Albert Einstein's theory of general relativity is a century old next year and, as far as the test of time is concerned, it seems to have done rather well. For many, indeed, it doesn't merely hold up: it is the archetype for what a scientific theory should look like. Einstein's achievement was to explain gravity as a geometric phenomenon: a force that results from the distortion of space-time by matter and energy, compelling objects - and light itself - to move along particular paths, very much as rivers are constrained by the topography of their landscape. General relativity departs from classical Newtonian mechanics and from ordinary intuition alike, but its predictions have been verified countless times. In short, it is true.
Einstein himself seemed rather indifferent to the experimental tests, however. The first came in 1919, when the British physicist Arthur Eddington observed the Sun's gravity bending starlight during a solar eclipse. What if those results hadn't agreed with the theory? 'Then,' said Einstein, 'I would have been sorry for the dear Lord, for the theory is correct.'
That was Einstein all over. As the Danish physicist Niels Bohr commented at the time, he was a little too fond of telling God what to do. But this wasn't sheer arrogance, nor parental pride in his theory. The reason Einstein felt general relativity must be right is that it was too IMS Learning Resources Private Limited beautiful a theory to be wrong.
This sort of talk both delights today's physicists and makes them a little nervous. After all, isn't experiment - nature itself - supposed to determine truth in science? What does beauty have to do with it? 'Aesthetic judgments do not arbitrate scientific discourse,' the string theorist Brian Greene reassures his readers in his book The Elegant Universe. 'Ultimately, theories are judged by how they fare when faced with cold, hard, experimental facts.' Einstein, Greene insists, didn't mean to imply otherwise - he was just saying that beauty in a theory is a good guide, an indication that you are on the right track.
Einstein isn't around to argue, of course, but I think he would have done. It was Einstein, after all, who said that 'the only physical theories that we are willing to accept are the beautiful ones'. And if he were simply defending theory against too hasty a deference to experiment, there would be plenty of reason to side with him - for who is to say that, in case of a discrepancy, it must be the theory and not the measurement that is in error? But that's not really his point. Einstein seems to be asserting that beauty trumps experience come what may .
He wasn't alone. Here's the great German mathematician Hermann Weyl: 'My work always tries to unite the true with the beautiful; but when I had to choose one or the other, I usually chose the beautiful.' So much for John Keats's 'Beauty is truth, truth beauty.' And so much, you might be tempted to conclude, for scientists' devotion to truth: here were some of its greatest luminaries, pledging obedience to a different calling altogether.
What is the main point of this passage?
1) Beauty is not truth when it comes to scientific theories.
2) Scientists tend to prefer beautiful scientific theories over verifiable ones.
3) Some scientists, like Einstein, focus on the beauty rather than verification of scientific theories.
4) Einstein and other scientists have shown how beauty is an important quality of scientific theories.
Which of the following, if true, would not validate Einstein's views as stated in this passage?
1) Throughout history, the most successful and important scientific theories have been the most 'beautiful' ones.
2) Flawed experimental designs can sometimes invalidate scientific theories, in which case, the theories' beauty is a good guide to the truth.
3) The term 'beauty', as used by scientists, is merely another word for anything that throws light on the basic structure of the universe.
4) 'Beauty' in scientific terms merely means simplicity, and simple theories are more likely to be true.
Which of the following is true about Albert Einstein, as per this passage?
i] He thought himself superior to God.
ii] He and Niels Bohr were rivals.
iii] He and Brian Greene were friends.
iv] His theory of general relativity explained how gravity works.
v] His theory of general relativity suggested that light is bent by gravity.
1) [iv] and [v]
2) [i], [ii], [iv] and [v]
3) [ii], [iii], [iv] and [v]
4) [i] and [iv]
Elements of which of the following pairs do not belong together?
1) Niels Bohr - Denmark
2) Brian Greene - String theory
3) Arthur Eddington - Solar eclipse
4) Hermann Weyl - 'Beauty is truth, truth beauty.
If f(x) = min(3x-1,x-2 ), find the max value of f(x).
C : infinity
D: None of these
For people asking for percentiles, here's my speculation:
170 - 99.8%ile
150 - 99.5%ile
130 - 99%ile
120 - 98.5%ile
110 - 97%ile
100 - 95%ile
Be the change you wish to see in the world
(Nitin Gupta- Alpha Numeric)
These types of problems deal with athletes running on a circular track and questions being asked about when, where and how often would two or more athletes cross each other.
Meeting for the First Time When running in opposite directions:
Consider two friends, A and B, who are separated by 1500 meters and they move towards each other at speeds of 40 m/s and 10 m/s. After how much time would they meet?This is a straight question on relative speed and the time taken to meet = 1500/(40+10) = 30 sec
When running in the same direction:
Consider a police traveling at 40 m/s chasing a thief traveling at 10 m/s.
