CAT Quant Based Puzzles Questions
Master CAT Quant Based Puzzles Questions with practice questions and detailed solutions.
Instructions
Three participants - Akhil, Bimal and Chatur participate in a random draw competition for five days. Every day, each participant randomly picks up a ball numbered between 1 and 9. The number on the ball determines his score on that day. The total score of a participant is the sum of his scores attained in the five days. The total score of a day is the sum of participants’ scores on that day. The 2-day average on a day, except on Day 1, is the average of the total scores of that day and of the previous day. For example, if the total scores of Day 1 and Day 2 are 25 and 20, then the 2-day average on Day 2 is calculated as 22.5. Table 1 gives the 2-day averages for Days 2 through 5.
Participants are ranked each day, with the person having the maximum score being awarded the minimum rank (1) on that day. If there is a tie, all participants with the tied score are awarded the best available rank. For example, if on a day Akhil, Bimal, and Chatur score 8, 7 and 7 respectively, then their ranks will be 1, 2 and 2 respectively on that day. These ranks are given in Table 2.
The following information is also known.
1. Chatur always scores in multiples of 3. His score on Day 2 is the unique highest score in the competition. His minimum score is observed only on Day 1, and it matches Akhil’s score on Day 4.
2. The total score on Day 3 is the same as the total score on Day 4.
3. Bimal’s scores are the same on Day 1 and Day 3.
Three participants - Akhil, Bimal and Chatur participate in a random draw competition for five days. Every day, each participant randomly picks up a ball numbered between 1 and 9. The number on the ball determines his score on that day. The total score of a participant is the sum of his scores attained in the five days. The total score of a day is the sum of participants’ scores on that day. The 2-day average on a day, except on Day 1, is the average of the total scores of that day and of the previous day. For example, if the total scores of Day 1 and Day 2 are 25 and 20, then the 2-day average on Day 2 is calculated as 22.5. Table 1 gives the 2-day averages for Days 2 through 5.
Participants are ranked each day, with the person having the maximum score being awarded the minimum rank (1) on that day. If there is a tie, all participants with the tied score are awarded the best available rank. For example, if on a day Akhil, Bimal, and Chatur score 8, 7 and 7 respectively, then their ranks will be 1, 2 and 2 respectively on that day. These ranks are given in Table 2.
The following information is also known.
1. Chatur always scores in multiples of 3. His score on Day 2 is the unique highest score in the competition. His minimum score is observed only on Day 1, and it matches Akhil’s score on Day 4.
2. The total score on Day 3 is the same as the total score on Day 4.
3. Bimal’s scores are the same on Day 1 and Day 3.
Question 1.
What is Akhil's score on Day 1?
What is Akhil's score on Day 1?
Text Explanation:
Let the total score of day 1, day 2, day 3, day 4, and day 5 are d1, d2, d3, d4, and d5, respectively.
The table shows that d1+d2 = 30 ... eq (1), d2+d3 = 31 ... eq (2), d3+d4 = 32 .... eq(3), d4+d5 = 34 ... eq(4)
It is given that participants are ranked each day, with the person having the maximum score being awarded the minimum rank (1) on that day. All participants with a tied score are awarded the best available rank if there is a tie.
It is given that the total score on Day 3 is the same as the total score on Day 4.
Therefore, d3 = d4 => d3 = d4 = 16, which implies d2 = 15, d5 = 18, and d1 = 15.
The day-wise score is given below:
It is known that Chatur always scores in multiples of 3. His score on Day 2 is the unique highest score in the competition. His minimum score is observed only on Day 1, and it matches Akhil’s score on Day 4.
Hence, only Chatur scored 9 (one time) on Day 2, and no other person scored 9 on any of the given 5 days. Chatur scored 3 only one time, which was on Day 1. Therefore, the scores obtained by Chatur on Day 3, Day 4, and Day 5 are 6, 6, and 6, respectively. It is also known that Akhil's score on Day 4 is the same as the score obtained by Chatur on Day 1. Hence, Akhil's score on Day 4 is 3.
Hence, we get the following table:
From Table 2, we see that the rank of Bimal and Akhil is the same, which is 2. Hence, The score obtained by Akhil and Bimal is the same. Let the score be x. Therefore, 6+2x = 16 => x = 5
The rank of Chatur on Day 5 is 2, and the rank of Bimal is 1, which implies the score obtained by Bimal will be more than Chatur. Hence, Bimal can score either 7 or 8 on Day 5. Therefore, the score obtained by Akhil on Day 5 is either 5 or 4.
It is given that Bimal’s scores are the same on Day 1 and Day 3. Hence, the score obtained by Bimal on Day 1 is 5, which implies The score obtained by Akhil is 7 on Day 1.
From Table 2, we can see that the rank of Bimal is 3 on Day 2, and the rank of Akhil is 2 on Day 2. Hence, the score of Bimal will be lower than Akhil on Day 2.
Let the score of Akhil be a, and the score of Bimal be b. Then 9+a+b = 15, and a > b
=> a+b =6, and a> b
Hence, the value of a can be 4/5, and the value of b can be 2/1
Therefore, the final table is given below:
From the table, we can see that the score of Akhil is 7 on day 1.
The correct option is B
Question 2.
Who attains the maximum total score?
Who attains the maximum total score?
Text Explanation:
Let the total score of day 1, day 2, day 3, day 4, and day 5 are d1, d2, d3, d4, and d5, respectively.
The table shows that d1+d2 = 30 ... eq (1), d2+d3 = 31 ... eq (2), d3+d4 = 32 .... eq(3), d4+d5 = 34 ... eq(4)
It is given that participants are ranked each day, with the person having the maximum score being awarded the minimum rank (1) on that day. All participants with a tied score are awarded the best available rank if there is a tie.
It is given that the total score on Day 3 is the same as the total score on Day 4.
Therefore, d3 = d4 => d3 = d4 = 16, which implies d2 = 15, d5 = 18, and d1 = 15.
The day-wise score is given below:
It is known that Chatur always scores in multiples of 3. His score on Day 2 is the unique highest score in the competition. His minimum score is observed only on Day 1, and it matches Akhil’s score on Day 4.
Hence, only Chatur scored 9 (one time) on Day 2, and no other person scored 9 on any of the given 5 days. Chatur scored 3 only one time, which was on Day 1. Therefore, the scores obtained by Chatur on Day 3, Day 4, and Day 5 are 6, 6, and 6, respectively. It is also known that Akhil's score on Day 4 is the same as the score obtained by Chatur on Day 1. Hence, Akhil's score on Day 4 is 3.
Hence, we get the following table:
From Table 2, we see that the rank of Bimal and Akhil is the same, which is 2. Hence, The score obtained by Akhil and Bimal is the same. Let the score be x. Therefore, 6+2x = 16 => x = 5
The rank of Chatur on Day 5 is 2, and the rank of Bimal is 1, which implies the score obtained by Bimal will be more than Chatur. Hence, Bimal can score either 7 or 8 on Day 5. Therefore, the score obtained by Akhil on Day 5 is either 5 or 4.
It is given that Bimal’s scores are the same on Day 1 and Day 3. Hence, the score obtained by Bimal on Day 1 is 5, which implies The score obtained by Akhil is 7 on Day 1.
From Table 2, we can see that the rank of Bimal is 3 on Day 2, and the rank of Akhil is 2 on Day 2. Hence, the score of Bimal will be lower than Akhil's on Day 2.
Let the score of Akhil be a, and the score of Bimal be b. Then 9+a+b = 15, and a > b
=> a+b =6, and a> b
Hence, the value of a can be 4/5, and the value of b can be 2/1
Therefore, the final table is given below:
From the table, we can see that the maximum score is obtained by Chatur.
The correct option is D
Question 3.
What is the minimum possible total score of Bimal?
What is the minimum possible total score of Bimal?
Text Explanation:
Let the total score of day 1, day 2, day 3, day 4, and day 5 are d1, d2, d3, d4, and d5, respectively.
The table shows that d1+d2 = 30 ... eq (1), d2+d3 = 31 ... eq (2), d3+d4 = 32 .... eq(3), d4+d5 = 34 ... eq(4)
It is given that participants are ranked each day, with the person having the maximum score being awarded the minimum rank (1) on that day. All participants with a tied score are awarded the best available rank if there is a tie.
It is given that the total score on Day 3 is the same as the total score on Day 4.
Therefore, d3 = d4 => d3 = d4 = 16, which implies d2 = 15, d5 = 18, and d1 = 15.
