Solutions for LRPE1002101

Solutions for questions 1 to 3: Solutions for questions 7 to 10:

As it is a knock out tournament for eliminating any single player, It is given that
one match is needed. Now only one of the 64 players is to be the (a) Kamal’s tokens are perfect squares [any of 1, 4 and 9].
winner, hence remaining 63 players are to be eliminated. Hence, (b) Manish’s tokens are multiples of 3 [any of 3, 6 and 9].
63 matches are required. (c) Nihar’s tokens are even numbers, one of which is a multiple
of 3 [6 and any even number].
1. If each match is an upset, we will get a maximum of (d) Lohit gets exactly one token [assume that the number of this
63 upsets. Choice (B) token is A].
Also assume that the number on second token of Nihar is B.
2. As there are no upsets, all the top 8 players will reach quarter Then A = 6 + B
finals. In the quarterfinals, seed 1 plays seed 8, seed 2 plays  8  A  10
seed 7, seed 3 plays seed 6, seed 4 plays seed 5.  A = 8 or 10
Choice (A) [since sum of two even numbers is an even number.]
 B = 2 or 4
3. In the first round, seed 43 plays seed 22. In the second
Let us assume that
round, seed 43 plays with winner of the match between seed
B=4
11 and seed 54. In the third round, seed 43 and the winner,
Then,
of the matches between 59th and 6th seeded players or the
Kamal’s tokens = 1 and 9
winner of 27th and 38th seeded players are played.
Manish’s tokens = only 3
 43rd seeded player would have played with 6th or 27th Because 9 is Kamal’s token and 6 is Nihar’s token.
or 38th, or 59th seeded player in the third round. But it is given that Lohit is the only person to get exactly one token.
Choice (D)
B4
B=2
Solutions for questions 4 to 6:
A=8
The following table shows the changes in the pocket money,  Manish’s tokens = 3 and 9
expenditure and savings of each boy in this month.  Kamal’s tokens = 1 and 4
“+” indicates increase. “–” indicates decrease. The following table shows the numbers of tokens:

Name of Pocket money Expenditure Savings Name of the boy Tokens

the boy P Q P–Q Jayaraj 5, 7 and 10
A +10 +5 +5 Kamal 1 and 4
B –5 –5 0 Lohit 8
C Note (a) +25 +35 –10 Manish 3 and 9
D Note (b) +5 0 +5 Nihar 6 and 2
E Note (c) –20 0 –20
7. 5 + 7 + 10 is the required sum = 22. Choice (A)
Note (a): A’s saving is `5 more than the normal saving 8. 8 is Lohit’s token. Choice (B)
[say z + 5; where z = normal saving]. C’s saving is `15 less than
A’s savings. 9. 1 and 4 are the numbers on Kamal’s tokens.
 z + 5 – 15 = z – 10 Choice (C)
 `10 less than the normal saving.
Now P = 25; P – Q = –10 10. 3 and 9 are the numbers on Manish’s tokens.
 Q = 35 Choice (B)
Note (b): B’s pocket money = x – 5 [where x = normal pocket
money] Solutions for questions 11 to 14:
 D’s = x – 5 + 10 = x + 5
P = +5 and P – Q = +5  Q = 0 From the information, the maximum number of persons travelling
Note (c): C = x + 25 by any car is three, in each car there are at least one male and
E = x + 25 – 45 = x – 20 one female. We can say that the ratio of the persons travelling by
E’s expenditure did not change  Q = 0 all cars is 2 : 3 : 3.
P = –20; Q = 0 From the information only G and F’s sister are travelling to
 P – Q = – 20 Bangalore, we can say that G is a male and F is not travelling to
 Saving is reduced by `20. Bangalore and only two persons are travelling to Bangalore. From
It is given that E could not save any money. the above information, B and E are travelling by the same car, but
 z – 20 = 0 [where z is the normal saving.] it is not a Wagon R, we can say that B and E are not travelling to
 z = 20. Bangalore i.e., either they are travelling to Chennai or Hyderabad.
From the information, C who is a female, and A are travelling to
4. From the table, it can be observed that C’s expenditure is the the same place but not to Hyderabad, and A is a female. We can
maximum, i.e. `35 more than the normal expenditure. say that they must be travelling to Chennai.
Choice (C) Hence B and E are travelling to Hyderabad.
From the information, A and A’s brother H are travelling by two
5. `20 is the normal saving. Choice (C) different cars, we can say that H is travelling to Hyderabad since
H is a male.
6. C received `25 more than the normal pocket money, which Hence D is travelling to Bangalore and F is travelling to Chennai.
is the maximum. Choice (C)
 Triumphant Institute of Management Education Pvt. Ltd. (T.I.M.E.), 95B, Siddamsetty Complex, Park Lane, Secunderabad – 500 003.
All rights reserved. No part of this material may be reproduced, in any form or by any means, without permission in writing.
This course material is only for the use of bonafide students of Triumphant Institute of Management Education Pvt. Ltd. and its
licensees/franchisees and is not for sale. (4 pages) (ancu/ancv) LRPE1002101.Sol/1
From the information, two females are travelling by Honda city and D Thus A, C and F are in Wagon R
is not travelling by Wagon R, we can say that the persons travelling to Since two females are there in Honda City, B and E must be
Bangalore are not travelling by Wagon R and Honda city. females.
Hence D and G are travelling by Swift. Since at least one male is in each car, we can say that F is a male.
From the information B and E are not travelling by Wagon R, we
can say that they are travelling by Honda City.
 B, E and H are travelling by Honda City.

