Set 1
According to a coding scheme the sentence
Peacock is designated as the national bird of India is coded as
5688999 35 1135556678 56 458 13666689 1334 79 13366
This coding scheme has the following rules:
The scheme is case-insensitive (does not distinguish between upper case and lower case letters).
Each letter has a unique code which is a single digit from among 1,2,3, …, 9.
The digit 9 codes two letters, and every other digit codes three letters.
The code for a word is constructed by arranging the digits corresponding to its letters in a non-decreasing sequence.
Explanation:
Solution
In the question, we have a statement PEACOCK IS DESIGNATED AS THE NATIONAL BIRD OF INDIA
Which is coded as 5688999 35 1135556678 56 458 13666689 1334 79 13366
Also what we know is
each one letter is only coded as only one character.
Number 1 to 8 are assigned to 3 letters while 9 is assigned to only 2.
This set can be approached by comparing letters and code of one word to other.
To start, IS is coded as 35 and AS is coded as 56. S is common between these two words and so as 5 so S must be assigned to 5, I to 3 and 6 to A.
Similarly, If we compare OF with PEACOCK, we get O is coded as 9 and so F must be coded as 7.
Comparing THE with DESIGNATED, H must be 4.
Till now, we have
If we cross the letters and numbers we have found then we are left with PECCK DEGNTED TE NTNL BRD ND coded as 58899 1155678 58 1668 134 16.
TE is coded as 58 and NTONL is coded as 16689 that means T must be 8 as it is the common one and so E must be 5.
PECCK is coded as 58899. If C is coded as 8 then two letters will be coded as 9. As we know O is already 9, 9 will have 3 letters that means C must be 9. E is coded as 5 so P and K must be 8.
Now the table looks like
And DGND NNL BRD ND is coded as 1167 166 134 16
Moving on with similar approach of comparing, BRD is 134 and ND is 16 so D must be 1 and N must be 6.
NNL is coded as 166 and N is 6 so L must be 1.
DGND is 1167 so D must be 1 and G must be 7.
The final table is
B and R can be 3 or 4 in any order.
Question 1
What best can be concluded about the code for the letter L?
6
8
1 or 8
1
Answer: L is coded as 1.
Question 2
What best can be concluded about the code for the letter B?
3
3 or 4
1
1 or 3 or 4
Answer: 3 or 4.
Question 3
For how many digits can the complete list o letters associated with that digit be identified?
1
2
0
3
Answer: Only for 8 and 9 can the complete list be identified so 2.
Question 4
Which set of letters CANNOT be coded with the same digit?
I, B, M
X, Y, Z
S, U, V
S, E, Z
Answer: we can see that S is associated with 5 and E is also associated with 5. That means only 1 more letter can be associated with 5 so in Option 3, U and V cannot be associated with 5 so 3. S, U, V.
Set 2
Each visitor to an amusement park needs to buy a ticket. Tickets can be Platinum, Gold, or Economy. Visitors are classified as Old, Middle-aged, or Young. The following facts are known about visitors and ticket sales on a particular day:
140 tickets were sold.
The number of Middle-aged visitors was twice the number of Old visitors, while the number of Young visitors was twice the number of Middle-aged visitors.
Young visitors bought 38 of the 55 Economy tickets that were sold, and they bought half the total number of Platinum tickets that were sold.
The number of Gold tickets bought by Old visitors was equal to the number of Economy tickets bought by Old visitors.
Explanation:
Solution
In this, we need to make a table.
By point 1, we get 140 tickets were sold in total and by point 2, Old, Middle-aged and young visitors were in the ratio of 1:2:4. Hence they were 20, 40 and 80 respectively.
The table we get
By point 3, we get total number of economy tickets as 55 and economy bought by young as 38.
Further we assume that X Young people bought Platinum tickets that means 2X in total bought Platinum Tickets.
By point 4, Gold tickets bought by Old are equal to economy tickets bought by them so we can assume them to be Y.
The table we get
Now we can fill others blanks in terms of X and Y as
Question 1
If the number of Old visitors buying Platinum tickets was equal to the number of Middle-aged visitors buying Platinum tickets, then which among the following could be the total number of Platinum tickets sold?
