After successfully completing the CAT 2016 sprint series and the SNAP 2016 sprint series, we are back with the XAT 2017 sprint preparation series – Quant 3 to boost your prep. This series will consist of 10 sets of questions from past year XAT papers, leading to XAT 2017 and covered almost all the question types that you needed to know come the 8th of January.
XAT 2017 sprint preparation series – Quant 3
1. The football league of a certain country is played according to the following rules:
-Each team plays exactly one game against each of the other teams.
-The winning team of each game is awarded 1 point and the losing team gets 0 points.
– If a match ends in a draw, both the teams get 1/2 points.
After the league was over, the teams were ranked according to the points that they earned at the end of the tournament. Analysis of the points table revealed the following:
Exactly half of the points earned by each team were earned in games against the ten teams which finished at the bottom of the table. Each of the bottom ten teams earned half of their total points against the other nine teams in the bottom ten. How many teams participated in the league?
A.16
B.18
C.19
D.25
E.30
2. In a city, there is a circular road. There are four points of entry into the park, namely – P, Q, R and S. Three paths were constructed which connected the points PQ, RS, and PS. The length of the path PQ is 10 units, and the length of the path RS is 7 units. Later, the municipal corporation extended the paths PQ and RS past Q and R respectively, and they meet at a point T on the main road outside the park. The path from Q to T measures 8 units, and it was found that the angle PTS is 60 . Find the area (in square units) enclosed by the paths PT, TS, and PS.
A.36 √3
B.54 √3
C.72 √3
D.90 √3
E.None of the above
Answer question numbers 3 and 4 based on the following information.
From a group of 545 contenders, a party has to select a leader. Even after holding a series of meetings, the politicians and the general body failed to reach a consensus. It was then proposed that all 545 contenders be given a number from 1 to 545. Then they will be asked to stand on a podium in a circular arrangement, and counting would start from the contender numbered 1. The counting would be done in a clockwise fashion. The rule is that every alternate contender would be asked to step down as the counting continued, with the circle getting smaller and smaller, till only one person remains standing. Therefore the first person to be eliminated would be the contender numbered 2.
3.Which position should a contender choose if he has to be the leader?
A.3
B.67
C.195
D.323
E.451
4. One of the contending politicians, Chanaya, was quite proficient in calculations and could correctly figure out the exact position. He was the last person remaining in the circle. Sensing foul play the politicians decided to repeat the game. However, this time, instead of removing every alternate person, they agreed on removing every 300th person from the circle. All other rules were kept intact. Mr. Chanaya did some quick calculations and found that for a group of 542 people the right position to become a leader would be 437. What is the right position for the whole group of 545 as per the modified rule?
A.3
B.194
C.249
D.437
E.543
5. Little Pika who is five and half years old has just learnt However, he does not know how to carry. For example, he can add 14 and 5, but he does not know how to add 14 and 7. How many pairs of consecutive integers between 1000 and 2000 (both 1000 and 2000 included) can Little Pika add?
A.150
B.155
C.156
D.258
E.None of the above
6. In the country of Twenty, there are exactly twenty cities, and there is exactly one direct road between any two cities. No two direct roads have an overlapping road segment. After the election dates are announced, candidates from their respective cities start visiting the other cities. Following are the rules that the election commission has laid down for the candidates:
-Each candidate must visit each of the other cities exactly once.
-Each candidate must use only the direct roads between two cities for going from one city to another.
-The candidate must return to his own city at the end of the campaign.
-No direct road between two cities would be used by more than one candidate. The maximum possible number of candidates is
A.5
B.6
C.7
D.8
E.9
7. The micro manometer in a certain factory can measure the pressure inside the gas chamber from 1 unit to 999999 Lately this instrument has not been working properly. The problem with the instrument is that it always skips the digit 5 and moves directly from 4 to 6. What is the actual pressure inside the gas chamber if the micro manometer displays 0030l6?
A.2201
B.2202
C.2600
D.2960
E.None of the above options
8. Consider a square ABCD of side 60 lt contains arcs BD and AC drawn with centres at A and D respectively. A circle is drawn such that it is tangent to side AB, and the arcs BD and AC. What is the radius of the circle?