If the distance between the police and thief when the chase starts is 1500 meters, find the time after which the police catches up with the thief.
Again, the solution should be known by now as 1500/(40-10) = 50sec
Frequency of meeting:
Say two athletes start running from a same point on a circular track and meet after t units of time. When they meet they are again together at a point (not necessarily the same point where they started but nevertheless they are together). And if they continue running in the direction they were running and at their respective speeds, then this instance can be thought of as a new beginning, two athlete starting to run from a same point. And thus, from this instant onwards, they would again meet after t units of time. And the reasoning could be again argued similarly at their 2nd meeting. So considering the 2nd meeting as a fresh start, they will again meet after t duration of time. Thus, the time after which they meet for the first time is also the frequency with which they keep meet if they continue running at their respective speeds and respective directions.
If the first meeting takes place after t duration after start, the nth meeting will take place after n x t duration after start.
*Where would the first meeting take place?
Once the time is known after which two athletes, running on a circular track, meet, we can find the distance run individually by them (rather
by any one of them) and find the exact position on the track where the meeting takes place. Rather than measuring this distance in absolute terms, it is more beneficial to measure this distance in terms of the track length.
*Meeting for the First Time at Start :
The question of identifying after how much time would two or more athletes meet at the starting point is an application of LCM rather than a problem of Time Speed Distance. The key word here is meeting at the starting point. Please realize that this may not be the first time that they meet. The could have crossed each other (met) at some other point on the track but then that would not be counted as a meeting point for this question as it has not occurred at the starting point.
Consider two athletes, A and B running on a circular track and taking x and y units of time to complete one full circle.A would reach the starting point for the first time after x units and thereafter would be at the starting point after every x units. Similarly B would reach the starting point for the first time after y units and thereafter would be at the starting point after every y units.Thus, A and B would both be at the starting time after common multiple of x and y and the first time that this would occur would be the LCM of x and y.For these types of problems, it does not matter in which direction the two athletes are running. Even if both are running clockwise or if one is running clockwise and other anticlockwise, the time when they would be at the starting point would remain the same.
Another Important Approach: 'Relative' Distances or Rounds run
When running in opposite directions ............ they meet whenever together they have covered one full round ............ and hence at the nth meeting, together they would have covered n rounds.
This is easy to understand. Since both start from common point and run in opposite directions, at 1st meeting, if one runs f fraction of the track, the other has to run (1 - f ) fraction of the track so that he is at same point. Thus, sum of distances run = f + (1 - f ) i.e. 1 full round.
This is new beginning, and from now onwards, till 2nd meeting they would together again cover 1 full round i.e. they would together cover 2 rounds since start. And so on.
When running in same directions ......
...... they meet whenever the faster one has covered 1 round more than the slower one ............ and hence at the nth meeting, the faster one would have covered n rounds more than the slower one.
Consider both start from same point and run in same direction. Focus on the gap between them. The faster one will race ahead of the slower one and a gap will start emerging between them. As time passes the gap will start increasing.
Three or more people meeting
To find time when three or more people meet, find the time after which pairs of athletes meet, such that at least one athlete is common to all the pairs (say, A & B, A & C, A & D, ......). The respective pairs will keep meeting after any multiple of the time found. At the LCM of these durations of time, we are sure that A & B have met, that A and C have met, that A & D have met and so on. But the only way that A could have met all these people at the same instant is when all of them are together.
Position of the meeting points:
These types of questions do not deal with when or how often do the
athletes meet when they are running on a circular track. These questions pertain to the number of points and their placement on the circular track where the athletes can possibly meet.
When running in the opposite directions:
If the ratio of speeds of two athletes (in the most reducible form) is a : b, the number of distinct meeting points on the track would be would be a + b.
Position of the meeting points :
When running in the same direction:
If the ratio of speeds of two athletes (in the most reducible form) is a : b, the number of distinct meeting points on the track would be would be |a - b|.
1. Two men , Jain Abhishek & Sood Saransh ,walk round a circle 1200 metres in circumference. Jain walks at the rate 150 meters/miute , and sood walks @80 minutes per minute. if both start at the same time from the same point & walk in the same direction.
(a) when will they 1st be together again at starting point.