38
34
32
36
Answer: Old buying Platinum is equal to Middle-Aged buying platinum that is
20 – 2Y = X + 2Y – 20
40 = X + 4Y
Or 80 – 8Y = 2X
That is 80 – 8Y people could have bought platinum ticket.
Out of the options, only 32 is possible as 80 – 8*6.
Question 2
If the number of Old visitors buying Platinum tickets was equal to the number of Middle-aged visitors buying Economy tickets, then the number of Old visitors buying Gold tickets was
Answer: Old visitors buying platinum tickets is 20 – 2Y
Middle aged visitors buying Economy tickets is 17 – Y
As given in the question, these are equal so 20 – 2Y = 17 – Y
20 – 17 = 2Y – Y
Y = 3
Old visitors buying Gold tickets, from the table, are Y.
So the answer is 3.
Question 3
If the number of Old visitors buying Gold tickets was strictly greater than the number of Young visitors buying Gold tickets, then the number of Middle-aged visitors buying Gold tickets was
Answer: Old vistors buying Gold tickets = Y
Young Visitors buying Gold tickets = 42 – X
As per the question, Y > 42 – X
That is X + Y > 42
Now Middle-aged visitors buying Gold tickets is 43 – X – Y = 43 – (X + Y)
Now if X + Y is greater than 42 then minimum value it can take is 43.
So possible value of 43 – (X + Y) = 43 – 43 = 0
Zero is the answer.
Question 4
Which of the following statements MUST be FALSE?
- The numbers of Old and Middle-aged visitors buying Economy tickets were equal
- The numbers of Gold and Platinum tickets bought by Young visitors were equal
- The numbers of Middle-aged and Young visitors buying Gold tickets were equal
- The numbers of Old and Middle-aged visitors buying Platinum tickets were equal
Answer: Check all the options.
In option 1, Old and Middle-Aged visitors buying Economy tickets are Y and 17 – Y. If they are equal then Y would 8.5. But Number of visitors cannot be in decimal so this is not possible.
Set 3
An agency entrusted to accredit colleges looks at four parameters: faculty quality (F), reputation (R), placement quality (P), and infrastructure (I). The four parameters are used to arrive at an overall score, which the agency uses to give an accreditation to the colleges. In each parameter, there are five possible letter grades given, each carrying certain points: A (50 points), B (40 points), C (30 points), D (20 points), and F (0 points). The overall score for a college is the weighted sum of the points scored in the four parameters. The weights of the parameters are 0.1, 0.2, 0.3 and 0.4 in some order, but the order is not disclosed. Accreditation is awarded based on the following scheme:
Eight colleges apply for accreditation, and receive the following grades in the four parameters (F, R, P, and I):
It is further known that in terms of overall scores:
High Q is better than Best Ed;
Best Ed is better than Cosmopolitan; and
Education Aid is better than A-one.
Solution
Explanation:
In this what we have to find is the weightage of the given parameters in the total score.
We start using point 2, Best Ed is better than Cosmopolitan. Comparing the rating of these two, we see that they are rated equal in two parameters while equally different (one higher and one lower) in other two parameters. As they are rated similarly, there scores must be same too but Best Ed is rated better than Cosmopolitan that means that the parameter in which Best Ed is rated higher, must carry more weight than the parameter in which Cosmopolitan is rated higher.
And so R > I
Similarly using point 3 that is education aid is better than A-One, we can see I > P.
Now using point 1 that is High Q is better than Best Ed, We see High Q is rated better than Best Ed in only 1 parameter that is I while Best Ed is rated better in 2 these are F and R.
Now we know R > I > P. F and R both cannot be better than I because then Best will have higher overall score definitely so we have two possible cases R>I>F>P and R>I>P>F.
Calculating their scores as per the table 2, In first case, High Q’s overall rating is 8+12+6+2=28 and Best Ed’s Rating is 12+6+8+3=29.
In Second case, High Q’s rating is 8+12+4+3=27 and Best Ed’s score is 12+6+4+4=26.
So R>I>P>F is true. That means weight of R is 0.4, I is 0.3, P is 0.2, F is 0.1.