The day-wise score is given below:
It is known that Chatur always scores in multiples of 3. His score on Day 2 is the unique highest score in the competition. His minimum score is observed only on Day 1, and it matches Akhil’s score on Day 4.
Hence, only Chatur scored 9 (one time) on Day 2, and no other person scored 9 on any of the given 5 days. Chatur scored 3 only one time, which was on Day 1. Therefore, the scores obtained by Chatur on Day 3, Day 4, and Day 5 are 6, 6, and 6, respectively. It is also known that Akhil's score on Day 4 is the same as the score obtained by Chatur on Day 1. Hence, Akhil's score on Day 4 is 3.
Hence, we get the following table:
From Table 2, we see that the rank of Bimal and Akhil is the same, which is 2. Hence, The score obtained by Akhil and Bimal is the same. Let the score be x. Therefore, 6+2x = 16 => x = 5
The rank of Chatur on Day 5 is 2, and the rank of Bimal is 1, which implies the score obtained by Bimal will be more than Chatur. Hence, Bimal can score either 7 or 8 on Day 5. Therefore, the score obtained by Akhil on Day 5 is either 5 or 4.
It is given that Bimal’s scores are the same on Day 1 and Day 3. Hence, the score obtained by Bimal on Day 1 is 5, which implies The score obtained by Akhil is 7 on Day 1.
From Table 2, we can see that the rank of Bimal is 3 on Day 2, and the rank of Akhil is 2 on Day 2. Hence, the score of Bimal will be lower than Akhil on Day 2.
Let the score of Akhil be a, and the score of Bimal be b. Then 9+a+b = 15, and a > b
=> a+b =6, and a> b
Hence, the value of a can be 4/5, and the value of b can be 2/1
Therefore, the final table is given below:
From the table, we can see that the minimum score obtained by Bimal is 25.
Instructions
An air conditioner (AC) company has four dealers - D1, D2, D3 and D4 in a city. It is evaluating sales performances of these dealers. The company sells two variants of ACs - Window and Split. Both these variants can be either Inverter type or Non-inverter type. It is known that of the total number of ACs sold in the city, 25% were of Window variant, while the rest were of Split variant. Among the Inverter ACs sold, 20% were of Window variant.
The following information is also known:
1. Every dealer sold at least two window ACs.
2. D1 sold 13 inverter ACs, while D3 sold 5 Non-inverter ACs.
3. A total of six Window Non-inverter ACs and 36 Split Inverter ACs were sold in the city.
4. The number of Split ACs sold by D1 was twice the number of Window ACs sold by it.
5. D3 and D4 sold an equal number of Window ACs and this number was one-third of the number of similar ACs sold by D2.
6. D2 and D3 were the only ones who sold Window Non-inverter ACs. The number of these ACs sold by D2 was twice the number of these ACs sold by D3.
7. D3 and D4 sold an equal number of Split Inverter ACs. This number was half the number of similar ACs sold by D2.
An air conditioner (AC) company has four dealers - D1, D2, D3 and D4 in a city. It is evaluating sales performances of these dealers. The company sells two variants of ACs - Window and Split. Both these variants can be either Inverter type or Non-inverter type. It is known that of the total number of ACs sold in the city, 25% were of Window variant, while the rest were of Split variant. Among the Inverter ACs sold, 20% were of Window variant.
The following information is also known:
1. Every dealer sold at least two window ACs.
2. D1 sold 13 inverter ACs, while D3 sold 5 Non-inverter ACs.
3. A total of six Window Non-inverter ACs and 36 Split Inverter ACs were sold in the city.
4. The number of Split ACs sold by D1 was twice the number of Window ACs sold by it.
5. D3 and D4 sold an equal number of Window ACs and this number was one-third of the number of similar ACs sold by D2.
6. D2 and D3 were the only ones who sold Window Non-inverter ACs. The number of these ACs sold by D2 was twice the number of these ACs sold by D3.
7. D3 and D4 sold an equal number of Split Inverter ACs. This number was half the number of similar ACs sold by D2.
Question 4.
How many Split Inverter ACs did D2 sell?
How many Split Inverter ACs did D2 sell?
Text Explanation:
Let us assume, A is the total number of AC's sold=> From the information that the total number of ACs sold in the city, 25% were of Window variant => Window AC's = A/4 and Split AC's = 3A/4Now, let us assume B is the total number of inverter ACs => From the information that among the Inverter ACs sold, 20% were of Window variant.=> Window Inverter AC's = B/5 and Window Non-Inverter AC's = 4B/5
From - Condition-3=> A/4 - B/5 = 6 and 4B/5 = 36 => B = 46 and A = 60.
Now, from condition-6a) D1 & D4 sold "0" window Non-inverter ACs => D2 & D3 sold 6 window non-inverter ACs, it is given that D2 sold twice as many as D3 => D2 sold 4 and D3 sold 2 ACs of this type.From condition-2b) Let us assume, D1 sold "x" window inverter ACs => Number of split inverter ACs sold is 13-xFrom condition-4c) Number of split ACs sold by D1 will be "2x"From condition-5d) Let us assume 'y' is the number of window ACs sold by D3 & D4 => D2 sold 3y ACs of this type.From condition-7e) Let us assume 'z' is the number of split inverter ACs sold by D3 and D4 => D2 sold 2z ACs of this type.Let us use a, b, c, d, and e make a table:
We know that the total number of window ACs is 15=> x + 3y + y + y = 15 => x + 5y = 15, also x and y should be greater than or equal to 2 from condition-1=> x = 5 and y = 2 is the only solution.Filling this in the table:
Now, Number of split inverter ACs is 36=> 8 + 2z + z + z = 36 => 4z = 28 => z = 7.Filling this and using (5), the number of split AC's sold by D1 is 2*5 = 10.
From the table, we see that 14 split inverter ACs are sold.
Question 5.
Which of the following statements is necessarily false?
Which of the following statements is necessarily false?
Text Explanation:
Let us assume, A is the total number of AC's sold=> From the information that the total number of ACs sold in the city, 25% were of Window variant => Window AC's = A/4 and Split AC's = 3A/4Now, let us assume B is the total number of inverter ACs => From the information that among the Inverter ACs sold, 20% were of Window variant.=> Window Inverter AC's = B/5 and Window Non-Inverter AC's = 4B/5From - Condition-3=> A/4 - B/5 = 6 and 4B/5 = 36 => B = 46 and A = 60.Now, from condition-6a) D1 & D4 sold "0" window Non-inverter ACs => D2 & D3 sold 6 window non-inverter ACs, it is given that D2 sold twice as many as D3 => D2 sold 4 and D3 sold 2 ACs of this type.From condition-2b) Let us assume, D1 sold "x" window inverter ACs => Number of split inverter ACs sold is 13-xFrom condition-4c) Number of split ACs sold by D1 will be "2x"From condition-5d) Let us assume 'y' is the number of window ACs sold by D3 & D4 => D2 sold 3y ACs of this type.From condition-7e) Let us assume 'z' is the number of split inverter ACs sold by D3 and D4 => D2 sold 2z ACs of this type.Let us use a, b, c, d, and e make a table:We know that the total number of window ACs is 15=> x + 3y + y + y = 15 => x + 5y = 15, also x and y should be greater than or equal to 2 from condition-1=> x = 5 and y = 2 is the only solution.Filling this in the table:Now, Number of split inverter ACs is 36=> 8 + 2z + z + z = 36 => 4z = 28 => z = 7.Filling this and using (5), the number of split AC's sold by D1 is 2*5 = 10.We see that D1 & D3 sold 27 ACs together which is less than 60 - 27 = 33 sold by D2 & D4.=> Option-D is definitely false.
Question 6.
If D3 and D4 sold an equal number of ACs, then what was the number of Non-inverter ACs sold by D2?
If D3 and D4 sold an equal number of ACs, then what was the number of Non-inverter ACs sold by D2?
