CAT 2017 LRDI Slot 2 | Previous Year CAT Paper
Join our Telegram Group for CAT/MBA Preparation.
Directions for next 4 questions.
Funky Pizzaria was required to supply pizzas to three different parties. The total number of pizzas it had to deliver was 800, 70% of which were to be delivered to party 3 and the rest equally divided between Party 1 and Party 2.
Pizzas could be of Thin Crust (T) or Deep Dish (D) variety and come in either Normal Cheese (NC) or Extra Cheese (EC) versions. Hence, there are four types of pizzas: T-NC, T-EC, D-NC and D-EC. Partial information about proportions of T and NC pizzas ordered by the three parties is given below:
How many Thin Crust pizzas were to be delivered to Party 3?
- A.
398
- B.
162
- C.
196
- D.
364
Answer: Option B
Join our Telegram Group for CAT/MBA Preparation.
Explanation :
Now 70% of the pizzas were delivered to party 3 and the balance 30% of the pizzas were delivered to the remaining 2 parties i.e., Party 1 and Party 2. So each of Party 1 and Party 2 will receive
$\Rightarrow \frac{100-70}{100}\times \frac{1}{2}$ × 800 = 120 pizzas.
Party 3 will receive $\Rightarrow \frac{70}{100}$ × 800
= 560 pizzas.
Now number of Thin crust pizzas received by Party 1 = 0.6 × 120 = 72
So number of Deep dish pizzas received by Party 1 = 120 – 72 = 48
Similarly, number of Thin crust pizzas received by Party 2 = 0.55 × 120 = 66 and number of deep dish pizzas received by party 2 = 120 – 60 = 54
Number of normal cheese pizzas received by party 2 = 0.3 × 120 = 36
Number of extra cheese pizzas received by party 2 = 120 – 36 = 84
Now number of normal cheeze pizzas received by party 3 = 0.65 × 560 = 364
So number of extra cheese pizzas received by party 3 = 560 – 364 = 196
Now total number of Thin crust pizzas delivered to the 3 parties = 0.375 × 800
= 300
So, number of Deep Dish Pizzas delivered to the 3 parties = 800 – 416 = 384
So number of thin crust pizzas received by party B = 300 – (72 + 66) = 162 and number of deep dish pizzas received by party 3 = 500 – (48 + 54) = 398
Let number of thin crust normal cheese pizzas received by party 1, 2 and 3 be a, b and c.
Now number of deep dish normal cheese pizzas received by party 2 = 36 – b
So, the number of deep dish extra cheese pizzas received by party 2 = 54 – (36 – b) = 18 + b
Now total number of extra cheese deep dish pizzas received by party 3 = 196 – (162 – c) = 34 + c
Let number of deep pan normal cheese pizzas received by party 1 be ‘d’.
∴ Number of deep dish extra cheeze pizzas will be ‘48-d’
Now let us represent all this information in a table given below:
As can be seen from the table, number of thin crust pizzas delivered to party 3 is 162.
Hence, option (b).
Workspace:
How many Normal Cheese pizzas were required to be delivered to Party 1?
- A.
104
- B.
84
- C.
16
- D.
196
Answer: Option C
Join our Telegram Group for CAT/MBA Preparation.
Explanation :
Now 70% of the pizzas were delivered to party 3 and the balance 30% of the pizzas were delivered to the remaining 2 parties i.e., Party 1 and Party 2. So each of Party 1 and Party 2 will receive
$\Rightarrow \frac{100-70}{100}\times \frac{1}{2}$ × 800 = 120 pizzas.
Party 3 will receive $\Rightarrow \frac{70}{100}$ × 800
= 560 pizzas.
Now number of Thin crust pizzas received by Party 1 = 0.6 × 120 = 72
So number of Deep dish pizzas received by Party 1 = 120 – 72 = 48
Similarly, number of Thin crust pizzas received by Party 2 = 0.55 × 120 = 66 and number of deep dish pizzas received by party 2 = 120 – 60 = 54
Number of normal cheese pizzas received by party 2 = 0.3 × 120 = 36
Number of extra cheese pizzas received by party 2 = 120 – 36 = 84
Now number of normal cheeze pizzas received by party 3 = 0.65 × 560 = 364
So number of extra cheese pizzas received by party 3 = 560 – 364 = 196
Now total number of Thin crust pizzas delivered to the 3 parties = 0.375 × 800
= 300
So, number of Deep Dish Pizzas delivered to the 3 parties = 800 – 416 = 384
So number of thin crust pizzas received by party B = 300 – (72 + 66) = 162 and number of deep dish pizzas received by party 3 = 500 – (48 + 54) = 398
Let number of thin crust normal cheese pizzas received by party 1, 2 and 3 be a, b and c.
Now number of deep dish normal cheese pizzas received by party 2 = 36 – b
So, the number of deep dish extra cheese pizzas received by party 2 = 54 – (36 – b) = 18 + b
Now total number of extra cheese deep dish pizzas received by party 3 = 196 – (162 – c) = 34 + c
Let number of deep pan normal cheese pizzas received by party 1 be ‘d’.
∴ Number of deep dish extra cheeze pizzas will be ‘48-d’
Now let us represent all this information in a table given below:
Now, number of normal cheese pizzas delivered to all parties is 416. Also total number of normal cheese pizzas delivered to party 1 = a + d
∴ 416 = a + b + c + d + 36 – b + 364 – c
416 = a + d + 400
⇒ a + d = 16.
Hence, option (c).
Workspace:
For Party 2, if 50% of the Normal Cheese pizzas were of Thin Crust variety, what was the difference between the numbers of T-EC and D-EC pizzas to be delivered to party 2?
- A.
18
- B.
12
- C.
30
- D.
24
Answer: Option B
Join our Telegram Group for CAT/MBA Preparation.
Explanation :
Now 70% of the pizzas were delivered to party 3 and the balance 30% of the pizzas were delivered to the remaining 2 parties i.e., Party 1 and Party 2. So each of Party 1 and Party 2 will receive
$\Rightarrow \frac{100-70}{100}\times \frac{1}{2}$ × 800 = 120 pizzas.
Party 3 will receive ⇒ $\frac{70}{100}$ × 800
= 560 pizzas.
Now number of Thin crust pizzas received by Party 1 = 0.6 × 120 = 72
So number of Deep dish pizzas received by Party 1 = 120 – 72 = 48
Similarly, number of Thin crust pizzas received by Party 2 = 0.55 × 120 = 66 and number of deep dish pizzas received by party 2 = 120 – 60 = 54
Number of normal cheese pizzas received by party 2 = 0.3 × 120 = 36
Number of extra cheese pizzas received by party 2 = 120 – 36 = 84
Now number of normal cheeze pizzas received by party 3 = 0.65 × 560 = 364
So number of extra cheese pizzas received by party 3 = 560 – 364 = 196
Now total number of Thin crust pizzas delivered to the 3 parties = 0.375 × 800
= 300
So, number of Deep Dish Pizzas delivered to the 3 parties = 800 – 416 = 384
So number of thin crust pizzas received by party B = 300 – (72 + 66) = 162 and number of deep dish pizzas received by party 3 = 500 – (48 + 54) = 398
Let number of thin crust normal cheese pizzas received by party 1, 2 and 3 be a, b and c.
Now number of deep dish normal cheese pizzas received by party 2 = 36 – b
So, the number of deep dish extra cheese pizzas received by party 2 = 54 – (36 – b) = 18 + b
Now total number of extra cheese deep dish pizzas received by party 3 = 196 – (162 – c) = 34 + c
Let number of deep pan normal cheese pizzas received by party 1 be ‘d’.
∴ Number of deep dish extra cheeze pizzas will be ‘48-d’
Now let us represent all this information in a table given below:
If 50% of the normal cheese pizzas delivered to party 2 are of thin crust variety, then b : 36 – b
= 1 : 1
$\frac{b}{36-b}$ = 1 ⇒ 2b = 36 or b = 18
So number if T-EC pizzas = 66 – b = 66 – 18 = 48
Also, number of D-EC pizzas = 18 + b = 18 + 18 = 36
So difference between number of T-EC and D-EC pizzas = 48 – 36 = 12
Hence, option (b).
Workspace:
Suppose that a T-NC pizza cost as much as a D-NC pizza, but 3/5th of the price of a D-EC pizza. A D-EC pizza costs Rs. 50 more than a T-EC pizza, and the latter costs Rs. 500.
If 25% of the Normal Cheese pizzas delivered to Party 1 were of Deep Dish variety, what was the total bill for Party 1?
- A.
Rs. 59480
- B.
Rs. 59840
- C.
Rs. 42520
- D.
Rs. 45240
Answer: Option A
Join our Telegram Group for CAT/MBA Preparation.
Explanation :
Now 70% of the pizzas were delivered to party 3 and the balance 30% of the pizzas were delivered to the remaining 2 parties i.e., Party 1 and Party 2. So each of Party 1 and Party 2 will receive
$\Rightarrow \frac{100-70}{100}\times \frac{1}{2}$ × 800 = 120 pizzas.
Party 3 will receive ⇒ $\frac{70}{100}$ × 800
= 560 pizzas.
Now number of Thin crust pizzas received by Party 1 = 0.6 × 120 = 72
So number of Deep dish pizzas received by Party 1 = 120 – 72 = 48
Similarly, number of Thin crust pizzas received by Party 2 = 0.55 × 120 = 66 and number of deep dish pizzas received by party 2 = 120 – 60 = 54
Number of normal cheese pizzas received by party 2 = 0.3 × 120 = 36
Number of extra cheese pizzas received by party 2 = 120 – 36 = 84
Now number of normal cheeze pizzas received by party 3 = 0.65 × 560 = 364
So number of extra cheese pizzas received by party 3 = 560 – 364 = 196
Now total number of Thin crust pizzas delivered to the 3 parties = 0.375 × 800
= 300
So, number of Deep Dish Pizzas delivered to the 3 parties = 800 – 416 = 384
So number of thin crust pizzas received by party B = 300 – (72 + 66) = 162 and number of deep dish pizzas received by party 3 = 500 – (48 + 54) = 398
Let number of thin crust normal cheese pizzas received by party 1, 2 and 3 be a, b and c.
