Grouping and Selection: CAT Previous Year Questions

Ans: D

Explanation:The problem states that the testing device detects impurity if the percentage of impurity in the tested content is 10% or more.

We are given that 5 ml of content from bottle A is mixed with 5 ml of content from bottle B. The resultant mixture, when tested, detects the presence of I.
We are also given that bottle A contains only P (pure content). This means bottle A has 0% impurity.

Let the volume of impurity in bottle B be $V_I$ (in ml) and the volume of pure content in bottle B be $V_P$ (in ml). The total volume of liquid in bottle B is 50 ml, so $V_I + V_P = 50$.

When 5 ml from bottle A (which is 100% P) is mixed with 5 ml from bottle B, the total volume of the mixture is 10 ml.
The amount of impurity in the mixture comes solely from bottle B.
The amount of impurity from 5 ml of bottle B is $5 \times \frac{V_I}{50}$ ml.
The total volume of the mixture is 10 ml.

The test detects impurity, which means the percentage of impurity in the mixture is 10% or more.
Percentage of impurity in the mixture = $\frac{\text{Volume of impurity in mixture}}{\text{Total volume of mixture}} \times 100$

The volume of impurity in the mixture is the amount of impurity contributed by the 5 ml taken from bottle B.
Volume of impurity in the mixture = $5 \times \frac{V_I}{50}$

So, the percentage of impurity in the mixture is:
$\frac{5 \times \frac{V_I}{50}}{10} \times 100 \ge 10$

Let’s simplify this inequality:
$\frac{5V_I}{500} \times 100 \ge 10$
$\frac{5V_I}{5} \ge 10$
$V_I \ge 10$

This means the total volume of impurity in bottle B must be 10 ml or more.

Now let’s consider the options provided for the volume of I in bottle B. The question asks what BEST can be concluded about the volume of I in bottle B. Our calculation shows that $V_I \ge 10$ ml.

Option A: 1 ml. This is incorrect because $V_I$ must be at least 10 ml.
Option B: Less than 1 ml. This is incorrect because $V_I$ must be at least 10 ml.
Option C: 10 ml. This is a possible value for $V_I$, but it doesn’t encompass all possibilities (e.g., $V_I$ could be 15 ml).
Option D: 10 ml or more. This conclusion directly matches our derived inequality $V_I \ge 10$.

Therefore, the best conclusion is that the volume of impurity in bottle B is 10 ml or more.

Correct_Option: D

Ans: 1

Bottles are classified as either P (pure) or I (impure). The potential combinations are:
1. (P, P, P, P)
2. (P, P, P, I)
3. (P, P, I, I)
4. (P, I, I, I)
5. (I, I, I, I)

In scenario 1, where all four solutions are pure, combining equal volumes from each bottle will result in a determination for dispatch.

In scenario 2, with three pure bottles and one impure, mixing equal volumes from all four and testing will reveal the impurity, preventing dispatch.

In scenario 3, with two pure and two impure bottles, combining equal volumes from all four and testing will confirm the impurity, thus preventing dispatch.

In scenario 4, where only one bottle is pure, mixing equal volumes from all four bottles and testing will identify the impurity, leading to non-dispatch.

In scenario 5, if all four bottles are impure, combining equal volumes from all four bottles and testing will confirm the impurity, preventing dispatch.

Across all these scenarios, a single test suffices to decide whether the batch can be dispatched.

