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605-655 Level|   Math Related|         
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GMAT Club Olympics 2024 (Day 8): An athlete training for a cross-count 17 Jul 2024, 14:27
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GMAT Club Olympics 2024 (Day 8): An athlete training for a cross-count 17 Jul 2024, 14:32
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­x = distance ran on packet dirt
4x = distance ran on grass

Total time = x/201 + 4x/100.5 = 21

x/201 + 8x/201 = 21
9x = 21*201 = 4221
x = 469

distance ran on packet dirt = 469
distance ran on grass = 1876
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GMAT Club Olympics 2024 (Day 8): An athlete training for a cross-count 17 Jul 2024, 14:52
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Bunuel
­An athlete training for a cross-country race covered three types of terrain in three distinct segments of a practice run; she ran across stretches of packed dirt, rocky scree, and grass, in that order. On packed dirt, she ran at an average speed of 201 meters per minute; she then rapidly slowed on the rocky scree, and covered the grass segment at an average speed of 100.5 meters per minute. The distance she covered on grass was four times the distance she covered on packed dirt, and it took her 21 minutes to cover both. On the table, select the distance she covered on both these terrains, in meters. Make only two selections, one in each column.


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Let's say she covered D meters on packed dirt, so she covered 4D meters on grass (we don't care about the rocky scree part for this problem).

Knowing her speeds too, and that Time=Distance/Speed, we can set up the equation as:

\(\frac{D}{201} + \frac{4D}{100.5} = 21\)

\(\frac{D}{201} + \frac{8D}{201} = 21\)

\(\frac{9D}{201}= 21\)

\(D=469 meters\) -> Distance she covered on packed dirt.

\(4D=1876 meters\) -> Distance she covered on grass.

The answers are:

Distance covered on packed dirt (meters): 469
Distance covered on grass (meters): 1876 ­
 ­
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GMAT Club Olympics 2024 (Day 8): An athlete training for a cross-count 17 Jul 2024, 17:26
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Let's donate the distances, times and velocities to cross the packed dirt, rocky scree, and grass respectively \((D_{pd}, T_{pd}, V_{pd})\),  \((D_r, T_r, V_r)\) and \((D_g, T_g, V_g)\).
According to the stem :

\(V_{pd} = \frac{D_{pd}}{ T_{pd}} = 201\ m/min\\
V_g = \frac{D_g}{ T_g} = 100.5\ m/min­\)­

and 

\(T_g + T_{pd} = 21\ min­\)­ 
\(D_g = 4*D_{pd} \) 

from the equations above:

\(\frac{D_g}{ V_g} + \frac{D_{pd}}{V_{pd}} = \frac{4*D_{pd}}{100.5} + \frac{D_{pd}}{201} = 21\)

After simplification we find ­
\( D_{pd} = 469\ m \) and \( D_g= 4*D_{pd} = 1876\ m\)­
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GMAT Club Olympics 2024 (Day 8): An athlete training for a cross-count 17 Jul 2024, 18:31
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­For this question we have that:

Speed dirt = 201 = D/td  and Speed grass = 100,5 = 4D/tg
We also know that td  + tg = 21

So we can put the equation as: D/200 + 4D/100 = 21

Since the answers are far apart, we can estimate

(D+8D)/200 = 21
9D = 4200
D= 469, this is the distance for dirt

And for grass is 4*469 = 1,876
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GMAT Club Olympics 2024 (Day 8): An athlete training for a cross-count 17 Jul 2024, 18:43
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­As per the question,

Packed dirt:
Speed: 201 m/min

Grass:
Speed: 100.5 m/min

Distance on Grass (y) = 4*distance on packed dirt (x)

Time to cover:
Packed dirt: \(\frac{x}{201}\) m
Grass: \(\frac{4x}{100.5}\) m

Time to cover these two tracks: 21 min

So,
\(\frac{x}{201} + \frac{4x}{100.5} = 21 \)
\(9x = 21*201 \)
\(x = 469 m \)