Now number of deep dish normal cheese pizzas received by party 2 = 36 – b
So, the number of deep dish extra cheese pizzas received by party 2 = 54 – (36 – b) = 18 + b
Now total number of extra cheese deep dish pizzas received by party 3 = 196 – (162 – c) = 34 + c
Let number of deep pan normal cheese pizzas received by party 1 be ‘d’.
∴ Number of deep dish extra cheeze pizzas will be ‘48-d’
Now let us represent all this information in a table given below:
Let the price of a T-NC pizza and D-NC pizza be Rs ‘3x’. So, the price of a D-EC pizza will be ‘5x’ and the price of T-EC pizza will be Rs. ‘5x – 50’
Now 5x – 50 = 500 ⇒ 5x = 550 or x = 110
So the price of T-NC and D-NC pizza will be 3 × 110 or Rs. 330
Further, the price of D-NC pizza will be Rs. 550 and that of T-EC pizza will be Rs. 500
Now $\frac{d}{a+d}=\frac{1}{4}$⇒ 3d = a
Given a + d = 16 ⇒ 4d = 16 or d = 4
∴ a = 16 – 4 = 12
So total number of T-NC, T-EC, D-NC and D-EC pizzas, delivered to party 1 is 12, 60, 4 and 44 respectively.
Cost of pizzas to party 1 is calculated as below
Cost of T-NC pizzas = 12 × 330 = 3960
Cost of D-NC pizzas = 4 × 300 = 1320
Cost of T-EC pizzas = 60 × 500 = 30000
Cost of D-EC pizzas = 44 × 550 = 24200
Total cost = 3960 + 1320 + 30000 + 24200 = Rs. 59480.
Hence, option (a).
Workspace:
Answer the following question based on the information given below.
There were seven elective courses - E1 to E7 – running in a specific term in a college. Each of the 300 students enrolled had chosen just one elective from among these seven. However, before the start of the term, E7 was withdrawn as the instructor concerned had left the college. The students who had opted for E7 were allowed to join any of the remaining electives. Also, the students who had chosen other electives were given one chance to change their choice. The table below captures the movement of the students from one elective to another during this process. Movement from one elective to the same elective simply means no movement. Some numbers in the table got accidentally erased; however, it is known that these were either 0 or 1.
Further, the following are known:
- Before the change process there were 6 more students in E1 than in E4, but after the reshuffle, the number of students in E4 was 3 more than that in E1.
- The number of students in E2 increased by 30 after the change process.
- Before the change process, E4 and 2 more students than E6, while E2 had 10 more students than E3.
How many elective courses among E1 to E6 had a decrease in their enrollments after the change process?
- A.
4
- B.
1
- C.
2
- D.
3
Answer: Option C
Join our Telegram Group for CAT/MBA Preparation.
Explanation :
Let us first try and tabulate the number of people for each of the electives before and after the change in elective
ow we know that after change of electives number of people choosing E2 increases by 30. This means that before change of electives, number of people choosing E2 is 76 – 30 = 46.
Now if we check row E2, the values 34, 8, 2 and 2 add upto 46. Since the values of blank cells can be 0 or 1, it means that the remaining 2 cells in row E2 will be 0.
Now before the change, E1 had 6 more students than E4, which means E4 had 31 – 6 = 25 students. If we look at the row E4, there are 2 blank cells. Also, the total of the remaining 4 cells adds upto 3 + 2 + 14 + 4 = 23. As total of all 6 cells has to be 25, it would mean that remaining 2 cells can only have a value of 1. Also the number of students before change of elective in E6 is 25 – 2 = 23. Again in row E6, sum of 4 occupied cells = 7 + 3 + 2 + 9 = 21. Here too with only 2 vacant cells, the balance 2 cells will have a value of 1 each. Now in E2 there are 10 students more than E3 before change of electives, which means in row E3 the total will be 36. This further implies that in row E5, total will be 300 – (31 + 46 + 36 + 25 + 23 + 101) = 38
So far, the total of columns E1 to E6 is 17 + 76 + 78 + 21 + 44 + 60 = 296
This means that the remaining 6 vacant cells in the table will be filled up four ‘1’s and two 0’s. Since the number of students after the reshuffle in E1 is 3 less than the number of students in E4 (which is 21 without the 2 blank cells), the total of column E1 has to be at least 18 which means the vacant cell in column E1 will be filled with 1. This further implies that the balance 2 cells in column E4 will be filled with 0. So then the remaining cells in column E3, E5 and E6 will be filled with 1. Our final table will appear as below.
Now using this table let us answer each of the questions
If we look at the table there is a decrease in the number of students for electives E1 (which decreases from 31 to 18) and E4 (which decreases from 25 to 21).
Hence, option (c).
Workspace:
After the change process, which of the following is the correct sequence of number of students in the six electives E1 to E6?
- A.
19, 76, 79, 21, 45, 60
- B.
19, 76, 78, 22, 45, 60
- C.
18, 76, 79, 23, 43, 61
- D.
18, 76, 79, 21, 45, 61
Answer: Option D
Join our Telegram Group for CAT/MBA Preparation.
Explanation :
Let us first try and tabulate the number of people for each of the electives before and after the change in elective
ow we know that after change of electives number of people choosing E2 increases by 30. This means that before change of electives, number of people choosing E2 is 76 – 30 = 46.
Now if we check row E2, the values 34, 8, 2 and 2 add upto 46. Since the values of blank cells can be 0 or 1, it means that the remaining 2 cells in row E2 will be 0.
Now before the change, E1 had 6 more students than E4, which means E4 had 31 – 6 = 25 students. If we look at the row E4, there are 2 blank cells. Also, the total of the remaining 4 cells adds upto 3 + 2 + 14 + 4 = 23. As total of all 6 cells has to be 25, it would mean that remaining 2 cells can only have a value of 1. Also the number of students before change of elective in E6 is 25 – 2 = 23. Again in row E6, sum of 4 occupied cells = 7 + 3 + 2 + 9 = 21. Here too with only 2 vacant cells, the balance 2 cells will have a value of 1 each. Now in E2 there are 10 students more than E3 before change of electives, which means in row E3 the total will be 36. This further implies that in row E5, total will be 300 – (31 + 46 + 36 + 25 + 23 + 101) = 38
So far, the total of columns E1 to E6 is 17 + 76 + 78 + 21 + 44 + 60 = 296
This means that the remaining 6 vacant cells in the table will be filled up four ‘1’s and two 0’s. Since the number of students after the reshuffle in E1 is 3 less than the number of students in E4 (which is 21 without the 2 blank cells), the total of column E1 has to be at least 18 which means the vacant cell in column E1 will be filled with 1. This further implies that the balance 2 cells in column E4 will be filled with 0. So then the remaining cells in column E3, E5 and E6 will be filled with 1. Our final table will appear as below.
Now using this table let us answer each of the questions
As can be seen from the table, the correct sequence of the number of students in the 6 electives E1 to E6 is 18, 76, 79, 21, 45 and 61.
Hence, option (d).
Workspace:
After the change process, which course among E1 to E6 had the largest change in its enrollment as a percentage of its original enrollment?
- A.
E1
- B.
E2
- C.
E3
- D.
E6
Answer: Option D
Join our Telegram Group for CAT/MBA Preparation.
Explanation :
Let us first try and tabulate the number of people for each of the electives before and after the change in elective
ow we know that after change of electives number of people choosing E2 increases by 30. This means that before change of electives, number of people choosing E2 is 76 – 30 = 46.
Now if we check row E2, the values 34, 8, 2 and 2 add upto 46. Since the values of blank cells can be 0 or 1, it means that the remaining 2 cells in row E2 will be 0.
Now before the change, E1 had 6 more students than E4, which means E4 had 31 – 6 = 25 students. If we look at the row E4, there are 2 blank cells. Also, the total of the remaining 4 cells adds upto 3 + 2 + 14 + 4 = 23. As total of all 6 cells has to be 25, it would mean that remaining 2 cells can only have a value of 1. Also the number of students before change of elective in E6 is 25 – 2 = 23. Again in row E6, sum of 4 occupied cells = 7 + 3 + 2 + 9 = 21. Here too with only 2 vacant cells, the balance 2 cells will have a value of 1 each. Now in E2 there are 10 students more than E3 before change of electives, which means in row E3 the total will be 36. This further implies that in row E5, total will be 300 – (31 + 46 + 36 + 25 + 23 + 101) = 38
So far, the total of columns E1 to E6 is 17 + 76 + 78 + 21 + 44 + 60 = 296
This means that the remaining 6 vacant cells in the table will be filled up four ‘1’s and two 0’s. Since the number of students after the reshuffle in E1 is 3 less than the number of students in E4 (which is 21 without the 2 blank cells), the total of column E1 has to be at least 18 which means the vacant cell in column E1 will be filled with 1. This further implies that the balance 2 cells in column E4 will be filled with 0. So then the remaining cells in column E3, E5 and E6 will be filled with 1. Our final table will appear as below.
Now using this table let us answer the questions
As can be seen from the table above, the highest percentage is for E6.
Hence, option (d).
Note: We do not need to calculate the exact value. If we compare the value of number of students in electives before and after change, we will see by observation that the ratio for E6 will be highest.
Workspace:
Later, the college imposed a condition that if after the change of electives, the enrollment in any elective (other than E7) dropped to less than 20 students, all the students who had left that course will be required to re-enroll for that elective.
Which of the following is a correct sequence of electives in decreasing order of their final enrollments?
- A.
E2, E3, E6, E5, E1, E4
- B.
E3, E2, E6, E5, E4, E1
- C.
E2, E5, E3, E1, E4, E6
- D.