Ans: 2

The impure solution has an 80 percent concentration of impurity.
When equal quantities of all four solutions are combined.
If we take 10 ml from each, the impurity is 2 ml out of a total of 40 ml. This impurity level is below 10 percent, making it undetectable.
Similarly, when equal quantities of one impure solution and two pure solutions are mixed.
The impurity in this mixture is 2 ml out of 30 ml, which is less than 10 percent and therefore cannot be detected.
Thus, to identify the impure solution, we must test two solutions at a time.
Let’s represent the three pure solutions as ‘P’ and the impure solution as ‘I’.
The bottles are P, P, P, I.
If we combine equal quantities from one bottle of ‘P’ and one bottle of ‘I’, the impurity will be detectable.
Following this, consider one of the remaining two ‘P’ bottles and test it against one of the already tested ‘P’ or ‘I’ solutions.
If the solution chosen for this second test is ‘I’, the impurity will be detected, confirming that bottle as ‘I’.
If the solution chosen is ‘P’, the impurity will not be detected. In this case, the remaining untested bottle must be ‘I’.
Therefore, a minimum of two tests are necessary to pinpoint the bottle containing the impurity.

Ans: D

The potential compositions of the bottles are:
Scenario A: One bottle is pure, and three are impure.
Scenario B: Two bottles are pure, and two are impure.

Given that the impurity concentration in any impure bottle is 85 percent.

In Scenario A, when equal volumes are taken from all four bottles and combined, the test will indicate the presence of impurity. This is because the overall impurity level exceeds 10 percent.
For instance, if 10 ml is taken from each of the four bottles (total volume 40 ml):
The total impurity would be 30 ml (from the three impure bottles) * 0.85 = 25.5 ml.
The impurity concentration would be 25.5 ml / 40 ml, which is significantly greater than 10 percent. (Impurity is detected).

In Scenario B, with two pure and two impure bottles, the outcome differs when equal volumes are mixed.
When equal volumes, such as 10 ml from each of the four bottles, are combined:
The total impurity would be 20 ml (from the two impure bottles) * 0.85 = 17 ml.
The impurity concentration would be 17 ml / 40 ml, which is less than 10 percent. (Impurity is not detected).

Therefore, a single test can differentiate between these two possibilities. The outcome of the test will reveal whether the impurity is detected or not, allowing for the determination of the number of pure and impure bottles.

Ans: 6

Explanation:Let $n_x$ be the number of prizes of type $x$. We are given that the prizes are of types a, b, c, … in decreasing order of value.
There is exactly one prize of type a, so $n_a = 1$.
The problem states that the number of items at least doubles as you move to the next type. This means:
$n_b \ge 2n_a$
$n_c \ge 2n_b$
$n_d \ge 2n_c$
and so on.

We want to maximize the number of different types of prizes. Let’s say there are $k$ types of prizes: $a_1, a_2, …, a_k$, where $a_1$ is the most expensive (type a), $a_2$ is the next most expensive (type b), and so on.
Let $n_i$ be the number of prizes of type $a_i$.
We have $n_1 = 1$.
The condition is $n_{i+1} \ge 2n_i$ for $i = 1, 2, …, k-1$.

To maximize the number of types ($k$), we should assume the minimum possible number for each type, which is exactly double the number of the previous type.
So, we assume:
$n_1 = 1$
$n_2 = 2n_1 = 2 \times 1 = 2$
$n_3 = 2n_2 = 2 \times 2 = 4$
$n_4 = 2n_3 = 2 \times 4 = 8$
$n_i = 2^{i-1}$

The total number of boxes is 100. So, the sum of the number of prizes of each type must be less than or equal to 100.
$n_1 + n_2 + n_3 + … + n_k \le 100$

Using the minimum counts:
$1 + 2 + 4 + 8 + … + 2^{k-1} \le 100$

This is a geometric series with first term 1 and common ratio 2. The sum of the first $k$ terms is given by $\frac{1(2^k – 1)}{2-1} = 2^k – 1$.
So, we need to find the largest $k$ such that:
$2^k – 1 \le 100$
$2^k \le 101$

Let’s check powers of 2:
$2^1 = 2$
$2^2 = 4$
$2^3 = 8$
$2^4 = 16$
$2^5 = 32$
$2^6 = 64$
$2^7 = 128$

The largest integer $k$ for which $2^k \le 101$ is $k=6$.
If $k=6$, the total number of prizes is $2^6 – 1 = 64 – 1 = 63$. This is less than or equal to 100.
If $k=7$, the total number of prizes would be $2^7 – 1 = 128 – 1 = 127$, which is greater than 100.