Now we can remake the table as
Question 1
What is the weight of the faculty quality parameter?
0.3
0.4
0.1
0.2
Answer: As per our solution, 0.1 is the weight of Faculty.
Question 2
How many colleges receive the accreditation of AAA?
Answer: By the table, 3 colleges got more than 45 that is AAA rating.
Question 3
What is the highest overall score among the eight colleges?
Answer: By the table, 48 is the highest score.
Question 4
How many colleges have overall scores between 31 and 40, both inclusive?
3
1
0
2
Answer: No colleges scored between 31 and 40. Option 3
Set 4
Each of the 23 boxes in the picture below represents a product manufactured by one of the following three companies: Alfa, Bravo and Charlie. The area of a box is proportional to the revenue from the corresponding product, while its centre represents the Product popularity and Market potential scores of the product (out of 20). The shadings of some of the boxes have got erased.
The companies classified their products into four categories based on a combination of scores (out of 20) on the two parameters – Product popularity and Market potential as given below:
The following facts are known:
Alfa and Bravo had the same number of products in the Blockbuster category.
Charlie had more products than Bravo but fewer products than Alfa in the No-hope category.
Each company had an equal number of products in the Promising category.
Charlie did not have any product in the Doubtful category, while Alfa had one product more than Bravo in this category.
Bravo had a higher revenue than Alfa from products in the Doubtful category.
Charlie had a higher revenue than Bravo from products in the Blockbuster category.
Bravo and Charlie had the same revenue from products in the No-hope category.
Alfa and Charlie had the same total revenue considering all products.
Solution
Explanation:
In this, we make a table or graph with 4 zones naming Promising, Blockbuster, Doubtful and No-Hope. In each zone, we see how many products they have (Number of blocks) and how much revenue they generate (area of the blocks)
Now point 1, Alfa Bravo had same number of products in Blockbuster category. As there are two unknown blocks in that category, Alfa and Bravo cannot have 3 each because for that we need 3 blank blocks. So they must have 2 each and the other blank block must belong to Charlie.
Also by point 6, Charlie had higher revenue than Bravo in this category. This is possible if the block of area 4 goes to Bravo and block with area 9 goes to Charlie.
Similarly by point 2, Alfa has more products than Charlie while Charlie had more products than Bravo. We have 3 blank blocks in the No-Hope zone. They can be distributed as 1 to Alfa and 2 to Charlie to satisfy the condition.
By point 7, Bravo and Charlie had same revenue in No-Hope category. As we know Bravo had a revenue of 4 and no new addition is done to him, the products Charlie receive must be 3+1.
Similarly by the combination of 4 and 5, we can figure out that in doubtful category, Alfa gets 2 blocks of 1+6 and Bravo gets 1 of 9.
By Point 3, Alfa, Bravo and Charlie must have 1 product each in Promising category.
Further by point 8, Alfa had a revenue of 29 in other 3 blocks and Charlie had revenue of 21. So in promising category, Alfa must get block of area 1 and Charlie must get a block of area 9.
The final table is like this.
Question 1
Considering all companies’ products, which product category had the highest revenue?
No-Hope
Doubtful
Promising
Blockbuster
Answer: By the table above, Blockbuster had the highest revenue that is 36.
Question 2
Which of the following is the correct sequence of numbers of products Bravo had in No-hope, Doubtful, Promising and Blockbuster categories respectively?
1, 3, 1, 3
2, 3, 1, 2
3, 3, 1, 2
1, 3, 1, 2
Answer: In No-Hope, Doubtful, Promising and Blockbuster categories, Bravo had 1, 3, 1, 2 products respectively so option 4.
Question 3
Which of the following statements is NOT correct?
- The total revenue from No-hope products was less than the total revenue from Doubtful products
- Bravo’s revenue from Blockbuster products was greater than Alfa’s revenue from Doubtful products
- Alfa’s revenue from Blockbuster products was the same as Charlie’s revenue from Promising products
- Bravo and Charlie had the same revenues from No-hope products
Answer: Option 2, Bravo’s revenue from Blockbuster was 10 which was less than Alfa’s revenue from Doubtful products which is 12.
