Previous CAT Questions  Solved (Logical Reasoning)  Set 5

Sixteen teams have been invited to participate in the ABC Gold Cup cricket tournament. The tournament is conducted in two stages. In the first stage, the teams are divided into two groups. Each group consists of eight teams, with each team playing every other team in its group exactly once. At the end of the first stage, the top four teams from each group advance to the second stage while the rest are eliminated. The second stage comprises of several rounds. A round involves one match for each team. The winner of a match in a round advances to the next round, while the loser is eliminated; the team that remains undefeated in the second stage is declared the winner and claims the Gold Cup. The tournament rules are such that each match results in a winner and a loser with no possibility of a tie. In the first stage a team earns one point for each win and no points for a loss. At the end of the first stage teams in each group are ranked on the basis of total points to determine the qualifiers advancing to the next stage. Ties are resolved by a series of complex tiebreaking rules so that exactly four teams from each group advance to the next stage.
Q1) What is the total number of matches played in the tournament?
a. 28
b. 55
c. 63
d. 35Q2) The minimum number of wins needed for a team in the first stage to guarantee its advancement to the next stage is:
a. 5
b. 6
c. 7
d. 4Q3) What is the highest number of wins for a team in the first stage in spite of which it would be eliminated at the end of first stage?
a. 1
b. 2
c. 3
d. 4Q4) What is the number of rounds in the second stage of the tournament?
a. 1
b. 2
c. 3
d. 4Q5) Which of the following statements is true?
a. The winner will have more wins than any other team in the tournament.
b. At the end of the first stage, no team eliminated from the tournament will have more wins than any of the teams qualifying for the second stage.
c. It is possible that the winner will have the same number of wins in the entire tournament as a team eliminated at the end of the first stage.
d. The number of teams with exactly one win in the 2nd stage of the tournament is 4.Q#1,
Group A & B has 8C2matches in the 1st round. Hence total 56 in 1st round. (1st stage)
4 teams from each group qualify to 2nd round, and play against each other once, eliminating 4 losers out of this. This round has 4 matches. (2nd stage)
Next round has 4 teams, and hence 2 matches and 2 losers. (2nd stage)
Next round has 2 teams, hence 1 match. (2nd stage)
Hence, there are totally, 56 + 4 + 2 + 1 = 63 matches.Q#2,
Let the teams be 1, 2, 3, 4, 5, 6, 7, 8.
Let us assume
1 can defeat 2, 3, 4, 5, 6.
2 can defeat 3, 4, 5, 6, 7.
3 can defeat 4, 5, 6, 7, 8.
Now we do not consider 4, 5, 6 as it is already defeated by 1,2 & 3.
7 can defeat 1, 4, 5, 6, 8
8 can defeat 1, 2, 4, 5, 6.
Here, we see that 5 teams can win 5 games each, and still 1 team will get eliminated.
Let us consider 6 games now:
1 can defeat 2, 3, 4, 5, 6, 7
2 can defeat 3, 4, 5, 6, 7, 8.
We will not consider 3, 4, 5, 6, 7 as they are defeated twice.
8 can defeat 1, 3, 4, 5, 6, 7.
No other team can have 6 wins. So these 3 teams qualify for sure.
Hence 6 wins are required for a guaranteed position.Q#3,
There are 28 matches in the 1st round, and hence 28 wins.
To find this, we have to assign maximum wins for 3 teams, and find the result for the 4th team. 1 team can win 7 matches, the next team a maximum of 6 matches, and the 3rd team can win a maximum of 5 matches. This means 18 wins for these 3 teams together.
There are 10 wins left, and 5 teams left. Hence, it is possible that each team can have 2 wins, and 1 out of this team can qualify for the next stage.
However, if a team wins only 1 game, there is no chance that it qualifies for the next stage.Q#4,
As explained in Q#1, 2nd stage has 3 rounds, and 7 matches.Q#5,
Consider 1st option. From Q#3, it is seen that a team with 2 wins can qualify for the 2nd round. 2nd round has 3 wins, and a team can win the tournament with as less as 5 games. It is possible that, in the opposite group a team can win 7 matches in the 1st round. Hence, the 1st option can be ruled out.