Name of the person CAR CITY Gender

A Wagon R Chennai Female

B Honda City Hyderabad Female
C Wagon R Chennai Female
D Swift Bangalore Female
E Honda City Hyderabad Female
F Wagon R Chennai Male
G Swift Bangalore Male
H Honda City Hyderabad Male

11. There are 5 females. Choice (C) 13. F is travelling to Chennai. Choice (B)

12. A, C and F are travelling in Wagon R. Choice (C) 14. Choice (C) is true. Choice(C)

Solutions for questions 15 to 18:

Dancing Singing Drama Total

Mall X Y (2X + 1) – X – Y + 1 = X – Y + 1 2X + 1
Stadium A 2A – 1
Theatre 5(2A – 1)/3 2(2A – 1)
Total 2Y 2(2A – 1) 250

15. The given data can be tabulated in the following way:We Solutions for questions 19 to 21:
know that 2A-1 is odd and divisible by 3(Since it’s 5/3 times
the students in Drama at theatre). First, let us observe the second statements made by Madhuri and
And 2(2A-1) is even. All these conditions are satisfied only Mithu. Clearly, both of them cannot be true simultaneously. First,
in choice (D). Choice (D) let us assume the second statement made by Madhuri is true, so
the second statement of .Mithu is false, which implies the first
16. In the given situation, the numbers are Y/2 each. Therefore, statement of Mithu is true.
2Y is definitely divisible by 4.  The second statement of Mamta is true.
We know that 2(2A-1) (Product of 2 and an odd number) is  We get
never divisible by 4.
As 250 is not divisible by 4, the total number of students Case (1):
participating in the dancing should definitely be divisible by
4. (For clarity take 2 as common from Singing, Drama and Name   Belongs to Working in
250. This makes, singing even, dancing odd and 250 as 125 Mamta F T Hyderabad Delhi
which is odd. Odd + even + something = odd  something
= even) Madhuri F T Delhi Mumbai
Therefore, the only option can be 84. Choice (B) Mithu T F Mumbai Hyderabad

17. From the total students we get the equation, 2X+1 +2A- Now, let us assume the second statement of Madhuri is falso, so,
1+2(2A-1) = 250 her first statement is true.
X+3A = 126.  The second statement of Mamta is true and also, the first
Given, 2X+1 = 61. X =30. A =32. statement of Mithu is true. So, we get,
Also, as Singing = Dancing, 4Y+2(2A-1) =250.
Y = 31  2Y = 62. Choice (A) Case (2):

18. From the above question, we got that X = 30, Y = 31and Name   Belongs to Working in
A =32. From this, the students participating in dancing at the
theatre is zero. If that is zero, the students participating in Mamta F T Hyderabad Mumbai
singing at the theatre = 2(2A-1)-5(2A-1)/3 =(2A-1)/3 = 21.
As the total students who participated in the singing is 62, Madhuri T F Mumbai Delhi
the people who participated in singing at stadium is
Mithu T F Delhi Hyderabad
62-31-21 = 10. Choice (C)

Triumphant Institute of Management Education Pvt. Ltd. (T.I.M.E.) HO: 95B, 2nd Floor, Siddamsetty Complex, Secunderabad – 500 003.
Tel : 040–40088400 Fax : 040–27847334 email : info@time4education.com website : www.time4education.com LRPE1002101.Sol/2
19. In case (1), Mithu belongs to Mumbai. Choice (C) 21. Mamta belongs to Hyderabad. Choice (B)

