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GMAT 1: 650 Q49 V30
Posts: 216
Updated on: 12 Jun 2024, 00:20
Originally posted by imhimanshu on 15 Sep 2012, 23:25.
Last edited by Bunuel on 12 Jun 2024, 00:20, edited 9 times in total.
Last edited by Bunuel on 12 Jun 2024, 00:20, edited 9 times in total.
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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Question Stats:
68% (02:30) correct
32%
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A car is traveling on a straight stretch of roadway, and the speed of the car is increasing at a constant rate with respect to time. At time 0 seconds, the speed of the car is \(v_0\) meters per second; 10 seconds later, the front bumper of the car has traveled 125 meters, and the speed of the car is \(v_{10}\) meters per second.
Select values of \(v_0\) and \(v_{10}\) that are together consistent with the information provided. Make only two selections, one in each column.
Select values of \(v_0\) and \(v_{10}\) that are together consistent with the information provided. Make only two selections, one in each column.
ID: 100401
| \(v_{0} \) | \(v_{10} \) | |
| 5 | ||
| 18 | ||
| 20 | ||
| 36 | ||
| 72 |
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imhimanshu
Responding to a pm:
Your best friend in GMAT is your reasoning skill. You can solve this question easily using brute force. Of course there is nothing wrong with the complete AP approach but I would use it only partially...
The car travels 125 m in 10 sec. It's average speed is 12.5 m/s
vo can only be 5. If it were 18, the car would have traveled more than 180m in 10 sec since the minimum speed is 18. The speed keeps increasing thereafter. Same logic goes for all other values.
Now, with vo = 5, which value will give the average as 12.5? Obviously 20. (recall that average of an AP is the average of its first and last terms)
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Posts: 4,475
danzig
Dear Danzig & Mbmanoj & Chakdum,
This is tricky. When a car is increasing at constant acceleration, then it is NOT a sequence, either arithmetic or geometric. Thinking about the motion in terms of a sum of what happens in each second is not helpful, and in particular, thinking of it as a sum of constant-motion chunks each second is DEAD WRONG. If it goes from, say 5 m/s to 15 m/s in 10 seconds, then the acceleration is 2 m/s^2, but that does not mean: 5 m/s for the duration of the 1st second, 7 m/s for the duration of the 2nd second, etc. That is a complete misunderstanding of the nature of acceleration.
Instead, the formula given in the OE, average velocity = (vo + vf)/2, is always correct for constant acceleration. Here's one way to think about that formula. Think about the graph of speed vs. time. The speed is continuously increasing from (vo) to (vf).
Attachment:
constant acceleration v vs. t.JPG [ 16.64 KiB | Viewed 56862 times ]
Area of a Trapezoid = (average of the parallel bases)*(height)
Here, the parallel bases are the two vertical segments --- the one on the left has a length of (vo) and the one on the right has a length of (vf), so we average those two. The trapezoid is flipped on its edge, so the "height" (i.e. the distance between the two parallel segments) is the horizontal length at the bottom --- 10 s. Thus
Area = (10 s)*(vo + vf)/2 = distance traveled
Now, think about average velocity (AV). We know that this relates total distance (DT) and total time (TT) of any trip.
(DT) = (AV)*(TT)
DT = 10*(AV)
But from the equation above, from the area of the trapezoid, we know
DT = 10*(vo + vf)/2
Comparing those two makes immediately clear:
AV = (vo + vf)/2
This formula has nothing to do with a sequence of any kind. It comes from the area of a trapezoid! For this problem, it is a very useful shortcut:
DT = 125
TT = 10
125 = 10*(vo + vf)/2
125 = 5*(vo + vf)
25 = (vo + vf)
So we just need to numbers that have a sum of 25.
Does all this make sense?
Mike
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Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)
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Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)
this question is a application of AP. and yes the previous answers are correct.
Speed at time '0' = v0
Speed at time '10' = v10
Lets assume the speed is increasing at a rate of 'd' m/s
Note:- Distance traveled in any 1 second is equal to the speed during that second.
