Quantitative Reasoning
Directions for Questions 1 to 5:
A quiz competition was organised in a school and the performance
of students was recorded on piece of paper with ink. But somehow some water fell on the paper and
the information remained incomplete.
However the scorer has same clues which are:
| Average | Good | Excellent | Total | |
|---|---|---|---|---|
| Male | 12 | |||
| Female | 36 | |||
| Total | 33 |
(i) Half the students were either excellent or good.
(ii) 40% of the students were females.
(iii) One-third of the males students were average.
Ans .
a
- Explanation :
From Clue No. (iii), 40% of the total students are female 40% = 36 100% = 90 Total students = 90
Average Good Excellent Total Male 18 24 12 54 Female 27 9 0 36 Total 45 33 12 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
Ans .
a
- Explanation :
From Clue No. (iii), 40% of the total students are female 40% = 36 100% = 90 Total students = 90
Average Good Excellent Total Male 18 24 12 54 Female 27 9 0 36 Total 45 33 12 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
Ans .
a
- Explanation :
From Clue No. (iii), 40% of the total students are female 40% = 36 100% = 90 Total students = 90
Average Good Excellent Total Male 18 24 12 54 Female 27 9 0 36 Total 45 33 12 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
Ans .
b
- Explanation :
From Clue No. (iii), 40% of the total students are female 40% = 36 100% = 90 Total students = 90
Average Good Excellent Total Male 18 24 12 54 Female 27 9 0 36 Total 45 33 12 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
Ans .
b
- Explanation :
From Clue No. (iii), 40% of the total students are female 40% = 36 100% = 90 Total students = 90
Average Good Excellent Total Male 18 24 12 54 Female 27 9 0 36 Total 45 33 12 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
Directions for Questions 6 to 10:
A, B, C and D are four friends living together in a flat and they
have an agreement that whatever edible comes they will share equally among themselves. One day
A’s uncle came to him and gave him a box of laddoos. Since no one was around, A divided the
laddoos in four equal parts and ate his share after which he put the rest in the box. As he was closing
the box, B walked in, took the Box from A & divided the ladoos in 4 equal parts & A & B took one
part each and ate it. Suddenly C appeared and snatched the box. He again divided the laddoos in four
equal parts, the three of them ate one part each and kept the remaining laddoos in the box. Later when
D came he again divided the laddoos in four equal parts and all four ate their respective share. In
total D ate 3 laddoos.
Ans .
b
- Explanation :
The following structure would work:
A B C D Total Left Last 3 3 3 3 0 Second last 12 12 24 12 12 Third last 24 48 First 32 96 Start 128
3 + 12 = 15 (b)
Ans .
c
- Explanation :
The following structure would work:
A B C D Total Left Last 3 3 3 3 0 Second last 12 12 24 12 12 Third last 24 48 First 32 96 Start 128
3 + 12 + 24 = 39 (c)
Ans .
c
- Explanation :
The following structure would work:
A B C D Total Left Last 3 3 3 3 0 Second last 12 12 24 12 12 Third last 24 48 First 32 96 Start 128
3 + 12 + 28 + 32 = 71 (c)
Ans .
a
- Explanation :
The following structure would work:
A B C D Total Left Last 3 3 3 3 0 Second last 12 12 24 12 12 Third last 24 48 First 32 96 Start 128
128 (a)
Ans .
a
- Explanation :
The following structure would work:
A B C D Total Left Last 3 3 3 3 0 Second last 12 12 24 12 12 Third last 24 48 First 32 96 Start 128
32 (a)
Directions for Questions 11 to 14: Rajeev planted some plants in his lawn but in certain fixed
pattern:
i. In most of the rows there are neither Roses nor Marigolds.
ii. There are two more rows of Orchids than Tulips and two more rows of Roses than Orchids.
iii. There are four more rows of Roses than Tulips.
iv. There aren’t as many rows of Lilly as Fireball.
v. There is one less Marigold row than Rose.
vi. There is just one row of Tulips.
vii. The maximum number of rows he planted is six.
Ans .
b
- Explanation :
From clues 1, 2, 3, 4, 5 we get: Orchids = Tulips + 2, Rose = Tulips + 4 Marigold = Tulips + 3 and since Lily < Fireball. If Tulips is 1 (Clue 6), we get Tulip = 1, Orchids = 3, Marigold = 4, Rose = 5 Hence, Lily = 2 and Fireball = 6
Five (b)
Ans .
b
- Explanation :
From clues 1, 2, 3, 4, 5 we get: Orchids = Tulips + 2, Rose = Tulips + 4 Marigold = Tulips + 3 and since Lily < Fireball. If Tulips is 1 (Clue 6), we get Tulip = 1, Orchids = 3, Marigold = 4, Rose = 5 Hence, Lily = 2 and Fireball = 6
Statement (iii) is redundant. Hence (b)
Ans .
c
- Explanation :
From clues 1, 2, 3, 4, 5 we get: Orchids = Tulips + 2, Rose = Tulips + 4 Marigold = Tulips + 3 and since Lily < Fireball. If Tulips is 1 (Clue 6), we get Tulip = 1, Orchids = 3, Marigold = 4, Rose = 5 Hence, Lily = 2 and Fireball = 6
3 + 4 = 7 (c)
Ans .
b
- Explanation :
From clues 1, 2, 3, 4, 5 we get: Orchids = Tulips + 2, Rose = Tulips + 4 Marigold = Tulips + 3 and since Lily < Fireball. If Tulips is 1 (Clue 6), we get Tulip = 1, Orchids = 3, Marigold = 4, Rose = 5 Hence, Lily = 2 and Fireball = 6
Six (b)
Directions for Questions 15 to 20:
In a class of 540 students, for every 9 girls these are 11 boys.