20. In case (2), the first statement of the person from Delhi is
true. So, Mamta works in Mumbai. Choice (A)

Solutions for questions 22 to 25:

Given the pointer is initially pointed at 10. Then in the next spin, the pointer can be pointed to the immediate left or the immediate right.
The two possible cases are
Case -1 Case - 2
11th 11th
10th 10th
12th 12th
3 3th 9th 3 9th 13th
th th th
8 7 7 8th
6th 5th 5th 6th
4th 3rd 3rd 4th
nd st
2 1 2nd
1st

10 10
Note: 1st , …….represent
2nd , 3rd, spin , 1st
spin , ---- 2nd 3rdspin
Given the person wins if the sum of two consecutive spins is 2(1+1) This is possible only if 11th and 12th spins pointed to 1

By point II case 1 can be eliminated, as the 13th spin cannot be to immediate right of 8. Therefore, in case 2,8th spin results number 8.

By point IV the least sum of three numbers can be 6 only when the numbers are 1,2 & 3

Therefore, 2 is in between one and three the arrangement is

1
11th
10th
2
9th 12th 13th 8
3 7 th
8th
5th 6th

3rd 4th
1 st 2nd

10
From I, spins 4th, 5th , 6th do not result in composite number.

The left, non – composite are 5,7 & 11

Also form II we get the sum of three adjacent numbers as 31. This can be possible only if the numbers adjacent to 10 are 9 & 12.
Therefore, the final arrangement is 1
1
11th
10th 6/4
2
th 12th
9
3 13th 8
7th 8th
5/7/11 5th 6th 5/7/11

3rd 4th
4/6 5/7/11
nd
1 st 2
9/12 12/9
10

22. In that case, 9 will be to the left of 10. Therefore, the second 24. In this scenario, 1st spin cannot result in 12. Therefore, it
spin will be 12. Choice (D) results in 9. Choice (C)

23. In this scenario, neither of the 5th or 6th spin can result in 11. 25. In 11th and 12th spins the super human scores 1 and 1
Therefore, only 4th spin can result in 11. Choice (D) consecutively. Therefore, he wins after 12th spin.
Choice (B)

Triumphant Institute of Management Education Pvt. Ltd. (T.I.M.E.) HO: 95B, 2nd Floor, Siddamsetty Complex, Secunderabad – 500 003.
Tel : 040–40088400 Fax : 040–27847334 email : info@time4education.com website : www.time4education.com LRPE1002101.Sol/3
Solutions for questions 26 to 30: 29. If a student enrols for Microbiology and Forensic Medicine,
then he must also enrol for Biochemistry, Physiology and
26. If a student enrols for Forensic Medicine, he must enrol for Pathology.
Physiology and Pathology. Since Pathology is taken, Anatomy, and hence Pharmacology,
Pathology enrolled  No Anatomy and hence, no cannot be taken.
Pharmacology. Since he has to take Physiology, he cannot take both
Since Physiology & Cardiology are taken  cannot enrol for Ophthalmology and Cardiology but only one of these two.
Ophthalmology. Choice (D)
Hence, choices (B), (C) and (D) are eliminated Choice (A) 30. If D chooses Ophthalmology and Cardiology, then he cannot
does not violate any condition. Choice (A) choose Physiology.
 he cannot choose Pharmacology or Forensic Medicine.
27. If a student takes up Microbiology, then he must also take He cannot enrol for 3 courses. Also, out of Anatomy and
Pathology and Biochemistry. This implies Anatomy cannot Pathology, he can enrol only for one course. Hence, he
be chosen. Since Anatomy cannot be chosen, even cannot enrol totally for 4 courses.
Pharmacology cannot be selected. Choice (D)  he can take up a maximum of 5 courses.
Choice (B)
28. Since Pathology and Anatomy cannot be selected together,
those courses, requiring these two as prerequisites, cannot
be selected together. Hence, Microbiology or Forensic
Medicine cannot be taken with Pharmacology.
Choice (A)

Triumphant Institute of Management Education Pvt. Ltd. (T.I.M.E.) HO: 95B, 2nd Floor, Siddamsetty Complex, Secunderabad – 500 003.
Tel : 040–40088400 Fax : 040–27847334 email : info@time4education.com website : www.time4education.com LRPE1002101.Sol/4