relative to A is 5 mph. What is A’s speed relative to B? Is it 0? No! A’s speed relative to Earth is 0. A’s speed relative to B is 5 mph since distance between A and B is increasing at a rate of 5 mph. Confused? When we say ‘relative to B’, we assume that B is stationary. Since distance between the two is increasing at a rate of 5 mph, we say A’s speed relative to B is 5 mph.
Case 2: A is walking due east at a speed of 5 mph and B is walking due east at a speed of 2 mph.
What is A’s speed relative to B? Distance between A and B is increasing by 3 miles every hour. So A’s speed relative to B is only 3 mph, not 5 mph. Think of it this way – A is moving fast but not as fast when compared with B since B is also moving. Similarly, B’s speed relative to A is also 3 mph. Notice that when they move in the same direction, their relative speed is the difference of their speeds.
Case 3: A is walking due east at a speed of 5 mph and B is walking due west, away from A, at a speed of 2 mph.
What is A’s speed relative to B? Distance between A and B is increasing by 7 miles every hour. So A’s speed relative to B is 7 mph, not 5 mph. Think of it this way – A is moving fast and even faster as compared with B since B is moving in the opposite direction. Similarly, B’s speed relative to A is also 7 mph. Notice that when they move in the opposite directions, their relative speed is the sum of their speeds. It doesn’t matter whether they are moving toward each other or away from each other.
Jane can paint the wall in J hours, and Bill can paint the same wall in B hours. They begin at noon together. If J and B are both even numbers is J=B?
(1) Jane and Bill finish at 4:48 p.m.
(2) (J+B)^2=400
Jane can paint the wall in J hours, and Bill can paint the same wall in B hours. They begin at noon together. If J and B are both even numbers is J=B?
Jane and Bill working together will paint the wall in T=JB/J+B hours. Now suppose that J=B-> T=J^2/2J, as J and B are even J=2n–> T=2n T=n, as n is an integer, working together Jane and Bill will paint the wall in whole number of hours, meaning that in any case TT must be an integer.
(1) They finish painting in 4 hours and 48 minutes, TT is not an integer, –> JJ and BB are not equal. Sufficient.
(2) J+B=20J+B=20, we can even not consider this one, clearly insufficient. JJ and BB can be 1010 and 1010 or 1212 and 88.
Answer: A.
Machines X and Y produced identical bottles at different constant rates. Machine X, operating alone for 4 hours, filled part of a production lot; then Machine Y, operating alone for 3 hours, filled the rest of this lot. How many hours would it have taken Machine X operating alone to fill the entire production lot?
(1) Machine X produced 30 bottles per minute.
(2) Machine X produced twice as many bottles in 4 hours as Machine Y produced in 3 hours.
Ans = B
Machine X produced twice as many bottles in 4 hours as Machine Y did in 3 hrs. Then I can say that Machine X filled 2/3 rd of the lot in 4 hrs. (If this is unclear, think Machine Y made ‘b’ bottles in 3 hrs, then Machine X made ‘2b’ bottles in 4 hrs and together the lot contained ‘3b’ bottles. So Machine X filled 2/3 of the lot in 4 hours.) In how much time will Machine X fill the rest of the 1/3 rd lot? In 2 hours. Hence, it takes a total of 6 hours to fill the lot on its own. This is sufficient to answer the question.
It takes 6 technicians a total of 10 hours to build and program a new server from Direct Computer, with each working at the same rate. If six technicians start to build the server at 11:00 AM, and one technician per hour is added beginning at 5:00 PM, at what time will the server be complete?
A. 7:30 PM B. 7:45 PM C. 8:00 PM D. 9:00 AM E. 10:00 PM
To complete the job 610=60 man/hours are needed. By 5:00 PM 66=36 man/hour done. 24 is left.
5: 00 PM - 6:00 PM 7 man/hour;
6: 00 PM - 7:00 PM 8 man/hour;
7: 00 PM - 8:00 PM 9 man/hour.
7+8+9=24.
We are given that 6 technicians can complete a job in 10 hours. Since rate = work/time, the rate of the 6 technicians is 1/10. Thus, the rate of 1 technician is (1/10)/6 = 1/60.
We see that 6 technicians work from 11:00 AM to 5:00 PM, or 6 hours. Since rate x time = work, those technicians complete 1/10 x 6 = 6/10 = 3/5 of the job.
If another technician is added to the job at 5:00 PM, the new rate for the technicians is (1/60) x 7 = 7/60, and thus all 7 technicians complete 7/60 of the job in the hour from 5:00 PM to 6:00 PM.
Thus, 3/5 + 7/60 = 36/60 + 7/60 = 43/60 of the job will be completed.
If another technician is added to the job at 6:00 PM, the new rate for the technicians is (1/60) x 8 = 8/60, and thus all 8 technicians complete 8/60 of the job in the hour from 6:00 PM to 7:00 PM.
