Best Preparation Strategy - 30 Days to CMAT

The countdown has begun. With just 30 days to go for CMAT  2016, this is probably going to be the most important phase of your CMAT preparations. This is quite a tricky phase, because almost all of you are already done with solving all the modules at least twice and are pretty much confident about your concepts but, still there is a that stupid feeling of unpreparedness or fear of exam that keeps lurking around in some corner of your mind. 

Believe me when I say this, the only way to overcome your fear of exams is to face it as Dale Carnegie aptly said “Do the thing you fear, and continue to do so. This is the quickest and surest way of all victory over fear”.

Read Complete Strategy at - http://www.endeavorcareers.com/gurus-speak/761-30-days-to-cmat-preparation-strategy

Arithmetic & Algebraic Problems - A LEGACY PUZZLE

Question: A man left legacies to his three sons and to a hospital, amounting in all to $1,320.00. If he had left the hospital legacy also to his first son, that son would have received as much as the other two sons together. If he had left it to his second son, he would have received twice as much as the other two sons together. If he had left the hospital legacy to his third son, he would have received then thrice as much as the first son and second son together. Find the amount of each legacy.

Answer: The legacy to the first son was $55.00, to the second son $275.00, to the third son $385.00, and to the hospital $605.00, making $1,320.00 in all.

Arithmetic & Algebraic Problems - THE STAIRCASE RACE

Question: This is a rough sketch of the finish of a race up a staircase in which three men took part. Ackworth, who is leading, went up three steps at a time, as arranged; Barnden, the second man, went four steps at a time, and Croft, who is last, went five at a time. Undoubtedly Ackworth wins. But the point is, how many steps are there in the stairs, counting the top landing as a step?

I have only shown the top of the stairs. There may be scores, or hundreds, of steps below the line. It was not necessary to draw them, as I only wanted to show the finish. But it is possible to tell from the evidence the fewest possible steps in that staircase. Can you do it?

Answer: If the staircase were such that each man would reach the top in a certain number of full leaps, without taking a reduced number at his last leap, the smallest possible number of steps would, of course, be 60 (that is, 3 X 4 X 5). But the sketch showed us that A. taking three steps at a leap, has one odd step at the end; B. taking four at a leap, will have three only at the end and C. taking five at a leap, will have four only at the finish. Therefore, we have to find the smallest number that, when divided by 3, leaves a remainder I, when divided by 4 leaves 3, and when divided by 5 leaves a remainder 4. This number is 19. So there were 19 steps in all, only 4 being left out in the sketch.

Arithmetic & Algebraic Problems - TIMING THE CAR

Question: "I was walking along the road at three and a half miles an hour," said Mr. Pipkins, "when the car dashed past me and only missed me by a few inches." "Do you know at what speed it was going?" asked his friend. "Well, from the moment it passed me to its disappearance round a corner.

I took twenty-seven steps and walking on reached that corner with one hundred and thirty-five steps more." "Then, assuming that you walked, and the car ran, each at a uniform rate, we can easily work out the speed."

Answer: As the man can walk 27 steps while the car goes 162, the car is clearly going six times as fast as the man. The man walks 3 1/2 miles an hour: therefore the car was going at 21 miles an hour.

Arithmetic & Algebraic Problems - HILL CLIMBING

Question: Weary Willie went up a certain hill at the rate of one and a half miles per hour and came down at the rate of four and a half miles per hour, so that it took him just six hours to make the double journey. How far was it to the top of the hill?

Answer: It must have been 6 3/4 miles to the top of the hill. He would go up in 4 1/2 hours and descend in 1 1/2 hours.

Arithmetic & Algebraic Problems - WESTMINSTER CLOCK

Question: A man crossed over Westminster Bridge one morning between eight and nine o'clock by the tower clock (often mistakenly called Big Ben, which is the name of the large bell only, but this by the way). On his return between four and five o'clock he noticed that the hands were exactly reversed. What were the exact times that he made the two crossings?

Answer: The times were 8 hours 23 71/143 minutes, and 4 hours 41 137/143 minutes. We are always allowed to assume that these fractional times can be indicated in clock puzzles.

Arithmetic & Algebraic Problems - AT RIGHT ANGLES

Question: Rackbrane asked his young friends at the breakfast table one morning this little question:

"How soon between the hours of five and six will the hour and minute hands of a clock be exactly at right angles?"

Answer: To be at right angles the minute hand must always be exactly fifteen minutes either behind or ahead of the hour hand. Each case would happen eleven times in the twelve hours-i.e., every 1 hour 5 5/11 minutes. Starting from nine o'clock, the eighth addition will give the case 5 hours 43 7/11 minutes. In the other case, starting from three o'clock, the second addition gives 5 hours 10 10/11 minutes.

These are the two cases between five and six, and the latter will, of course, be the sooner.

Arithmetic & Algebraic Problems - EQUAL DISTANCES

Question: A few mornings ago the following clock puzzle was sprung on his pupils by Professor Rackbrane. At what time between three and four o'clock is the minute hand the same distance from VIII as the hour hand is from XII?

Answer: At 23 1/13 minutes past three o'clock.

Arithmetic & Algebraic Problems - MISTAKING THE HANDS

Question: "Between two and three o'clock yesterday," said Colonel Crackham, "I looked at the clock and mistook the minute hand for the hour hand, and consequently the time appeared to be fifty-five minutes earlier than it actually was. What was the correct time?"

Answer: The time must have been 5 5/11 minutes past two o'clock.

Arithmetic & Algebraic Problems - WHEN DID THE DANCING BEGIN?

Question: "The guests at that ball the other night," said Dora at the breakfast table, "thought that the clock had stopped, because the hands appeared in exactly the same position as when the dancing began. But it was found that they had really only changed places. As you know, the dancing commenced between ten and eleven oclock. What was the exact time of the start?"

Answer: The dancing must have begun at 59 83/143 minutes past ten, and the hands were noticed to have changed places at 54 l38/l43 minutes past eleven.

Arithmetic & Algebraic Problems - THE AMBIGUOUS CLOCK

Question: A man had a clock with an hour hand and minute hand of the same length and indistinguishable. If it was set going at noon, what would be the first time that it would be impossible, by reason of the similarity of the hands, to be sure of the correct time?

Readers will remember that with these clock puzzles there is the convention that we may assume it possible to indicate fractions of seconds. On this assumption an exact answer can be given.

Answer: The first time would be 5 5/143 minutes past twelve, which might also (the hands being similar) indicate 60/143 minutes past one o'clock.

Arithmetic & Algebraic Problems - WHAT IS THE TIME?

Question: At what time are the two hands of a clock so situated that, reckoning as minute points past XII, one is exactly the square of the distance of the other?

Answer: The time is 6 3/4 minutes past IX, when the hour hand is 45 9/16 (the square of 6 3/4) minutes past XII. If we allow fractions less than a minute point, there is also the solution, five seconds (one-twelfth of a minute) past XII o'clock.