Question 4
If the smallest box on the grid is equivalent to revenue of Rs.1 crore, then what approximately was the total revenue of Bravo in Rs. crore?
24
30
40
34
Answer: In this, we have taken the area of 1 as revenue of 1 crore. So total revenue of Bravo becomes 17 + 3 + 4 + 10 = 34 that is the option 4.
Set 5
Fun Sports (FS) provides training in three sports – Gilli-danda (G), Kho-Kho (K), and Ludo (L). Currently it has an enrollment of 39 students each of whom is enrolled in at least one of the three sports. The following details are known:
The number of students enrolled only in L is double the number of students enrolled in all the three sports.
There are a total of 17 students enrolled in G.
The number of students enrolled only in G is one less than the number of students enrolled only in L.
The number of students enrolled only in K is equal to the number of students who are enrolled in both K and L.
The maximum student enrollment is in L.
Ten students enrolled in G are also enrolled in at least one more sport.
Solution
This is a question of venn diagram.
By point 1, number of students enrolled only in L is double of the number of students enrolled in all 3 sports. So we assume those playing all three as X then those playing only Ludo becomes 2X.
By point 3, the number of students enrolled in G is one less than those playing only L. By previous point, only L is 2X and hence number of students playing only G becomes 2X – 1.
By point 2, 17 students are playing G in total.
The venn that we have now is
Gilli-Danda (17)
Kho-Kho
Ludo
X
2X
By point 6, we get that those who are enrolled in G and one more sport are 10. That as we know that total number of students in G is 17, the number of those playing only G becomes 17 – 10=7
Now as those playing only G were 2X – 1. Value of X turns out to be 4 (2X – 1 = 7).
Hence, those playing all three are 4 and only L are 8.
By point 4, we get the number of students playing K and L (includes all three) is equal to the number of students playing only K. If we assume those playing only K as Y then those playing K and L but not G becomes Y – 4 (4 is all three).
The venn now is
Gilli-Danda (17)
Kho-Kho
Ludo
4
8
9
5 In this, we see that Y, Y-4 and 8 constitute 22 students as this area is total minus G that is 39 – 17=22.
Equating we get the value of Y=9.
Now, we see that the two remaining sections of the Venn must be equal to 6 in total.
Also, point 5 says that maximum students are in L. Hence those playing L and G but not K must be 4, 5 or 6 while those playing G and K but not L can be 0, 1 or 2. Otherwise those playing will be more than L. So number of students in L can be 21-23.
Question 1
What is the minimum number of students enrolled in both G and L but not in K?
Answer: As we see in the end, minimum value of L and G but not K is 4.
Question 2
If the numbers of students enrolled in K and L are in the ratio 19:22, then what is the number of students enrolled in L?
18
22
19
17
Answer: students enrolled in K and L are in the ratio of 19:22. That means the numbers of students in K will be 19 and L will be 22.
Question 3
Due to academic pressure, students who were enrolled in all three sports were asked to withdraw from one of the three sports. After the withdrawal, the number of students enrolled in G was six less than the number of students enrolled in L, while the number of students enrolled in K went down by one. After the withdrawal, how many students were enrolled in both G and K?
Answer: Students enrolled in K went down by 1. That means that 1 students joined L and G but not K section. Other 3 might have gone to G and K or L and K section.
Total number of students in L and G is 39 – 9=28 as 9 is the number of people enrolled only in K.
As we know that number of students enrolled in G is 6 less than L and their sum is 28 while number of students liking both of them is 5, 6 or 7 (+1 to original possible numbers).
This is only possible if the common number of students is 6 otherwise number of students in G and L will be in decimals. So number of students in G becomes 15 and number of students in L becomes 21.
So the number of students in G and K previously becomes 1 and 1 added after the transfer so answer is 2.
Question 4
8
7
5
6
Answer: By the above explanation, Students in G and L will be 6.
Set 6
There are only four brands of entry-level smartphones called Azra, Bysi, Cxqi, and Dipq in a country.