Text Explanation:
Let us assume, A is the total number of AC's sold=> From the information that the total number of ACs sold in the city, 25% were of Window variant => Window AC's = A/4 and Split AC's = 3A/4Now, let us assume B is the total number of inverter ACs => From the information that among the Inverter ACs sold, 20% were of Window variant.=> Window Inverter AC's = B/5 and Window Non-Inverter AC's = 4B/5From - Condition-3=> A/4 - B/5 = 6 and 4B/5 = 36 => B = 46 and A = 60.Now, from condition-6a) D1 & D4 sold "0" window Non-inverter ACs => D2 & D3 sold 6 window non-inverter ACs, it is given that D2 sold twice as many as D3 => D2 sold 4 and D3 sold 2 ACs of this type.From condition-2b) Let us assume, D1 sold "x" window inverter ACs => Number of split inverter ACs sold is 13-xFrom condition-4c) Number of split ACs sold by D1 will be "2x"From condition-5d) Let us assume 'y' is the number of window ACs sold by D3 & D4 => D2 sold 3y ACs of this type.From condition-7e) Let us assume 'z' is the number of split inverter ACs sold by D3 and D4 => D2 sold 2z ACs of this type.Let us use a, b, c, d, and e make a table:We know that the total number of window ACs is 15=> x + 3y + y + y = 15 => x + 5y = 15, also x and y should be greater than or equal to 2 from condition-1=> x = 5 and y = 2 is the only solution.Filling this in the table:Now, Number of split inverter ACs is 36=> 8 + 2z + z + z = 36 => 4z = 28 => z = 7.Filling this and using (5), the number of split AC's sold by D1 is 2*5 = 10.If D3 and D4 sold equal number of AC's, the table will look as follows:
Number of non-inverter ACs sold is 1 + 4 = 5
Instructions
A train travels from Station A to Station E, passing through stations B, C, and D in that order. The train has a seating capacity of 200.
A ticket may be booked from any station to any other station ahead on the route, but not to any earlier station. A ticket from one station to another reserves one seat on every intermediate segment of the route. For example, a ticket from B to E reserves a seat in the intermediate segments B–C, C–D, and D–E.
The occupancy factor for a segment is the total number of seats reserved in the segment as a percentage of the seating capacity. The total number of seats reserved for any segment cannot exceed 200.
The following information is known:
- The occupancy factor for segment C–D was 95%, and only segment B–C had a higher occupancy factor.
- Exactly 40 tickets were booked from B to C, and 30 tickets were booked from B to E.
- Among the seats reserved on segment D–E, exactly four-sevenths were from stations before C.
- The number of tickets booked from A to C was equal to the number booked from A to E, and this number was higher than the number booked from B to E.
- No tickets were booked from A to B, from B to D, and from D to E.
- The number of tickets booked for any segment was a multiple of 10.
A train travels from Station A to Station E, passing through stations B, C, and D in that order. The train has a seating capacity of 200.
A ticket may be booked from any station to any other station ahead on the route, but not to any earlier station. A ticket from one station to another reserves one seat on every intermediate segment of the route. For example, a ticket from B to E reserves a seat in the intermediate segments B–C, C–D, and D–E.
The occupancy factor for a segment is the total number of seats reserved in the segment as a percentage of the seating capacity. The total number of seats reserved for any segment cannot exceed 200.
The following information is known:
- The occupancy factor for segment C–D was 95%, and only segment B–C had a higher occupancy factor.
- Exactly 40 tickets were booked from B to C, and 30 tickets were booked from B to E.
- Among the seats reserved on segment D–E, exactly four-sevenths were from stations before C.
- The number of tickets booked from A to C was equal to the number booked from A to E, and this number was higher than the number booked from B to E.
- No tickets were booked from A to B, from B to D, and from D to E.
- The number of tickets booked for any segment was a multiple of 10.
Question 7.
What was the occupancy factor for segment D - E?
What was the occupancy factor for segment D - E?
Text Explanation:
Putting all the given information in the table, we get,
From the above table, the tickets in the segments A - B, B - C, C - D and D - E can be calculated as,A - B $=a+b+c=2a+b$
B - C $=a+b+c+30+40=2a+b+70$
C - D $=a+b+x+y+30$
D - E $=c+y+30=a+y+30$
We are given that segment C - D had an occupancy factor of 95%. We can calculate the number of seats occupied in the segment C - D as,
Seats occupied in segment C - D $=\dfrac{95}{100}\times\ 200\ =\ 190$
We are given that the B - C segment had more occupancy than the C - D segment. We are also given that a, b, c, x and y are multiples of 10 as per clue 6.
We also know that the people on board in a segment cannot be more than 200. The only value greater than 190 that is a multiple of 10 is 200. So, B - C segment had 200 seats occupied.
Equating it with the above equation, we get,
$2a+b+70=200$
$2a+b=130$ --(1)
$a+b+x+y+30 = 190$
$a+b+x+y=160$ --(2)
We are also given that among the seats reserved on segment D - E, exactly four-sevenths were from stations before C. The seats reserved on segment D - E before station C are a + 30.
Equating the expressions, we get,
$\dfrac{4}{7}\left(a\ +\ y\ +\ 30\right)\ =\ a\ +\ 30$
$a\ +\ y\ +\ 30\ =\ \dfrac{7}{4}\left(a\ +\ 30\right)$ ---(3)
So, the seats occupied during D - E $=a\ +\ y\ +\ 30\ =\ \dfrac{7}{4}\left(a\ +\ 30\right)$
We know that this value has to be an integer.
We are given that the number of tickets booked from A to C was higher than that from B to E. This means that the value of a > 30.
We know that a and b are positive multiples of 10, so the possible values of a that are greater than 30 and satisfy the equation (1) are 40, 50 and 60.
The only value of a at which the equation (3) is an integer is when $a+30$ is a multiple of 4, and the only value out of the above values that satisfies the condition is when $a=50$, as at $a=40$ and 60, the value of $a+30$ is not a multiple of 4.
We can conclude that the value of $a=50$, and substituting in (1), we get the value of $b=30$.
Substituting in (3), we get,
$50\ +\ y\ +\ 30\ =\ \frac{7}{4}\left(50\ +\ 30\right)$
$\ y\ +\ 80\ =\ 140$
$\ y\ =\ 60$
Substituting all the values in (2), we get,
$50\ +\ 30\ +\ x\ +\ 60\ =\ 160$
$x\ =\ 20$
Putting all the values in the table, we get,
A - B $=a+b+c=2a+b=130$B - C $=a+b+c+30+40=2a+b+70=200$
C - D $=a+b+x+y+30=190$
D - E $=c+y+30=a+y+30=140$
Occupancy factor of segment D - E can be calculated as,
Occupancy factor $=\dfrac{140}{200}\times\ 100\ =\ 70\%$
Hence, the correct answer is option B.
Question 8.
How many tickets were booked from Station A to Station E?
How many tickets were booked from Station A to Station E?
Text Explanation:
Putting all the given information in the table, we get,
From the above table, the tickets in the segments A - B, B - C, C - D and D - E can be calculated as,A - B $=a+b+c=2a+b$
B - C $=a+b+c+30+40=2a+b+70$
C - D $=a+b+x+y+30$
D - E $=c+y+30=a+y+30$
We are given that segment C - D had an occupancy factor of 95%. We can calculate the number of seats occupied in the segment C - D as,
Seats occupied in segment C - D $=\dfrac{95}{100}\times\ 200\ =\ 190$
We are given that the B - C segment had more occupancy than the C - D segment. We are also given that a, b, c, x and y are multiples of 10 as per clue 6.
We also know that the people on board in a segment cannot be more than 200. The only value greater than 190 that is a multiple of 10 is 200. So, B - C segment had 200 seats occupied.
Equating it with the above equation, we get,
$2a+b+70=200$
$2a+b=130$ --(1)
$a+b+x+y+30 = 190$
$a+b+x+y=160$ --(2)
We are also given that among the seats reserved on segment D - E, exactly four-sevenths were from stations before C. The seats reserved on segment D - E before station C are a + 30.
Equating the expressions, we get,
$\dfrac{4}{7}\left(a\ +\ y\ +\ 30\right)\ =\ a\ +\ 30$
$a\ +\ y\ +\ 30\ =\ \dfrac{7}{4}\left(a\ +\ 30\right)$ ---(3)
So, the seats occupied during D - E $=a\ +\ y\ +\ 30\ =\ \dfrac{7}{4}\left(a\ +\ 30\right)$
We know that this value has to be an integer.
We are given that the number of tickets booked from A to C was higher than that from B to E. This means that the value of a > 30.
We know that a and b are positive multiples of 10, so the possible values of a that are greater than 30 and satisfy the equation (1) are 40, 50 and 60.
The only value of a at which the equation (3) is an integer is when $a+30$ is a multiple of 4, and the only value out of the above values that satisfies the condition is when $a=50$, as at $a=40$ and 60, the value of $a+30$ is not a multiple of 4.
We can conclude that the value of $a=50$, and substituting in (1), we get the value of $b=30$.
