TECHNOTHLON SAMPLE TEST PAPER - 3 ACADEMIC SESSION : 2012-13 CLASS - IX & X Time : 2 Hr.
Maximum Marks : 210
GENERAL INSTRUCTIONS 1.
This booklet is your Question Paper.
2.
The question paper contains 42 questions.
3.
Blank papers, clip boards, log tables, slide rule, calculators, mobile or any other electronic gadgets in any form are not allowed to be used.
4.
Write your Name & Application Form Number in the space provided in the bottom of this booklet.
5.
No rough sheets will be provided by the invigilators. All the rough work is to be done in the blank space provided in the question paper.
6.
Marking scheme Question 1 to 30 are objective type : +5 for correct answer and –2 for wrong answer. Question 31 to 42 are subjective type : +5 for correct answer and no negative marking.
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OBJECTIVE QUESTIONS Directions : (1 to 2) Read the following information carefully and answer the questions given below it. My house has a number. There are three conditions. I. If my house number is a multiple of 3, then it is a number from 50 to 59. II. If my house number is not a multiple of 4, then it is a number from 60 to 69. III. If my house number is not a multiple of 6, then it is a number from 70 to 79. 1.
What could be my house number ? (A) 72
2.
(B) 76
(C) 68
(D) None of these
Which of conditions is not necessary in order to get my house number ? (A) I
(B) II
(C) III
(D) None of these
Directions : (3 to 4) Read the following information carefully and answer the questions given below. PQR is a three-digit number. P3 + Q3 + R3 = PQR 3.
If PQR < 200, then Q is (A) 4
4.
(B) 2
(C) 1
(D) None of these
(C) 3
(D) Data insufficient
If 300 < PQR < 400, then R is (A) 1
(B) 2
Directions : (5 to 9) Answer the questions based on the given information. Three football teams played in a tournament where they meet each other once. But some of the information of the matches has been erased. Only the following is known : Teams
Played
Won
Lost
Draw
Goals For
Goals Against
Points
A
2
2
–
–
–
1
6
B
2
–
–
1
2
4
–
C
2
–
–
–
3
7
–
If the point awarded for a draw is 1 and for a defeat is 0, then answer the following questions. 5.
What was the final points tally for B ? (A) 6
6.
(B) 3
(C) 1
(D) 2
What was the score line of the match between B & C ? (Read as : the goals scored by B – the goals scored by C) (A) 3-1
(B) 2-2
(C) 1-1
(D) Cannot be determined
Space For Rough Work
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11 1
7.
What was the score line of the match between A & B ? (Read as : the goals scored by A – the goals scored by B) (A) 3-0
8.
(B) 2-0
(C) 4-1
(D) Cannot be determined
What was the score line of the match between A & C ? (Read as : the goals scored by A – the goals scored by C) (A) 5-1
9.
(B) 7-0
(C) 2-0
(D) Cannot be determined
If ranking is done based on points and then goal difference = (goals for – goals against) in case of a tie, then who stood third ? (A) A
(B) C
(C) B
(D) None of these
Direction : (10 to 13) In a target shooting competition, a person is allowed to shoot at four targets successively, then followed by the next competitor. When all have finished one such round, the process is repeated. If a target is hit, the shooter is answered two points. If he misses the target, the others are awarded one point each. The first person who gets 60 points wins the award. In a contest between A, B and C, the final score card is A = 60, B = 53, C = 43. Out of a total of 78 shots fired, 43 hit the target. 10.
Who was the first to shoot ? (A) A
11.
(B) B
(B) B
(C) C
(D) Cannot be determined
(C) 17
(D) Cannot be determined
(C) 12
(D) Cannot be determined
How many targets did A hit ? (A) 42
13.
(D) Cannot be determined
Who was the second to shoot ? (A) A
12.
(C) C
(B) 34
How many targets did B miss ? (A) 6
(B) 10
Directions : (14 to 18) Answer the questions based on the given information. In the following table : A B C
D G
E F
The letters A, B, C, D, E, F and G represent distinct digits chosen from (0, 1, 2, 3, 4, 5, 6, 7, 8 and 9) such that A*B*C = B*G*E = D*E*F, where (*) means multiplication. 14.