Distance traveled in 1st sec = v0
Distance traveled in 2nd sec = v0+d
Distance traveled in 3rd sec = v0+2d
Distance traveled in 10th sec = v0+9d
So Total Distance traveled in 10 second can be given by
v0 + (v0+d) + (v0+2d)+.........+(v0+8d)+(v0+9d) = 10/2[2v0 + (10-1)d]=10/2[2v0 + 9d]
& as per question the bumper has traveled 125 m at the end of 10th second. So
10/2[2v0 + 9d]= 125
[2v0 + 9d]= 25
v0 + (v0+9d) = 25
v0 + v10 = 25......(1)
We are asked the value for speed at 0th & 10th second. From equation (1) we can say that the sum of speed at these moments is equal to 25.
The only options available are 5 , 20.
The speed v0 must be 5 & v10 must 20 because the speed is increasing at constant rate.
Note:- If the speed was decreasing at a constant rate, then v0=20 & v10=5
Hope it helps.
Speed at time '0' = v0
Speed at time '10' = v10
Lets assume the speed is increasing at a rate of 'd' m/s
Note:- Distance traveled in any 1 second is equal to the speed during that second.
Distance traveled in 1st sec = v0
Distance traveled in 2nd sec = v0+d
Distance traveled in 3rd sec = v0+2d
Distance traveled in 10th sec = v0+9d
So Total Distance traveled in 10 second can be given by
v0 + (v0+d) + (v0+2d)+.........+(v0+8d)+(v0+9d) = 10/2[2v0 + (10-1)d]=10/2[2v0 + 9d]
& as per question the bumper has traveled 125 m at the end of 10th second. So
10/2[2v0 + 9d]= 125
[2v0 + 9d]= 25
v0 + (v0+9d) = 25
v0 + v10 = 25......(1)
We are asked the value for speed at 0th & 10th second. From equation (1) we can say that the sum of speed at these moments is equal to 25.
The only options available are 5 , 20.
The speed v0 must be 5 & v10 must 20 because the speed is increasing at constant rate.
Note:- If the speed was decreasing at a constant rate, then v0=20 & v10=5
Hope it helps.
imhimanshu
Wow it looks like a different Question. I am trying to solve it with a physics average speed formula: Avg Velocity : (u+v)/2, where u=initial velocity, V=final velocity.
From Question We understand the Car travels 125 m in 10 sec. Therefore the average velocity = 125/10 m/s or 12.5 or 25/2.
Now from the formula (u+v)/2 = 25/2, Hence u+v=25. By scanning the answer choices answer choice u=5 & v=20 fits. Hence it may be the answer.
General Discussion
SOURH7WK
In this case, your reasoning is correct. The average speed is the mean between the initial and the final speed because the sequence of the speeds for every minute is an arithmetic progression (each speed is the previous one increased by the same amount).
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Love GMAT Quant questions and running.
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Wow this appears to be more of a physics question :D
Can be solved by
s =(v+u)*t /2
or combining
v=u+at and
v^2 =u^2+2as
Can be solved by
s =(v+u)*t /2
or combining
v=u+at and
v^2 =u^2+2as
imhimanshu
danzig
t=0 speed V0
t=10 " V10
speed is raised at const rate. SO, V0, V1 ,....V10 are in AM
Let V10= V0 + (10-1)X .....X is increment for every sec
Since d=SXt
125= (V0(1sec) +V1(1sec) +...V10(1sec
So, 125= V0+ V0+1X + V0+2X +...V0+9X = 10V0 + (1+2+3+..9)X= 10V0 + (9*5)X
But, 9X=V10-V0
So, 125 = 5V0+5V10.................V0+v10=25 (5+20)
chakdum
Joined: 02 Oct 2012
Last visit: 06 Aug 2014
Posts: 12
Location: United States
Concentration: Entrepreneurship, General Management
Schools: Kellogg '16 (D) Cambridge (D) Sloan '16 (D) Duke '16 (D) Ross '16 (WL) Anderson '16 (D) Marshall '16 (D) HEC Jan'16 (D) Oxford (D) Babson '16 (A)
GMAT 1: 720 Q49 V38
GPA: 2.77
WE:General Management (Education)
Schools: Kellogg '16 (D) Cambridge (D) Sloan '16 (D) Duke '16 (D) Ross '16 (WL) Anderson '16 (D) Marshall '16 (D) HEC Jan'16 (D) Oxford (D) Babson '16 (A)
GMAT 1: 720 Q49 V38
Posts: 12
mbmanoj
Doubt :
I agree with
"Speed is raised at const rate. SO, V0, V1 ,....V10 are in AM"
But not with
"Let V10= V0 + (10-1)X .....X is increment for every sec," which will give V1=V0 for t=1 seconds
i.e. V1=V0+X, V2=V0+2X, V3 = V0+3X, so.. V10 = V0+10X
Please comment.