Distance on packed dirt = 469 m
Distance on Grass = 4*469 = 1876 m­
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GMAT Club Olympics 2024 (Day 8): An athlete training for a cross-count 17 Jul 2024, 18:59
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Bunuel
­An athlete training for a cross-country race covered three types of terrain in three distinct segments of a practice run; she ran across stretches of packed dirt, rocky scree, and grass, in that order. On packed dirt, she ran at an average speed of 201 meters per minute; she then rapidly slowed on the rocky scree, and covered the grass segment at an average speed of 100.5 meters per minute. The distance she covered on grass was four times the distance she covered on packed dirt, and it took her 21 minutes to cover both. On the table, select the distance she covered on both these terrains, in meters. Make only two selections, one in each column.


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­Let the distance covered grass is 4d; 
Distance covered on dirt is d;

Approximating as options are not close.  

4d/100 +d/200 = 21;

9d = 21*200

d = 1400/3 = 466..

So the distance covered on dirt is ~466
distance covered grass is 4d is ~1864

Hence best choices are 469,1876
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GMAT Club Olympics 2024 (Day 8): An athlete training for a cross-count 17 Jul 2024, 19:22
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Bunuel
­An athlete training for a cross-country race covered three types of terrain in three distinct segments of a practice run; she ran across stretches of packed dirt, rocky scree, and grass, in that order. On packed dirt, she ran at an average speed of 201 meters per minute; she then rapidly slowed on the rocky scree, and covered the grass segment at an average speed of 100.5 meters per minute. The distance she covered on grass was four times the distance she covered on packed dirt, and it took her 21 minutes to cover both. On the table, select the distance she covered on both these terrains, in meters. Make only two selections, one in each column.


­
 


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­Let the distance covered in Dirt=a , Rocky scree=b , Grass=c

Speed in dirt=201 m/min  
Speed in grass=100.5 m/min  

Given c=4a

time to cross dirt+ time to cross grass=21
a/201+c/100.5=21
a/201+4a/100.5=21
So a= 469
And c=4*469=1876

Ans is 
Distance covered on packed dirt= 469
Distance covered on grass =1876
 
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GMAT Club Olympics 2024 (Day 8): An athlete training for a cross-count 17 Jul 2024, 20:49
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Let d denote the distance on packed dirt at 201 meters per minute. The equation we get from the given info is:
\(\frac{d}{201}+\frac{4d}{100.5}=21\)
Since the distance travelled is 4 times for the grass. 
Now, we just need to solve the equation to get 
d=469
and 4d = 1876. 
 ­
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GMAT Club Olympics 2024 (Day 8): An athlete training for a cross-count 17 Jul 2024, 21:10
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­An athlete training for a cross-country race covered three types of terrain in three distinct segments of a practice run; she ran across stretches of packed dirt, rocky scree, and grass, in that order. On packed dirt, she ran at an average speed of 201 meters per minute; she then rapidly slowed on the rocky scree and covered the grass segment at an average speed of 100.5 meters per minute. The distance she covered on grass was four times the distance she covered on packed dirt, and it took her 21 minutes to cover both.

On the table, select the distance she covered on both these terrains, in meters. Make only two selections, one in each column.



Solution: Let's assume
Average speed on packed dirt as x meters/min
Average speed on grass as y meters/min
Distance covered on packed dirt as d1 meters
Distance covered on grass as d2 meters
Time taken to cover packed dirt as t1 minutes
Time taken to cover grass as t2 minutes

Given that,
x = 201 meters/min
y = 100.5 meters/min
d2 = 4d1
t1 + t2 = 21  ------- (1)

Now, t1 = \(\frac{d1}{x}\)
t1 = \(\frac{d1}{201}\)

Similarly, t2 = \(\frac{d2}{y}\)
t2 = \(\frac{4d1}{100.5}\)

t2 = \(\frac{8d1}{201}\)

From (1)
\(\frac{d1}{201}\) + \(\frac{8d1}{201}\) = 21

\(\frac{9d1}{201}\) = 21
d1 = \(\frac{21 * 201}{9}\)
d1 = 469 meters
and d2 = 4 * 469 meters = 1876 meters