From Q#2, it is seen that a team can be eliminated even with 5 wins. At the same time, in the other group, it can happen as in Q#3, where a team can move to the next round with as less than 2 games. Hence this option can be ruled out.Consider the 1st option of this question, where we proved that a team can win the tournament with 5 games. Also, from question# 2, a team with 5 wins can be eliminated. Hence this is true.
Consider the last option before the 3rd option, as it is very simple. There are 3 rounds in the 2nd stage the one with 8 teams. 4 teams win and qualify to the next part of the 2nd round. Here 2 teams win and 2 teams lose. So, there are only 2 teams with exactly 1 win in the 2nd round. Hence this option can also be ruled out.
(CAT 2002)
Four students (Ashish, Dhanraj, Felix and Sameer) sat for the Common Entrance Exam for Management (CEEM). One student got admission offers from three National Institutes of Management (NIM), another in two NIMs, the third in one NIM, while the fourth got none. Below are some of the facts about who got admission offers from how many NlMs and what is their educational background.
The one who is an engineer didn’t get as many admissions as Ashish.
The one who got offer for admissions in two NIMs isn’t Dhanraj nor is he a chartered accountant.
Sameer is an economist.
Dhanraj isn’t an engineer and received more admission offers than Ashish.
The medical doctor got the most number of admission offers.Which one of the following statements is necessarily true?
a. Ashish is a chartered accountant and got offer for admission in three NIMs.
b. Dhanraj is a medical doctor and got admission offer in one NIM.
c. Sameer is an economist who got admission offers in two NIMs.
d. Felix who is not an engineer did not get any offer for admission.Read the entire question, and form a table.
Student Specialization Offers Ashish Dhanraj Dhanraj Economist Dhanraj Now we can deduce the following:
Ashish is not an engineer from clue 1, and Dhanraj is not an engineer from clue 4. Hence Felix is the engineer.
Ashish has more offers than the engineer, and Dhanraj has more offer than Ashish. i.e. Dhanraj > Ashish > Engineer. So, Dhanraj can either have 2 or 3 offers. But from clue 2, Dhanraj does not have 2 offers. Hence he has 3 offers, and he is the medical doctor.
So we know the profession of everyone now, Ashish is C.A
Clue 2 says that, C.A/Ashish does not have 2 offers, but he has more offers than engineer. So Ashish has 1 offer, engineer has 0, and economist 2.
The table looks like:
Student Specialization Offers Ashish CA 1 Dhanraj Doctor 3 Sameer Economist 2 Felix Engineer 0 (CAT 2002)
Five boys went to a store to buy sweets. One boy had Rs. 40. Another boy had Rs. 30. Two other boys had Rs. 20 each. The remaining boy had Rs. 10. Below are some more facts about the initial and final cash positions.
Alam started with more than Jugraj.
Sandeep spent Rs. 1.50 more than Daljeet.
Ganesh started with more money than just only one other person.
Daijeet started with 2/3 of what Sandeep started with.
Alam spent the most, but did not end with the least.
Jugraj spent the least and ended with more than Alam or Daljeet.
Alam spent 10 times more than what Ganesh did.
Ganesh spent Rs 3.50In the choices given below, all statements except one are false. Which one of the following statements can be true?
a. Alam started with Rs.40 and ended with Rs.9.50.
b. Sandeep started with Rs.30 and ended with Rs. 1 .00.
c. Ganesh started with Rs.20 and ended with Rs.4.00.
d. Jugraj started with Rs.10 and ended with Rs.7.00.Read the question fully, and thoroughly. Work on the fixed details first, before moving to the conditional/confusing/’notfixed’ statements.
So Ganesh started with more money than just 1 person. So Ganesh has 20Rs.
Daljeet is 2/3 of Sandeep. The only number divisible by 3 here is 30. So Sandeep has 30 and Daljeet has 20.
