**Set 1**

The rock band, “The linkin park” during one of its iteniaries of India,staged exactly five concerts one in each of the five major cities – Delhi, Mumbai,Kolkata, Hyderabad and chennai. In each of the concerts they performed all their five biggest hit numbers ‘Living on the edge’,’falling in love’,’walk this way’, ‘Taste of India’ and ‘love in the elavator’. Though they performed each of the 5 concerts no number was performed in the same position at any two concerts.

(1) Taste of India was performed as the third number in hyderabad and as fourth in Delhi.

(2) In all the cities except chennai, falling in love before love in elevator.

(3) In Kolkata ‘Living on the edge’ was performed as the third no. whereas ‘Taste of India’ was performed before ‘living on the edge’.

Q.1. which of the number was performed as the first number in Mumbai.

Q.2. In which city was living on the edge performed as the fifth number.

Q.3. In which city was walk this way performed as the second number ?

**Solution** :- We know that no number was performed in the same position at any two concerts. By case (1) Taste of India was performed as the third in Hyderabad and fourth in Delhi. By case (3) ‘living on the edge’ was performed at third no. at kolkata. By case (2), ‘falling in love’ was performed before ‘love in elevator’ except at chennai. Therefore in Hyderabad, 1st ‘Falling in love’ & 2nd ‘Love in elevator’. In Delhi 2nd ‘Falling in love’ & ‘Love in elevator’. In Mumbai 3rd ‘Falling in love’ 4th ‘Love in the elevator’.In Kolkata 4th ‘Falling in love’ & 5th ‘Love in the elevator’. Rest can be made by table.

Numbers | First | Second | Third | Fourth | Fifth |
---|---|---|---|---|---|

Living on the edge | Delhi | Mumbai | Kolkata | Chennai | Hyderabad |

Falling in love | Hyderabad | Delhi | Mumbai | Kolkata | Chennai |

Walk this way | Mumbai | Kolkata | Chennai | Hyderabad | Delhi |

Taste of India | Kolkata | Chennai | Hyderabad | Delhi | Mumbai |

Love in the elevator | Chennai | Hyderabad | Delhi | Mumbai | Kolkata |

Answers can be given easily now.

Q1 - Walk this way.

Q2 - Hyderabad.

Q3 - Kolkata.

**Set 2**

Four candidates P,Q,R,S applied for a post for which the selection is based on the performance of the candidates in four test I,II,III & IV conducted in that order. A candidate can clear a test, only if he attempts at least one question correctly in that test. Only the candidates who have cleared a test are eligible to take the next test. Initially, each of the four test is sheduled to have exactly three questions in it.However, for each candidate based on his performance in each of the tests,the number of questions in each of his subsequent tests are revised as follows.

I. In any test, if the number of questions attempted correctly by a candidate is at least half, but not all,of the total number of questions available, he gets a bonus question in each of his subsequent tests.

II. In any test, if a candidate correctly attempts all the questions available,then he/she gets two bonus questions in each of his/her student tests.

In any test, for each correct answer, a candidate is awarded three marks and each wrong answer he panellised one mark. The same scheme of marking is followed for all all questions,including the bonus questions. The candidate who obtained the maximum net marks, in all 4 tests put together, will be given the post.

Q1. If P scored a net of 7 marks in test II and attempted not more than 20 questions in all the four tests put together then the maximum possible net score of P in test III is

A. 2 marks

B. 8 marks

C. 14 marks

D. 18 marks

**Solution :-** 7 marks can be get only in the case while he solved 3 question correctly and 2 wrong. It means he solved 5 question in test II. He solved 3 questions in test I.

He solved all 3 questions correctly therefore he got 5 questions in test II. Test II he solved 3 questions correctly and 2 were incorrect. In test III he will get 6 questions.

As in the question, it is given he didn’t solve more than 20 hence in test IV he got only 6 questions.

20 = 3+5+6+6.

In test III, he solved maximum 2 correct and 4 incorrect.