The weight of students varies from 40 to 50 kg. There are as many44 kg girls as there are 46 kg boys
and as many 40 kg boys as 50 kg girls. The number of 50 kg boys is 35 more than that of 44 kg girls
while there are as many 44 kg boys as 46 kg girls. The ratio of 40 kg boys and girls is 4:3 while that
of 50 kg girls and boys is 1:3.
Ans .
b
- Explanation :
Weight Boys Girls 40 (a+35)/3 (a+35)/4 44 a a 46 b b 50 a + 35 (a+35)/3 Total 297 243
Start from the fourth line and take ‘a’ as the number of girls of 44 kgs.
Ans .
a
- Explanation :
Weight Boys Girls 40 (a+35)/3 (a+35)/4 44 a a 46 b b 50 a + 35 (a+35)/3 Total 297 243
Start from the fourth line and take ‘a’ as the number of girls of 44 kgs.
Ans .
b
- Explanation :
Weight Boys Girls 40 (a+35)/3 (a+35)/4 44 a a 46 b b 50 a + 35 (a+35)/3 Total 297 243
Start from the fourth line and take ‘a’ as the number of girls of 44 kgs.
Ans .
a
- Explanation :
Weight Boys Girls 40 (a+35)/3 (a+35)/4 44 a a 46 b b 50 a + 35 (a+35)/3 Total 297 243
Start from the fourth line and take ‘a’ as the number of girls of 44 kgs.
Ans .
b
- Explanation :
Weight Boys Girls 40 (a+35)/3 (a+35)/4 44 a a 46 b b 50 a + 35 (a+35)/3 Total 297 243
Start from the fourth line and take ‘a’ as the number of girls of 44 kgs.
Ans .
a
- Explanation :
Weight Boys Girls 40 (a+35)/3 (a+35)/4 44 a a 46 b b 50 a + 35 (a+35)/3 Total 297 243
Start from the fourth line and take ‘a’ as the number of girls of 44 kgs.
Directions for Questions 21 to 26:
A, B, C, D and E are five different integer. When written in the
ascending order of values, the difference between any two adjacent integers is 4. D is the greatest and
A the least. B is greater than E but less than C. The sum of the integers is equal to E.
Ans .
b
- Explanation :
First Statement A, B, C, D & E Second Statement The numbers are in A.P. with common difference 4. Third Statement D _ _ _ A
(Descending order from left to right)Fourth Statement D C B E A
For the sum to be equal to E, the sum of A, B, C and D shall be zero.
(Using options in Q21 the value of A should be 9.) Then the series becomes:
D C B E A
7 3 –1 –5 –9
(b)
Ans .
a
- Explanation :
First Statement A, B, C, D & E Second Statement The numbers are in A.P. with common difference 4. Third Statement D _ _ _ A
(Descending order from left to right)Fourth Statement D C B E A
For the sum to be equal to E, the sum of A, B, C and D shall be zero.
(Using options in Q21 the value of A should be 9.) Then the series becomes:
D C B E A
7 3 –1 –5 –9
–10 (a)
Ans .
d
- Explanation :
First Statement A, B, C, D & E Second Statement The numbers are in A.P. with common difference 4. Third Statement D _ _ _ A
(Descending order from left to right)Fourth Statement D C B E A
For the sum to be equal to E, the sum of A, B, C and D shall be zero.
(Using options in Q21 the value of A should be 9.) Then the series becomes:
D C B E A
7 3 –1 –5 –9
D is ‘7’. Hence (d).
Ans .
d
- Explanation :
First Statement A, B, C, D & E Second Statement The numbers are in A.P. with common difference 4. Third Statement D _ _ _ A
(Descending order from left to right)Fourth Statement D C B E A
For the sum to be equal to E, the sum of A, B, C and D shall be zero.
(Using options in Q21 the value of A should be 9.) Then the series becomes:
D C B E A
7 3 –1 –5 –9
The value of E is –5. Hence (d).
Ans .
a
- Explanation :
First Statement A, B, C, D & E Second Statement The numbers are in A.P. with common difference 4. Third Statement D _ _ _ A
(Descending order from left to right)Fourth Statement D C B E A
For the sum to be equal to E, the sum of A, B, C and D shall be zero.
(Using options in Q21 the value of A should be 9.) Then the series becomes:
D C B E A
7 3 –1 –5 –9
7 ¥ 3 ¥ 1 ¥ –5 ¥ –9 = –945 (a).