A.9 cm
B.10 cm
C.12 cm
D.I5 cm
E.None of the above
9. There are 240 second year students in a B-School. The Finance area offers 3 electives in the second year. These are Financial Derivatives, Behavioural Finance, and Security Analysis. Four students have taken all the three electives, and 48 students have taken Financial Derivatives. There are twice as many students who study Financial Derivatives and Security Analysis but not Behavioural Finance, as those who study both Financial Derivatives and Behavioural Finance but not Security Analysis, and 4 times as many who study all the three. 124 students study Security Analysis. There are 59 students who could not muster courage to take up any of these subjects. The group of students who study both Financial Derivatives and Security Analysis but not Behavioural Finance, is exactly the same as the group made up of students who study both Behavioural Finance and Security Analysis. How many students study Behavioural Finance only?
A.29
B.30
C.32
D.35
E.None of the above
10. ln a plane rectangular coordinate system, points L, M, N and O are represented by the coordinates (-5, 0), (1, -1), (0, 5), and (-1, 5) respectively. Consider a variable point P in the same plane. The minimum value of PL + PM + PN + PO is
A.1+ √37
B.5√2 + 2√10
C.√41 + √37
D.√41 +1
E.None of the above
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Solutions
1. Let there be top x teams and bottom 10 teams in the competition. So, we have 10C2 = 45 matches between the bottom 10 teams and xC2 matches between the top x teams. Also, we have 10x matches that involved a top team and a bottom 10 team.
Points won by the bottom 10 teams against the top x teams will be 45.
Points won by the top x teams against the bottom 10 teams will be 10x – 45
Exactly half the points scored by the top x teams came in matches against the bottom 10 teams.
So, the remaining half came in the matches that they played among themselves
xC2 = 10x – 45
x(x – 1) = 20x – 90
x^2 – 21x + 90 = 0
So, x = 15 or x = 6.
If x = 6, the total points scored by the top 6 teams would be 30 which is not possible as the average of the top 6 teams would then be lesser than that of the bottom 10 teams.
So, option d is correct.
2. Using the intersecting chords theorem, we will get x(x + 7) = 8*18 and so, RT = 9 units. So, area of triangle PTS will be ½ * 16 * 18 * sin60 = 72√3 square units. Option c.
3. Classic Josephus’ puzzle. 545 in binary is 1000100001. So, the safe number is the decimal equivalent of 1000011 which is 67. Option b.
4. It would be interesting to note that if every alternate person is removed, for a set of 1, 2, 3, 4, 5, 6, 7, 8 … people, the eventual winners will be 1, 1, 3, 1, 3, 5, 1, 3 and so on. So, for sets that have 2^n elements, we get a 1 followed by consecutive odd numbers from 1 till n. Now, if every third person is removed, we get this series in the form of 1, 4, 1, 4, 7, 1, 4, 7, 10 and so on for sets of 4, 5, 6, 7, 8, 9, 10, 11, 12 people and so on. So, if the 300th person is eliminated, we will get an arithmetic progression in terms of 300. Now they have given us that 542 corresponds to 437. So, 543 will correspond to 437 + 300 – 543 = 194. Similarly, 544 will correspond to 194 + 300 = 494. And 545 will correspond to 494 + 300 – 545 = 249. Option c.
5. Let the numbers be in the form of 1abc and 1ab(c+1) or 1999 and 2000.
Considering the first scenario, we get that, a, b, c have to be less than 5. So, 5*5*5 = 125 ways.
If c = 9, c + 1 will end in a 0 and so, we can have b take values from 0 to 4 and a can take values from 0 to 4. So, 25 cases.
Now, if b and c are both 9, we get 5 cases.
If a, b and c are all 9, we get 1 case.
Total of 156 cases.
6. 20c2 roads i.e. 190 roads in total. Every candidate uses 20 roads. So, if there are x candidates, we get 20x <= 190. x < 10. Option e.
7. We have to check for 3014 and add 1 to it. So, 3014 is in base 9 which is the equivalent of 2200. So, 3016 will correspond to 2201. Option a.
8. Let O be the center of the circle. AO = 60 – r.
(60 – r)^2 – r^2 = (60 + r)^2 – (60 – r)^2
3600 – 120r = 240r
r = 10 cm. Option b.
9. Option a.
10. The point should be on the intersection of the two diagonals and so, total lengths of the segments will equal the sum of the lengths of the diagonals. So, option b.
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You can follow the entire sprint series here: XAT 2017 Sprint Preparation Series by Learningroots
