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In a 1000 m race Usha beats Shiny by 50 m. In the same race,

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In a 1000 m race Usha beats Shiny by 50 m. In the same race, [#permalink] New post 05 Apr 2013, 10:38
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In a 1000 m race Usha beats Shiny by 50 m. In the same race, by what time margin Shiny beats Mercy
who runs at 4 m/s

a) 100 sec.
b) 50 sec
c) 25 sec
d) Data not sufficient
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Two persons C & D started traveling from A and B which are 3 [#permalink] New post 05 Apr 2013, 12:28
Two persons C & D started traveling from A and B which are 300 km apart, towards B and A
respectively at 1.00 p.m. C travels at a constant speed of 30 kmph whereas D doubles his speed every
hour. If D reaches A in 4 5/8 hours, at what time did C and D meet each other?

a) 4:30 p.m.
b) 4:40 p.m.
c) 5:00 p.m.
d) 5:10 p.m.
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Two trains T1 and T2 start simultaneously from two stations [#permalink] New post 05 Apr 2013, 12:48
Two trains T1 and T2 start simultaneously from two stations X and Y respectively towards each
other. If they are 70 km apart both 3 and 6 hours after start, then find the distance between the two
stations.

a) 210 km
b)240 km
c)220km
d)180km
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In a class of 150, students can enroll for Judo or Taekwondo [#permalink] New post 12 Apr 2013, 14:08
In a class of 150, students can enroll for Judo or Taekwondo or both or none.The number of students enrolling for Taekwondo only is twice the number of students enrolling for Judo. The number of students opting for both is ten more than the number of students who opt out of both. The number of students opting at least 1 course is nine times the number of students opting for none. find -

1) the number of students who enrolled for Taekwondo
A. 110
B. 115
C. 90
D. 125
E. NONE

2) the number of students who enrolled for Judo only
A. 20
B. 25
C. 15
D. 40
E. 45
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Re: In a 1000 m race Usha beats Shiny by 50 m. In the same race, [#permalink] New post 12 Apr 2013, 14:41
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Two persons C & D started traveling from A and B which are 300 km apart, towards B and A
respectively at 1.00 p.m. C travels at a constant speed of 30 kmph whereas D doubles his speed every
hour. If D reaches A in 4 5/8 hours, at what time did C and D meet each other?

a) 4:30 p.m.
b) 4:40 p.m.
c) 5:00 p.m.
d) 5:10 p.m.

To calculate the speed of D we use this formula: (it travels 1h at v, 1h at 2v and so on till it reach 300 km)
v+2v+4v+8v+5/8*16v=300 , 25v=300 , v=12
D started at 12kmph. C travels at 30 kmph. They meet after 4 hours.
Distance of C=30*4=120km
Distence of D=12+24+48+96=180km. 120+180=300km.
They meet at 1pm + 4 h = 5pm. C
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Re: In a 1000 m race Usha beats Shiny by 50 m. In the same race, [#permalink] New post 12 Apr 2013, 16:38
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Quote:
Two trains T1 and T2 start simultaneously from two stations X and Y respectively towards each
other. If they are 70 km apart both 3 and 6 hours after start, then find the distance between the two
stations.

a) 210 km
b)240 km
c)220km
d)180km


Variables -
a the speed of train T1
b the speed of train T2
d the distance between A and B

a + b = (d - 70) /3 - trains are traveling towards each other
a + b = (d + 70) /6 - trains are traveling away from each other

(d-70)/3 = (d + 70)/6 ==> d = 210

Answer is A

What is the source of these problems? This problem took me long time... more than 15 mins.. :-( Not good...

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Re: In a 1000 m race Usha beats Shiny by 50 m. In the same race, [#permalink] New post 12 Apr 2013, 18:14
Please explain how to get the working for the first answer
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Re: In a 1000 m race Usha beats Shiny by 50 m. In the same race, [#permalink] New post 13 Apr 2013, 02:25
Expert's post
iwantto wrote:
In a 1000 m race Usha beats Shiny by 50 m. In the same race, by what time margin Shiny beats Mercy
who runs at 4 m/s

a) 100 sec.
b) 50 sec
c) 25 sec
d) Data not sufficient


Usha beats Shiny by 50 m means Usha reaches the finish line when Shiny is 50 m away from the finish line i.e. Usha covers 1000m in the time in which Shiny covers 950m.
We can say that the Ratio of their speeds, speed of Usha : speed of Shiny = 1000:950 = 20:19
We don't know their actual speeds.
Usha's speed could be 20 m/s and Shiny's will be 19 m/s.
or
Usha's speed could be 40 m/s and Shiny's will be 38 m/s.
or
Usha's speed could be 10 m/s and Shiny's will be 9.5 m/s.
etc

We know Mercy's speed is 4 m/s so we know Mercy takes 1000/4 = 250 s to complete the 1000m but we don't know Shiny's speed. We don't know how long Shiny takes to complete the 1000 m hence data is not sufficient here.
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Re: Two persons C & D started traveling from A and B which are 3 [#permalink] New post 13 Apr 2013, 02:34
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Expert's post
iwantto wrote:
Two persons C & D started traveling from A and B which are 300 km apart, towards B and A
respectively at 1.00 p.m. C travels at a constant speed of 30 kmph whereas D doubles his speed every
hour. If D reaches A in 4 5/8 hours, at what time did C and D meet each other?

a) 4:30 p.m.
b) 4:40 p.m.
c) 5:00 p.m.
d) 5:10 p.m.