Hadrienlbb
Joined: 21 Oct 2017
Last visit: 23 Mar 2020
Posts: 67
Location: France
Concentration: Entrepreneurship, Technology
GMAT 1: 750 Q48 V44
GPA: 4
WE:Project Management (Internet and New Media)
mikemcgarry,
Would you mind briefly commenting my following logic?
We are told that the speed of the car is increasing at a constant rate with respect to time, that time is 10 seconds and that the car has traveled 125 meters.
Therefore, I conclude that my Speed at V0 will be lower than 12.5 meters (125/10 = 12.5 meters / second). Then there is only one choice for V0 in the table. V0 = 5.
Then I know I need the rate to constantly increase until V10 and need to get to 125 meters.
(V10 + V0)/2 = 12.5 meters/s
(V10 + 5)/2 = 12.5
V10 + 5 = 12.5*2
V10 = 25 - 5 = 20
Thanks!
Would you mind briefly commenting my following logic?
We are told that the speed of the car is increasing at a constant rate with respect to time, that time is 10 seconds and that the car has traveled 125 meters.
Therefore, I conclude that my Speed at V0 will be lower than 12.5 meters (125/10 = 12.5 meters / second). Then there is only one choice for V0 in the table. V0 = 5.
Then I know I need the rate to constantly increase until V10 and need to get to 125 meters.
(V10 + V0)/2 = 12.5 meters/s
(V10 + 5)/2 = 12.5
V10 + 5 = 12.5*2
V10 = 25 - 5 = 20
Thanks!
mikemcgarry
Posts: 4,475
Hadrienlbb
Dear Hadrienlbb,
My friend, I would say that yours is a brilliant and wonderfully elegant solution! Kudos!
Mike
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Mike McGarry
Magoosh Test Prep
Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)
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Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)
There is a simple formula to calculate the distance traveled by object with consistently increasing speed:
S=V0t+at²/2
where S is the distance, a is the rate of speed increase, t is the time
So we get 125 = 10*V0 + a*(10)^2/2 = 10*V0 + 50a
--> 25=2V0 + 10a
And we know that V10 = V0+10a (speed after 10 seconds)
So S = V0 + V0 + 10a = V0+V10.
Only 5 and 20 will satisfy this equation.
S=V0t+at²/2
where S is the distance, a is the rate of speed increase, t is the time
So we get 125 = 10*V0 + a*(10)^2/2 = 10*V0 + 50a
--> 25=2V0 + 10a
And we know that V10 = V0+10a (speed after 10 seconds)
So S = V0 + V0 + 10a = V0+V10.
Only 5 and 20 will satisfy this equation.
Posts: 16,401
Official Explanation
Since the speed of the car increases at a constant rate, the car's average speed, in meters per second, over the 10-second period is equal to \(\frac{1}{2} (v_{0} + v_{10})\). Thus, the distance the car traveled, in meters, over that period is equal to \(\frac{1}{2} (v_{0} + v_{10})(10)\), or \(5(v_{0} + v_{10})\). But this value is given to be 125, so it must be true that \(v_{0} + v_{10} = \frac{125}{5}\), or 25. Because speed must be positive, both \(v_{0}\) and \(v_{10}\) must be less than 25, ruling out 36 and 72 as possible values.
Among the available alternatives—5, 18, and 20—the only pair that sum to 25 are 5 and 20. Since the speed is increasing, it must also be true that \(v_{0} ≤ v_{10}.\) Therefore \(v_{0} = 5.\)
The correct answer is 5.
Following the analysis above, the value of \(v_{10}\) must correspond to \(v_{0} = 5\), and it is given by \(v_{10} = 20.\)
The correct answer is 20.
Since the speed of the car increases at a constant rate, the car's average speed, in meters per second, over the 10-second period is equal to \(\frac{1}{2} (v_{0} + v_{10})\). Thus, the distance the car traveled, in meters, over that period is equal to \(\frac{1}{2} (v_{0} + v_{10})(10)\), or \(5(v_{0} + v_{10})\). But this value is given to be 125, so it must be true that \(v_{0} + v_{10} = \frac{125}{5}\), or 25. Because speed must be positive, both \(v_{0}\) and \(v_{10}\) must be less than 25, ruling out 36 and 72 as possible values.