What does the letter G represent ? (A) 6
(B) 2
(C) 1
(D) Cannot be determined
Space For Rough Work
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22 2
15.
What does the letter A represent if the letter B represent 4 ? (A) 3
16.
17.
18.
(B) 6
(C) 1
(D) Either (A) or (B)
Which three digits are not represented by any of the seven letters ? (A) 1, 3, 7 (B) 0, 1, 5 (C) 0, 5,7
(D) 0, 3, 8
What is the value of ABC ? (A) 60 (B) 72
(D) Cannot be determined
(C) 36
What is the value of B if the letter D represents 8 ? (A) 4 (B) 9 (C) 6
(D) Cannot be determined
Direction : (19 to 23) Read the following paragraph carefully. Seven digits are chosen from among 0,1,......, 9 and each of the seven is represented by a different letter form among A, B, C, D, E, F, G, such that A × B × C = B ×G × E = D × E × F. 19.
20.
21.
22.
23.
Which digit does G represent (A) 2 (B) 3
(C) 4
(D) 5
Which digit could B represent ? (A) 2 (B) 3
(C) 4
(D) 5
Which digit could E represent ? (A) 2 (B) 3
(C) 4
(D) 5
Which digit could A not represent ? (A) 3 (B) 5
(C) 6
(D) 8
Which digit could D not represent ? (A) 3 (B) 6
(C) 7
(D) 8
Directions : (24 to 26) Read the following statement. Seven digits are chosen from among 0,1,........., 9 and each is represented by a different letter in the subtraction problem below : S L I D E – D EA N 3 6 5 1 24.
25.
26.
What digit does S represent ? (A) 0 (B) 1
(C) 2
(D) 3
What digit does N represent ? (A) 0 (B) 1
(C) 2
(D) 3
(C) 2
(D) 3
What digit does E represent ? (A) 0
(B) 1
Space For Rough Work
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33 3
Directions for questions 27 to 30 : Study the given algorithm & answer the questions that follow. STEP 1 : a = 2, B = 1, C = 0 STEP 2 : A = B + 2*C STEP 3 : B = 2*A – C STEP 4 : C = 2*A + B STEP 5 : If C < 1000, then goto STEP 2. STEP 6 : Stop. 27.
What is the number of iterations when the program terminates ? (A) 2
28.
(D) 8
(B) A
(C) C
(D) None of these
Which of the following relations hold at the end of each iteration ? (A) A > B < C
30.
(C) 6
Which of the following gets a negative value during iterations ? (A) B
29.
(B) 4
(B) C > B < A
(C) B > C < A
(D) C > B >A
(C) 36
(D) 4
What is the value of C at the end of 2nd iteration ? (A) 10
(B) 16
SUBJECTIVE QUESTIONS 31.
Acting on an anonymous phone call, the police raid a house to arrest a suspected murderer. They don’t know what he looks like but they know his name is John and that he is inside the house. The police bust in on a carpenter, a lorry driver, a mechanic and a fireman all playing poker. Without hesitation or communication of any kind, they immediately arrest the fireman. How do they know they’ve got their man?
32.
A man lives in the penthouse of an apartment building. Every morning he takes the elevator down to the lobby and leaves the building. Upon his return, however, he can only travel halfway up in the lift and has to walk the rest of the way - unless it’s raining. What is the explanation for this?
33.
Many shops have prices set just under a round figure, e.g. $9.99 instead of $10.00 or $99.95 instead of $100.00 . It is assumed that this is done because the price seems lower to the consumer. But this is not the reason the practice started. What was the original reason for this pricing method?
34.
Assume there are approximately 5,000,000,000 (5 billion) people on Earth. What would you estimate to be the result, if you multiply together the number of fingers on every person’s left-hands? (For the purposes of this exercise, thumbs count as fingers, for five fingers per hand.) If you cannot estimate the number then try to guess how long the number would be. Space For Rough Work
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44 4
35.
An American nightclub called ‘The Coconut Grove’ had a terrible fire in which over 400 people died. A simple design flaw in the building led to the death toll being so high. Subsequently, regulations were changed to ensure that all public buildings throughout the country eliminated this one detail which proved so deadly. What was it?