So, the maximum possible number of different types of prizes is 6.

Let’s verify the number of prizes for k=6:
Type a: $n_a = 1$
Type b: $n_b \ge 2 \times 1 = 2$. Let $n_b = 2$.
Type c: $n_c \ge 2 \times 2 = 4$. Let $n_c = 4$.
Type d: $n_d \ge 2 \times 4 = 8$. Let $n_d = 8$.
Type e: $n_e \ge 2 \times 8 = 16$. Let $n_e = 16$.
Type f: $n_f \ge 2 \times 16 = 32$. Let $n_f = 32$.
Total prizes = $1 + 2 + 4 + 8 + 16 + 32 = 63$.
We have 100 boxes, so we can accommodate these 6 types. The remaining $100 – 63 = 37$ boxes can contain prizes of any of these 6 types (to maintain the “at least doubles” condition, we would have had to increase the counts beyond the minimum, but we are looking for the maximum number of types).

If we try to have 7 types:
$n_a = 1$
$n_b \ge 2$
$n_c \ge 4$
$n_d \ge 8$
$n_e \ge 16$
$n_f \ge 32$
$n_g \ge 2 \times 32 = 64$
The sum of the minimum number of prizes for 7 types is $1+2+4+8+16+32+64 = 127$, which exceeds the 100 boxes.

Therefore, the maximum possible number of different types of prizes is 6.

Final_Answer:6

Ans: C

Option A: Exactly 75 items are of type ‘e’.
If we assign values: a=1, b=2, c=4, d=8, then ‘e’ can be 85. The maximum possible value for ‘e’ is 85, which means it can certainly take the value of 75. A valid scenario is a=1, b=2, c=4, d=18, e=75.

Option B: Exactly 30 items are of type ‘b’.
Consider a=1, b=30, and c=69. This combination is feasible.

Option C: Exactly 45 items are of type ‘c’.
If ‘d’ must be at least 90, then the total count (45 for ‘c’ + at least 90 for ‘d’) would exceed 100 (the maximum total items). This implies ‘d’ cannot be present. With only ‘a’, ‘b’, and ‘c’ present, the maximum value for ‘b’ could be 22, and ‘a’ could be 1. However, 45 (for ‘c’) + 22 (max ‘b’) + 1 (max ‘a’) = 68, which is less than 100. Therefore, this scenario is not possible.

Option D: Exactly 60 items are of type ‘d’.
A possible distribution is: d=60, c=30, b=9, and a=1. The sum a+b+c+d equals 100. This scenario is possible.

Option C is the correct answer.

Ans: A

The count of items from 1 to 100 that share the same characteristic as those in box 45 is found to be 31 + 1 + 43 = 75.

To achieve the highest possible number of items, we assign the following quantities: a=1, b=2, c=4, d=18, and e=75 (as provided).

It’s possible to have a maximum of five distinct categories of items.

Suppose we are considering a scenario with six categories of items. In this case, the smallest possible quantity for the fifth category would be 16. The sum of items (1 + 2 + 4 + 8 + 16 + 75 = 106) exceeds the total limit of 100.

Ans: 7

It is understood that the preferred denomination for the customer is the 500 rupee note.
Consequently, the quantity of 500 rupee notes distributed must exceed the quantity of notes of any other denomination.

Consider a scenario where Rs.3500 is disbursed using 500 rupee notes (which amounts to 7 notes). The remaining Rs.1500 must then be distributed using Rs.100 and Rs.200 notes. The least number of other denomination notes required in this instance is 8 (achieved with 7 Rs.200 notes and 1 Rs.100 note). This implies that a minimum of Rs.4000 must be dispensed in 500 rupee notes.

Scenario 1:
If Rs.4000 is disbursed in 500 rupee notes, this accounts for 8 five hundred rupee notes.
The remaining Rs.1000 cannot be entirely distributed using only 100 rupee notes, as this would necessitate 10 notes.