(b) when will they be together again
(c) no. of distinct meeting points.
a) lcm (time taken by jain to complete 1 round,time taken by sood to complete 1 round) = lcm (1200/150,1200/80) = 120 min.
c) since ratio of speed is 15:8 ===> no. of distinct point = 15-8 = 7
b) from the previous 2 question we can say that they are meeting 7 times in 120 min==> 1st meeting will take after 120/7min( no need to use relative speed funda-- always work on ratio of speed)
(e) when they meet for 100th time what is the distance from the starting point to the point at which they will meet..in anticlockwise direction.
d) ratio of speed = ratio of no. of rounds ===> no. of rounds covered by jain to sood = 15:8 ==> here there is agap of 7 but for first meeting there must be a gap of 1 round so distance covered by jain & sood when they meet for the first time = 15/7 & 8/7( in terms of rounds) so at 100th meeting distance covered is 15/7*100 & 8/7&*100 ( in terms of round 1 round =1200m) & time required = 120/7* 100.( avoid using relative speed work on ratio)
(e) since there are 7 distinct points : 100 = 7k+2; so 100th meeting point will coincide with 2nd meeting point & when they meet for 2nd time no. of rounds covered covered by sood = 8/7 *2 = 16/7=2 +2/7==> 2/7 clockwise or anticlockwise depending upon whether they started clockwise or anticlockwise respectively. CBD
Q. For opposite direction: 2. two men , Jain Abhishek & Sood Saransh ,walk round a circle 1200 metres in circumference. Jain walks at the rate 150 meters/minute , and sood walks @80 minutes per minute. if both start at the same time from the same point & walk in the opposite direction.
(a) when will they 1st be together again at starting point.
(b) when will they be together again
(c) no. of distinct meeting points.
(d) find the distance travelled & time required by Sood Saransh & Jain Abhishek when they meet for 100th time. (e) when they meet for 100th time what is the distance from the starting point to the point at which they will meet..in anticlockwise direction
2. a) lcm (time taken by jain to complete 1 round,time taken by sood to complete 1 round) = lcm (1200/150,1200/80) = 120 min.c) since ratio of speed is 15:8 ===> no. of distinct point = 15+8 = 23 b) from the previous 2 question we can say that they are meeting 23 times in 120 min==> 1st meeting will take after 120/23min( no need to use relative speed funda-- always work on ratio of speed) d) ratio of speed = ratio of no. of rounds ===> no. of rounds covered by jain to sood = 15:8 ==> for first meeting distance covered by jain & sood when they meet for the first time = 15/23 & 8/23( in terms of rounds) so at 100th meeting distance covered is 15/23*100 & 8/23&*100 ( in terms of round 1 round =1200m) & time required = 120/23* 100.( avoid using relative speed work on ratio) e) (e) since there are 23 distinct points : 100 = 23k+7; so 100th meeting point will coincide with 8th meeting point & when they meet for 12nd time no. of rounds covered covered by sood = 8/23 *8 = 64/23=2 +18/23==> 18/23 clockwise or anticlockwise depending upon whether they started clockwise or anticlockwise respectively
Q.Three athletes A, B and C are running on a
circular track of length 1200 meters with speeds 30 m/s, 50 m/s and 80 m/s. A is running clockwise and B and C are running anticlockwise.
3: Find the time after which A and B will meet for the first time and
the frequency (in seconds) after which they will keep meeting.
Also find the sum of the distance (in fraction of the track length)
run by them till their first meeting.
no. of meeting points for a & b = 3+5 = 8 ( as direction is opposite so at first meeting a & b will travel 3/8 & 5/8 ; at 2nd meeting 3/*2 & 5/8*2; at third meeting 3/8*3& 5/8*3 & so on.................( i hope that concept is clear now)
Q. Which of the following cannot be the ratio of speeds of two joggers running on a circular jogging track if while running they meet at a diametrically opposite point to the point from where both of them started?
(1) 3 : 5 (2) 1 : 3 (3) 1 : 5 (4) 2 : 5
5. Consider athletes A , B & C running at speeds of 150 m/s , 70 m/s &110 m/s respectively on a circular track of 1000 meters, A & B running clockwise and C anti-clockwise. If they keep running indefinitely, at how many distinct point on the circle would they meet? what is the distance covered by each when they meet for 100th time.
a:b = 15:7 so A&B will meet @ 8 points(as direction is same). b:c =7:11 so B&C will meet at 18 points ( as direction is opposite) so no. of distinct meeting points for all 3 = hcf(8,18) = 2, QED
6. In a circular track, there are two points P and Q which are diametrically opposite.C starts running clockwise with a speed of 4m/s from P. At the same time, from Q, A starts running clockwise with a speed of 3 m/s and B starts running anti clockwise with a speed of 5 m/s. If the length of the track is 300m, then after how much time will C be equidistant from A and B for the first
6. B and C meets after 150/(4 + 5) = 50/3 seconds.
At this point distance between A and C is 150 + 50 - 200/3 = 400/3
So, suppose that from this moment on wards it take them t time to reach the situation when C is equidistant from A and B
After time t, distance between A and C will be 400/3 + 3t - 4t = 400/3 - t and distance between B and C will be 9t
9t = 400/3 - t
=> t = 40/3
So, after 50/3 + 40/3 = 30 seconds C will be equidistant from A and B for the first time
7. Three boys A, B and C start running at constant speeds from the same point P along the circumference of a circular track. The speed of A, B and C are in the ratio 5:1:1. A and B run clockwise while C runs in the anticlockwise direction. Each time A meets B or C on the track he gives them a card.What is the difference in the number of cards received by B and C if A distributes 33 cards in all?