Thus, 43/60 + 8/60 = 51/60 of the job will be completed.
If another technician is added to the job at 7:00 PM, the new rate for the technicians is (1/60) x 9 = 9/60, and thus all 9 technicians complete 9/60 of the job in the hour from 7:00 PM to 8:00 PM.
Thus, 51/60 + 9/60 = 60/60 = 1 whole job will be completed at 8:00 PM.
Pipe A fills a tank of capacity 700 liters at the rate of 40 liters a minute. Another pipe B fills the same tank at the rate of 30 liters a minute. A pipe at the bottom of the tank drains the tank at the rate of 20 liters a minute. If pipe A is kept open for a minute and then closed and pipe B is open for a minute and then closed and then pipe C is open for a minute and then closed and the cycle is repeated, when will the tank be full?
A) 42 minutes B) 14 minutes C) 39 minutes D) 40 minutes 20 seconds E) None of these
Why answer is not A ?
In three minutes net gain is 40+30-20=50 liters;
After 13 cycles (133=39 minutes) net gain will be 1350=650 liters.
Then in 1 minute pipe A will add 40 liters, 10 liters to be filled;
Then to fill 10 liters pipe B will need 10/30=1/3 min.
So total time 39+1+13=401339+1+13=4013 min.
if we have 13 cycles (with 1 cycle = A for 1 minute + B for 1 minutes - C for 1 minute), we will have filled 650 out of 700 litres.
After this, for the next minute (39+1)th minute, A will add another 40, leading to 650+40=690 litres and then B will have added another 10/30 litres in 1/3 of a minute. So technically we dont need the entire 14th cycle as we have reached the capacity fo 700 litres. If you keep B on for any longer than 1/3rd of a minute, the water will start to overflow from the tank.
If you go the entire 14 cycles, the water would have overflowed by the time entire of B was done and with C draining water after B was done, there would be less water than 700 litres. This is the reason why we did not do 14*50=700. This is a classical trap in GMAT.
A furniture manufacturer has two machines, but only one can be used at a time. Machine A is utilized during the first shift and Machine B during the second shift, while both work half of the third shift. If Machine A can do the job in 12 days working two shifts and Machine B can do the job in 15 days working two shifts, how many days will it take to do the job with the current work schedule?
(A) 14 (B) 13 (C) 11 (D) 9 (E) 7
Machine A needs 12 days * 2 shifts = 24 shifts to do the whole job;
Machine B needs 15 days * 2 shifts = 30 shifts to do the whole job;
In one day each machine works 1.5 shifts (3/2 shifts), and together, in one day, they are doing (3/2)/24+(3/2)/30=9/80 th of the whole, thus with the current work schedule they’ll need 80/9=~9 days to do the whole job.
Answer: D.
Al and Ben are drivers for SD Trucking Company. One snowy day, Ben left SD at 8:00 a.m. heading east and Al left SD at 11:00 a.m. heading west. At a particular time later that day, the dispatcher retrieved data from SD’s vehicle tracking system. The data showed that, up to that time, Al had averaged 40 miles per hour and Ben had averaged 20 miles per hour. It also showed that Al and Ben had driven a combined total of 240 miles. At what time did the dispatcher retrieve data from the vehicle tracking system?
A. 1:00 p.m. B. 2:00 p.m. C. 3:00 p.m. D. 5:00 p.m. E. 6:00 p.m.
Let t represent the time passed between the departure of Al (11:00 a.m.) and the retrieval of data by the dispatcher. Since Ben left at 8 a.m., his time is (t + 3). Since they both traveled for a total 240 miles, we can use the formula distance = rate x time and create the following equation:
Ben’s distance + Al’s distance = 240
20(t + 3) + 40t = 240
20t + 60 + 40t = 240
60t = 180
t = 3 hours
Thus, it was 11 a.m. + 3 hours = 2 p.m. when the dispatcher retrieved data from the vehicle tracking system.
Answer: B
During a trip, Francine traveled x percent of the total distance at an average speed of 40 miles per hour and the rest of the distance at an average speed of 60 miles per hour. In terms of x, what was Francine’s average speed for the entire trip?
A. (180−x)2(180−x)2
B. (x+60)4(x+60)4
C. (300−x)5(300−x)5
D. 600(115−x)600(115−x)
E. 12,000(x+200)
https://gmatclub.com/forum/during-a-trip-francine-traveled-x-percent-of-the-total-distance-at-an-94933.html
IF YOU COVER SAME DISTANCE AT DIFFERENT SPEEDS, THEN YOU WILL SPEN MORE TIME WITH LOWER SPEED, SO AVERAGE SPEED WILL BE CLOSER TO THE LOWER SPEED
think of extremes if X = 0, X=100 .
Also smart numbers d= 100 , X = 40
ANS = E