Substituting in (3), we get,
$50\ +\ y\ +\ 30\ =\ \frac{7}{4}\left(50\ +\ 30\right)$
$\ y\ +\ 80\ =\ 140$
$\ y\ =\ 60$
Substituting all the values in (2), we get,
$50\ +\ 30\ +\ x\ +\ 60\ =\ 160$
$x\ =\ 20$
Putting all the values in the table, we get,
A - B $=a+b+c=2a+b=130$B - C $=a+b+c+30+40=2a+b+70=200$
C - D $=a+b+x+y+30=190$
D - E $=c+y+30=a+y+30=140$
The number of tickets booked from station A to E from the above table is 50.
Hence, the correct answer is 50.
Question 9.
How many tickets were booked from Station C?
How many tickets were booked from Station C?
Text Explanation:
Putting all the given information in the table, we get,
From the above table, the tickets in the segments A - B, B - C, C - D and D - E can be calculated as,A - B $=a+b+c=2a+b$
B - C $=a+b+c+30+40=2a+b+70$
C - D $=a+b+x+y+30$
D - E $=c+y+30=a+y+30$
We are given that segment C - D had an occupancy factor of 95%. We can calculate the number of seats occupied in the segment C - D as,
Seats occupied in segment C - D $=\dfrac{95}{100}\times\ 200\ =\ 190$
We are given that the B - C segment had more occupancy than the C - D segment. We are also given that a, b, c, x and y are multiples of 10 as per clue 6.
We also know that the people on board in a segment cannot be more than 200. The only value greater than 190 that is a multiple of 10 is 200. So, B - C segment had 200 seats occupied.
Equating it with the above equation, we get,
$2a+b+70=200$
$2a+b=130$ --(1)
$a+b+x+y+30 = 190$
$a+b+x+y=160$ --(2)
We are also given that among the seats reserved on segment D - E, exactly four-sevenths were from stations before C. The seats reserved on segment D - E before station C are a + 30.
Equating the expressions, we get,
$\dfrac{4}{7}\left(a\ +\ y\ +\ 30\right)\ =\ a\ +\ 30$
$a\ +\ y\ +\ 30\ =\ \dfrac{7}{4}\left(a\ +\ 30\right)$ ---(3)
So, the seats occupied during D - E $=a\ +\ y\ +\ 30\ =\ \dfrac{7}{4}\left(a\ +\ 30\right)$
We know that this value has to be an integer.
We are given that the number of tickets booked from A to C was higher than that from B to E. This means that the value of a > 30.
We know that a and b are positive multiples of 10, so the possible values of a that are greater than 30 and satisfy the equation (1) are 40, 50 and 60.
The only value of a at which the equation (3) is an integer is when $a+30$ is a multiple of 4, and the only value out of the above values that satisfies the condition is when $a=50$, as at $a=40$ and 60, the value of $a+30$ is not a multiple of 4.
We can conclude that the value of $a=50$, and substituting in (1), we get the value of $b=30$.
Substituting in (3), we get,
$50\ +\ y\ +\ 30\ =\ \frac{7}{4}\left(50\ +\ 30\right)$
$\ y\ +\ 80\ =\ 140$
$\ y\ =\ 60$
Substituting all the values in (2), we get,
$50\ +\ 30\ +\ x\ +\ 60\ =\ 160$
$x\ =\ 20$
Putting all the values in the table, we get,
A - B $=a+b+c=2a+b=130$B - C $=a+b+c+30+40=2a+b+70=200$
C - D $=a+b+x+y+30=190$
D - E $=c+y+30=a+y+30=140$
Number of tickets booked from station C = 20 + 60 = 80.
Hence, the correct answer is 80.
Question 10.
What is the difference between the number of tickets booked to Station C and the number of tickets booked to Station D?
What is the difference between the number of tickets booked to Station C and the number of tickets booked to Station D?
Text Explanation:
Putting all the given information in the table, we get,
From the above table, the tickets in the segments A - B, B - C, C - D and D - E can be calculated as,A - B $=a+b+c=2a+b$
B - C $=a+b+c+30+40=2a+b+70$
C - D $=a+b+x+y+30$
D - E $=c+y+30=a+y+30$
We are given that segment C - D had an occupancy factor of 95%. We can calculate the number of seats occupied in the segment C - D as,
Seats occupied in segment C - D $=\dfrac{95}{100}\times\ 200\ =\ 190$
We are given that the B - C segment had more occupancy than the C - D segment. We are also given that a, b, c, x and y are multiples of 10 as per clue 6.
We also know that the people on board in a segment cannot be more than 200. The only value greater than 190 that is a multiple of 10 is 200. So, B - C segment had 200 seats occupied.
Equating it with the above equation, we get,
$2a+b+70=200$
$2a+b=130$ --(1)
$a+b+x+y+30 = 190$
$a+b+x+y=160$ --(2)
We are also given that among the seats reserved on segment D - E, exactly four-sevenths were from stations before C. The seats reserved on segment D - E before station C are a + 30.
Equating the expressions, we get,
$\dfrac{4}{7}\left(a\ +\ y\ +\ 30\right)\ =\ a\ +\ 30$
$a\ +\ y\ +\ 30\ =\ \dfrac{7}{4}\left(a\ +\ 30\right)$ ---(3)
So, the seats occupied during D - E $=a\ +\ y\ +\ 30\ =\ \dfrac{7}{4}\left(a\ +\ 30\right)$
We know that this value has to be an integer.
We are given that the number of tickets booked from A to C was higher than that from B to E. This means that the value of a > 30.
We know that a and b are positive multiples of 10, so the possible values of a that are greater than 30 and satisfy the equation (1) are 40, 50 and 60.
The only value of a at which the equation (3) is an integer is when $a+30$ is a multiple of 4, and the only value out of the above values that satisfies the condition is when $a=50$, as at $a=40$ and 60, the value of $a+30$ is not a multiple of 4.
We can conclude that the value of $a=50$, and substituting in (1), we get the value of $b=30$.
Substituting in (3), we get,
$50\ +\ y\ +\ 30\ =\ \frac{7}{4}\left(50\ +\ 30\right)$
$\ y\ +\ 80\ =\ 140$
$\ y\ =\ 60$
Substituting all the values in (2), we get,
$50\ +\ 30\ +\ x\ +\ 60\ =\ 160$
$x\ =\ 20$
Putting all the values in the table, we get,
A - B $=a+b+c=2a+b=130$B - C $=a+b+c+30+40=2a+b+70=200$
C - D $=a+b+x+y+30=190$
D - E $=c+y+30=a+y+30=140$
The number of tickets booked to station C = 50 + 40 = 90
The number of tickets booked to station D = 30 + 20 = 50
Difference = 90 - 50 = 40.
Hence, the correct answer is 40.
Question 11.
How many tickets were booked to travel in exactly one segment?
How many tickets were booked to travel in exactly one segment?
Text Explanation:
Putting all the given information in the table, we get,
From the above table, the tickets in the segments A - B, B - C, C - D and D - E can be calculated as,A - B $=a+b+c=2a+b$
B - C $=a+b+c+30+40=2a+b+70$
C - D $=a+b+x+y+30$
D - E $=c+y+30=a+y+30$
We are given that segment C - D had an occupancy factor of 95%. We can calculate the number of seats occupied in the segment C - D as,
Seats occupied in segment C - D $=\dfrac{95}{100}\times\ 200\ =\ 190$
We are given that the B - C segment had more occupancy than the C - D segment. We are also given that a, b, c, x and y are multiples of 10 as per clue 6.
We also know that the people on board in a segment cannot be more than 200. The only value greater than 190 that is a multiple of 10 is 200. So, B - C segment had 200 seats occupied.
Equating it with the above equation, we get,
$2a+b+70=200$
$2a+b=130$ --(1)
$a+b+x+y+30 = 190$
$a+b+x+y=160$ --(2)
We are also given that among the seats reserved on segment D - E, exactly four-sevenths were from stations before C. The seats reserved on segment D - E before station C are a + 30.
Equating the expressions, we get,
$\dfrac{4}{7}\left(a\ +\ y\ +\ 30\right)\ =\ a\ +\ 30$
$a\ +\ y\ +\ 30\ =\ \dfrac{7}{4}\left(a\ +\ 30\right)$ ---(3)
So, the seats occupied during D - E $=a\ +\ y\ +\ 30\ =\ \dfrac{7}{4}\left(a\ +\ 30\right)$
We know that this value has to be an integer.
We are given that the number of tickets booked from A to C was higher than that from B to E. This means that the value of a > 30.
We know that a and b are positive multiples of 10, so the possible values of a that are greater than 30 and satisfy the equation (1) are 40, 50 and 60.
The only value of a at which the equation (3) is an integer is when $a+30$ is a multiple of 4, and the only value out of the above values that satisfies the condition is when $a=50$, as at $a=40$ and 60, the value of $a+30$ is not a multiple of 4.