If Rs.800 is disbursed using 100 rupee notes, then 9 notes are required to cover the Rs.1000 (8 notes of Rs.100 and 1 note of Rs.200). Therefore, this possibility can be excluded.

If Rs.600 is disbursed using 100 rupee notes, then a minimum of 8 notes are needed to cover Rs.1000 (6 notes of Rs.100 and 2 notes of Rs.200). This scenario can also be disregarded.

If Rs.400 is disbursed using 100 rupee notes, then 7 notes are required (4 notes of Rs.100 and 3 notes of Rs.200). This is a viable option.

If Rs.200 is disbursed using 100 rupee notes, then 6 notes are required (2 notes of Rs.100 and 4 notes of Rs.200). This is also a viable option.

Rs.1000 can be disbursed using 5 notes of Rs.200.
Thus, there are 3 viable sub-cases within this scenario.

Scenario 2:
If Rs.4500 is disbursed in 500 rupee notes, this involves 9 five hundred rupee notes.
The remaining Rs.500 can be disbursed as 100 rupee notes (requiring 5 notes) or a combination of 100 rupee and 200 rupee notes.
The equation for this is: 200*a + 100*b = 500, where ‘a’ represents the count of Rs.200 notes and ‘b’ represents the count of Rs.100 notes.
The possible values for ‘a’ are 0, 1, and 2.
Therefore, there are 3 viable sub-cases in this scenario.

Scenario 3:
If Rs.5000 is disbursed using 500 rupee notes, this means 10 five hundred rupee notes are used.
In this case, there is only 1 viable sub-case.

The total count of viable scenarios is the sum of the viable cases from each scenario: 3 + 3 + 1 = 7.
Hence, the correct answer is 7.

Ans: 6

If a customer’s preferred denomination is 500 rupee notes, the quantity of 500 rupee notes provided must exceed the count of notes of any other denomination.

Consider an instance where Rs.3500 is dispensed in 500 rupee notes (comprising 7 notes). The remaining Rs.1500 must then be dispensed using Rs.100 and Rs.200 notes. In this scenario, the minimum number of other denomination notes needed is 8 (achieved by dispensing 7 notes of Rs.200 and 1 note of Rs.100, totaling Rs.1400 + Rs.100 = Rs.1500). Consequently, at least Rs.4000 must be dispensed using 500 rupee notes for this condition to be met.

Scenario (1):
If Rs.4000 is dispensed using 500 rupee notes, then 8 five-hundred rupee notes are dispensed.
The remaining Rs.1000 cannot be fully dispensed using only 100 rupee notes (as this would require 10 notes).

* If Rs.800 is dispensed using 100 rupee notes (8 notes), then 9 notes are required to dispense the remaining Rs.1000 (8 notes of Rs.100 plus 1 note of Rs.200). This case is therefore invalid.
* If Rs.600 is dispensed using 100 rupee notes (6 notes), then a minimum of 8 notes are required to dispense the remaining Rs.1000 (6 notes of Rs.100 plus 2 notes of Rs.200). This case is also invalid.
* If Rs.400 is dispensed using 100 rupee notes (4 notes), then 7 notes are required to dispense the remaining Rs.1000 (4 notes of Rs.100 plus 3 notes of Rs.200). This is a valid combination.
* If Rs.200 is dispensed using 100 rupee notes (2 notes), then 6 notes are required to dispense the remaining Rs.1000 (2 notes of Rs.100 plus 4 notes of Rs.200). This is also a valid combination.
* The Rs.1000 can also be dispensed using 5 notes of Rs.200. This is a valid combination.

Therefore, within Scenario (1), there are 3 valid ways to dispense the remaining amount.

Scenario (2):
If Rs.4500 is dispensed using 500 rupee notes, then 9 five-hundred rupee notes are dispensed.
The remaining Rs.500 can be dispensed as 100 rupee notes (5 notes) or a combination of 100 rupee and 200 rupee notes, following the equation 200\*a + 100\*b = 500. The possible values for ‘a’ (number of 200 rupee notes) are 0, 1, and 2, leading to 3 valid combinations.