Details about their market share, unit selling price, and profitability (defined as the profit as a percentage of the revenue) for the year 2016 are given in the table below:
In 2017, sales volume of entry-level smartphones grew by 40% as compared to that in 2016. Cxqi offered a 40% discount on its unit selling price in 2017, which resulted in a 15% increase in its market share. Each of the other three brands lost 5% market share. However, the profitability of Cxqi came down to half of its value in 2016. The unit selling prices of the other three brands and their profitability values remained the same in 2017 as they were in 2016.
Solution
Explanation:
We remake the table including the data of 2017.
Let the initial market equal to 100. That means market in 2017 became 140.
Cxqi gained 15% market share while others lost 5% each that means Cxqi’s market became 15+15=30% and now that total market is 140, Cxqi’s sale would be 30% of 140 that is 42.
Similarly, the sale of others can be known.
Selling Price of Cxqi became 60% of 2016 that means 60% of 30,000 = 18,000.
Profitability of Cxqi became half that is half of 40% = 20%.
Now we make the table to find profits of both year (Profit=saleSPProfitability)
Question 1
The brand that had the highest revenue in 2016 is:
Dipq
Cxqi
Azra
Bysi
Answer: By the data, we can see that Azra had the highest Revenue that is 40*15,000= 6,00,000.
Question 2
The brand that had the highest profit in 2016 is:
Bysi
Azra
Dipq
Cxqi
Answer: By the table, we can see that Cxqi had the highest profit in 2016 that is 1,80,000.
Question 3
The brand that had the highest profit in 2017 is:
Azra
Bysi
Cxqi
Dipq
Answer: BY the Table, Bysi had the highest profit in 2017 that is 1,68,000
Question 4
The complete list of brands whose profits went up in 2017 from 2016 is:
- Azra, Bysi, Cxqi
- Azra, Bysi, Dipq
- Cxqi, Azra, Dipq
- Bysi, Cxqi, Dipq
Answer: By the table, Only Cxqi’s Profits went down in 2017 hence Option 2.
Set 7
The base exchange rate of a currency X with respect to a currency Y is the number of units of currency Y which is equivalent in value to one unit of currency X. Currency exchange outlets buy currency at buying exchange rates that are lower than base exchange rates, and sell currency at selling exchange rates that are higher than base exchange rates.
A currency exchange outlet uses the local currency L to buy and sell three international currencies A, B, and C, but does not exchange one international currency directly with another. The Base Exchange rates of A, B and C with respect to L are in the ratio 100:120:1. The buying exchange rates of each of A, B, and C with respect to L are 5% below the corresponding Base Exchange rates, and their selling exchange rates are 10% above their corresponding Base Exchange rates.
The following facts are known about the outlet on a particular day:
The amount of L used by the outlet to buy C equals the amount of L it received by selling C.
The amounts of L used by the outlet to buy A and B are in the ratio 5:3.
The amounts of L the outlet received from the sales of A and B are in the ratio 5:9.
The outlet received 88000 units of L by selling A during the day.
The outlet started the day with some amount of L, 2500 units of A, 4800 units of B, and 48000 units of C.
The outlet ended the day with some amount of L, 3300 units of A, 4800 units of B, and 51000 units of C.
Solution
Explanation:
The exchange rates as per given in the question are
We are assuming x as base values are not absolute values but ratios.
Now using the points the table for outlets is
Now in outlet C, we know that the amount of L used to buy and sell currency C is equal. Buying and Selling exchange rates of Currency C are 0.95x and 1.1x that is in the ratio 0f 19:22. As they are equal, the amount of currency C bought and sold must be in 22:19 so that the amount of L will be equal.
We also know that the currency C bought was 3,000 more than C sold there is 3,000 difference between initial and final amount of C.
This means that Currency C bought and sold must be 22,000 and 19,000. And the values of these will be 20,900x and 20,900x that is amount of C * exchange rate.
We can also see in the table that amount of Currency A bought is 800 more than sold (final amount minus initial amount) and amount of B bought is equal to sold. We assume these values as ‘a’ and ‘b’.
By the points in the question, we also know that the amount of L used in Buying A and B are in the ratio 5:3 and Selling A and B are in the ratio 5:9.