E2, E3, E5, E6, E1, E3
Answer: Option A
Join our Telegram Group for CAT/MBA Preparation.
Explanation :
Let us first try and tabulate the number of people for each of the electives before and after the change in elective
After Change in electives | ||||||||
Before Change in Elective |
E1 | E2 | E3 | E4 | E5 | E6 | Total | |
E1 | 9 | 5 | 10 | 1 | 4 | 2 | 31 | |
E2 | 34 | 8 | 2 | 2 | ||||
E3 | 2 | 6 | 25 | 2 | ||||
E4 | 3 | 2 | 14 | 4 | ||||
E5 | 5 | 30 | ||||||
E6 | 7 | 3 | 2 | 9 | ||||
E7 | 4 | 16 | 30 | 5 | 5 | 41 | 101 | |
Total | 76 |
ow we know that after change of electives number of people choosing E2 increases by 30. This means that before change of electives, number of people choosing E2 is 76 – 30 = 46.
Now if we check row E2, the values 34, 8, 2 and 2 add upto 46. Since the values of blank cells can be 0 or 1, it means that the remaining 2 cells in row E2 will be 0.
Now before the change, E1 had 6 more students than E4, which means E4 had 31 – 6 = 25 students. If we look at the row E4, there are 2 blank cells. Also, the total of the remaining 4 cells adds upto 3 + 2 + 14 + 4 = 23. As total of all 6 cells has to be 25, it would mean that remaining 2 cells can only have a value of 1. Also the number of students before change of elective in E6 is 25 – 2 = 23. Again in row E6, sum of 4 occupied cells = 7 + 3 + 2 + 9 = 21. Here too with only 2 vacant cells, the balance 2 cells will have a value of 1 each. Now in E2 there are 10 students more than E3 before change of electives, which means in row E3 the total will be 36. This further implies that in row E5, total will be 300 – (31 + 46 + 36 + 25 + 23 + 101) = 38
After Change in electives | ||||||||
Before Change in Elective |
E1 | E2 | E3 | E4 | E5 | E6 | Total | |
E1 | 9 | 5 | 10 | 1 | 4 | 2 | 31 | |
E2 | 0 | 34 | 8 | 0 | 2 | 2 | 46 | |
E3 | 2 | 6 | 25 | 2 | 36 | |||
E4 | 1 | 3 | 2 | 14 | 4 | 25 | ||
E5 | 5 | 30 | 38 | |||||
E6 | 1 | 7 | 3 | 1 | 2 | 9 | 23 | |
E7 | 4 | 16 | 30 | 5 | 5 | 41 | 101 | |
Total | 76 |
So far, the total of columns E1 to E6 is 17 + 76 + 78 + 21 + 44 + 60 = 296
This means that the remaining 6 vacant cells in the table will be filled up four ‘1’s and two 0’s. Since the number of students after the reshuffle in E1 is 3 less than the number of students in E4 (which is 21 without the 2 blank cells), the total of column E1 has to be at least 18 which means the vacant cell in column E1 will be filled with 1. This further implies that the balance 2 cells in column E4 will be filled with 0. So then the remaining cells in column E3, E5 and E6 will be filled with 1. Our final table will appear as below.
After Change in electives | ||||||||
Before Change in Elective |
E1 | E2 | E3 | E4 | E5 | E6 | Total | |
E1 | 9 | 5 | 10 | 1 | 4 | 2 | 31 | |
E2 | 0 | 34 | 8 | 0 | 2 | 2 | 46 | |
E3 | 2 | 6 | 25 | 0 | 1 | 2 | 36 | |
E4 | 1 | 3 | 2 | 14 | 1 | 4 | 25 | |
E5 | 1 | 5 | 1 | 0 | 30 | 1 | 38 | |
E6 | 1 | 7 | 3 | 1 | 2 | 9 | 23 | |
E7 | 4 | 16 | 30 | 5 | 5 | 41 | 101 | |
Total | 18 | 76 | 79 | 21 | 45 | 61 | 300 |
Now using this table let us answer the questions
It is only for elective E1 that the enrollments have fallen to below 20. If the students who had left the course E1, now re-enroll for it, the final number of students for each of the electives will be as follows
After Change in electives | ||||||||
Before Change in Elective |
E1 | E2 | E3 | E4 | E5 | E6 | Total | |
E1 | 31 | 0 | 0 | 0 | 0 | 0 | 31 | |
E2 | 0 | 34 | 8 | 0 | 2 | 2 | 46 | |
E3 | 2 | 6 | 25 | 0 | 1 | 2 | 36 | |
E4 | 1 | 3 | 2 | 14 | 1 | 4 | 25 | |
E5 | 1 | 5 | 1 | 0 | 30 | 1 | 38 | |
E6 | 1 | 7 | 3 | 1 | 2 | 9 | 23 | |
E7 | 4 | 16 | 30 | 5 | 5 | 41 | 101 | |
Total | 40 | 71 | 79 | 20 | 41 | 59 | 300 |
As can be seen from the table the correct sequence of the electives in decreasing order of their electives is E2, E3, E6, E5, E1 and E4.
Hence, option (a).
Workspace:
Answer the following question based on the information given below.
An old woman had the following assets:
(a) Rs. 70 lakh in bank deposits
(b) 1 house worth Rs. 50 lakh
(c). 3 flats, each worth Rs. 30 lakh
(d) Certain number of gold coins, each worth Rs. 1 lakh
She wanted to distribute her assets among her three children; Neeta, Seeta and Geeta.
The house, any of the flats or any of the coins were not to be split. That is, the house went entirely to one child; a flat went to one child and similarly, a gold coin went to one child.
Among the three, Neeta received the least amount in bank deposits, while Geeta received the highest. The value of the assets was distributed equally among the children, as were the gold coins.How much did Seeta receive in bank deposits (in lakhs of rupees)?
- A.
30
- B.
40
- C.
20
- D.
10
Answer: Option C
Join our Telegram Group for CAT/MBA Preparation.
Explanation :
As per the data given in the question the gold coins were equally distributed amongst the 3 children. We are also told that the remaining assets (apart from the gold coins) are also equally distributed and that no flat or house is split or divided between 2 or more people. Only the bank deposit of Rs.70 lakhs can be split. Apart from the gold coins, the total value of the assets is 70 + 50 + 30 + 30 + 30 = 210 lakhs
So 210 lakhs split equally would means that each of the 3 daughters would get Rs.70 lakhs. This would mean that the person receiving the Rs.50 lakh house would get Rs.20 lakhs from the bank deposit. Now there are 3 flats of Rs.30 lakhs each and a balance of 50 lakhs in the bank deposit. The only way these assets can be split equally is if one of the 2 remaining people gets one of the flats of Rs.30 lakhs and receives 40 lakhs from the bank deposit. The second person would receive 70-40-20 = 10 Lakhs as bank deposit and 2 flats of 30 lakhs each, the sum of which adds upto 70 lakhs.
As Neeta receives the least amount in bank deposits, she will receive 10 lakhs. Also, as Geeta receives the highest amount in bank deposits she wil receive 40 lakhs. Therefore, Seeta will receive 20 lakhs from the bank deposit.
Hence, option (c).
Workspace:
Among the three, Neeta received the least amount in bank deposits, while Geeta received the highest. The value of the assets was distributed equally among the children, as were the gold coins. How many flats did Neeta receive?
Answer: 2
Join our Telegram Group for CAT/MBA Preparation.
Explanation :
As per the data given in the question the gold coins were equally distributed amongst the 3 children. We are also told that the remaining assets (apart from the gold coins) are also equally distributed and that no flat or house is split or divided between 2 or more people. Only the bank deposit of Rs.70 lakhs can be split. Apart from the gold coins, the total value of the assets is 70 + 50 + 30 + 30 + 30 = 210 lakhs
So 210 lakhs split equally would means that each of the 3 daughters would get Rs.70 lakhs. This would mean that the person receiving the Rs.50 lakh house would get Rs.20 lakhs from the bank deposit. Now there are 3 flats of Rs.30 lakhs each and a balance of 50 lakhs in the bank deposit. The only way these assets can be split equally is if one of the 2 remaining people gets one of the flats of Rs.30 lakhs and receives 40 lakhs from the bank deposit. The second person would receive 70-40-20 = 10 Lakhs as bank deposit and 2 flats of 30 lakhs each, the sum of which adds upto 70 lakhs.
As Neeta receives the least amount in bank deposits, she will receive 10 lakhs. Also, as Geeta receives the highest amount in bank deposits she wil receive 40 lakhs.
Since Neeta is the person who receives the least amount in bank deposits i.e., 10 lakhs, he will receive 2 flats of 30 lakhs each.
Answer: 2
Workspace:
The value of the assets distributed among Neeta, Seeta and Geeta was in the ratio of 1 : 2 : 3, while the gold coins were distributed among them in the ratio of 2 : 3 : 4. One child got all three flats and she did not get the house. One child, other than Geeta, got Rs. 30 lakh in bank deposits.
How many gold coins did the old woman have?
- A.
72
- B.
90
- C.
180
- D.
216
Answer: Option B
Join our Telegram Group for CAT/MBA Preparation.
Explanation :
As per the data given in the question the gold coins were equally distributed amongst the 3 children. We are also told that the remaining assets (apart from the gold coins) are also equally distributed and that no flat or house is split or divided between 2 or more people. Only the bank deposit of Rs.70 lakhs can be split. Apart from the gold coins, the total value of the assets is 70 + 50 + 30 + 30 + 30 = 210 lakhs
So 210 lakhs split equally would means that each of the 3 daughters would get Rs.70 lakhs. This would mean that the person receiving the Rs.50 lakh house would get Rs.20 lakhs from the bank deposit. Now there are 3 flats of Rs.30 lakhs each and a balance of 50 lakhs in the bank deposit. The only way these assets can be split equally is if one of the 2 remaining people gets one of the flats of Rs.30 lakhs and receives 40 lakhs from the bank deposit. The second person would receive 70-40-20 = 10 Lakhs as bank deposit and 2 flats of 30 lakhs each, the sum of which adds upto 70 lakhs.