Alam has more money than Jugraj. So Alam has 40 and Jugraj 10.
A = 40, S = 30, D = 20, G = 20, J = 10Now, we know the money spent by G = 3.5.
So A spent 3.5 * 10 = 35.
Let D spend x Rs.
So S spent x + 1.5.Now we have the following after the money spent:
A has 5. Spent 35
G has 16.5. Spent 3.5
D has 20 – x and this is definitely lesser than 10.
S has 28.5 – x.
J has 10 – y. (y is the amount spent by J).So, now we know Option 1 and 3 are wrong.
If S ended with 1Re, that means he spent 29rs, which means, D spent,
29  1.5 = 27.5 Rs. But that is not possible as D has only 20Rs with him.
So option 2 is also wrong.
Hence the correct one is option 4.(CAT 2002)
In a hospital there were 200 Diabetes, 150 Hyperglycaemia and 150 Gastroenteritis patients. Of these, 80 patients were treated for both Diabetic and Hyperglycaemia. Sixty patients were treated for Gastroenteritis and Hyperglycaemia, while 70 were treated for Diabetes and Gastroenteritis. Some of these patients have all the three diseases. Doctor Dennis treats patients with only Diabetes. Doctor Hormis treats patients with only Hyperglycaemia and Doctor Gerard treats patients with only Gastroenteritis. Doctor Paul is a generalist. Therefore, he can treat patients with multiple diseases. Patients always prefer a specialist for their disease. If Dr. Dennis had 80 patients the other three doctors can be arranged in terms of the number of patients treated as:
a. Paul > Gerard > Hormis
b. Paul > Hormis > Gerard
c. Gerard > Paul > Hormis
d. none of theseDraw a pie chart.
Let x be the common portion of all the 3 – DGH = x.
DG = 70 – x
DH = 80 – x
HG = 60 – x
D = 80. And Total diabetes = 200.
So 80 + x + 70 – x + 80 – x = 200. x = 30. i.e. DGH = 30
DG = 40
DH = 50
GH = 30
D = 80
G = 50
H = 40.
Dennis = 80, Hormis = 40, Gerard = 50, Paul = 150.(CAT 2002)
Three children won the prizes in the Bournvita Quiz contest. They are from the schools; Loyola, Convent, Little Flowers, which are located at different cities. Below are some of the facts about the schools, the children the city they are from:
One of the children is Bipin.
Loyola School’s contestant did not come first.
Little Flower’s contestant was named Riaz.
Convent School is not in Hyderabad.
The contestant from Pune took third place.
The contestant from Pune is not from Loyola School.
The contestant from Bangalore did not come first.
Convent School’s contestant’s name is not Balbir.Which of the following statements is true?
a. 1st prize: Riaz (Little Flowers), 2nd prize: Bipin (Convent), 3rd prize: Balbir (Loyola).
b. 1st prize: Bipin (Convent), 2nd prize: Riaz (Little Flowers), 3rd prize: Balbir (Loyola)
c. 1st prize: Riaz (Little Flowers), 2nd prize: Balbir(Loyola), 3rd prize: Bipin (Convent).
d. 1st prize: Bipin (Convent), 2nd prize: Balbir (Loyola), 3rd prize: Riaz(Little Flowers)Read the question fully and thoroughly, form a table, with the “fixed” data available, taking 1 data at a time. i.e. either school, else place, else prize.
We can easily map the student with the school, as Riaz’s school is given, and Balbir is not from Convent. So, he will be from Loyola. Bipin is from Convent, hence.
Now, Loyola is not 1st (Clue 2). So Loyola has to be 2nd or 3rd.
Pune is 3rd (Clue 5), Loyola is not in Pune (Clue 6). So Loyola can only be 2nd.
Bangalore is not 1st, and Pune is 3rd, so Bangalore is 2nd, and Loyola is in Bangalore hence.
The rest of the table can be filled easily, Convent is not in Hyderabad, hence LF is in Hyderabad, and Convent is in Pune with 3rd place.