Responding to a pm:

Case of C is simple.
Distance he needs to cover is 300 km at constant speed of 30 kmph so he takes 10 hrs and covers 30 km every hour.

Case of D.
He starts with a speed of s and covers s km in the first hr. Then his speed becomes 2s and he covers 2s km in the next hr. Then his speed becomes 4s and he covers 4s km in the next hour and so on...
s + 2s + 4s + 8s + 16s*(5/8) = 300
25s = 300
s = 12 kmph
So distance covered by D every hr: 12, 24, 48, 96, 120

Together, to meet, they need to cover a total distance of 300 km.
Now just try to sum up the distance covered by them in each hr till you get 300.

(30+12) + (30+24) + (30+ 48) + (30+96) = 300

So they take a total of 4 hrs together to cover 300 km.

So they meet at 1+4 = 5:00 pm
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Re: Two trains T1 and T2 start simultaneously from two stations [#permalink] New post 13 Apr 2013, 02:43
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Expert's post
iwantto wrote:
Two trains T1 and T2 start simultaneously from two stations X and Y respectively towards each
other. If they are 70 km apart both 3 and 6 hours after start, then find the distance between the two
stations.

a) 210 km
b)240 km
c)220km
d)180km


Responding to a pm:

What information do you get from this: If they are 70 km apart both 3 and 6 hours after start?
Say, they started at 12:00 noon. This means that distance between them at 3:00 pm was 70 km. Then they met. Then they traveled further and the distance between them at 6:00 pm was again 70 km.

It means that in 3 hrs (from 3:00 pm to 6:00 pm), they together covered 140 km (70 km between them, then they met and then they traveled further to create a distance of 70 km between them again so total 140 km)

Now, to have a distance of 70 km between them for the first time i.e. at 3:00 pm, how much must have they traveled since 12:00 noon?
They both traveled for 3 hrs. How much distance do they cover together in 3 hrs? We already know that they together cover 140 km in 3 hrs.

Hence the distance between the stations must be 140 + 70 = 210 km
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Re: In a class of 150, students can enroll for Judo or Taekwondo [#permalink] New post 13 Apr 2013, 02:54
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Expert's post
iwantto wrote:
In a class of 150, students can enroll for Judo or Taekwondo or both or none.The number of students enrolling for Taekwondo only is twice the number of students enrolling for Judo. The number of students opting for both is ten more than the number of students who opt out of both. The number of students opting at least 1 course is nine times the number of students opting for none. find -

1) the number of students who enrolled for Taekwondo
A. 110
B. 115
C. 90
D. 125
E. NONE

2) the number of students who enrolled for Judo only
A. 20
B. 25
C. 15
D. 40
E. 45


Total number of students = 150

Given: The number of students opting at least 1 course is nine times the number of students opting for none

This means no of students opting for at least 1 course: No of students opting for no course = 9:1
Since total number of students is 150, the number of students opting for at least 1 course = 135
the number of students opting for none = 15

Given: The number of students opting for both is ten more than the number of students who opt out of both.

This means no of students opting for both = 15 + 10 = 25

Given: The number of students enrolling for Taekwondo only is twice the number of students enrolling for Judo.
If no of students enrolling for Judo is J, number of students enrolling for Taekwondo only = 2J
2J + J = 135
J = 45

No of students who enrolled for Taekwondo only = 2J = 2*45 = 90
Total no of students who enrolled for Taekwondo (Taekwondo only + Both) = 90 + 25 = 115

No of students who enrolled for Judo = J = 45
No of students who enrolled for Judo only (Judo - Both) = 45 - 25 = 20

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http://www.veritasprep.com/blog/2012/09 ... ping-sets/
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AD:DC = 2:3, then angle ABC is equal to? [#permalink] New post 13 Jun 2013, 12:44
Attachment:
Untitled.jpg
Untitled.jpg [ 5.55 KiB | Viewed 1055 times ]
AD:DC = 2:3, then angle ABC is equal to?

A. 30
B. 40
C. 45
D. 60
E. 110

Hi,
This may be very easy ... I couldn't solve ,, or may be this is missing some information.

Please help...

Answer:
[Reveal] Spoiler:
B
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Re: AD:DC = 2:3, then angle ABC is equal to? [#permalink] New post 14 Jun 2013, 02:37
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iwantto wrote:
Attachment:
The attachment Untitled.jpg is no longer available
AD:DC = 2:3, then angle ABC is equal to?
A. 30
B. 40
C. 45
D. 60
E. 110
Hi,
This may be very easy ... I couldn't solve ,, or may be this is missing some information.
Please help...


Well the easy way to look at it would be with the angle-bisector theorem, which states that if the angle bisector divides the opposite side by a given ratio such as given in the diagram. The ratio:
AD/DC = AB/BC always holds. Now, the reverse can be assumed as true as well.

Now we know that DBC = 20 degree from the traingle property. Now it can also be seen, AB/BC = 10/15 = 2/3 which is equal to AD/DC as per the question. Hence it can be assumed that BD is the angle bisector. Hence the total angle B is 2*20 = 40 degrees. Hence the answer is [B].

Hope it helps! :)

Regards,
A
Attachments

traingle.png
traingle.png [ 5.57 KiB | Viewed 1003 times ]


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Re: AD:DC = 2:3, then angle ABC is equal to?   [#permalink] 14 Jun 2013, 02:37
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