Solutions

Questions 1 to 5

From Clue No. (iii),

40% of the total students are female

40% = 36

100% - 90

Total students = 90

Half of the students are either good or excellent means that total of (good + excellent) students = 45

∖ Number of excellent =12

1/3 of male = average male students = 18

Questions 6 to 10

The following structure would work:

Hence, the answers are:

6.

7. 3 + 12+ 24 = 39 (c)

8. 3 + 12 + 28 + 32 = 71 (c)

9. 128 (a)

10. 32(a)

Questions 11 to 14

Fromclues 1, 2, 3, 4, 5 we get:

Orchids = Tulips + 2, Rose = Tulips + 4 Marigold = Tulips + 3 and since Lily

Fireball.

IfTulips is 1 (Clue 6), we get

Tulip = 1, Orchids = 3, Marigold = 4, Rose = 5

Hence, Lily = 2 and Fireball = 6

The answer are:

11. Five (b)

12. Statement (iii) is redundant. Hence (b)

13. 3 + 4-7 (c)

14. Six(b)

Questions 15 to 20

Start from the fourth line and take ‘a’ as the number of girls of 44 kgs.

Questions 21 to 26

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Hence the answer are:

21. (b)

22. -10 (a)

23. D is ‘7’. Hence (d).

24. The value of E is -5. Hence (d).

25. 7¥3¥ 1 ¥-5 ¥-9--945 (a).

26. 7-(-9) - 16. Hence (c).

Questions 27 to 31

At this point the table would look like:

Q’s Answer Key: He must have had T and V as his sources. In such a case his answer to Questions 25, 46 and 90 would remain blank and he would introduce the wrong answer to Q. 96.

We are now left with R, U & X and need to think how these answer keys could have been created.

It is evident that X must have been R’s only source (since the blanks are just carried forward by R and he has introduced the wrong answer to Q. 56).

Further, X’s answer key could have been formed if he had his sources as S and T.

U’s answer key could have been created only if his sources were T and W. In that case he would introduce the blank answers 90 and 92, copy the incorrect answer 46 as it is (since there would be no mismatch in that answer) and introduce the wrong answer to Q. 14. Thus the final table would look like:

Thus the answers to the questions are:

27. (b)

28. P, T, S, X. Hence 4, Option (b).

29. U. Option (a)

ÇÎ. (ñ)

Dz- (ñ)

Questions 32 to 35

Hence answers are:

32. (b)

33. (a)

34. (a)

35. (b)

Questions 36 to 41

The alphabets PQRSTUVWX represent the nine digit number with the given property. Deductionl: Since all numbers having even digits have to be divisible by even numbers (i.e., PQ is divisible by 2, PQRS is divisible by 4, PQRSTU is divisible by 6 and PQRSTUVW is divisible by 8), Q, S, U & W must be the 4 even numbers viz. 2, 4, 6 and 8 in some random order.

Consequently P, R, T, V & X (the odd placed digits) must be sharing the odd digits 1,3, 5,7 and 9 in some order.

Deduction 2: Since PQRST is divisible by 5, T must be equal to 5. Thus, the number must be PQRS5UVWX.

Deduction 3: PQR is divisible by 3 and PQRSTU is divisible by 6, hence STU must be divisible by 3. i.e., S+T+U=S+5+U must be divisible by 3. Also, S and U are even number. Through trial and error you should realise that there are only two ways this could happen:

(a) IfSTU represents 258

or (b) If STU represents 654

Note: S can only be 2 or 6 since for PQRS to be divisible by 4, RS should be divisible by 4 and R being an odd number S can only be 2 or 6.

Deduction 4: The following 4 options of filling in the other two even numbers (4 and 6 in case STU is 258 or 2 and 8 in case STU is 654) emerge:

36.

Hence (b)

37. Digit sum = 4 + 5 = 9. Hence (a)

38. R = l. Hence (c)

39. 2 = W. Hence(b)

Now solve Questions 40 and 41 through options.

40. The first five digits could only be 38165 (no other option fits). Hence (a)

41. If 38165 are the first 5 digits, the last 5 emerge out of possibility 4 in the table above and it gives us 54729. Hence (a)

Questions 42 to 45

You will get the following equations:

A + G = 40___________ (1)

A + D = 5____________ (2)

J + D = 8_____________ (3)

This means that A is 3 less than J.

D + R = 37________ (4)

A-G = 8__________ (5)

Using (1) and (5) we get A = 24 and G = 16,

Hence D = 19, J = 27 and R = 18.

Thus the answer are:

42. R= 18 (a).

43. 27+ 19 = 46 (a)

44. 19-18 = 1(a)

45. 24 + 16 + 19+ 27 + 18 → 104 ¥2 = 208 (a)

Questions 46 to 49

Runs scored by top three batsmen against Pakistan

-40+130 + 28

- 198

90% - 198

100% - 220 runs ‘

The rest of the players made only 22 runs.

Runs scored by top three batsmen against South Africa

-51+75 + 49

- 175

70% - 175

100%-250

The rest of the players made only 75 runs

Runs scored by top three batsmen against Australia

-55 + 87 + 50

- 192

80% - 192

100-240

The rest of the players made only 48 runs Y→ 40 + South Africa + 87 - 127 + runs scored against South Africa.

V → 130 + South Africa + Australia -130 + runs scored against South Africa + Australia

K→28 + 51 + Australia - 79 + Runs scored against Australia.

S → Pakistan + 75 + 50 - 125 + Runs scored against Pakistan.

R → Pakistan + 49 + 55- 104 + Runs scored against Pakistan.

46. (c) For Yuvraj, middle index cannot be determined as we don’t know the exact runs scored by him against South Africa.

Same is the case with Virendra and Kaif

For Saurav the middle index will be 50 because whatever runs he scores against

Pakistan, they could be maximum 22 or minimum zero.