Ans .
c
- Explanation :
First Statement A, B, C, D & E Second Statement The numbers are in A.P. with common difference 4. Third Statement D _ _ _ A
(Descending order from left to right)Fourth Statement D C B E A
For the sum to be equal to E, the sum of A, B, C and D shall be zero.
(Using options in Q21 the value of A should be 9.) Then the series becomes:
D C B E A
7 3 –1 –5 –9
7–(–9) = 16. Hence (c).
Directions for Questions 27 to 31:
In November the answers of a prestigious test held nationwide
were leaked to a group of unscrupulous people. The CBI has arrested the Don, the mastermind behind
it and nine other people—P, Q, R, S, T, U, V, W and X in this matter. On interrogation, certain facts
came into light:
Their modus operandi consisted of the Don initially obtaining the answer key, then the other nine
persons created their answer keys in the following manner:
They obtained the answer key from one or two sources, then he/she compares the answer keys to a
question from both sources. If the key to a question from both sources is identical, it is copied,
otherwise it is left blank. If the person has only one source, he/she copies the source’s answer into
his/her copy. Finally, each person compulsorily replaces one of the answers (not a blank one) with a
wrong answer in his/her answer key.
The paper contained 150 questions. So the CBI has ruled out the possibility of two or more of them
introducing wrong answer to the same question. The CBI has a copy of correct answer key and
tabulated the following data. The data represents question numbers.
| Name | Wrong Answer(s) | Blank Answer(s) |
|---|---|---|
| P | 46 | ------ |
| Q | 96 | 46,90,25 |
| R | 27,56 | 17,46,90 |
| S | 17 | ------ |
| T | 46,90 | ------ |
| U | 14,46 | 92,90 |
| V | 25 | ------ |
| W | 46,92 | ------ |
| X | 27 | 17,46,90 |
Ans .
b
- Explanation :
Paragraph 1 Ten people P, Q, R, S, T, U, V, W, X and the Mastermind 3 Paragraph 2 & 3 Two ways of creating an answer key: ONE SOURCE: Copy entire answer key and introduce 1 wrong answer. Deduction: If you have one source, then you introduce only 1 wrong answer and carry over any wrong answers as well as any blanks from the answer key of the source. TWO SOURCES: Introduce blank if the two answer keys differ on one particular question. [This means that if one answer key has the answer correct and the other answer key has it wrong then we introduce a blank]. Note: If both answer keys are correct or if both answer keys are incorrect on a particular question, then they will give us the same answer and hence will not differ. Consequently their answer will be copied into the answer key ‘under construction’. Paragraph 4 Note also that since it is given that two or more people have not introduced a wrong answer to the same question, we can deduce that if two answer keys have the same answer wrong, they will be showing the same incorrect answer to that question. Consequently if someone has two sources who have the wrong answer to thesame question both of them willshow the same incorrect answer to that particular question and that answer will get replicated as it is, into the answer key which is constructed using them both. With this understanding we move to the information contained in the table. Deductions from the table The first thing we see when we see the table is that P, S and V have only 1 incorrect answer and no blanks. A little thought will give you that this can happen only if there is a single all-correct answer key as the source. Hence, P, S & V must have had the Don as their source and further that P must have introduced the wrong answer to Q 46. S must have introduced the wrong answer to Q 17 & V must have introduced the wrong answer to Q 25. At this point we also know that— In case of 1 source, there will be the introduction of only 1 extra wrong answer and that blanks can only be introduced if there are two sources. From this point on move ahead in the question using two main objectives— (a) Decoding the answer key patterns of the remaining six people (Q, R, T, U, W & X) and (b) Trying to decode the introduction of the remaining wrong answers and the blanks. From the table, T and W are pretty easy to decode. P must have been the source for both of them and T must have introduced the wrong answer to Q 90, while W must have introduced the wrong answer to Q 92. 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:
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(b)
Ans .
b
- Explanation :
Paragraph 1 Ten people P, Q, R, S, T, U, V, W, X and the Mastermind 3 Paragraph 2 & 3 Two ways of creating an answer key: ONE SOURCE: Copy entire answer key and introduce 1 wrong answer. Deduction: If you have one source, then you introduce only 1 wrong answer and carry over any wrong answers as well as any blanks from the answer key of the source. TWO SOURCES: Introduce blank if the two answer keys differ on one particular question. [This means that if one answer key has the answer correct and the other answer key has it wrong then we introduce a blank]. Note: If both answer keys are correct or if both answer keys are incorrect on a particular question, then they will give us the same answer and hence will not differ. Consequently their answer will be copied into the answer key ‘under construction’. Paragraph 4 Note also that since it is given that two or more people have not introduced a wrong answer to the same question, we can deduce that if two answer keys have the same answer wrong, they will be showing the same incorrect answer to that question. Consequently if someone has two sources who have the wrong answer to thesame question both of them willshow the same incorrect answer to that particular question and that answer will get replicated as it is, into the answer key which is constructed using them both. With this understanding we move to the information contained in the table. Deductions from the table The first thing we see when we see the table is that P, S and V have only 1 incorrect answer and no blanks. A little thought will give you that this can happen only if there is a single all-correct answer key as the source. Hence, P, S & V must have had the Don as their source and further that P must have introduced the wrong answer to Q 46. S must have introduced the wrong answer to Q 17 & V must have introduced the wrong answer to Q 25. At this point we also know that— In case of 1 source, there will be the introduction of only 1 extra wrong answer and that blanks can only be introduced if there are two sources. From this point on move ahead in the question using two main objectives— (a) Decoding the answer key patterns of the remaining six people (Q, R, T, U, W & X) and (b) Trying to decode the introduction of the remaining wrong answers and the blanks. From the table, T and W are pretty easy to decode. P must have been the source for both of them and T must have introduced the wrong answer to Q 90, while W must have introduced the wrong answer to Q 92. 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:
P, T, S, X. Hence 4, Option (b).