As Neeta receives the least amount in bank deposits, she will receive 10 lakhs. Also, as Geeta receives the highest amount in bank deposits she wil receive 40 lakhs.
Let the total value of assets held by Neeta, Geeta and Seeta be x, 2x and 3x and the total number of gold coins held by each of them be 2y, 3y and 4y. So the total assets (apart from the gold coins) held by Neeta, Geeta and Seeta will be x – 2y, 2x – 3y and 3x – 4y.
Now total assets apart from gold coin = 210 lakhs
∴ (x – 2y) + (2x -3y) + (3x – 4y) = 210
⇒ 6x – 9y = 210
On simplifying we get 2x – 3y = 70
This means that the total assets held by Geeta apart from the gold coins = 70 lakhs We are further told that one child got all 3 flats and one person other than Geeta got 30 lakhs in bank deposits. So then the only way Geeta could have got assets (other than gold coins) worth 70 lakhs is by getting the house of 50 lakhs and 20 lakhs from the bank deposit. This is also because all the 3 flats of Rs. 30 lakh each have gone to one person and obviously that person cannot be Geeta. This means that the 3rd sister has got 20 lakhs from the bank deposit (since one more sister has got 30 lakhs). Since Geeta cannot receive 30 lakhs from the bank deposit, it would imply that Neeta will receive 30 lakhs and Geeta will receive 20 lakhs from the bank deposit.
Now, all 3 flats go to either Neeta or Geeta.
If all 3 flats are received by Neeta, then the ratio of value of assets of Neeta and Geeta
(in terms of y) will be (30 + 90 + 2y): (20 + 4y)
⇒ (120 + 2y) : (20 + 4y) = $\frac{120+2y}{20+4y}$
$=\frac{1}{3}$
⇒ 360 + 6y = 20 + 4y ⇒ 2y = -340 or y = -170
However ‘y’ cannot be a negative value. So all 3 flats will go to Geeta
∴ Ratio of value of assets of Neeta and Geeta (in terms of y) will be (30 + 2y) : (20 + 90 + 4y)
$\Rightarrow \frac{30+2y}{110+4y}=\frac{1}{3}$
⇒ 90 + 6y = 110 + 4y ⇒ y = 10
Total number of gold coins = 2y + 3y + 4y = 9y = 90.
Hence, option (b).
Workspace:
The value of the assets distributed among Neeta, Seeta and Geeta was in the ratio of 1 : 2 : 3, while the gold coins were distributed among them in the ratio of 2 : 3 : 4. One child got all three flats and she did not get the house. One child, other than Geeta, got Rs. 30 lakh in bank deposits.
How much did Geeta get in bank deposits (in lakhs of rupees)?
Answer: 20
Join our Telegram Group for CAT/MBA Preparation.
Explanation :
As per the data given in the question the gold coins were equally distributed amongst the 3 children. We are also told that the remaining assets (apart from the gold coins) are also equally distributed and that no flat or house is split or divided between 2 or more people. Only the bank deposit of Rs.70 lakhs can be split. Apart from the gold coins, the total value of the assets is 70 + 50 + 30 + 30 + 30 = 210 lakhs
So 210 lakhs split equally would means that each of the 3 daughters would get Rs.70 lakhs. This would mean that the person receiving the Rs.50 lakh house would get Rs.20 lakhs from the bank deposit. Now there are 3 flats of Rs.30 lakhs each and a balance of 50 lakhs in the bank deposit. The only way these assets can be split equally is if one of the 2 remaining people gets one of the flats of Rs.30 lakhs and receives 40 lakhs from the bank deposit. The second person would receive 70-40-20 = 10 Lakhs as bank deposit and 2 flats of 30 lakhs each, the sum of which adds upto 70 lakhs.
As Neeta receives the least amount in bank deposits, she will receive 10 lakhs. Also, as Geeta receives the highest amount in bank deposits she wil receive 40 lakhs.
Geeta gets 20 lakhs from bank deposits.
Answer: 20
Workspace:
Answer the following question based on the information given below.
At a management school, the oldest 10 dorms, numbered 1 to 10, need to be repaired urgently. The following diagram represents the estimated repair costs (in Rs. Crores) for the 10 dorms. For any dorm, the estimated repair cost (in Rs. Crores) in an integer. Repairs with estimated cost Rs. 1 or 2 Crores are considered light repairs, repairs with estimated cost Rs. 3 or 4 are considered moderate repairs and repairs with estimated cost Rs. 5 of 6 Crores are considered extensive repairs.
Further, the following are known:
- Odd – numbered dorms do not need light repair; even-numbered dorms do not need moderate repair and dorms, whose numbers are divisible by 3, do not need extensive repair.
- Dorms 4 to 9 all need different repair costs, with Dorm 7 needing the maximum and Dorm 8 needing the minimum.
Which of the following is NOT necessarily true?
- A.
Dorm 1 needs a moderate repair
- B.
Dorm 5 repair will cost no more than Rs. 4 Crores
- C.
Dorm 7 needs an extensive repair
- D.
Dorm 10 repair will cost no more than Rs. 4 Crores
Answer: Option D
Join our Telegram Group for CAT/MBA Preparation.
Explanation :
Let us tabulate the information given in the bar chart. As per condition (i) we get the following information
From this we can conclude that Dorm 6 would require light repairs and that Dorm 3 and 9 would require moderate repairs. Now following condition (2), we get to know that Dorm 7 would require 6 cr for repair. and Dorm 8 would required 1 cr. This means that in light repairs, Dorm 8 requires 1 crore for repair and Dorm 2 requires 6 crores for repair. Now as each of Dorms to 9 requires a different amount for repairs, and no even numbered dorm requires ,moderate repair, it would imply that Dorm no 4 requires heavy repairs. Already Dorm 7 needs the maximum amount for repairs, (i.e, 6 crores) which means that Dorm4 would require 5 crores for repair. Now Dorm 9 would require 3 or 4 cr for repair. This leads us to 2 possible cases.
In case 1, as 4 cr is required for Dorm no 9, 3 cr will be required for Dorm no. 5. Also, as no even numbered dorm can have a moderate expenditure, out of the remaining 4 dorms, 1 and 3 (which are also the only odd numbered dorms out of the remaining 4 dorms) will have an expenditure of 3 crores (also since there is only 1 dorm which has a expense of 4 cr which in this case is the expense for Dorm 9). This leaves out only Dorm 2 and Dorm 10, one of these 2 dorms will have an expense of 1 cr and the other an expense of 6 crores.
In case 2, 3 cr is the expense for Dorm No. 9 and 4 cr will be the expense for Dorm no 5. Like in case 1, Dorm no 1 and 3 will have an expense of 1 crore and the other an expense of 6 crores. So our final table for both cases will be as below
Now using these tables let us answer each of the questions.
Let us examine each of the statements individually
1] Statement 1 is true as Dorm no 1 needs a moderate repair of 3 crores
2] Statement 2 is also true as Dorm no 5 needs a repair of 3 or 4 crores
3] Statement 3 is also true as Dorm 7 requires a extensive repair of 6 crores
4] Statement 4 may or may not be true as Dorm 10 has a repair cost of 1 crore or 6 crores
Hence, option (d).
Workspace:
What is the total cost of repairing the odd-numbered dorms (in Rs. Crores)?
Answer: 19
Join our Telegram Group for CAT/MBA Preparation.
Explanation :
Let us tabulate the information given in the bar chart. As per condition (i) we get the following information
From this we can conclude that Dorm 6 would require light repairs and that Dorm 3 and 9 would require moderate repairs. Now following condition (2), we get to know that Dorm 7 would require 6 cr for repair. and Dorm 8 would required 1 cr. This means that in light repairs, Dorm 8 requires 1 crore for repair and Dorm 2 requires 6 crores for repair. Now as each of Dorms to 9 requires a different amount for repairs, and no even numbered dorm requires ,moderate repair, it would imply that Dorm no 4 requires heavy repairs. Already Dorm 7 needs the maximum amount for repairs, (i.e, 6 crores) which means that Dorm4 would require 5 crores for repair. Now Dorm 9 would require 3 or 4 cr for repair. This leads us to 2 possible cases.
In case 1, as 4 cr is required for Dorm no 9, 3 cr will be required for Dorm no. 5. Also, as no even numbered dorm can have a moderate expenditure, out of the remaining 4 dorms, 1 and 3 (which are also the only odd numbered dorms out of the remaining 4 dorms) will have an expenditure of 3 crores (also since there is only 1 dorm which has a expense of 4 cr which in this case is the expense for Dorm 9). This leaves out only Dorm 2 and Dorm 10, one of these 2 dorms will have an expense of 1 cr and the other an expense of 6 crores.
In case 2, 3 cr is the expense for Dorm No. 9 and 4 cr will be the expense for Dorm no 5. Like in case 1, Dorm no 1 and 3 will have an expense of 1 crore and the other an expense of 6 crores. So our final table for both cases will be as below
Dorms 1 and 3 cost 3 crores each to repair.
One of Dorm 5 or Dorm 9 costs 3 crores to repair and the other costs 4 crores repair.
Dorm 7 costs 6 crores to repair.
Total cost of repairing odd numbered dorms = 3 + 3 + 3 + 4 + 6 = 19 crores
Answer: 19
Workspace:
Suppose further that:
- 4 of the 10 dorms needing repair are women’s dorms and need a total of Rs. 20 Crores for repair.
- Only one of Dorms 1 to 5 is a women’s dorm.
What is the cost for repairing Dorm 9 (in Rs. Crores)?
Answer: 3
Join our Telegram Group for CAT/MBA Preparation.