Maximum marks = 2 x 3 + 4 x (-1) = 2.

Hence Option A.

Q2. If Q’s net score in test III was 10 marks, then his net score in test II cannot be

A. 7 marks

B. 8 marks

C.11 marks

D.12 marks.

**Solution :-**

10 = 3 x 4 + (-1) x 2.

Test III he attempted 6 questions.

In test II, his attempted questions could be 4 or 5 questions.

If he attempted 4 questions in test II, he needed to attempt all 4 questions correctly.

Marks = 3 x 4 = 12.

If he attempted, 5 questions in test II, he had two possibilities.

- 3 correct & 2 incorrect.

Marks = 3 x 3 + (-1) x 2 = 7 marks. - 4 correct & 1 incorrect.

Marks = 3 x 4 + (-1) x 1 = 11 marks.

He can’t get 8 marks in any case.

Hence option B.

Q3. If R attempted total 21 questions in all the four tests put together,then what is the maximum possible net marks that he could have scored well in all the four tests put together.

A.51

B.55

C.59

D.63.

**Solution :-**

21 = 3 + 4 + 6 + 8.

In test I, he solved 2 correctly and 1 incorrectly.Then all the questions solved correctly in next two stages.

Test I marks = 2 x 3 + 1 x (-1) = 5.

Test II marks = 3 x 4 = 12.

Test III marks = 3 x 6 = 18.

Test IV marks = 3 x 8 = 24.

Total = 5 + 12 + 18 + 24 = 59.

Hence option C.

Q4. If the number of correct attempts of P in test I is the as same as that of S in test III and P scored a net of 5 marks in test II, then what is the maximum possible net score of S in test IV ?

A. 12 marks

B. 15 marks

C. 18 marks

D. 21 marks.

**Solution :-**

P got 5 marks in test II.

5 = 2 x 3 + 1 x (-1).

P attempted 3 questions in test II.P will attempt 4 questions in test III.

P have got 3 questions in test II, it means he solved only one correctly inj test I.

S solved 1 correctly in test III. To get maximum questions S solved all 3 correctly in test I then got 5 questions in test II. He solved all 5 questions correctly again.Now he gets 7 questions in test III, he solved only 1 correct.

Now he will get 7 questions in test IV, where he will solve all 7 correctly.

Marks = 3 x 7 = 21.

Hence option D.

**Set 3**

Two of A, B, C are fighting.

- The shorter of A & B is the older of the fighter.
- The younger of B & C is the shorter of the two fighters.
- The taller of A & C is the younger of the two fighters.

Q1. Who is taller.

Q2. Who is younger.

Q3. Who is not fighting.

**Solution :-**

Case I

(i) A -- shorter -- older

(ii) B -- shorter -- older

Case II

(iii) B -- shorter -- younger

(iv) C -- shorter -- younger

Case III

(v) A -- Taller -- younger

(vi) C -- Taller -- younger

From ii & iii, we get B is shorter.

From iv & Vi, we get C is younger.

It means (ii) & (v) are correct.

A & B are fighting.

Q1. A

Q2. A

Q3. C

**Set 4**

A, B, C often eat dinner out.Each orders either coffee or tea after dinner.

- If A orders coffee then B orders the drink that C orders.
- If B orders coffee then A orders the drink that C doesn’t order.
- If C orders tea then A orders the drink that B orders.

Who do you know always orders the same drink after dinner ?

Solution :-

Case | 1(i) | 1(ii) | 2(i) | 2(ii) | 3(i) | 3(ii) |
---|---|---|---|---|---|---|

A | coffee | Coffee | Tea | Coffee | Coffee | Tea |

B | Tea | Coffee | Coffee | Coffee | Coffee | Tea |

C | Tea | Coffee | Coffee | Tea | Tea | Tea |

From case 1, 2 (ii) & 3 (i) can’t be possible.

From case 2, 1(ii) can’t be possible.

From case 3, 1(i) can’t be possible.

The only cases left 2(i) & 3(ii).

A is only there who orders tea everytime.