Ans .
a
- Explanation :
Paragraph 1 Ten people P, Q, R, S, T, U, V, W, X and the Mastermind 3 Paragraph 2 & 3 Two ways of creating an answer key: ONE SOURCE: Copy entire answer key and introduce 1 wrong answer. Deduction: If you have one source, then you introduce only 1 wrong answer and carry over any wrong answers as well as any blanks from the answer key of the source. TWO SOURCES: Introduce blank if the two answer keys differ on one particular question. [This means that if one answer key has the answer correct and the other answer key has it wrong then we introduce a blank]. Note: If both answer keys are correct or if both answer keys are incorrect on a particular question, then they will give us the same answer and hence will not differ. Consequently their answer will be copied into the answer key ‘under construction’. Paragraph 4 Note also that since it is given that two or more people have not introduced a wrong answer to the same question, we can deduce that if two answer keys have the same answer wrong, they will be showing the same incorrect answer to that question. Consequently if someone has two sources who have the wrong answer to thesame question both of them willshow the same incorrect answer to that particular question and that answer will get replicated as it is, into the answer key which is constructed using them both. With this understanding we move to the information contained in the table. Deductions from the table The first thing we see when we see the table is that P, S and V have only 1 incorrect answer and no blanks. A little thought will give you that this can happen only if there is a single all-correct answer key as the source. Hence, P, S & V must have had the Don as their source and further that P must have introduced the wrong answer to Q 46. S must have introduced the wrong answer to Q 17 & V must have introduced the wrong answer to Q 25. At this point we also know that— In case of 1 source, there will be the introduction of only 1 extra wrong answer and that blanks can only be introduced if there are two sources. From this point on move ahead in the question using two main objectives— (a) Decoding the answer key patterns of the remaining six people (Q, R, T, U, W & X) and (b) Trying to decode the introduction of the remaining wrong answers and the blanks. From the table, T and W are pretty easy to decode. P must have been the source for both of them and T must have introduced the wrong answer to Q 90, while W must have introduced the wrong answer to Q 92. 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:
U. Option (a)
Ans .
c
- Explanation :
Paragraph 1 Ten people P, Q, R, S, T, U, V, W, X and the Mastermind 3 Paragraph 2 & 3 Two ways of creating an answer key: ONE SOURCE: Copy entire answer key and introduce 1 wrong answer. Deduction: If you have one source, then you introduce only 1 wrong answer and carry over any wrong answers as well as any blanks from the answer key of the source. TWO SOURCES: Introduce blank if the two answer keys differ on one particular question. [This means that if one answer key has the answer correct and the other answer key has it wrong then we introduce a blank]. Note: If both answer keys are correct or if both answer keys are incorrect on a particular question, then they will give us the same answer and hence will not differ. Consequently their answer will be copied into the answer key ‘under construction’. Paragraph 4 Note also that since it is given that two or more people have not introduced a wrong answer to the same question, we can deduce that if two answer keys have the same answer wrong, they will be showing the same incorrect answer to that question. Consequently if someone has two sources who have the wrong answer to thesame question both of them willshow the same incorrect answer to that particular question and that answer will get replicated as it is, into the answer key which is constructed using them both. With this understanding we move to the information contained in the table. Deductions from the table The first thing we see when we see the table is that P, S and V have only 1 incorrect answer and no blanks. A little thought will give you that this can happen only if there is a single all-correct answer key as the source. Hence, P, S & V must have had the Don as their source and further that P must have introduced the wrong answer to Q 46. S must have introduced the wrong answer to Q 17 & V must have introduced the wrong answer to Q 25. At this point we also know that— In case of 1 source, there will be the introduction of only 1 extra wrong answer and that blanks can only be introduced if there are two sources. From this point on move ahead in the question using two main objectives— (a) Decoding the answer key patterns of the remaining six people (Q, R, T, U, W & X) and (b) Trying to decode the introduction of the remaining wrong answers and the blanks. From the table, T and W are pretty easy to decode. P must have been the source for both of them and T must have introduced the wrong answer to Q 90, while W must have introduced the wrong answer to Q 92. 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:
(c)
Ans .