Explanation :
Let us tabulate the information given in the bar chart. As per condition (i) we get the following information
From this we can conclude that Dorm 6 would require light repairs and that Dorm 3 and 9 would require moderate repairs. Now following condition (2), we get to know that Dorm 7 would require 6 cr for repair. and Dorm 8 would required 1 cr. This means that in light repairs, Dorm 8 requires 1 crore for repair and Dorm 2 requires 6 crores for repair. Now as each of Dorms to 9 requires a different amount for repairs, and no even numbered dorm requires ,moderate repair, it would imply that Dorm no 4 requires heavy repairs. Already Dorm 7 needs the maximum amount for repairs, (i.e, 6 crores) which means that Dorm4 would require 5 crores for repair. Now Dorm 9 would require 3 or 4 cr for repair. This leads us to 2 possible cases.
In case 1, as 4 cr is required for Dorm no 9, 3 cr will be required for Dorm no. 5. Also, as no even numbered dorm can have a moderate expenditure, out of the remaining 4 dorms, 1 and 3 (which are also the only odd numbered dorms out of the remaining 4 dorms) will have an expenditure of 3 crores (also since there is only 1 dorm which has a expense of 4 cr which in this case is the expense for Dorm 9). This leaves out only Dorm 2 and Dorm 10, one of these 2 dorms will have an expense of 1 cr and the other an expense of 6 crores.
In case 2, 3 cr is the expense for Dorm No. 9 and 4 cr will be the expense for Dorm no 5. Like in case 1, Dorm no 1 and 3 will have an expense of 1 crore and the other an expense of 6 crores. So our final table for both cases will be as below
Now the only way the cost of repairing 4 dorms adds upto 20 crores is when the repair cost of 2 of the dorms is 6 crores each, the repair cost of the 3rd dorm is 5 crores and the repair cost of the 4th dorm is 3 crores.
This means that dorms 7 and 4 are definitely women’s dorms. Now if we follow condition. Now if we follow condition (2), we get to know that from Dorm 1 to 5 only Dorm 4 is a women’s dorm. So, the remaining 3 women’s dorms have to be from Dorm 6 to 10. Now from Dorm 6 to Dorm 10, we already know that Dorm 7 is a women’s dorm. Now the Dorm for which the cost of repair is 3 crore has to be a dorm where number is between 6 to 10. This only happens in case 2, where Dorm 9 is one of the 3 dorm that costs Rs. 3 crore to repair. Also in case 2, the second dorm that cost Rs.6 crores to repair has to be Dorm No. 10 to satisfy condition (2).
Hence the 4 women’s dorms that Rs.20 crores to repair are Dorms 7, 10, 4 and 9. So the cost of repairing Dorm 9 is Rs. 3 crores.
Answer: 3
Workspace:
Suppose further that:
- 4 of the 10 dorms needing repair are women’s dorms and need a total of Rs. 20 Crores for repair.
- Only one of Dorms 1 to 5 is a women’s dorm.
Which of the following is a women’s dorm?
- A.
Dorm 2
- B.
Dorm 5
- C.
Dorm 8
- D.
Dorm 10
Answer: Option D
Join our Telegram Group for CAT/MBA Preparation.
Explanation :
Let us tabulate the information given in the bar chart. As per condition (i) we get the following information
From this we can conclude that Dorm 6 would require light repairs and that Dorm 3 and 9 would require moderate repairs. Now following condition (2), we get to know that Dorm 7 would require 6 cr for repair. and Dorm 8 would required 1 cr. This means that in light repairs, Dorm 8 requires 1 crore for repair and Dorm 2 requires 6 crores for repair. Now as each of Dorms to 9 requires a different amount for repairs, and no even numbered dorm requires ,moderate repair, it would imply that Dorm no 4 requires heavy repairs. Already Dorm 7 needs the maximum amount for repairs, (i.e, 6 crores) which means that Dorm4 would require 5 crores for repair. Now Dorm 9 would require 3 or 4 cr for repair. This leads us to 2 possible cases.
In case 1, as 4 cr is required for Dorm no 9, 3 cr will be required for Dorm no. 5. Also, as no even numbered dorm can have a moderate expenditure, out of the remaining 4 dorms, 1 and 3 (which are also the only odd numbered dorms out of the remaining 4 dorms) will have an expenditure of 3 crores (also since there is only 1 dorm which has a expense of 4 cr which in this case is the expense for Dorm 9). This leaves out only Dorm 2 and Dorm 10, one of these 2 dorms will have an expense of 1 cr and the other an expense of 6 crores.
In case 2, 3 cr is the expense for Dorm No. 9 and 4 cr will be the expense for Dorm no 5. Like in case 1, Dorm no 1 and 3 will have an expense of 1 crore and the other an expense of 6 crores. So our final table for both cases will be as below
we can see that out of the given options only Dorm 10 is a women’s dorm.
Hence, option (d).
Workspace:
Answer the following question based on the information given below.
A tea taster was assigned to rate teas from six different locations – Munnar, Wayanand, Ooty, Darjeeling, Assam and Himachal. These teas were placed in six cups, numbered 1 to 6, not necessarily in the same order. The tea taster was asked to rate these teas on the strength of their flavour on a scale of 1 to 10. He gave a unique integer rating to each tea. Some other information is given below:
- Cup 6 contained tea from Himachal.
- Tea from Ooty got the highest rating, but it was not in Cup 3.
- The rating of tea in Cup 3 was double the rating of the tea in Cup 5.
- Only two cups got ratings in even numbers.
- Cup 2 got the minimum rating and this rating was an even number.
- Tea in Cup 3 got a higher rating than that in Cup 1.
- The rating of tea from Wayanad was more than the rating of tea from Munnar, but less than that from Assam.
What was the second highest rating given?
Answer: 7
Join our Telegram Group for CAT/MBA Preparation.
Explanation :
Following condition (1) and (2) we can form the table as below
Now rating of Tea in cup 3 is twice the rating of tea in cup 5. So the rating of tea in cup 3 will be an even number. Now following condition. (4) and (5) we know that cup 2 and cup 3 are the only cups that have an even numbered rating. This means that the rating of tea in cup 1, 4, 5 and 6 will be an odd number. Now since the rating of cup 3 is twice the rating of cup 5 which has an odd numbered rating, rating of cup 5 will be 3. The reason for this is that the since the rating of cup 2 is the least and an even number, the rating of cup 5 has to be higher than that of cup 2 (which will be atleast 2) and hence it’s rating cannot be 1. Further, since only the rating of cup 2 and cup 3 are even numbers and neither of them contains the highest rated tea, so then rating of Tea in Ooty (which is not in cup 3) will be less than 10. So this means the only possibility for rating of tea in cup 3 has to be 6, as that is the only even number which is twice that of an odd number. This further implies that the rating of tea in cup 5 is 3. This also means that rating of Tea in cup 2 is 2. We know need to find the ratings of tea in cups 1, 4 and 6. Further, as all these cups will be 5, 7 and 9 (not essentially in that order). Since the rating of tea in cup 3 (which is 6) is higher than the rating of tea in cup 1, the rating of tea in cup 1 has to be 5. Further, tea from Ooty is placed in cup 4 and since the tea from Ooty has the highest rating it’s rating will be 9. This would mean that the rating of tea from Himachal in cup 6 will be 7. The only other information we have is that Tea from Assam has a better rating than Tea from Wayanand which in turn is better than Tea from Munnar.
We have no other information about the location of tea in cups 1, 2, 3 and 5. So using the table above let us answer the questions.
As can be seen from the table, 2nd highest rating given was 7.
Answer: 7
Workspace:
What was the number of the cup that contained tea from Ooty?
Answer: 4
Join our Telegram Group for CAT/MBA Preparation.
Explanation :
Following condition (1) and (2) we can form the table as below
Now rating of Tea in cup 3 is twice the rating of tea in cup 5. So the rating of tea in cup 3 will be an even number. Now following condition. (4) and (5) we know that cup 2 and cup 3 are the only cups that have an even numbered rating. This means that the rating of tea in cup 1, 4, 5 and 6 will be an odd number. Now since the rating of cup 3 is twice the rating of cup 5 which has an odd numbered rating, rating of cup 5 will be 3. The reason for this is that the since the rating of cup 2 is the least and an even number, the rating of cup 5 has to be higher than that of cup 2 (which will be atleast 2) and hence it’s rating cannot be 1. Further, since only the rating of cup 2 and cup 3 are even numbers and neither of them contains the highest rated tea, so then rating of Tea in Ooty (which is not in cup 3) will be less than 10. So this means the only possibility for rating of tea in cup 3 has to be 6, as that is the only even number which is twice that of an odd number. This further implies that the rating of tea in cup 5 is 3. This also means that rating of Tea in cup 2 is 2. We know need to find the ratings of tea in cups 1, 4 and 6. Further, as all these cups will be 5, 7 and 9 (not essentially in that order). Since the rating of tea in cup 3 (which is 6) is higher than the rating of tea in cup 1, the rating of tea in cup 1 has to be 5. Further, tea from Ooty is placed in cup 4 and since the tea from Ooty has the highest rating it’s rating will be 9. This would mean that the rating of tea from Himachal in cup 6 will be 7. The only other information we have is that Tea from Assam has a better rating than Tea from Wayanand which in turn is better than Tea from Munnar.
We have no other information about the location of tea in cups 1, 2, 3 and 5. So using the table above let us answer the questions.
The number of the cup that contained the tea from Ooty is 4.
Answer: 4
Workspace:
If the tea from Munnar did not get the minimum rating, what was the rating of the tea from Wayanad?
- A.
3
- B.
5
- C.
1
- D.
6
Answer: Option B
Join our Telegram Group for CAT/MBA Preparation.