We can conclude that the value of $a=50$, and substituting in (1), we get the value of $b=30$.
Substituting in (3), we get,
$50\ +\ y\ +\ 30\ =\ \frac{7}{4}\left(50\ +\ 30\right)$
$\ y\ +\ 80\ =\ 140$
$\ y\ =\ 60$
Substituting all the values in (2), we get,
$50\ +\ 30\ +\ x\ +\ 60\ =\ 160$
$x\ =\ 20$
Putting all the values in the table, we get,
A - B $=a+b+c=2a+b=130$B - C $=a+b+c+30+40=2a+b+70=200$
C - D $=a+b+x+y+30=190$
D - E $=c+y+30=a+y+30=140$
The number of tickets booked for exactly one segment can be calculated using values from the above table as,
Tickets booked for exactly one segment = A - B + B - C + C - D + D - E = 0 + 40 + 20 + 0 = 60.
Hence, the correct answer is 60.
Instructions
The charts depict details of research papers written by four authors, Arman, Brajen, Chintan, and Devon. The papers were of four types, single-author, two-author, three-author, and four-author, that is, written by one, two, three, or all four of these authors, respectively. No other authors were involved in writing these papers.
The following additional facts are known:
1. Each of the authors wrote at least one of each of the four types of papers.
2. The four authors wrote different numbers of single-author papers.
3. Both Chintan and Devon wrote more three-author papers than Brajen.
4. The number of single-author and two-author papers written by Brajen were the same.
The charts depict details of research papers written by four authors, Arman, Brajen, Chintan, and Devon. The papers were of four types, single-author, two-author, three-author, and four-author, that is, written by one, two, three, or all four of these authors, respectively. No other authors were involved in writing these papers.
The following additional facts are known:
1. Each of the authors wrote at least one of each of the four types of papers.
2. The four authors wrote different numbers of single-author papers.
3. Both Chintan and Devon wrote more three-author papers than Brajen.
4. The number of single-author and two-author papers written by Brajen were the same.
Question 12.
What was the total number of two-author and three-author papers written by Brajen?
What was the total number of two-author and three-author papers written by Brajen?
Text Explanation:
If all the three-author-type books are counted for all three authors separately, then we get the sum to be 3 * 3 = 9.
If all the four-author-type books are counted for all four authors separately, then we get the sum to be 4 * 2 = 8.
We are given that there are 2 four-author books, so there are exactly two books for each that are of the four-author type.
Putting the values in the table, we get,
We are given that each author has at least one book of each type. In the case of Arman, there are three books left to be assigned, and there are 3 types of books left, so each of them has to be equal to 1 for the above condition to be satisfied.In the case of Brajen, we are told that he had an equal number of single-author and two-author books.
Case 1: If the equal number is 1, then the three-author type books become 4, and we are given that both Chintan and Devon had more three-author books than Brajen, which is not possible in this case, as both of them combined will be left with 5 books together and whaterver we split 5 books there wont be a case of both of them having more books than Brajen.
Case 2: If the equal number is 2, then the three-author type books become 2, and we are given that both Chintan and Devon had more three-author books than Brajen, which is possible in this case, as both of them combined will be left with 6 books together and the only way to split 6 books such that both of them are greater than 2 is to split them into 3, 3 and it is the only possible way.
Case 3: If the equal number is greater than or equal to 3, then the three-author type books become 0 or negative, which can directly be eliminated, as we are given that all authors wrote at least 1 book of each type.
So, we can eliminate both cases 1 and 3, and we will be left with the case of Brajen having 2 books each of Single-author, Two-author, and Three-author types. Chintan and Devon have three books each of the three-author type.
Filling up the table with these values, we get,
We are given that each author has a different number of single-author type books. The only possibilities for the single-author books of Chintan and Devon are 3 and 4 in some order. If the single-author books of Chintan are 3, then the double-author books of Chintan become 4, the single-author books of Devon become 4, and the double-author books of Devon become 1.
If the single-author books of Chintan are 4, then the double-author books of Chintan become 3, the single-author books of Devon become 3, and the double-author books of Devon become 2.
Placing both possibilities in the table, we get,
The total number of two-author and three-author papers written by Brajen = 2 + 2 = 4.Hence, the correct answer is 4.
Question 13.
Which of the following statements is/are NECESSARILY true?
i. Chintan wrote exactly three two-author papers.
ii. Chintan wrote more single-author papers than Devon.
Which of the following statements is/are NECESSARILY true?
i. Chintan wrote exactly three two-author papers.
ii. Chintan wrote more single-author papers than Devon.
Text Explanation:
If all the three-author-type books are counted for all three authors separately, then we get the sum to be 3 * 3 = 9.
If all the four-author-type books are counted for all four authors separately, then we get the sum to be 4 * 2 = 8.
We are given that there are 2 four-author books, so there are exactly two books for each that are of the four-author type.
Putting the values in the table, we get,
We are given that each author has at least one book of each type. In the case of Arman, there are three books left to be assigned, and there are 3 types of books left, so each of them has to be equal to 1 for the above condition to be satisfied.In the case of Brajen, we are told that he had an equal number of single-author and two-author books.
Case 1: If the equal number is 1, then the three-author type books become 4, and we are given that both Chintan and Devon had more three-author books than Brajen, which is not possible in this case, as both of them combined will be left with 5 books together and whaterver we split 5 books there wont be a case of both of them having more books than Brajen.
Case 2: If the equal number is 2, then the three-author type books become 2, and we are given that both Chintan and Devon had more three-author books than Brajen, which is possible in this case, as both of them combined will be left with 6 books together and the only way to split 6 books such that both of them are greater than 2 is to split them into 3, 3 and it is the only possible way.
Case 3: If the equal number is greater than or equal to 3, then the three-author type books become 0 or negative, which can directly be eliminated, as we are given that all authors wrote at least 1 book of each type.
So, we can eliminate both cases 1 and 3, and we will be left with the case of Brajen having 2 books each of Single-author, Two-author, and Three-author types. Chintan and Devon have three books each of the three-author type.
Filling up the table with these values, we get,
We are given that each author has a different number of single-author type books. The only possibilities for the single-author books of Chintan and Devon are 3 and 4 in some order. If the single-author books of Chintan are 3, then the double-author books of Chintan become 4, the single-author books of Devon become 4, and the double-author books of Devon become 1.
If the single-author books of Chintan are 4, then the double-author books of Chintan become 3, the single-author books of Devon become 3, and the double-author books of Devon become 2.
Placing both possibilities in the table, we get,
Statement 1: Chintan wrote exactly three two-author papers.This is not necessarily true, as he could have written four books as well.
Statement 2: Chintan wrote more single-author papers than Devon.
This is not necessarily true, as the opposite can be possible.
So, neither I nor 2 is necessarily true.
Hence, the correct answer is option A.
Question 14.
Which of the following statements is/are NECESSARILY true?
i. Arman wrote three-author papers only with Chintan and Devon.
ii. Brajen wrote three-author papers only with Chintan and Devon.
Which of the following statements is/are NECESSARILY true?
i. Arman wrote three-author papers only with Chintan and Devon.
ii. Brajen wrote three-author papers only with Chintan and Devon.
Text Explanation:
If all the three-author-type books are counted for all three authors separately, then we get the sum to be 3 * 3 = 9.
If all the four-author-type books are counted for all four authors separately, then we get the sum to be 4 * 2 = 8.
We are given that there are 2 four-author books, so there are exactly two books for each that are of the four-author type.
Putting the values in the table, we get,
We are given that each author has at least one book of each type. In the case of Arman, there are three books left to be assigned, and there are 3 types of books left, so each of them has to be equal to 1 for the above condition to be satisfied.In the case of Brajen, we are told that he had an equal number of single-author and two-author books.
Case 1: If the equal number is 1, then the three-author type books become 4, and we are given that both Chintan and Devon had more three-author books than Brajen, which is not possible in this case, as both of them combined will be left with 5 books together and whaterver we split 5 books there wont be a case of both of them having more books than Brajen.
Case 2: If the equal number is 2, then the three-author type books become 2, and we are given that both Chintan and Devon had more three-author books than Brajen, which is possible in this case, as both of them combined will be left with 6 books together and the only way to split 6 books such that both of them are greater than 2 is to split them into 3, 3 and it is the only possible way.
Case 3: If the equal number is greater than or equal to 3, then the three-author type books become 0 or negative, which can directly be eliminated, as we are given that all authors wrote at least 1 book of each type.
So, we can eliminate both cases 1 and 3, and we will be left with the case of Brajen having 2 books each of Single-author, Two-author, and Three-author types. Chintan and Devon have three books each of the three-author type.