c
- Explanation :
Paragraph 1 Ten people P, Q, R, S, T, U, V, W, X and the Mastermind 3 Paragraph 2 & 3 Two ways of creating an answer key: ONE SOURCE: Copy entire answer key and introduce 1 wrong answer. Deduction: If you have one source, then you introduce only 1 wrong answer and carry over any wrong answers as well as any blanks from the answer key of the source. TWO SOURCES: Introduce blank if the two answer keys differ on one particular question. [This means that if one answer key has the answer correct and the other answer key has it wrong then we introduce a blank]. Note: If both answer keys are correct or if both answer keys are incorrect on a particular question, then they will give us the same answer and hence will not differ. Consequently their answer will be copied into the answer key ‘under construction’. Paragraph 4 Note also that since it is given that two or more people have not introduced a wrong answer to the same question, we can deduce that if two answer keys have the same answer wrong, they will be showing the same incorrect answer to that question. Consequently if someone has two sources who have the wrong answer to thesame question both of them willshow the same incorrect answer to that particular question and that answer will get replicated as it is, into the answer key which is constructed using them both. With this understanding we move to the information contained in the table. Deductions from the table The first thing we see when we see the table is that P, S and V have only 1 incorrect answer and no blanks. A little thought will give you that this can happen only if there is a single all-correct answer key as the source. Hence, P, S & V must have had the Don as their source and further that P must have introduced the wrong answer to Q 46. S must have introduced the wrong answer to Q 17 & V must have introduced the wrong answer to Q 25. At this point we also know that— In case of 1 source, there will be the introduction of only 1 extra wrong answer and that blanks can only be introduced if there are two sources. From this point on move ahead in the question using two main objectives— (a) Decoding the answer key patterns of the remaining six people (Q, R, T, U, W & X) and (b) Trying to decode the introduction of the remaining wrong answers and the blanks. From the table, T and W are pretty easy to decode. P must have been the source for both of them and T must have introduced the wrong answer to Q 90, while W must have introduced the wrong answer to Q 92. 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:
(c)
Directions for Questions 32 to 35:
Three classmates—X, Y and Z live on the AN Jha Marg, yet they
do not know the house number of each other. The houses are numbered from 1 to 99. Since Z is a
regular student and attends every class sincerely, his notes are very good and updated. X and Y are
not so regular, therefore they desire to meet Z at his house individually.
One day X asks Z, “The number of your house in which you reside is a perfect square or not?” Z
replies. Then X asks, “Is it greater than 50?” He again replies. X thinks that he has got the address
and decides to visit Z. When X reaches at the address he realises that he is wrong. He then thinks
over it again and is not surprised as Z answered only the second question honestly.
Y not aware of X’s conversation, asks Z two questions of his own. Y asks “Is your house number a
perfect cube?” Z replies. Then Y asks “Is it greater than 25?”
He answers again. Y thinks that he has got the address but upon reaching there he finds the address
incorrect and realises that Z answered only the second question honestly.
If Z’s house number is less than the house number of X and Y and the sum of all three of their house
numbers is twice the perfect square of some number then answer the following question:
Ans .
b
- Explanation :
Paragraph 2 Since X has thought that he has the answer to the house number of Z, he must have got ‘yes’ as an answer to both questions he asked. In such a case he would think that Z’s number is one of 64 or 81 (the two 2 digit perfect squares which are greater than 50). Also, we can deduce that since he thinks that he knows the answer he must be living in one of the two houses. Hence, X’s house number is either 64 or 81. Paragraph 3 By a similar logic Y’s house number is either 27 or 64. Paragraph 4 We know Z’s house number is greater than 50 but less than Y’s and X’s house numbers. Hence Y’s number must be 64 and X’s number is 81. Also from options to Q. No. 34 we get that Z’s number must be 55 since (81+64+55) is the only addition that satisfies the condition of the sum of the three numbers to be double of a perfect square.
(b)
Ans .
a
- Explanation :
Paragraph 2 Since X has thought that he has the answer to the house number of Z, he must have got ‘yes’ as an answer to both questions he asked. In such a case he would think that Z’s number is one of 64 or 81 (the two 2 digit perfect squares which are greater than 50). Also, we can deduce that since he thinks that he knows the answer he must be living in one of the two houses. Hence, X’s house number is either 64 or 81. Paragraph 3 By a similar logic Y’s house number is either 27 or 64. Paragraph 4 We know Z’s house number is greater than 50 but less than Y’s and X’s house numbers. Hence Y’s number must be 64 and X’s number is 81. Also from options to Q. No. 34 we get that Z’s number must be 55 since (81+64+55) is the only addition that satisfies the condition of the sum of the three numbers to be double of a perfect square.