Explanation :
Following condition (1) and (2) we can form the table as below
Now rating of Tea in cup 3 is twice the rating of tea in cup 5. So the rating of tea in cup 3 will be an even number. Now following condition. (4) and (5) we know that cup 2 and cup 3 are the only cups that have an even numbered rating. This means that the rating of tea in cup 1, 4, 5 and 6 will be an odd number. Now since the rating of cup 3 is twice the rating of cup 5 which has an odd numbered rating, rating of cup 5 will be 3. The reason for this is that the since the rating of cup 2 is the least and an even number, the rating of cup 5 has to be higher than that of cup 2 (which will be atleast 2) and hence it’s rating cannot be 1. Further, since only the rating of cup 2 and cup 3 are even numbers and neither of them contains the highest rated tea, so then rating of Tea in Ooty (which is not in cup 3) will be less than 10. So this means the only possibility for rating of tea in cup 3 has to be 6, as that is the only even number which is twice that of an odd number. This further implies that the rating of tea in cup 5 is 3. This also means that rating of Tea in cup 2 is 2. We know need to find the ratings of tea in cups 1, 4 and 6. Further, as all these cups will be 5, 7 and 9 (not essentially in that order). Since the rating of tea in cup 3 (which is 6) is higher than the rating of tea in cup 1, the rating of tea in cup 1 has to be 5. Further, tea from Ooty is placed in cup 4 and since the tea from Ooty has the highest rating it’s rating will be 9. This would mean that the rating of tea from Himachal in cup 6 will be 7. The only other information we have is that Tea from Assam has a better rating than Tea from Wayanand which in turn is better than Tea from Munnar.
We have no other information about the location of tea in cups 1, 2, 3 and 5. So using the table above let us answer the questions.
If tea from Munnar did not get the least rating of 2, it would imply that the tea from Munnar got the 2nd least rating of 3, since the tea from Assam and Wayanad have a better rating than the tea from Munnar. As we already know that tea from Ooty and Himachal are the ones with the top 2 ratings, it would mean that tea from Assam is rated 3rd highest with a rating of 6 and tea from Wayanad is rated 4th highest with a rating of 5.
Hence, option (b).
Workspace:
If cups containing teas from Wayanad and Ooty had consecutive numbers, which of the following statements may be true?
- A.
Cup 5 contains tea from Assam
- B.
Cup 1 contains tea from Darjeeling
- C.
Tea from Wayanad has got a rating of 6
- D.
Tea from Darjeeling got the minimum rating
Answer: Option B
Join our Telegram Group for CAT/MBA Preparation.
Explanation :
Following condition (1) and (2) we can form the table as below
Now rating of Tea in cup 3 is twice the rating of tea in cup 5. So the rating of tea in cup 3 will be an even number. Now following condition. (4) and (5) we know that cup 2 and cup 3 are the only cups that have an even numbered rating. This means that the rating of tea in cup 1, 4, 5 and 6 will be an odd number. Now since the rating of cup 3 is twice the rating of cup 5 which has an odd numbered rating, rating of cup 5 will be 3. The reason for this is that the since the rating of cup 2 is the least and an even number, the rating of cup 5 has to be higher than that of cup 2 (which will be atleast 2) and hence it’s rating cannot be 1. Further, since only the rating of cup 2 and cup 3 are even numbers and neither of them contains the highest rated tea, so then rating of Tea in Ooty (which is not in cup 3) will be less than 10. So this means the only possibility for rating of tea in cup 3 has to be 6, as that is the only even number which is twice that of an odd number. This further implies that the rating of tea in cup 5 is 3. This also means that rating of Tea in cup 2 is 2. We know need to find the ratings of tea in cups 1, 4 and 6. Further, as all these cups will be 5, 7 and 9 (not essentially in that order). Since the rating of tea in cup 3 (which is 6) is higher than the rating of tea in cup 1, the rating of tea in cup 1 has to be 5. Further, tea from Ooty is placed in cup 4 and since the tea from Ooty has the highest rating it’s rating will be 9. This would mean that the rating of tea from Himachal in cup 6 will be 7. The only other information we have is that Tea from Assam has a better rating than Tea from Wayanand which in turn is better than Tea from Munnar.
We have no other information about the location of tea in cups 1, 2, 3 and 5. So using the table above let us answer the questions.
Given that cups containing tea from Wayanad and Ooty have consecutive numbers, tea from Wayanad can either be in cup 3 or cup 5. However, as tea in cup 3 has the 3rd highest rating it cannot be from Wayanad. This is because the tea from Assam is better than the tea from Wayanad. So then tea from Wayanad will be in cup no 5 and will have a rating of 3 and tea in cup 2 which has the lowest rating of 2 is from Munnar. This means that in cup 1 and cup 3 we can have tea from Assam and Darjeeling (not essentially in that order). If we now examine the statements in each of the 4 options we will see that statements in options (1), (3) and (4) are not true. However statement in option (2) may be true.
Hence, option (b).
Workspace:
Answer the following question based on the information given below.
In an 8 × 8 chessboard a queen placed anywhere can attack another piece if the piece is present in the same row, or in the same column or in any diagonal position in any possible 4 directions, provided there is no other piece in between in the path from the queen to that piece.
The columns are labelled a to h (left to right) and the rows are numbered 1 to 8 (bottom to top). The position of a piece given by the combination of column and row labels. For example, position c5 means that the piece is cth column and 5th row.
If the queen is at c5, and the other pieces at positions c2, g1, g3, g5 and a3, how many are under attack by the queen? There are no other pieces on the board.
- A.
2
- B.
3
- C.
4
- D.
5
Answer: Option C
Join our Telegram Group for CAT/MBA Preparation.
Explanation :
In the diagram given above let Q represent the queen at cell C_{5}. Now let us suppose pieces P_{1}, P_{2}, P_{3}, P_{4} and P_{5} are at cells. C_{2}, G_{1}, G_{3}, G_{5} and A_{3}. As can seen that Q can attack P_{1} as they are is the same column and can attack P_{4} as both are in the same row. Further Q can attack P_{5} and P_{2} as both are in a diagonal position to Q. So it cannot be attacked by Q. So 4 pieces i.e., P_{1}, P_{2}, P_{4} and P_{5} can attacked by Q.
Hence, option (c).
Workspace:
If the other pieces are only at positions a1, a3, b4, d7, h7 and h8, then which of the following positions of the queen results in the maximum number of pieces being under attack?
- A.
f8
- B.
a7
- C.
c1
- D.
d3
Answer: Option D
Join our Telegram Group for CAT/MBA Preparation.
Explanation :
Let P_{1}, P_{2}, P_{3}, P_{4}, P_{5} and P_{6} be the 6 pieces at positions A_{1}, A_{3}, B_{4}, D_{7}, H_{7} and H_{8} respectively.
1] If the queen is at position F_{8} it can attack P_{6} as it is in the same row and P_{3} as it in a diagonal position. However it cannot attack P_{2}, which is in a diagonal position as P_{3} is placed between P_{2} and Q along the same diagonal. So a total of 2 pieces can be attacked if the queen is at position F_{8}.
2] If the queen is at position A_{7} it can directly attack P_{4} as it is in the same row and P_{2} as it is in the same column. However it cannot attack P_{5} because P_{4} is in between the queen and P_{5}. Similarly it cannot attack P_{1} because P_{2} is in between the Queen and P_{1}. So a total of 2 pieces can be attacked by the queen if it is at position A_{7}.
3] If the queen is at position C_{1}, then it can only attack P_{1} which is in the same row and P_{2} which is along the same diagonal as the queen. So at positions C_{1}, the queen can attack only 2 pieces.
4] If the queen is at position D_{3} then it can attack P_{2} along the same row, P_{4} along the same column and P_{5} along the same diagonal i.e., a total of 3 pieces. So out of the given options, the queen at position D_{3} can attack maximum number of pieces.
Hence, option (d).
Workspace:
If the other pieces are only at positions a1, a3, b4, d7, h7 and h8, then from how many positions the queen cannot attack any of the pieces?
- A.
0
- B.
3
- C.
4
- D.
6
Answer: Option C
Join our Telegram Group for CAT/MBA Preparation.
Explanation :
All cells in rows 1, 3, 4, 7 and 8 and columns A, B, D and H will be eliminated as one of the ways the queen can the attack us from the same row or column. If we check for cell E_{2}, F_{2}, G_{2} and G_{5}, queen will not be able to attack any of the 6 pieces. So, from a total of 4 positions, the queen cannot attack any of the 6 pieces.
Hence, option (d).
Workspace:
Suppose the queen is the only piece on the board and it is at position d5.
In how many positons can another piece be placed on the board such that it is safe from attack from the queen?
- A.
32
- B.
35
- C.
36
- D.
37
Answer: Option C
Join our Telegram Group for CAT/MBA Preparation.
Explanation :
Let Q represent the position of the queen at cell D_{5}. Now none of the pieces can be along column D or along row 5 i.e., a total of 8 + 8 – 1 = 15 cells (∵ D_{5} is common to column D and row 5 it is counted twice, hence we subtract 1) Now cells along the diagonal of D5 have to be eliminated since they can be directly attacked by the Queen from that cell. Therefore cells A_{8}, B_{7}, C_{6}, E_{4}, F_{3}, G_{2}, H_{1} and G_{8}, F_{7}, E_{6}, C_{4}, B_{3}, A_{2} i.e., a total of 13 cells are not to be considered. So a total of 64 – 28 = 36 positions are considered safe from attack if the queen is at position D_{5}
Hence, option (c).
Workspace:
Answer the following question based on the information given below.
Eight friends: Ajit, Byomkesh, Gargi, Jayanta, Kikira, Manik, Prodosh and Tapesh are goin to Delhi from Kolkatta by a flight operated by Cheap Air. In the flight, sitting is arranged in 30 rows, numbered 1 to 30, each consisting of 6 seats, marked by letters A to F from left to right, respectively. Seats A to C are to the left of the aisle (the passage running from the front of the aircraft to the back), and seats D to F are to the right of the aisle. Seats A and F are by the windows and referred to as Window seats, C and D are by the aisle and are referred to as Aisle seats while B and E are referred to as Middle seats. Seats marked by consecutive letters are called consecutive seats (or seats next to each other).A seat number is a combination of the row number, followed by the letter indicating the position in the row, e.g, 1A is the left window seat in the first row, while 12E is the right middle seat in the 12th row.