For Rahul middle index will be 49. (Same logic)

47. (d) R-Index → Difference between highest and lowest score.

For Yuvraj minimum R-index can be 87 - 40 - 47

For Virendra minimum R-index can be 130 - 48 = 82

For Kaif minimum R-index can be 51 - 28 = 23

For Sourav 75 - 22 = 53

For Rahul 55 - 22 = 33

∖ It is Kaif who can have minimum R-index.

48. (b) No. of players who definitely scored less than Yuvraj is 1.

In the worst case if we suppose that Yuvraj scored zero against South Africa, so we have to find how many players definitely scored less than 127.

This could be only Rahul who can score maximum 22 runs against Pakistan and will be able to get a score of 126, which is less than 127.

49. (b) Saurav’s middle index is the best.

Questions 50 to 52

From the given information it is clear that four numbers which must have been the values of money Spentwould be: 2517 (Clue 4), 2234 (Clue 1), 1340 (Clue 2) and 1193 (Clue 1 again).

We need to work out the fifth value.

Also, since Chellamma spent the least and Shahnaz the maximum and since one woman spent' 1378 more than Chellamma, a little bit of introspection will give you the following possibilities for the five numbers:

Possibility 1: If 1193 is the least value

The five numbers are:

1193, 1340, 2234, 2517 and 2571 (since 2571 -1193 + 1378)

Possibility 2: If 2517 is the maximum value, the five numbers are:

1139 (since 1139 - 2517- 1378), 1193, 1340, 2234 and 2517

Accordingly, we have the following possible arrangements for the five women and the amount they spent:

Note: The thought structure for placing the 5 values with 5 women in the case of

possibility 1, goes as follows:

Step 1:

After placing the least and maximum.

Step 2: 2234 should be before 1193 and Dhenuka cannot have spent' 1340.

A close look at the above table shows that clue 5 (H > D) is not obeyed by this arrangement. Hence, this solution is wrong.

We thus move into possibility 2, i.e.:

1139, 1193, 1340, 2234, 2517, are the five values. The thought structure for placing the five numbers for the five women goes as:

Step 1: Place the maximum and least values for G and C respectively.

This leaves us with 1340, 1193 and 2234 to place. ∙

Step 2: We need to keep 2 constraints in mind while doing this.

(a) 2234 has to come before 1193 (remember not immediately before). At the same time H > D. (Clue 5)

We can arrange 2234 before 1193 in 3 ways as shown below and then 1340 automatically falls into the vacant space.

Hence, the Onlypossible arrangement is as in (1) above. Hence the answers are:

50. (c) Dhenuka

51. (b) 1340

52. (a) 1139

53. (b) Each loaf of bread is divided into 3 parts. So we have 24 parts and each traveller gets 8 parts.

1st traveller has 15 parts. He ate 8 parts and gave his 7 parts to the Illrd traveller.

IInd traveller has 9 parts. He ate 8 parts and gave his 1 part to the IIIrd traveller. So 8 coins should be divided in the ratio 7: 1.

First traveller gets 7 coins.

54. (c) Maximum points of 22 can be achieved by taking (1 Management + 2 Fiction + 2 Maths + 5 Physics) books.

4 + 2 + 6+10-22

55. Option (b) is correct.

Questions 56 and 57

Elle 3y

Yogesh ó Wahida 2z

Zaheer z

56. (b) From the above table we can infer that Elle must be older than Wahida (as

she is thrice a higher value (y) while Wahida’s age is twice a lower value (z)).

57. (c) Using both pieces of information we get that if Zaheer = 10, then Wahida and Yogesh = 20 and hence Elle = 60 years.

Thus Option (c) is correct.

58. (b)

59. (b)

60. (b)

Questions 61 and 62

The following matrix will help you solve the problem.

61. '12. Hence (a)

62. The maximum negative he goes to is - 8 after the first game. Hence, Option (b) is correct.

63. Ghosh Babu’s net gain is ' 4. If after that he has ' 100 with him, he must have had ' 96 at the start. Hence, (d) is correct.

Puzzle Test

Reasoning puzzles are a favourite question type in all aptitude examinations. In the scheme of chapters on reasoning contained in this book, we have created separate chapters for specific kinds of puzzles which are often asked (like arrangements, rankings etc., which constituted the chapters prior to this one).

All other categories of puzzles—which cannot be specifically categorised as any of the foregoing chapters—you will get to see and practice in this chapter.

As the name suggests, questions on puzzles challenge you to match multiple factors (like name, colour of shirt, place of living, car model driven, etc.).The key skills involved in solving questions on puzzles include but are not limited to:

(i) The ability to make a relevant tabular structure for using the clues seamlessly: For example, suppose you have 5 people A, B, C, D, E wearing 5 colour of shirts red, yellow, green, blue and white drinking 5 kinds of soft drinks Coke, Pepsi, Mirinda, Thums Up and Seven Up—the solution table structure would look like this:

In the table above, you can see clearly that there is a direct correlation structure between each of the three ‘variables’ in the problem. More of this you would get to see and experience as you move through the solved illustrations and the exercises that follow it.

(ii) The ability to order the clues in the correct order of usage (as explained in the theory of logical reasoning): This includes the ability to perceive the direct clues and use them first to set up the ‘framework’ of problem solving.

(iii) The ability to perceive what indirect clues are talking about - and how to use them;

(iv) The ability to convert clues written in language form into visual cues so that you do not need to read the text again and again. Also, converting the language clues to visual cues is critical for the purpose of being able to ‘see’ all the clues at one go.

Illustrated below are the solutions to a few typical questions on puzzles. We would advise you to first have a look at the questions and try to solve the same on your own before looking at the step-by-step process of solving the same.