(a)
Ans .
a
- Explanation :
Paragraph 2 Since X has thought that he has the answer to the house number of Z, he must have got ‘yes’ as an answer to both questions he asked. In such a case he would think that Z’s number is one of 64 or 81 (the two 2 digit perfect squares which are greater than 50). Also, we can deduce that since he thinks that he knows the answer he must be living in one of the two houses. Hence, X’s house number is either 64 or 81. Paragraph 3 By a similar logic Y’s house number is either 27 or 64. Paragraph 4 We know Z’s house number is greater than 50 but less than Y’s and X’s house numbers. Hence Y’s number must be 64 and X’s number is 81. Also from options to Q. No. 34 we get that Z’s number must be 55 since (81+64+55) is the only addition that satisfies the condition of the sum of the three numbers to be double of a perfect square.
(a)
Ans .
b
- Explanation :
Paragraph 2 Since X has thought that he has the answer to the house number of Z, he must have got ‘yes’ as an answer to both questions he asked. In such a case he would think that Z’s number is one of 64 or 81 (the two 2 digit perfect squares which are greater than 50). Also, we can deduce that since he thinks that he knows the answer he must be living in one of the two houses. Hence, X’s house number is either 64 or 81. Paragraph 3 By a similar logic Y’s house number is either 27 or 64. Paragraph 4 We know Z’s house number is greater than 50 but less than Y’s and X’s house numbers. Hence Y’s number must be 64 and X’s number is 81. Also from options to Q. No. 34 we get that Z’s number must be 55 since (81+64+55) is the only addition that satisfies the condition of the sum of the three numbers to be double of a perfect square.
(b)
Directions for Questions 36 to 41: Study the following information and answer the following
questions.
It is very easy to remember the ID number of my ATM card which is a nine digit number and every
digit is distinct. If I tell you some clues then you will also be able to remember my ATM card ID
number. Let us say the number is PQRSTUVWX and the digit corresponding to it are 1 to 9 though not
respectively. The ID is divisible by 9.
If you delete the digit at its units place, the remaining 8-digit number of my ID is divisible by 8. If you
again delete the last digit of the 8-digit number the remaining 7-digit number is divisible by 7 and the
process goes on.
Ans .
b
- Explanation :
The alphabets PQRSTUVWX represent the nine digit number with the given property.
Deduction1: 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) lf STU represents 258 or (b) lf 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:
P Q R S T U V W X Possibility 1 4 2 5 8 6 Possibility 2 6 2 5 8 4 Possibility 3 2 6 5 4 8 Possibility 4 8 6 5 4 2
Sum of digits—1+2+3+4+5+6+7+8+9=45. Hence (b)
Ans .
a
- Explanation :
The alphabets PQRSTUVWX represent the nine digit number with the given property.
Deduction1: 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) lf STU represents 258 or (b) lf 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:
P Q R S T U V W X Possibility 1 4 2 5 8 6 Possibility 2 6 2 5 8 4 Possibility 3 2 6 5 4 8 Possibility 4 8 6 5 4 2
Digit sum = 4 + 5 = 9. Hence (a)
Ans .
c
- Explanation :
The alphabets PQRSTUVWX represent the nine digit number with the given property.
Deduction1: 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) lf STU represents 258 or (b) lf 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:
P Q R S T U V W X Possibility 1 4 2 5 8 6 Possibility 2 6 2 5 8 4 Possibility 3 2 6 5 4 8 Possibility 4 8 6 5 4 2
R = l. Hence (c)
Ans .
b
- Explanation :
The alphabets PQRSTUVWX represent the nine digit number with the given property.
Deduction1: 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) lf STU represents 258 or (b) lf 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:
P Q R S T U V W X Possibility 1 4 2 5 8 6 Possibility 2 6 2 5 8 4 Possibility 3 2 6 5 4 8 Possibility 4 8 6 5 4 2
2 = W. Hence (b) Now solve Questions 40 and 41 through options.
Ans .
a
- Explanation :
The alphabets PQRSTUVWX represent the nine digit number with the given property.
Deduction1: 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) lf STU represents 258 or (b) lf 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:
P Q R S T U V W X Possibility 1 4 2 5 8 6 Possibility 2 6 2 5 8 4 Possibility 3 2 6 5 4 8 Possibility 4 8 6 5 4 2
The first five digits could only be 38165 (no other option fits). Hence (a)
Ans .
a
- Explanation :
The alphabets PQRSTUVWX represent the nine digit number with the given property.
Deduction1: 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) lf STU represents 258 or (b) lf 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:
P Q R S T U V W X Possibility 1 4 2 5 8 6 Possibility 2 6 2 5 8 4 Possibility 3 2 6 5 4 8 Possibility 4 8 6 5 4 2
lf 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)
Directions for Questions 42 to 45:
Some friends went to Netram Sweets. Following is the
information about the number of rosogollas they ate:
i. Gimmy ate 8 less than Akshit.
ii. Dileep and Raj together ate 37.
iii. Jugal ate 8 more than Dileep.
iv. Akshit ate 5 more than Dileep.
v. Akshit and Gimmy together ate 40.
Ans .
a
- Explanation :
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.
R = 18 (a) .
Ans .
a
- Explanation :
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.
24 + 16 + 19 + 27 + 18 → 104 ¥ 2 = 208 (a)
Ans .
a
- Explanation :
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.