Scenario (3):
If Rs.5000 is dispensed using 500 rupee notes, then 10 five-hundred rupee notes are dispensed.
There is only 1 valid way to dispense the remaining amount in this case (which is zero).

It is stated that the automated teller machine (ATM) could serve only 10 customers with a stock of fifty 500 rupee notes. We need to determine the maximum number of customers who could have specified Rs.500 as their preferred denomination.

The minimum number of 500 rupee notes required to serve a customer who prefers this denomination is 8 (as determined from Scenario 1). With a stock of fifty 500 rupee notes, we can serve a maximum of [50 / 8] = 6 customers. Therefore, 6 is the correct answer.

Ans: A

It is stated that the maximum number of notes a customer can receive is 20. Additionally, there is a constraint on the quantity of 500 rupee notes (fifty), but no limitation on other denominations. To assist the largest possible number of customers, we must reduce the count of 500 rupee notes given out as much as feasible.

If zero 500 rupee notes are issued, at least 25 notes would be necessary (specifically, 25 notes of 200 rupee denomination).

If one 500 rupee note is issued, a minimum of one 100 rupee note and twenty-two 200 rupee notes are needed. The total count of notes in this scenario is 1 + 1 + 22 = 24. Thus, this scenario can be disregarded.

If two 500 rupee notes are issued, at least twenty 200 rupee notes would be required. This scenario can also be excluded as the total number of notes exceeds 20.

If three 500 rupee notes are issued, a minimum of one 100 rupee note and seventeen 200 rupee notes are necessary. The total notes required here is 3 + 1 + 17 = 21. Therefore, this scenario is also eliminated.

If four 500 rupee notes are issued, at least fifteen 200 rupee notes would be required. The total notes needed in this case is 4 + 15 = 19, which is less than 20. Consequently, this is a feasible scenario.

The smallest quantity of 500 rupee notes that allows serving a customer without exceeding 20 notes in total is 4. Therefore, with 50 five hundred rupee notes, a maximum of [50/4] = 12 customers can be served, making option A the correct choice.

Ans: A

It is stated that the overall quantity of notes issued is the smallest feasible. Consequently, our objective is to reduce the number of notes dispensed in each of the two scenarios presented.

For a customer requesting 500 rupees as their preferred denomination, the minimum number of notes needed is 10.
When serving 50 customers who have specified a preference for 500 rupee notes, we will need 50 multiplied by 10, totaling 500 notes.

Now, let’s analyze the situation where a customer’s preference is 100 rupees.
As previously observed, the fewest notes are required when we prioritize the use of five hundred rupee notes to the greatest extent possible.

If 4000 rupees are dispensed using 500 rupee notes, the remaining 1000 rupees can be issued with ten 100 rupee notes. In this specific configuration, the count of 100 rupee notes (10) is higher than the count of 500 rupee notes (8). This scenario is permissible. We must determine if further reductions in the required number of notes are achievable.

We cannot increase the allocation of 500 rupee notes to 9, as only five hundred rupee notes can be dispensed, which contradicts the customer’s stated preference for 100 rupee notes.

Should we substitute two 100 rupee notes with a single 200 rupee note, the quantity of 100 rupee notes will decrease to 6. The number of 500 rupee notes (8) then surpasses the number of 100 rupee notes (6). Therefore, distributing 4000 rupees using 500 rupee notes and the remainder using 100 rupee notes represents the most efficient distribution.

The lowest number of notes to serve one customer is 8 (five hundred notes) plus 10 (hundred notes), equalling 18 notes.
The quantity of five hundred rupee notes needed to serve 50 customers is 8 multiplied by 50, which equals 400.
Hence, the aggregate number of notes required is 400 plus 500, resulting in 900.

Thus, option A is the correct selection.