Filling the table with this Data
The amounts in brackets are the value of currency bought and sold in terms of Local currency.
By the points in the question, we also know that the amount of L used in Selling A and B are in the ratio 5:9.
That means 110ax and 132bx are in the ratio of 5:9.
Equating these we get 110ax/132bx=5/9
Solving we get a:b=2:3.
Taking a=2m and b=3m and filling the table again.
We also know that amount of L used in buying A and B is in the ratio 5:3.
Using this, (190mx+76000x)/342mx = 5/3
Solving we get m = 200.
Editing the table again.
By the 4th point of the set, we get the value of 44,000x as 88,000.
So X = 2.
Final table is looks like this now.
Question 1
How many units of currency A did the outlet buy on that day?
Answer: By the table, we get 1200.
Question 2
How many units of currency C did the outlet sell on that day?
3000
6000
22000
19000
Answer: By the table, 19,000.
Question 3
What was the base exchange rate of currency B with respect to currency L on that day?
Answer: The exchange rate of B to L was 120x and as we know x=2, the exchange rate becomes 240x.
Question 4
What was the buying exchange rate of currency C with respect to currency L on that day?
2.20
1.10
0.95
1.90
Answer: Buying exchange rate of C with respect to L was 0.95x. the value of X=2 then the answer would be 1.9
Set 8
Seven candidates, Akil, Balaram, Chitra, Divya, Erina, Fatima, and Ganeshan, were invited to interview for a position. Candidates were required to reach the venue before 8 am. Immediately upon arrival, they were sent to one of three interview rooms: 101, 102, and 103. The following venue log shows the arrival times for these candidates. Some of the names have not been recorded in the log and have been marked as ‘?’.
Additionally here are some statements from the candidates: Balaram: I was the third person to enter Room 101. Chitra: I was the last person to enter the room I was allotted to. Erina: I was the only person in the room I was allotted to. Fatima: Three people including Akil were already in the room that I was allotted to when I entered it. Ganeshan: I was one among the two candidates allotted to Room 102.
Solution
Explanation:
What we know from Balaram’s statement is that atleast 3 people were in room 101 and Balaram was the 3rd one to enter.
From Fatima’s Statement, we can understand that she was the 4th one in the room she entered.
From Erina’s statement we can say that she was alone in a room.
Now, if, Balaram and Fatima were in different rooms then these two rooms will have to have all 7 people. In which case, Erina cannot be alone in her room. So, we conclude that Balaram was the 3rd and Fatima was the 4th one in room 101. Also, Akil was the 1st one to enter room 101 as he is the 1st one to enter in all of them.
From Ganeshan’s Statement, we know that there were 2 people in room 102. And so Erina must be alone in room 103.
And from Chitra’s Statement, Chitra was the 2nd one in room 102 and Ganeshan must have been 1st one.
Divya has only one place left that is 2nd in room 101.
Till now, we have
As Chitra and Fatima were the last ones to enter in their respective rooms at 7:30 and 7:40, the only one who could have entered at 7:45 is Erina.
Divya could have entered at 7:10 or 7:15 and Balaram could have entered at 7:15 or 7:25 only.
Ganeshan could have entered at 7:10 or 7:15 or 7:25.
Question 1
What best can be said about the room to which Divya was allotted?
- Either Room 101 or Room 102
- Definitely Room 102
- Definitely Room 101
- Definitely Room 103
Answer: Divya was definitely allotted room 101
Question 2
Who else was in room 102 when Ganeshan entered?
Divya
Akil
No one
Chitra
Answer: Ganeshan was the first one to enter room 102 so no one.
Question 3
When did Erina reach the venue?
- 7:25 am
- 7:15 am
- 7:10 am
- 7:45 am
- Answer: By the table, Erina reached the venue at 7:45.
Question 4
If Ganeshan entered the venue before Divya, when did Balaram enter the venue?
- 7:15 am
- 7:25 am
- 7:45 am
- 7:10 am
Answer: Divya entered at 7:10 or 7:15. If Ganeshan entered before Divya then Divya must have entered at 7:15 and Ganeshan at 7:10. That means Balaram has only 7:25 as his possible entry time.