Filling up the table with these values, we get,
We are given that each author has a different number of single-author type books. The only possibilities for the single-author books of Chintan and Devon are 3 and 4 in some order.If the single-author books of Chintan are 3, then the double-author books of Chintan become 4, the single-author books of Devon become 4, and the double-author books of Devon become 1.
If the single-author books of Chintan are 4, then the double-author books of Chintan become 3, the single-author books of Devon become 3, and the double-author books of Devon become 2.
Placing both possibilities in the table, we get,
Statement 1: Arman wrote three-author papers only with Chintan and Devon.There are in total 3 three-author books, and Chintan and Devon are authors of all three of them according to the table. So, Arman wrote the three-author papers only with Chintan and Devon. So, the statement is necessarily true
Statement 2: Brajen wrote three-author papers only with Chintan and Devon.
There are in total 3 three-author books, and Chintan and Devon are authors of all three of them according to the table. So, Brajen wrote the three-author papers only with Chintan and Devon. So, the statement is necessarily true.
So, both I and 2 are necessarily true.
Hence, the correct answer is option B.
Question 15.
If Devon wrote more than one two-author papers, then how many two-author papers did Chintan write?
If Devon wrote more than one two-author papers, then how many two-author papers did Chintan write?
Text Explanation:
If all the three-author-type books are counted for all three authors separately, then we get the sum to be 3 * 3 = 9.
If all the four-author-type books are counted for all four authors separately, then we get the sum to be 4 * 2 = 8.
We are given that there are 2 four-author books, so there are exactly two books for each that are of the four-author type.
Putting the values in the table, we get,
We are given that each author has at least one book of each type. In the case of Arman, there are three books left to be assigned, and there are 3 types of books left, so each of them has to be equal to 1 for the above condition to be satisfied.In the case of Brajen, we are told that he had an equal number of single-author and two-author books.
Case 1: If the equal number is 1, then the three-author type books become 4, and we are given that both Chintan and Devon had more three-author books than Brajen, which is not possible in this case, as both of them combined will be left with 5 books together and whaterver we split 5 books there wont be a case of both of them having more books than Brajen.
Case 2: If the equal number is 2, then the three-author type books become 2, and we are given that both Chintan and Devon had more three-author books than Brajen, which is possible in this case, as both of them combined will be left with 6 books together and the only way to split 6 books such that both of them are greater than 2 is to split them into 3, 3 and it is the only possible way.
Case 3: If the equal number is greater than or equal to 3, then the three-author type books become 0 or negative, which can directly be eliminated, as we are given that all authors wrote at least 1 book of each type.
So, we can eliminate both cases 1 and 3, and we will be left with the case of Brajen having 2 books each of Single-author, Two-author, and Three-author types. Chintan and Devon have three books each of the three-author type.
Filling up the table with these values, we get,
We are given that each author has a different number of single-author type books. The only possibilities for the single-author books of Chintan and Devon are 3 and 4 in some order.If the single-author books of Chintan are 3, then the double-author books of Chintan become 4, the single-author books of Devon become 4, and the double-author books of Devon become 1.
If the single-author books of Chintan are 4, then the double-author books of Chintan become 3, the single-author books of Devon become 3, and the double-author books of Devon become 2.
Placing both possibilities in the table, we get,
If Devon wrote more than one two-author paper, he would have written 2 papers, making the number of two-author papers written by Chintan 3.Hence, the correct answer is 3.
Instructions
Aurevia, Brelosia, Cyrenia and Zerathania are four countries with their currencies being Aurels, Brins, Crowns, and Zentars, respectively. The currencies have different exchange values. Crown's currency exchange rate with Zentars $ = 0.5$, i.e., 1 Crown is worth 0.5 Zentars.
Three travelers, Jano, Kira, and Lian set out from Zerathania visiting exactly two of the countries. Each country is visited by exactly two travelers. Each traveler has a unique Flight Cost, which represents the total cost of airfare in traveling to both the countries and back to Zerathania. The Flight Cost of Jano was 4000 Zentars, while that of the other two travelers were 5000 and 6000 Zentars, not necessarily in that order.
When visiting a country, a traveler spent either 1000, 2000 or 3000 in the country's local currency. Each traveler had different spends (in the country's local currency) in the two countries he/she visited. Across all the visits, there were exactly two spends of 1000 and exactly one spend of 3000 (in the country's local currency).
The total "Travel Cost" for a traveler is the sum of his/her Flight Cost and the money spent in the countries visited.
The citizens of the four countries with knowledge of these travels made a few observations, with spends measured in their respective local currencies:
i. Aurevia citizen: Jano and Kira visited our country, and their Travel Costs were 3500 and 8000, respectively.
ii. Brelosia citizen: Kira and Lian visited our country, spending 2000 and 3000, respectively. Kira's Travel Cost was 4000.
iii. Cyrenia citizen: Lian visited our country and her Travel Cost was 36000.
Aurevia, Brelosia, Cyrenia and Zerathania are four countries with their currencies being Aurels, Brins, Crowns, and Zentars, respectively. The currencies have different exchange values. Crown's currency exchange rate with Zentars $ = 0.5$, i.e., 1 Crown is worth 0.5 Zentars.
Three travelers, Jano, Kira, and Lian set out from Zerathania visiting exactly two of the countries. Each country is visited by exactly two travelers. Each traveler has a unique Flight Cost, which represents the total cost of airfare in traveling to both the countries and back to Zerathania. The Flight Cost of Jano was 4000 Zentars, while that of the other two travelers were 5000 and 6000 Zentars, not necessarily in that order.
When visiting a country, a traveler spent either 1000, 2000 or 3000 in the country's local currency. Each traveler had different spends (in the country's local currency) in the two countries he/she visited. Across all the visits, there were exactly two spends of 1000 and exactly one spend of 3000 (in the country's local currency).
The total "Travel Cost" for a traveler is the sum of his/her Flight Cost and the money spent in the countries visited.
The citizens of the four countries with knowledge of these travels made a few observations, with spends measured in their respective local currencies:
i. Aurevia citizen: Jano and Kira visited our country, and their Travel Costs were 3500 and 8000, respectively.
ii. Brelosia citizen: Kira and Lian visited our country, spending 2000 and 3000, respectively. Kira's Travel Cost was 4000.
iii. Cyrenia citizen: Lian visited our country and her Travel Cost was 36000.
Question 16.
What is the sum of Travel Costs for all travelers in Zentars?
What is the sum of Travel Costs for all travelers in Zentars?
Text Explanation:
We are given the travel cost of Jano to be 3500A in Statement 1.
We are also given the travel cost of Kira to be 8000A in Statement 1 and 4000B in Statement 2.
We are also given the travel cost of Lian to be 36000C in statement 3.
We know that the travel cost has to be constant throughout, so by equating the travel costs of Kira, we get,
8000A = 4000B
B = 2A ---(1)
We are also given the value of C to be 0.5 Z.
C = 0.5Z ---(2)
So, the travel cost of Lian = 36000C = 36000*0.5Z = 18000Z
Let us put the known information in the table, and we get,
We are given that the spending cost of any individual is different in different cities.We are given that the spending amounts are amongst {1000, 2000, 3000} in their local currencies and also told that there were two spends of 1000 and one spend of 3000, which makes the total spending amount 2000, to be 3 because there are 6 spending amounts in total.
One 2000 and one 3000 are already assigned in the above table, so we will be left with two spends of 1000 and two spends of 2000 to be assigned.
Kira has already spent 2000, so the only amount that she must have spent is 1000, as she cannot spend 2000 twice in different countries.
Lian has already spent 3000, so the only amount that she must have spent is 2000. If she spent 1000, then Jano will be left with two 2000s, which is not possible.
We can also conclude that Jano spend 1000 and 2000 in some order in the two countries.
The flight costs are given as 4000Z, 5000Z, and 6000Z, out of which 4000Z is for Jano and 5000Z and 6000Z are for Kira and Lian, in some order.
Filling up the table with these values, we get,
Travel Cost = Spending Cost + Flight CostFor Kira,
Travel Cost = 8000A
Spending Cost = 1000A + 4000A = 5000A
Flight Cost = 5000Z/6000Z
CASE 1: Flight cost of Kira = 5000Z and Flight cost of Lian = 6000Z
If we assume the Flight cost to be 5000Z, then we get,
8000A = 5000A + 5000Z
3000A = 5000Z
A = $\frac{5}{3}$Z
For Lian,
Travel Cost = 18000Z
Spending Cost = 6000A + 1000Z = 6000$\times\ \frac{5}{3}$Z + 1000Z = 11000Z
Flight Cost = Travel Cost - Spending Cost = 18000Z - 11000Z = 7000Z
But in our assumption, the Flight Cost of Lian is 6000Z, which does not match the above answer.