Cheap Air charges Rs. 1000 extra for any seats in Rows 1, 12 and 13 as those have extra legroom. For Rows 2-10, it charges Rs. 300 extra for Window seats and Rs. 500 extra for Aisle seats. For Rows 11 and 14 to 20, it charges Rs. 200 extra for Window seats and Rs. 400 extra for Aisle seats. All other seats are available at no extra charge.
The following are known:
- The eight friends were seated in six different rows.
- They occupied 3 Window seats, 4 Aisle seats and 1 Middle seat.
- Seven of them had to pay extra amounts, totaling to Rs. 4600, for their choices of seat. One of them did not pay any additional amount of his/her choice of seat.
- Jayanta, Ajit and Byomkesh were sitting in seats marked by the same letter, in consecutive rows in increasing order of row numbers; but all of them paid different amounts for their choices of seat. One of these amounts may be zero.
- Gargi was sitting next to Kikira, and Manik was sitting next to Jayanta.
- Prodosh and Tapesh were sitting in seats marked by the same letter, in consecutive rows in increasing order of row numbers; but they paid different amounts for their choices of seat. One of these amounts may be zero.
In which row was Manik sitting?
- A.
10
- B.
11
- C.
12
- D.
13
Answer: Option A
Join our Telegram Group for CAT/MBA Preparation.
Explanation :
Using the information given in the questions let us tabulate the rates of the seats in the different rows.
Following condition (4) as Jayanta, Ajit and Byomkesh sit on seats marked by the same letter in increasing order of row in numbers and paid different amounts, the only way this would be possible is if they sat in rows 10, 11 and 12 respectively, either on the window seat or the aisle seat. So this would lead to 2 possible cases.
Case 1: Jayanta, Ajit and Byomkesh sit on the window seat in consecutive rows. In this case, as Manik sits adjacent to Jayanta, it would imply Manik sits on the middle seat next to Jayanta in row 10. This further implies that Manik did not pay any additional amount for his seat. Now Jayanta, Ajit and Byomkesh would pay an amount of 300 + 200 + 1000 = Rs. 1500. So the balance amount of 4600 – 1500 = 3100 must have been paid by Gargi, Kikira, Pradosh and Tapesh and all of them have to be occupying aisle seats. However, no contribution of amounts changeable for aisle seats adds upto 3100. So case is ruled out.
Case 2. Jayanta, Ajit and Byomkesh sit on the aisle seat in rows 10, 11 and 12 respectively. Here Manik can occupy either the middle seat next to Jayanta or the aisle seat next to Jayanta. Let us assume that Manik occupies the middle seat in row 10 next to Jayanta. In that case, cost of 3 aisle seats adds upto 4600 – (1000 + 500 + 400) = 2900. However, no contribution of amounts chargeable for the above mentioned seats adds upto 2900. So then Manik can only sit on the aisle seat adjacent to Jayanta in row 10. The total amount of seats chargeable to Jayanta, Ajit, Byomkesh and Manik adds upto 1000 + 500 + 400 + 500 = 2400. Now the balance amount of 4600 – 2400
= 2200 has to be accounted for by 4 people i.e., Gagi, Kikira, Pradosh and Tapesh. 3 of them sit on 3 window seats and one on the middle seat. Now out of Gargi and Kikira, one of them will sit on the middle seat and other on the window seat as they sit adjacent to each other. Which means that both Pradosh and Tapesh sit on the window seat. Since the amounts chargeable for the window/middle seats can only the amongst 1000, 300, 200 or 0 and the total we need to account for is 2200, at least one of the seats has to be a 1000 Rs. Seat. If we assume one of seats is Rs. 1000, then the balance amount of Rs 1200, cannot be accounted for. So there have to be at least 2 seats of Rs. 1000 each. Which means 2 out of Gargi, Kikira, Pradosh and Tapesh have to sit in the 13th row. Now these 2 people will have to be Gargi and Kikirsa as per condition (5). So each of Gargi and Kikira pay Rs. 1000 for their seat and these 2 seats will be a window and middle seat (both adjacent to each other)in the 13th row. The balance amount of Rs. 200 will be paid by Pradosh who sits on the window seat of the 20th row. This also implies that Tapesh will sit just behind Pradosh in the window seat of the 21st row. Let us represent all this information in the table given below
J – Jayanta, M – Manik, A – Ajit, B – Byomkesh, G – Gargi, K – Kikira, P – Pradosh, T – Tapesh
Using this data let us answer the question.
As can be seen from the table, Manik was sitting in row 10.
Hence, option (a).
Workspace:
How much extra did Jayanta pay for his choice of seat?
- A.
Rs. 300
- B.
Rs. 400
- C.
Rs. 500
- D.
Rs. 1000
Answer: Option C
Join our Telegram Group for CAT/MBA Preparation.
Explanation :
Using the information given in the questions let us tabulate the rates of the seats in the different rows.
Following condition (4) as Jayanta, Ajit and Byomkesh sit on seats marked by the same letter in increasing order of row in numbers and paid different amounts, the only way this would be possible is if they sat in rows 10, 11 and 12 respectively, either on the window seat or the aisle seat. So this would lead to 2 possible cases.
Case 1: Jayanta, Ajit and Byomkesh sit on the window seat in consecutive rows. In this case, as Manik sits adjacent to Jayanta, it would imply Manik sits on the middle seat next to Jayanta in row 10. This further implies that Manik did not pay any additional amount for his seat. Now Jayanta, Ajit and Byomkesh would pay an amount of 300 + 200 + 1000 = Rs. 1500. So the balance amount of 4600 – 1500 = 3100 must have been paid by Gargi, Kikira, Pradosh and Tapesh and all of them have to be occupying aisle seats. However, no contribution of amounts changeable for aisle seats adds upto 3100. So case is ruled out.
Case 2. Jayanta, Ajit and Byomkesh sit on the aisle seat in rows 10, 11 and 12 respectively. Here Manik can occupy either the middle seat next to Jayanta or the aisle seat next to Jayanta. Let us assume that Manik occupies the middle seat in row 10 next to Jayanta. In that case, cost of 3 aisle seats adds upto 4600 – (1000 + 500 + 400) = 2900. However, no contribution of amounts chargeable for the above mentioned seats adds upto 2900. So then Manik can only sit on the aisle seat adjacent to Jayanta in row 10. The total amount of seats chargeable to Jayanta, Ajit, Byomkesh and Manik adds upto 1000 + 500 + 400 + 500 = 2400. Now the balance amount of 4600 – 2400
= 2200 has to be accounted for by 4 people i.e., Gagi, Kikira, Pradosh and Tapesh. 3 of them sit on 3 window seats and one on the middle seat. Now out of Gargi and Kikira, one of them will sit on the middle seat and other on the window seat as they sit adjacent to each other. Which means that both Pradosh and Tapesh sit on the window seat. Since the amounts chargeable for the window/middle seats can only the amongst 1000, 300, 200 or 0 and the total we need to account for is 2200, at least one of the seats has to be a 1000 Rs. Seat. If we assume one of seats is Rs. 1000, then the balance amount of Rs 1200, cannot be accounted for. So there have to be at least 2 seats of Rs. 1000 each. Which means 2 out of Gargi, Kikira, Pradosh and Tapesh have to sit in the 13th row. Now these 2 people will have to be Gargi and Kikirsa as per condition (5). So each of Gargi and Kikira pay Rs. 1000 for their seat and these 2 seats will be a window and middle seat (both adjacent to each other)in the 13th row. The balance amount of Rs. 200 will be paid by Pradosh who sits on the window seat of the 20th row. This also implies that Tapesh will sit just behind Pradosh in the window seat of the 21st row. Let us represent all this information in the table given below
J – Jayanta, M – Manik, A – Ajit, B – Byomkesh, G – Gargi, K – Kikira, P – Pradosh, T – Tapesh
Using this data let us answer the question.
Jayanta paid Rs.500 for his choice of seat.
Hence, option (c).
Workspace:
How much extra did Gargi pay for her choice of seat?
- A.
0
- B.
Rs. 300
- C.
Rs. 400
- D.
Rs. 1000
Answer: Option D
Join our Telegram Group for CAT/MBA Preparation.
Explanation :
Using the information given in the questions let us tabulate the rates of the seats in the different rows.
Following condition (4) as Jayanta, Ajit and Byomkesh sit on seats marked by the same letter in increasing order of row in numbers and paid different amounts, the only way this would be possible is if they sat in rows 10, 11 and 12 respectively, either on the window seat or the aisle seat. So this would lead to 2 possible cases.
Case 1: Jayanta, Ajit and Byomkesh sit on the window seat in consecutive rows. In this case, as Manik sits adjacent to Jayanta, it would imply Manik sits on the middle seat next to Jayanta in row 10. This further implies that Manik did not pay any additional amount for his seat. Now Jayanta, Ajit and Byomkesh would pay an amount of 300 + 200 + 1000 = Rs. 1500. So the balance amount of 4600 – 1500 = 3100 must have been paid by Gargi, Kikira, Pradosh and Tapesh and all of them have to be occupying aisle seats. However, no contribution of amounts changeable for aisle seats adds upto 3100. So case is ruled out.