19 – 18 = 1 (a)
Ans .
a
- Explanation :
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.
27 + 19 = 46 (a)
Directions for Questions 46 to 49:
Coach Johan sat with the score cards of Indian players from the 3
games in a one-day cricket tournament where the same set of players played for India and all the
major batsmen go out. John summarised the batting performance through three figures, one for each
game. In each figure, the three outer triangles communicate the number of runs scored by the three top
scorers from India. K, R, S, V and Y represent Kaif, Rahul, Saurav, Virender and Yuvraj
respectively. The middle triangle in each diagram denotes the percentage of total score that was
scored by the top three Indian scorers in that game. No two players score the same number of runs in
the same game. John also calculated two batting indices for each players based on his scores it he
tournament; the R-index of a batsman is the difference between his highest and lowest scores in the 3
games while the M-index is the middle number, if his scores are arranged in a non-increasing order.
Ans .
c
- Explanation :
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.
(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. This will not affect the middle number 50. For Rahul middle index will be 49. (Same logic)
Ans .
d
- Explanation :
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.
(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.
Ans .
b
- Explanation :
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.
(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.
Ans .
b
- Explanation :
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.
(b) Saurav’s middle index is the best.
Directions for Questions 50 to 52:
Five women decided to go shopping to M.G. Road, Bangalore.
They arrived at the designated meeting place in the following order: 1. Archana, 2. Chellama, 3.
Dhenuka, 4. Helen and 5. Sahnaz.
Each woman spent at least ` 1000. Below are some additional facts about how much they spent during
their shopping spree.
i. The woman who spent ` 2234 arrive before the lady who spent ` 1193.
ii. One woman spent ` 1340 and she was not Dhenuka.
iii. One woman spent ` 1378 more than Chellamma.
iv. One woman spent ` 2517 and she was not Archana.
v. Helen spent more than Dhenuka.
vi. Shahnaz spent the largest amount and Chellamma the smallest.
Ans .
c
- Explanation :
From the given information it is clear that four numbers which must have been the values of money spent would 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:
Possibility 1 Possibility 2 A (x 2517) 2234 C (least) 1193 D (x 1340) 2517 H 1340 S (Max) 2571
Note: The thought structure for placing the 5 values with 5 women in the case of possibility 1, goes as follows:
Step 1:
A
C 1193
D
H
S 2571
After placing the least and maximum.
Step 2: 2234 should be before 1193 and Dhenuka cannot have spent ` 1340.
A 2234
C 1193
D 2517
H 1340
G 2571
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.
A (x 2517)
C (least, 1139
D (x 1340)
H
G (Maximum) 2517
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 only possible arrangement is as in (1) above.
(c) Dhenuka
Ans .
b
- Explanation :
From the given information it is clear that four numbers which must have been the values of money spent would 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:
Possibility 1 Possibility 2 A (x 2517) 2234 C (least) 1193 D (x 1340) 2517 H 1340 S (Max) 2571
Note: The thought structure for placing the 5 values with 5 women in the case of possibility 1, goes as follows:
Step 1:
A
C 1193
D
H
S 2571
After placing the least and maximum.
Step 2: 2234 should be before 1193 and Dhenuka cannot have spent ` 1340.
A 2234
C 1193
D 2517
H 1340
G 2571
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.
A (x 2517)
C (least, 1139
D (x 1340)
H
G (Maximum) 2517
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 only possible arrangement is as in (1) above.
(b) 1340
Ans .
a
- Explanation :
From the given information it is clear that four numbers which must have been the values of money spent would 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:
Possibility 1 Possibility 2 A (x 2517) 2234 C (least) 1193 D (x 1340) 2517 H 1340 S (Max) 2571
Note: The thought structure for placing the 5 values with 5 women in the case of possibility 1, goes as follows:
Step 1:
A
C 1193
D
H
S 2571
After placing the least and maximum.
Step 2: 2234 should be before 1193 and Dhenuka cannot have spent ` 1340.
A 2234
C 1193
D 2517
H 1340
G 2571
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.
A (x 2517)
C (least, 1139
D (x 1340)
H
G (Maximum) 2517
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 only possible arrangement is as in (1) above.
(a) 1139
Ans .
b
- Explanation :
From the given information it is clear that four numbers which must have been the values of money spent would 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:
Possibility 1 Possibility 2 A (x 2517) 2234 C (least) 1193 D (x 1340) 2517 H 1340 S (Max) 2571
Note: The thought structure for placing the 5 values with 5 women in the case of possibility 1, goes as follows:
Step 1:
A
C 1193
D
H
S 2571
After placing the least and maximum.
Step 2: 2234 should be before 1193 and Dhenuka cannot have spent ` 1340.
A 2234
C 1193
D 2517
H 1340
G 2571
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.
A (x 2517)
C (least, 1139
D (x 1340)
H
G (Maximum) 2517
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 only possible arrangement is as in (1) above.