Distance covered on packed dirt = 469 meters
Distance covered on grass = 1876 meters
 ­
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GMAT Club Olympics 2024 (Day 8): An athlete training for a cross-count 17 Jul 2024, 21:41
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Quote:
­An athlete training for a cross-country race covered three types of terrain in three distinct segments of a practice run; she ran across stretches of packed dirt, rocky scree, and grass, in that order. On packed dirt, she ran at an average speed of 201 meters per minute; she then rapidly slowed on the rocky scree, and covered the grass segment at an average speed of 100.5 meters per minute. The distance she covered on grass was four times the distance she covered on packed dirt, and it took her 21 minutes to cover both. On the table, select the distance she covered on both these terrains, in meters. Make only two selections, one in each column.
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Let Dp be the distance traveled in packed dirt, and Dg be that of grass.

It is mentioned that Dg = 4Dp
It is also given that Sp = 201, Sg = 100.5.

Total time = 21 minutes.

Therefore, 21 = Dp/201 + 4Dp/100.5

Solving this, we can get, Dp = 469, and Dg = 4*Dp = 4*469 => 1876
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GMAT Club Olympics 2024 (Day 8): An athlete training for a cross-count 17 Jul 2024, 21:55
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Let us collect all the information from the question and present it in a more clear way.

Let x be the distance for Dirt or D and so Grass or G will be 4x.

D = x
G = 4x 

Average speed of D = 201 meters per minute
Average speed of G = 100.5 meters per minute. 

Time to cover D and G = 21 minutes

Let us use the Distance Time and Speed formula for the area of D and G combined. 

Distance = 5x
Speed = ? or S
Time = 21 minutes

5x = 21*S

We also know the following two equations for D and G

x = 201*t1
4x = 100.5*t2
and that t1 +t2 = 21 

We can substitute x to get

804*t1 = 100.5*t2
t1 + t2 = 21

Now we can substitute either t1 or t2

804*t1 = 100.5*(21 - t1)
8*t1 = 21 - t1
9*t1 = 21
t1 = 7/3 

Now we subsitte t1 to this equation, x = 201*t1 to get

x = 201*7/3
x = 67*7
x = 469 which is the distance of Packed Dirt
Grass is 4x 

Therefore the answers are
Packed Dirt = 469; Grass = 1876

 
Bunuel
­An athlete training for a cross-country race covered three types of terrain in three distinct segments of a practice run; she ran across stretches of packed dirt, rocky scree, and grass, in that order. On packed dirt, she ran at an average speed of 201 meters per minute; she then rapidly slowed on the rocky scree, and covered the grass segment at an average speed of 100.5 meters per minute. The distance she covered on grass was four times the distance she covered on packed dirt, and it took her 21 minutes to cover both. On the table, select the distance she covered on both these terrains, in meters. Make only two selections, one in each column.


­
 


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GMAT Club Olympics 2024 (Day 8): An athlete training for a cross-count 17 Jul 2024, 22:02
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Bunuel
­An athlete training for a cross-country race covered three types of terrain in three distinct segments of a practice run; she ran across stretches of packed dirt, rocky scree, and grass, in that order. On packed dirt, she ran at an average speed of 201 meters per minute; she then rapidly slowed on the rocky scree, and covered the grass segment at an average speed of 100.5 meters per minute. The distance she covered on grass was four times the distance she covered on packed dirt, and it took her 21 minutes to cover both. On the table, select the distance she covered on both these terrains, in meters. Make only two selections, one in each column.


­
 


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­Packed dirt:
speed = 201 meters/minute
time = t1
distance = d

Grass:
speed = 100.5 meters/minute = 201/2 meters/minute
time = t2
distance = 4d

t1 + t2 = 21 minutes

Speed = Distance/Time => Time = Distance/Speed 

t1 + t2 = d/201 + 8d/201 = 21
=> 9d/201 = 21
=> d/67 = 7
=> d = 469

1st column: 469 meters
2nd column: 1,876 meters
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GMAT Club Olympics 2024 (Day 8): An athlete training for a cross-count 17 Jul 2024, 22:22
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\(\text{Total time taken=Time taken at Dirt + Time taken at Grass}\)