So, we can eliminate this case.
CASE 2: Flight cost of Kira = 6000Z and Flight cost of Lian = 5000Z
If we assume the Flight cost to be 6000Z, then we get,
8000A = 5000A + 6000Z
3000A = 6000Z
A = 2Z
For Lian,
Travel Cost = 18000Z
Spending Cost = 6000A + 1000Z = 6000*2Z + 1000Z = 13000Z
Flight Cost = Travel Cost - Spending Cost = 18000Z - 13000Z = 5000Z
In our assumption, the Flight Cost of Lian is 5000Z, which matches the above answer.
So, Flight cost of Kira = 6000Z, Flight cost of Lian = 5000Z and A = 2Z.
Filling the table with calculated values all in the currency of Z, we get,
In the case of Zano,Spending Cost = Travel Cost - Flight Cost = 7000Z - 4000Z = 3000Z
In the first case, the Spending Cost = 2000Z + 1000Z = 3000Z
In the second case, the Spending Cost = 4000Z + 500Z = 4500Z
We obtained a spending cost of 3000Z only in the first case, so we can eliminate the second case.
The final table looks like,
The total Travel Cost = 7000Z + 16000Z + 18000Z = 41000ZHence, the correct answer is 41000.
Question 17.
How many Zentars did Lian spend in the two countries he visited?
How many Zentars did Lian spend in the two countries he visited?
Text Explanation:
We are given the travel cost of Jano to be 3500A in Statement 1.
We are also given the travel cost of Kira to be 8000A in Statement 1 and 4000B in Statement 2.
We are also given the travel cost of Lian to be 36000C in statement 3.
We know that the travel cost has to be constant throughout, so by equating the travel costs of Kira, we get,
8000A = 4000B
B = 2A ---(1)
We are also given the value of C to be 0.5 Z.
C = 0.5Z ---(2)
So, the travel cost of Lian = 36000C = 36000*0.5Z = 18000Z
Let us put the known information in the table, and we get,
We are given that the spending cost of any individual is different in different cities.We are given that the spending amounts are amongst {1000, 2000, 3000} in their local currencies and also told that there were two spends of 1000 and one spend of 3000, which makes the total spending amount 2000, to be 3 because there are 6 spending amounts in total.
One 2000 and one 3000 are already assigned in the above table, so we will be left with two spends of 1000 and two spends of 2000 to be assigned.
Kira has already spent 2000, so the only amount that she must have spent is 1000, as she cannot spend 2000 twice in different countries.
Lian has already spent 3000, so the only amount that she must have spent is 2000. If she spent 1000, then Jano will be left with two 2000s, which is not possible.
We can also conclude that Jano spent 1000 and 2000 in some order in the two countries.
The flight costs are given as 4000Z, 5000Z, and 6000Z, out of which 4000Z is for Jano and 5000Z and 6000Z are for Kira and Lian, in some order.
Filling up the table with these values, we get,
Travel Cost = Spending Cost + Flight CostFor Kira,
Travel Cost = 8000A
Spending Cost = 1000A + 4000A = 5000A
Flight Cost = 5000Z/6000Z
CASE 1: Flight cost of Kira = 5000Z and Flight cost of Lian = 6000Z
If we assume the Flight cost to be 5000Z, then we get,
8000A = 5000A + 5000Z
3000A = 5000Z
A = $\frac{5}{3}$Z
For Lian,
Travel Cost = 18000Z
Spending Cost = 6000A + 1000Z = 6000$\times\ \frac{5}{3}$Z + 1000Z = 11000Z
Flight Cost = Travel Cost - Spending Cost = 18000Z - 11000Z = 7000Z
But in our assumption, the Flight Cost of Lian is 6000Z, which does not match the above answer.
So, we can eliminate this case.
CASE 2: Flight cost of Kira = 6000Z and Flight cost of Lian = 5000Z
If we assume the Flight cost to be 6000Z, then we get,
8000A = 5000A + 6000Z
3000A = 6000Z
A = 2Z
For Lian,
Travel Cost = 18000Z
Spending Cost = 6000A + 1000Z = 6000*2Z + 1000Z = 13000Z
Flight Cost = Travel Cost - Spending Cost = 18000Z - 13000Z = 5000Z
In our assumption, the Flight Cost of Lian is 5000Z, which matches the above answer.
So, Flight cost of Kira = 6000Z, Flight cost of Lian = 5000Z and A = 2Z.
Filling the table with calculated values all in the currency of Z, we get,
In the case of Zano,Spending Cost = Travel Cost - Flight Cost = 7000Z - 4000Z = 3000Z
In the first case, the Spending Cost = 2000Z + 1000Z = 3000Z
In the second case, the Spending Cost = 4000Z + 500Z = 4500Z
We obtained a spending cost of 3000Z only in the first case, so we can eliminate the second case.
The final table looks like,
Spending Cost of Lian = 12000Z + 1000Z = 13000ZHence, the correct answer is 13000.
Question 18.
What was Jano's total spend in the two countries he visited, in Aurels?
What was Jano's total spend in the two countries he visited, in Aurels?
Text Explanation:
We are given the travel cost of Jano to be 3500A in Statement 1.
We are also given the travel cost of Kira to be 8000A in Statement 1 and 4000B in Statement 2.
We are also given the travel cost of Lian to be 36000C in statement 3.
We know that the travel cost has to be constant throughout, so by equating the travel costs of Kira, we get,
8000A = 4000B
B = 2A ---(1)
We are also given the value of C to be 0.5 Z.
C = 0.5Z ---(2)
So, the travel cost of Lian = 36000C = 36000*0.5Z = 18000Z
Let us put the known information in the table, and we get,
We are given that the spending cost of any individual is different in different cities.We are given that the spending amounts are amongst {1000, 2000, 3000} in their local currencies and also told that there were two spends of 1000 and one spend of 3000, which makes the total spending amount 2000, to be 3 because there are 6 spending amounts in total.
One 2000 and one 3000 are already assigned in the above table, so we will be left with two spends of 1000 and two spends of 2000 to be assigned.
Kira has already spent 2000, so the only amount that she must have spent is 1000, as she cannot spend 2000 twice in different countries.
Lian has already spent 3000, so the only amount that she must have spent is 2000. If she spent 1000, then Jano will be left with two 2000s, which is not possible.
We can also conclude that Jano spend 1000 and 2000 in some order in the two countries.
The flight costs are given as 4000Z, 5000Z, and 6000Z, out of which 4000Z is for Jano and 5000Z and 6000Z are for Kira and Lian, in some order.
Filling up the table with these values, we get,
Travel Cost = Spending Cost + Flight CostFor Kira,
Travel Cost = 8000A
Spending Cost = 1000A + 4000A = 5000A
Flight Cost = 5000Z/6000Z
CASE 1: Flight cost of Kira = 5000Z and Flight cost of Lian = 6000Z
If we assume the Flight cost to be 5000Z, then we get,
8000A = 5000A + 5000Z
3000A = 5000Z
A = $\frac{5}{3}$Z
For Lian,
Travel Cost = 18000Z
Spending Cost = 6000A + 1000Z = 6000$\times\ \frac{5}{3}$Z + 1000Z = 11000Z
Flight Cost = Travel Cost - Spending Cost = 18000Z - 11000Z = 7000Z
But in our assumption, the Flight Cost of Lian is 6000Z, which does not match the above answer.
So, we can eliminate this case.
CASE 2: Flight cost of Kira = 6000Z and Flight cost of Lian = 5000Z
If we assume the Flight cost to be 6000Z, then we get,
8000A = 5000A + 6000Z
3000A = 6000Z
A = 2Z
For Lian,
Travel Cost = 18000Z
Spending Cost = 6000A + 1000Z = 6000*2Z + 1000Z = 13000Z
Flight Cost = Travel Cost - Spending Cost = 18000Z - 13000Z = 5000Z
In our assumption, the Flight Cost of Lian is 5000Z, which matches the above answer.
So, Flight cost of Kira = 6000Z, Flight cost of Lian = 5000Z and A = 2Z.
Filling the table with calculated values all in the currency of Z, we get,
In the case of Zano,Spending Cost = Travel Cost - Flight Cost = 7000Z - 4000Z = 3000Z
In the first case, the Spending Cost = 2000Z + 1000Z = 3000Z
In the second case, the Spending Cost = 4000Z + 500Z = 4500Z
We obtained a spending cost of 3000Z only in the first case, so we can eliminate the second case.