Among the available alternatives—5, 18, and 20—the only pair that sum to 25 are 5 and 20. Since the speed is increasing, it must also be true that \(v_{0} ≤ v_{10}.\) Therefore \(v_{0} = 5.\)
The correct answer is 5.
Following the analysis above, the value of \(v_{10}\) must correspond to \(v_{0} = 5\), and it is given by \(v_{10} = 20.\)
The correct answer is 20.
Elite097
Joined: 20 Apr 2022
Last visit: 07 May 2026
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Schools: Sloan '25 Columbia '27 Haas '27
GPA: 3.64
Schools: Sloan '25 Columbia '27 Haas '27
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KarishmaB @anishpassai pls explain this.
How do we know it is not in GP? Quest says it is increasing at 'constt growth rate'.
And since everyone has taken it in Ap, then does it mean there are a total of 11 terms from 0 secs to 10secs: v0,.......,v10. Thus shouldnt sum of ap be (11/2)(2v0+ 10d) and not (10/2)(2v0+ 9d)?
How do we know it is not in GP? Quest says it is increasing at 'constt growth rate'.
And since everyone has taken it in Ap, then does it mean there are a total of 11 terms from 0 secs to 10secs: v0,.......,v10. Thus shouldnt sum of ap be (11/2)(2v0+ 10d) and not (10/2)(2v0+ 9d)?
Elite097
"the speed of the car is increasing at a constant rate" means the speed is increasing by a fixed amount. This 'fixed amount' is the common diff and it has to be AP.
When it is a GP, the speed increases exponentially, not at a constant rate.
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Elite097
Joined: 20 Apr 2022
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KarishmaB yes but constant rate refers to constant percentage hence it could still increase exponentially and other questions also use growth rate to refer to GP so not clear
Elite097
No. If speed is increasing at a variable rate, how does "constant rate" make sense? It means acceleration is constant, say 2m/s^2.
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Dude Physics in GMAT.
Vt = speed at t
Vo = initial speed of car
a = acceleration
t = time
s = distance travel.
\(Vt = Vo + at\) and \(Vt^2 = Vo^2 + 2as\)
now,
\(Vt^2 - Vo^2 = 2as\)
\((Vt - Vo)*(Vt + Vo) = 2as\)
from \(Vt = Vo + at\)
\(Vt + Vo = 2*\frac{s}{t}\)
s=125 t=10
substittue and solve
\(Vt + Vo = 25\) and we know Vt>Vo
check the values in the table that will satisfy this.
Vo = 5 and V10= 20.
Vt = speed at t
Vo = initial speed of car
a = acceleration
t = time
s = distance travel.
\(Vt = Vo + at\) and \(Vt^2 = Vo^2 + 2as\)
now,
\(Vt^2 - Vo^2 = 2as\)
\((Vt - Vo)*(Vt + Vo) = 2as\)
from \(Vt = Vo + at\)
\(Vt + Vo = 2*\frac{s}{t}\)
s=125 t=10
substittue and solve
\(Vt + Vo = 25\) and we know Vt>Vo
check the values in the table that will satisfy this.
Vo = 5 and V10= 20.
Thanks for the helpful adive.
But I wonder why the speed at 2nd sec is V0+d not V0 x r when it is said to increase at a 'rate'.
Doesn't it meant that V1/V0 = V2/V1?
But I wonder why the speed at 2nd sec is V0+d not V0 x r when it is said to increase at a 'rate'.
Doesn't it meant that V1/V0 = V2/V1?
fameatop
Not an expert on this but; it worked for me thinking of it as an average,
So the average speed is 12.5 and I need a max and a min speeds so 3 speeds in total so 12.5* 3 is 37.5 minus the average speed= 25
The min speed + the max speed must be = 25 so the answer choices from there are straightforward.
So the average speed is 12.5 and I need a max and a min speeds so 3 speeds in total so 12.5* 3 is 37.5 minus the average speed= 25
The min speed + the max speed must be = 25 so the answer choices from there are straightforward.
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