36.
A man holidaying abroad fell off a yacht into deep water. He could not swim and he was not wearing anything to keep him afloat. It took 30 minutes for the people on the yacht to realize someone was missing. The missing man was rescued two hours later. Why didn’t he drown?
37.
A police officer saw a truck driver clearly going the wrong way down a one-way street, but did not try to stop him. Why not?
38.
A man buys rice at $1 a pound from American growers and sells them at $0.05 a pound in India. As a result of this he becomes a millionaire. How come ?
39.
Though Flow Case - 1 Lady and the Mathematician A lady and a mathematician are travelling in an airplane. The lady challenges the mathematician to solve her question. Here’s the sequence of events that follow. • Lady : “ guess my 3 daughter’s ages” • Mathematician : “ Alright ...... give me some clues“ • Lady : “Clue 1 : The product of their ages is 36” • Mathematician : “I can’t find their ages. I need one more clue” • Lady : “Clue 2 : Sum of their ages is your seat number” • Mathematician : “I still can’t find out” • Lady : “Clue 3 : My youngest daughter has blue eyes” • Mathematician : “I got their ages” • What is the seat no. ? What are the ages ?
40.
A truck is stuck at a road under a bridge. It’s just a couple of inches too high to pass under. Any other route, avoiding the bridge would add a couple of hours to the journey. A young boy comes along and again saves the day. How ?
41.
An American who has never been to another country sees the Great Wall of China with his own eyes. He’s standing on solid ground. How is this possible ?
42.
A man is lying dead in a field. Next to him there is an unopened package. There is no other creature in the field. How did he die ?
Space For Rough Work
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55 5
Sol. (10 to 13) : There are three shooters – A, B and C taking part in the competition. Each shooter will aim at 4 targets before the next gets his chance. Thus, one round is completed in every 12 shots. A total of 78 shots are fired of which 72 shots would complete 6 rounds. This means that the first to shoot will get (6 × 4 + 4) = 28 shots and the next shooter will get (6 × 4 + 2) = 26 shots and the third will get 24 shots. Hitting target A B C
a1 b1 c1
Missing target a2 (All numbers integers) b2 c2
Of the 78 shots, 43 hit the target and 35 miss the target. Thus, a1 + b1 + c1 = 43 ......(i) and a2 + b2 + c2 = 35 .....(ii) A scored 60 points, 2a1 + b2 + c2 = 60 ......(iii) B scored 53 points, 2b1 + a2 + c2 = 53 ......(iv) C scored 43 points, 2c1 + b2 + a2 = 43 ......(v) Let us assume that A was the first to shoot. Thus, a1 + a2 = 28 From (ii) and (iii), we have 2a1 – a2 = 25 3a1 = 53 a1 = 17.67 (Not possible, because a1 and a2 must be integers) Let us assume that B was the fist to shoot. Thus, b1 + b2 = 28 From (ii) and (iv), we have 2b1 – b2 = 18 3b1 = 46 b1 = 15.33 (Not possible, since b1 and b2 must also be integers) Let us assume that C was the first to shoot. Thus c1 + c2 = 28 From (ii) and (v), we have 2c1 – c2 = 8 or 3c1 = 36 c1 = 12 and c2 = 16 (Integral values, so possible) Let us assume that A was the second to shoot Thus a1 + a2 = 26 From (ii) and (iii) we have 2a1 – a2 = 25 3a1 = 51 a1 = 17 and a2 = 9 Thus, C was the first to shoot and A was the second to shoot. Also from the values above b1 =14 and b2 = 10. Alternative method : Each round = 12 shots Hence, for 78 shots to be fired, first shooter fired 28 shots (4 × 6 + 4 = 28) Second shooter fired 26 shots (ended the competition) Third shooter fired 24 shots. As per given conditions, for A to score 60 points, he must fire more than 