The final table looks like,
Total Spending Cost of Jano = 2000Z + 1000Z = 3000ZWe know that A = 2Z, so the above value in A can be calculated as,
3000Z = 3000*A/2 = 1500A
So, the spending of Jano in Aurels is 1500A.
Hence, the correct answer is 1500.
Question 19.
One Brin is equivalent to how many Crowns?
One Brin is equivalent to how many Crowns?
Text Explanation:
We are given the travel cost of Jano to be 3500A in Statement 1.
We are also given the travel cost of Kira to be 8000A in Statement 1 and 4000B in Statement 2.
We are also given the travel cost of Lian to be 36000C in statement 3.
We know that the travel cost has to be constant throughout, so by equating the travel costs of Kira, we get,
8000A = 4000B
B = 2A ---(1)
We are also given the value of C to be 0.5 Z.
C = 0.5Z ---(2)
So, the travel cost of Lian = 36000C = 36000*0.5Z = 18000Z
Let us put the known information in the table, and we get,
We are given that the spending cost of any individual is different in different cities.We are given that the spending amounts are amongst {1000, 2000, 3000} in their local currencies and also told that there were two spends of 1000 and one spend of 3000, which makes the total spending amount 2000, to be 3 because there are 6 spending amounts in total.
One 2000 and one 3000 are already assigned in the above table, so we will be left with two spends of 1000 and two spends of 2000 to be assigned.
Kira has already spent 2000, so the only amount that she must have spent is 1000, as she cannot spend 2000 twice in different countries.
Lian has already spent 3000, so the only amount that she must have spent is 2000. If she spent 1000, then Jano will be left with two 2000s, which is not possible.
We can also conclude that Jano spent 1000 and 2000 in some order in the two countries.
The flight costs are given as 4000Z, 5000Z, and 6000Z, out of which 4000Z is for Jano and 5000Z and 6000Z are for Kira and Lian, in some order.
Filling up the table with these values, we get,
Travel Cost = Spending Cost + Flight CostFor Kira,
Travel Cost = 8000A
Spending Cost = 1000A + 4000A = 5000A
Flight Cost = 5000Z/6000Z
CASE 1: Flight cost of Kira = 5000Z and Flight cost of Lian = 6000Z
If we assume the Flight cost to be 5000Z, then we get,
8000A = 5000A + 5000Z
3000A = 5000Z
A = $\frac{5}{3}$Z
For Lian,
Travel Cost = 18000Z
Spending Cost = 6000A + 1000Z = 6000$\times\ \frac{5}{3}$Z + 1000Z = 11000Z
Flight Cost = Travel Cost - Spending Cost = 18000Z - 11000Z = 7000Z
But in our assumption, the Flight Cost of Lian is 6000Z, which does not match the above answer.
So, we can eliminate this case.
CASE 2: Flight cost of Kira = 6000Z and Flight cost of Lian = 5000Z
If we assume the Flight cost to be 6000Z, then we get,
8000A = 5000A + 6000Z
3000A = 6000Z
A = 2Z
For Lian,
Travel Cost = 18000Z
Spending Cost = 6000A + 1000Z = 6000*2Z + 1000Z = 13000Z
Flight Cost = Travel Cost - Spending Cost = 18000Z - 13000Z = 5000Z
In our assumption, the Flight Cost of Lian is 5000Z, which matches the above answer.
So, Flight cost of Kira = 6000Z, Flight cost of Lian = 5000Z and A = 2Z.
Filling the table with calculated values all in the currency of Z, we get,
In the case of Zano,Spending Cost = Travel Cost - Flight Cost = 7000Z - 4000Z = 3000Z
In the first case, the Spending Cost = 2000Z + 1000Z = 3000Z
In the second case, the Spending Cost = 4000Z + 500Z = 4500Z
We obtained a spending cost of 3000Z only in the first case, so we can eliminate the second case.
The final table looks like,
We know that B = 2A and A = 2Z, so the value of B in terms of Z is B = 2*2Z = 4Z --(3)
We also know that
C = 0.5Z --(4)
Dividing (3) by (4), we get,
$\dfrac{B}{C}\ =\ \dfrac{4Z}{0.5Z}\ =\ 8$
B = 8C
So, one Brin is equivalent to 8 crowns.
Hence, the correct answer is option A.
Question 20.
Which of the following statements is NOT true about money spent in the local currency?
Which of the following statements is NOT true about money spent in the local currency?
Text Explanation:
We are given the travel cost of Jano to be 3500A in Statement 1.
We are also given the travel cost of Kira to be 8000A in Statement 1 and 4000B in Statement 2.
We are also given the travel cost of Lian to be 36000C in statement 3.
We know that the travel cost has to be constant throughout, so by equating the travel costs of Kira, we get,
8000A = 4000B
B = 2A ---(1)
We are also given the value of C to be 0.5 Z.
C = 0.5Z ---(2)
So, the travel cost of Lian = 36000C = 36000*0.5Z = 18000Z
Let us put the known information in the table, and we get,
We are given that the spending cost of any individual is different in different cities.We are given that the spending amounts are amongst {1000, 2000, 3000} in their local currencies and also told that there were two spends of 1000 and one spend of 3000, which makes the total spending amount 2000, to be 3 because there are 6 spending amounts in total.
One 2000 and one 3000 are already assigned in the above table, so we will be left with two spends of 1000 and two spends of 2000 to be assigned.
Kira has already spent 2000, so the only amount that she must have spent is 1000, as she cannot spend 2000 twice in different countries.
Lian has already spent 3000, so the only amount that she must have spent is 2000. If she spent 1000, then Jano will be left with two 2000s, which is not possible.
We can also conclude that Jano spend 1000 and 2000 in some order in the two countries.
The flight costs are given as 4000Z, 5000Z, and 6000Z, out of which 4000Z is for Jano and 5000Z and 6000Z are for Kira and Lian, in some order.
Filling up the table with these values, we get,
Travel Cost = Spending Cost + Flight CostFor Kira,
Travel Cost = 8000A
Spending Cost = 1000A + 4000A = 5000A
Flight Cost = 5000Z/6000Z
CASE 1: Flight cost of Kira = 5000Z and Flight cost of Lian = 6000Z
If we assume the Flight cost to be 5000Z, then we get,
8000A = 5000A + 5000Z
3000A = 5000Z
A = $\frac{5}{3}$Z
For Lian,
Travel Cost = 18000Z
Spending Cost = 6000A + 1000Z = 6000$\times\ \frac{5}{3}$Z + 1000Z = 11000Z
Flight Cost = Travel Cost - Spending Cost = 18000Z - 11000Z = 7000Z
But in our assumption, the Flight Cost of Lian is 6000Z, which does not match the above answer.
So, we can eliminate this case.
CASE 2: Flight cost of Kira = 6000Z and Flight cost of Lian = 5000Z
If we assume the Flight cost to be 6000Z, then we get,
8000A = 5000A + 6000Z
3000A = 6000Z
A = 2Z
For Lian,
Travel Cost = 18000Z
Spending Cost = 6000A + 1000Z = 6000*2Z + 1000Z = 13000Z
Flight Cost = Travel Cost - Spending Cost = 18000Z - 13000Z = 5000Z
In our assumption, the Flight Cost of Lian is 5000Z, which matches the above answer.
So, Flight cost of Kira = 6000Z, Flight cost of Lian = 5000Z and A = 2Z.
Filling the table with calculated values all in the currency of Z, we get,
In the case of Zano,Spending Cost = Travel Cost - Flight Cost = 7000Z - 4000Z = 3000Z
In the first case, the Spending Cost = 2000Z + 1000Z = 3000Z
In the second case, the Spending Cost = 4000Z + 500Z = 4500Z
We obtained a spending cost of 3000Z only in the first case, so we can eliminate the second case.
The final table looks like,
Statement 1) Jano spent 2000 in AureviaJano spent 2000Z in Aurevia = 1000A in Aurevia.
So, this option is incorrect.
Statement 2) Lian spent 2000 in Cyrenia
Lian spent 1000Z in Cyrenia = 2000C in Cyrenia.
So, this option is correct.
Statement 3) Jano spent 2000 in Cyrenia
Jano spent 1000Z in Cyrenia = 2000C in Cyrenia.
So, this option is correct.
Statement 4) Kira spent 1000 in Aurevia
Kira spent 2000Z in Aurevia = 1000A in Aurevia.
So, this option is correct.
Hence, the correct answer is option A.
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