(b) Each loaf of bread is divided into 3 parts. So we have 24 parts and each traveller gets 8 parts. Ist traveller has 15 parts. He ate 8 parts and gave his 7 parts to the lllrd traveller. IInd traveller has 9 parts. He ate 8 parts and gave his1 part to the IIIrd traveller. So 8 coins should be divided in the ratio 7: 1. First traveller gets 7 coins.
Ans .
c
- Explanation :
From the given information it is clear that four numbers which must have been the values of money spent would 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:
Possibility 1 Possibility 2 A (x 2517) 2234 C (least) 1193 D (x 1340) 2517 H 1340 S (Max) 2571
Note: The thought structure for placing the 5 values with 5 women in the case of possibility 1, goes as follows:
Step 1:
A
C 1193
D
H
S 2571
After placing the least and maximum.
Step 2: 2234 should be before 1193 and Dhenuka cannot have spent ` 1340.
A 2234
C 1193
D 2517
H 1340
G 2571
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.
A (x 2517)
C (least, 1139
D (x 1340)
H
G (Maximum) 2517
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 only possible arrangement is as in (1) above.
(c) Maximum points of 22 can be achieved by taking (1 Management + 2 Fiction + 2 Maths + 5 Physics) books. 4 + 2 + 6 + 10 = 22
Ans .
b
- Explanation :
From the given information it is clear that four numbers which must have been the values of money spent would 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:
Possibility 1 Possibility 2 A (x 2517) 2234 C (least) 1193 D (x 1340) 2517 H 1340 S (Max) 2571
Note: The thought structure for placing the 5 values with 5 women in the case of possibility 1, goes as follows:
Step 1:
A
C 1193
D
H
S 2571
After placing the least and maximum.
Step 2: 2234 should be before 1193 and Dhenuka cannot have spent ` 1340.
A 2234
C 1193
D 2517
H 1340
G 2571
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.
A (x 2517)
C (least, 1139
D (x 1340)
H
G (Maximum) 2517
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 only possible arrangement is as in (1) above.
Option (b) is correct.
Directions for Questions 56 to 57: Elle is three times older than Yogesh; Zaheer is half the age of Wahida. Yogesh is older than Zaheer.
Ans .
b
- Explanation :
Elle 3y Yogesh y Wahida 2z Zaheer z
(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)).
Ans .
c
- Explanation :
Elle 3y Yogesh y Wahida 2z Zaheer z
(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.
Ans .
b
- Explanation :
Elle 3y Yogesh y Wahida 2z Zaheer z
(b)
Ans .
b
- Explanation :
Elle 3y Yogesh y Wahida 2z Zaheer z
(b)
Ans .
b
- Explanation :
Elle 3y Yogesh y Wahida 2z Zaheer z
(b)
Directions for Questions 61 to 63: Recently, Ghosh Babu spent his winter vacation on Kyakya Island. During the vacation, he visited the local casino where he came across a new card game. Two players, using a normal deck of 52 playing cards, play this game. One player is called the ‘dealer’ and the other is called the ‘player’. First, the player picks a card at random from the deck. This is called the base card. The amount in rupees is equal to the face value of the base card and is called the base amount. The face values of Ace, King, Queen and Jack are ten. For other cards the face value is the number on the card. Once the ‘player’ picks a card from the deck, the ‘dealer’ pays him the base amount. Then the ‘dealer’ picks a card from the deck and this card is called the top card. If the top card is of the same suit as the base card, the ‘player’ pays twice the base amount to the ‘dealer’. If the top card is of the same colours as the base card (but not the same suit), then the ‘player’ pays the base amount to the ‘dealer’. lf the top card happens to be of a different colour than the base card the ‘dealer’ pays the base amount to the ‘player’. Ghosh Babu played the game four times. The first time he picked eight of clubs and the ‘dealer’ picked queen of clubs. Second time, he picked ten of hearts and the ‘dealer’ picked two of spades. Next time, Ghosh Babu picked six of diamonds and the ‘dealer’ picked ace of hearts. Lastly, he picked eight of spades and the ‘dealer’ picked Jack of spades. Answer the following questions based on these four games.
Ans .
a
- Explanation :
The following matrix will help you solve the problem.
Game Starting Money with Ghosh Babu Ghosh Babu Gives Gets Dealer Gets Gives End I 0 0 8 16 0 -8 II -8 0 10 0 10 12 III 12 0 6 6 0 12 IV 12 0 8 16 0 4
` 12. Hence (a)
Ans .
b
- Explanation :
The following matrix will help you solve the problem.
Game Starting Money with Ghosh Babu Ghosh Babu Gives Gets Dealer Gets Gives End I 0 0 8 16 0 -8 II -8 0 10 0 10 12 III 12 0 6 6 0 12 IV 12 0 8 16 0 4
The maximum negative he goes to is – 8 after the first game. Hence, Option (b) is correct.
Ans .
d
- Explanation :
The following matrix will help you solve the problem.
Game Starting Money with Ghosh Babu Ghosh Babu Gives Gets Dealer Gets Gives End I 0 0 8 16 0 -8 II -8 0 10 0 10 12 III 12 0 6 6 0 12 IV 12 0 8 16 0 4
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.