\(T=Td+Tg\)

\(\text{Time taken} = \frac{\text{Distance}}{\text{speed}}\)­

\(\text{Td} = \frac{\text{D}}{\text{Sd}}\)­ & \(\text{Tg} = \frac{\text{G}}{\text{Sg}}\)­

\(\frac{\text{D}}{201}+\frac{\text{G}}{100.5}=21\)­

\(\text{G=4*D}\)­

\(\frac{\text{D}}{201}+\frac{\text{4D}}{100.5}=21\)­

\(\frac{\text{D}}{201}+\frac{\text{8D}}{201}=21\)­

\(9D=201*21\)

\(D=469\)

\(G=4*469\)

\(G=1876\)­
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GMAT Club Olympics 2024 (Day 8): An athlete training for a cross-count 17 Jul 2024, 22:45
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Total time on PD+grass =21 minute
X/201+4X/100.5=21
rounding
x/200+8X/200=21
9x=21*200
x=4200/9
x=467

4x=1867

Ans-BD
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GMAT Club Olympics 2024 (Day 8): An athlete training for a cross-count 17 Jul 2024, 22:59
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­On packed dirt: Average speed = 201 m/min
On grass: Average speed = 100.5 m/min
Total time taken for both terrains = 21 min

D(dirt )=x
D(grass)=4x

td+tg=21
x/201+4x/100.5=21

x~467
4*x=1868
Ans B D
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GMAT Club Olympics 2024 (Day 8): An athlete training for a cross-count 17 Jul 2024, 23:06
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­So, we know the following:
\(g =  4d,\) where g is the length of grass patch and d stand for the dirt length.

And as for the time of 21 minutes, it is:
\(21 = \frac{d}{201} + \frac{g}{100.5 }= \frac{d}{201} + \frac{4d}{100.5} = d + \frac{8d}{201} = \frac{9d}{201}\)
From here, \(d = \frac{201*21}{9} = 469\)
and\( g = 4d = 1876\)

So, the answers are 469 and 1876.
 ­
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GMAT Club Olympics 2024 (Day 8): An athlete training for a cross-count 17 Jul 2024, 23:11
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Answer:
Dirt: 469
Grass: 1876

Distance on grass = 4*Distance on dirt.
Dg=4Dd

Speed on Dirt, Vd = 201m/min
Speed on Grass, Vg = 100.5m/min

Total time, Td + Tg = 21 mins

We know Total Time = Td + Tg = Dd/201 + Dg/100.5 = Dd/201 + 4Dd/100.5 = (Dd+8Dd)/201 = 9Dd/201 = 21

=> Dd = 21*201/9 = 469m

Also Dg = 4*Dd = 4*469 = 1876m
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GMAT Club Olympics 2024 (Day 8): An athlete training for a cross-count 17 Jul 2024, 23:32
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­An athlete training for a cross-country race covered three types of terrain in three distinct segments of a practice run; she ran across stretches of packed dirt (t1), rocky scree(t2), and grass(t3), in that order. On packed dirt, she ran at an average speed of 201 meters per minute; she then rapidly slowed on the rocky scree, and covered the grass segment at an average speed of 100.5 meters per minute. The distance she covered on grass was four times the distance she covered on packed dirt [t3*100.5=4*t1*201=>t3=8t1], and it took her 21 minutes to cover both[[b]t1+t3=21]. On the table, select the distance she covered on both these terrains, in meters. Make only two selections, one in each column.


[b]Solving both eqn, we get
t3= (21*8)/9
t1=t3/8

Distance on grass= t3*100.5=1876

Distance on dirt=1876/4=469­
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GMAT Club Olympics 2024 (Day 8): An athlete training for a cross-count 17 Jul 2024, 23:47
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Let's say the time taken for the travel on packed dirt = t1 
Let's say the time taken for the travel on grass segment = t2 

t1 + t2 = 21
(4d/201) + (d/100.5) = 21 
9d/201 = 21
d = 469 
and 4d = 1876 

Therfore, the right ans choice is 469 and 1876
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