8. Seven children A, B, C, D, E, F and G started walking from the same point at the same time, with speeds in the ratio of 1 : 2 : 3 : 4 : 5 : 6 : 7 respectively and they are running around a circular park. Each of them was carry flags of different colours. Whenever two or more children meet then they place their respective flag at that point, however nobody places more than 1 flag at a same point. They arerunning in anti-clockwise direction. How many flags will be there in total, when there will be no scope of putting more flags?
9. Sarah and Neha start running simultaneously from the diametrically opposite ends of a circular track towards each other at 15km/h and 25km/h respectively. After every 10 minutes their speed reduces to half of their current speeds. If the length of thecircular track is 1500 m, how many times will Sarah and Neha meet on the track? (1) 6 (2) 9 (3) 11 (4) 7 (5) 8
Total distance Shrikant can cover = 10/60*(15 + 15/2 + 15/4 + ...) = 5 km Total distance Sachin can cover = 10/60*(25 + 25/2 + 25/4 + ...) = 25/3 km First time they cover 750 m and subsequently they cover a distance of 1500 m to meet. The total distance they cover together is 40/3 km. Number of meetings possible is 1 + (40000/3 - 750)/1500 = 9.4 => choice (B) is the right answer
One of those days when everything went right.....
QA - 37/44 104
VA - 25/30 70
OA - 62/74 174 (Rank ~ 8)
Simcat's always boost my confidence.....!
QADI: 32C/41A = 87
VALR: 40C/47A = 113
Total: 72C/88A = 200
First ever CL double centuryy!! Yayyyeee
Found the center to be quiet for the first time, maybe the reason for my score
Be the change you wish to see in the world
An equilateral triangle is inscribed in a circle of radius 10 cm; and a square is inscribed in this triangle what is the area of this square ?
T.I.M.E. has changed the way the questions are presented to the candidate, the way u can navigate to the questions. I find it more user-friendly. Probably that's the way it is gonna be even in the main CAT exam. Hope so...
Yo YO ..
SIMCAT 113 ..
OA = 180 (84+96) with 88% accuracy
Percentile as shown = 99.96 :)
CAT 2014 STUDY MATERIAL http://www.mbarendezvous.com/study-material.php
A woman, her brother, her son and her daugher are chess players (all relations by birth). The worst player's twin (who is one of the four players) and the best player are of opposite sex. The worst player and the best player are the same age. Who is the worst player?
The probability of a bomb hitting a bridge when it is dropped from a plane is 0.5. At least two hits are required to destroy the bridge completely.
39.Find the least number of bombs that must be dropped so that the probability of destroying of the bridge completely is greater than 0.99. Ans=11
40.What is the probability that at least seven and at most nine bombs were needed to destroy the bridge completely? Ans= 0.09
Please explain the solution.
Hey guys,i didn't get my OBC certi yet...so if i apply as General candidate in CAT form,would i be eligible as OBC student while applying for colleges other than IIMs
Ok help me out here people,
I have to give AIMCAT 1508N today as it's the last day of the window, I couldn't give MOCK 10 as scheduled, due to various reasons and so I plan on giving it on Sunday...but the thing is CL people say you can only give it in unproctored mode for 3 days after the window gets over(i.e today,Friday,Saturday).
Which brings me to my question, is the 3 day thing true or is it just there to get maximum participation?
An Eternal Fire Burns Within.
QA: 84; VA:55; OA: 139 easy paper..similar to 1508....150+ is easy to score...
iCAT 4: QA 62 (98.47) / VA 76 (99.52) OA 138 (99.79) AIR 9
Was this iCAT really easy or am I getting good at it?
Sentence Completion: He was so ...............at tying fisherman's flies that he was asked to demonstrate his technique at sports fairs and exhibitions.
Bull CAT 29:
Section 1: 20A 17C 48
Section 2: 47A 36C 97
And I gave more time to Section 1.. :'(
Hello, does the 2 year PGDM programme also take the GMAT score for Indian Nationals, or is it valid for International students only
All incense are incantation.
No incantation is inept.
Some inept are indolent.
Some indolent are incisive.
I. 0.00002% incisive which are part of indolent, they have also a part of inept.
II. Atleast some incantation are not indolent.