Case 2. Jayanta, Ajit and Byomkesh sit on the aisle seat in rows 10, 11 and 12 respectively. Here Manik can occupy either the middle seat next to Jayanta or the aisle seat next to Jayanta. Let us assume that Manik occupies the middle seat in row 10 next to Jayanta. In that case, cost of 3 aisle seats adds upto 4600 – (1000 + 500 + 400) = 2900. However, no contribution of amounts chargeable for the above mentioned seats adds upto 2900. So then Manik can only sit on the aisle seat adjacent to Jayanta in row 10. The total amount of seats chargeable to Jayanta, Ajit, Byomkesh and Manik adds upto 1000 + 500 + 400 + 500 = 2400. Now the balance amount of 4600 – 2400
= 2200 has to be accounted for by 4 people i.e., Gagi, Kikira, Pradosh and Tapesh. 3 of them sit on 3 window seats and one on the middle seat. Now out of Gargi and Kikira, one of them will sit on the middle seat and other on the window seat as they sit adjacent to each other. Which means that both Pradosh and Tapesh sit on the window seat. Since the amounts chargeable for the window/middle seats can only the amongst 1000, 300, 200 or 0 and the total we need to account for is 2200, at least one of the seats has to be a 1000 Rs. Seat. If we assume one of seats is Rs. 1000, then the balance amount of Rs 1200, cannot be accounted for. So there have to be at least 2 seats of Rs. 1000 each. Which means 2 out of Gargi, Kikira, Pradosh and Tapesh have to sit in the 13th row. Now these 2 people will have to be Gargi and Kikirsa as per condition (5). So each of Gargi and Kikira pay Rs. 1000 for their seat and these 2 seats will be a window and middle seat (both adjacent to each other)in the 13th row. The balance amount of Rs. 200 will be paid by Pradosh who sits on the window seat of the 20th row. This also implies that Tapesh will sit just behind Pradosh in the window seat of the 21st row. Let us represent all this information in the table given below
J – Jayanta, M – Manik, A – Ajit, B – Byomkesh, G – Gargi, K – Kikira, P – Pradosh, T – Tapesh
Using this data let us answer the question.
Gargi paid Rs.1000 extra for her choice of seat.
Hence, option (d).
Workspace:
Who among the following did not pay any extra amount for his/her choice of seat?
- A.
Kikira
- B.
Manik
- C.
Gargi
- D.
Tapesh
Answer: Option D
Join our Telegram Group for CAT/MBA Preparation.
Explanation :
Using the information given in the questions let us tabulate the rates of the seats in the different rows.
Following condition (4) as Jayanta, Ajit and Byomkesh sit on seats marked by the same letter in increasing order of row in numbers and paid different amounts, the only way this would be possible is if they sat in rows 10, 11 and 12 respectively, either on the window seat or the aisle seat. So this would lead to 2 possible cases.
Case 1: Jayanta, Ajit and Byomkesh sit on the window seat in consecutive rows. In this case, as Manik sits adjacent to Jayanta, it would imply Manik sits on the middle seat next to Jayanta in row 10. This further implies that Manik did not pay any additional amount for his seat. Now Jayanta, Ajit and Byomkesh would pay an amount of 300 + 200 + 1000 = Rs. 1500. So the balance amount of 4600 – 1500 = 3100 must have been paid by Gargi, Kikira, Pradosh and Tapesh and all of them have to be occupying aisle seats. However, no contribution of amounts changeable for aisle seats adds upto 3100. So case is ruled out.
Case 2. Jayanta, Ajit and Byomkesh sit on the aisle seat in rows 10, 11 and 12 respectively. Here Manik can occupy either the middle seat next to Jayanta or the aisle seat next to Jayanta. Let us assume that Manik occupies the middle seat in row 10 next to Jayanta. In that case, cost of 3 aisle seats adds upto 4600 – (1000 + 500 + 400) = 2900. However, no contribution of amounts chargeable for the above mentioned seats adds upto 2900. So then Manik can only sit on the aisle seat adjacent to Jayanta in row 10. The total amount of seats chargeable to Jayanta, Ajit, Byomkesh and Manik adds upto 1000 + 500 + 400 + 500 = 2400. Now the balance amount of 4600 – 2400
= 2200 has to be accounted for by 4 people i.e., Gagi, Kikira, Pradosh and Tapesh. 3 of them sit on 3 window seats and one on the middle seat. Now out of Gargi and Kikira, one of them will sit on the middle seat and other on the window seat as they sit adjacent to each other. Which means that both Pradosh and Tapesh sit on the window seat. Since the amounts chargeable for the window/middle seats can only the amongst 1000, 300, 200 or 0 and the total we need to account for is 2200, at least one of the seats has to be a 1000 Rs. Seat. If we assume one of seats is Rs. 1000, then the balance amount of Rs 1200, cannot be accounted for. So there have to be at least 2 seats of Rs. 1000 each. Which means 2 out of Gargi, Kikira, Pradosh and Tapesh have to sit in the 13th row. Now these 2 people will have to be Gargi and Kikirsa as per condition (5). So each of Gargi and Kikira pay Rs. 1000 for their seat and these 2 seats will be a window and middle seat (both adjacent to each other)in the 13th row. The balance amount of Rs. 200 will be paid by Pradosh who sits on the window seat of the 20th row. This also implies that Tapesh will sit just behind Pradosh in the window seat of the 21st row. Let us represent all this information in the table given below
J – Jayanta, M – Manik, A – Ajit, B – Byomkesh, G – Gargi, K – Kikira, P – Pradosh, T – Tapesh
Using this data let us answer the question.
As Tapesh was sitting in row 21, he did not pay any extra amount for his seat.
Hence, option (d).
Workspace:
Answer the following question based on the information given below.
A high security research lab requires the researches to set a pass key sequence based on the scan of the five fingers of their left hands. When an employee first joins the lab, her fingers are scanned in an order of her choice, and then when she wants to re-enter the facility, she has to scan the five fingers in the same sequence.
The lab authorities are considering some relaxations of the scan order requirements of the scan order requirements, since it is observed that some employees often get locked-out because they forget the sequence.
The lab has decided to allow a variation in the sequence of scans of five fingers so that at most two scans (out of five) are out of place. For example, if the original sequence is Thumb (T), index finger (I), middle finger (M), ring finger (R) and little finger (L) then TLMRI is also allowed, but TMRLI is not.
How many different sequences of scans are allowed for any given person’s original scan?
Answer: 11
Join our Telegram Group for CAT/MBA Preparation.
Explanation :
As per the given information either the entire sequence of scans has to be correct or 3 out of the 5 scans have to be in place. So if we consider TIMRL as the original sequence, each of T, I, M, R and L has to be in the correct sequence which is only possible in 1 way. Now if the sequence of scans is out of place, then at least 2 scans will have to be out of place. Which implies that 3 scans are in place. So 3 out of T, I, M, R and L are in place. Now this is possible in ^{5}C_{3} or 10 ways. So number of different sequences of scans that are allowed = 1 + 10 = 11 ways.
Answer: 11
Workspace:
The lab has decided to allow variations of the original sequence so that input of the scanned sequence of five fingers is allowed to vary from the original sequence by one place for any of the fingers. Thus, for example, if TIMRL is the original sequence, then ITRML is also allowed, but LIMRT is not.
How many different sequences are allowed for any given person’s original scan?
- A.
7
- B.
5
- C.
8
- D.
13
Answer: Option C
Join our Telegram Group for CAT/MBA Preparation.
Explanation :
Let us suppose that the original sequence is TIMRL. As per the condition given in the question, for any of the given fingers, the scanned sequence is allowed to vary by one place for any of the given fingers. So for the sequence TIMRL, we can have the following possibilities of sequence of scans, so that it is accepted.
1] All the letters of original sequence are in place. This is possible in 1 way.
2] T & I get interchanged – 1 way
3] T & I and M & R get interchanged – 1 way
4] T & I and R & L interchanged – 1 way
5] Only R & L get interchanged – 1 way
6] Only M & R get interchanged – 1 way
7] I & M get interchanged – 1 way
8] I & M and R & L get interchanged – 1 way
So a total of 8 different sequences of scans will be accepted.
Hence, option (c).
Workspace:
The lab has now decided to require six scans in the pass key sequence, where exactly one finger is scanned twice, and the other fingers are scanned exactly once, which can be done in any order. For example, a possible sequence is TIMTRL.
Suppose the lab allows a variation of the original sequence (of six inputs) where at most two scans (out of six) are out of place, as long as the finger originally scanned twice is scanned twice and other fingers are scanned once.
How many different sequences if scans are allowed for any given person’s original scan?
Answer: 15
Join our Telegram Group for CAT/MBA Preparation.
Explanation :
Let us suppose original sequence is TIMTRL as indicated in the question. To answer this question, let us assume that all 6 scans in the sequence are distinct. As only 2 out of 6 scans can be out of place, it would imply that we choose 2 out of 6 scans and interchange their position. These can be done in ^{6}C_{2} or 15 ways. But since T occurs twice in the sequence, we eliminate the combination of T and T as interchanging these would not make any difference. So total number of ways possible where exactly 2 out of 6 scans can be in place and person can still be given access = 15 – 1 = 14
Another way the person can access is if the sequence of scans is as per the person’s original scan, which is possible in 1 way.
So total number of different scans allowed = 14 + 1 = 15
Answer: 15
Workspace:
The lab has now decided to require six scans in the pass key sequence, where exactly one finger is scanned twice, and the other fingers are scanned exactly once, which can be done in any order. For example, a possible sequence is TIMTRL.
Suppose the lab allows a variation of the original sequence (of six inputs) so that input in the form of scanned sequence of six fingers is allowed to vary from the original sequence by one place for any of the fingers, as long as the finger originally scanned twice is scanned twice and other fingers are scanned once.
How many different sequences of scans are allowed if the original scan sequence is LRLTIM?
- A.
8
- B.
11
- C.
13
- D.
14
Answer: Option C
Join our Telegram Group for CAT/MBA Preparation.
Explanation :
In the sequence LRLTIM, the different sequences of scans allowed are as follows
1] Original sequence
2] Interchange L & R
3] Interchange of L & R and L & T
4] Interchange of L & R, L & T and I & M
5] Interchange of R & L
6] Interchange of R & L and T & I
7] Interchange of L & T
8] Interchange of L & T and I & M
9] Interchange of T & I
10] Interchange of T & I and L & R
11] Interchange of T & I and R & L
12] Interchange of I & M
13] Interchange of I & M and R & L
So 13 different sequences of scan are allowed of the original sequence is LRLTIM.
Answer: 13
Workspace:
Feedback
Help us build a Free and Comprehensive Preparation portal for various competitive exams by providing us your valuable feedback about Apti4All and how it can be improved.