12 shots at target (since he can get a maximum of 35 from others’ misses and he needs to score balance 25 from hitting the target). Let A hit 15 targets, then A’s score = 15 × 2 = 30 + 30 (from others’ misses) Hence, A misses 5 shots only. So A’s total shots = 15 + 5 = 20 (not possible) Let A hit 17 targets, then A’s score = 17 × 2 = 34 + 26 (from others’ misses) Hence , A misses 9 shots only So A’s total shots = 17 + 9 = 26, i.e. A was the second shooter. Similarly, for B’s point to be 53, he must fire more than 9 shots at target (since he can get a maximum of 35 from others’ misses)
Let B hit 12 targets, then B’s score = 12 × 2 = 24 + 29 (from others’ misses) Hence, B misses 6 shots only So, B’s total shots = 12 + 6 = 18 (not possible) Let B hit 14 targets, then B’s score = 14 × 2 = 28 + 25 (from other’s misses) Hence, B misses 10 shots only. So, B’s total Shot = 14 + 10 = 24, i.e. B was the third shooter Accordingly, the following table can be complied First shooter C Targets Hit Misses Total shots Total points
12 16 28 43
Second Shooter A 17 9 26 60
Third Shooter B 14 10 24 53
Sol. (14 to 18) : As A*B*C = B*G*E = D*E*F –––(i) so each expression must possess those numbers whose integral multiple or sub-multiples is/are possessed by the other expressions. Thus 0, 5 & 7 are ruled out. Now, ‘9’ contains two 3’s while ‘3’ and ‘6’ contain one 3 each. As 9 would be present in some row or column, so that row or column has a factor 9, thus making it necessary for other row or column/s to have factor 9 to satisfy (i). 3 & 6 would suffice for the factor 9 for the other row or column. But the digit 9 should simultaneously be in one row & one column. So, either B = 9 or E = 9. If B = 9 then (D = 3 and F = 6) or (D = 6 and F =3) If E = 9 then (A = 3 and C = 6) or (A = 6 and C = 3) Now, ‘8’ contains three 2’s, ‘4’ contains two 2’s while ‘2’ and ‘6’ contain one 2 each. As 8 would be present in some row or column, so that row or column has a factor 8, thus making it necessary for other row or column/s to have factor 8 to satisfy (i). This can be done by allowing 8 to be present only in a single column while 4 would be present in the column containing 6 and should also be present in the row which should contain 2. Thus we have the following cases : 8
3
8
6
1
4
9
4
9
1
6
1
3
8
6
1
6
3
8
3
1
4
4
9
4
8
3
6
1
6
6
8
6
1
9
4
1
3
9
9
4
2
2
2
3
2
2
2
3 2
2
4
9 8
9 8
14.
G always represents ‘2’.
15.
When B = ‘4’ then A can take two values either ‘3’ or ‘6’
16.
0, 5, 7
17.
72
18.
4
Sol. (19 to 23) : No letter can be 0. No letter can be 5.No. letter can be 7. Hence the digits chosen are 1, 2, 3, 4, 6, 8, 9, Each product is a multiple of 1, 2, 3, 4, 6, 8, 9. The smallest such product is 72. 72 = 2 ×36 = 3 × 24. 1 × 8 × 9 = 2 × 4 × 9 = 3 × 4 × 6 . 4 and 9 are used twice. One of B and E is 4 and the other is 9. Hence G is 2. Sol. (24 to 26) : S has to 1. Then D ( under L ) must be greater than 6. D = 7, L = 0, A = 2 Or. D = 8 L = 1, A = 3. Or, D = 9 L = 2 or 3, A = 3 or 4. If L was carried over form D to E than N = 9 and, D ( Over A) would be 6 more then A , But this cannot be true ( from the above inference about D, L, A ), hence E is not 0. E is 1 more than N. E 1. E = 2 N = 1. E 2. E = 3 N = 2. For questions 27 to 30 : This is an example of algorithm which does not correspond to any known logic neither it can be simplified. This requires forming a table and performing operations as specified. The following table can be formed from the given algorithm : Iteration No.
A
B
C
Initial(0)
1
1
2
4
2
10
16
36
3
88
140
316
4
772
1228
2772
27. b 4 28. d No one got a negative value during any iterations. 29. d A < B< C. 30. c 36.
