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Here's the extracted text: --- Q.27) A BPO service provider engages male and female operators for answering 2600 calls per day. A male operator can handle 75 calls per day whereas a female operator can handle 110 calls per day. The male and the female operators get a fixed wage of ₹500 and ₹650 per day respectively. In addition, a male operator gets ₹3.50 per call he answers and a female operator gets ₹2.25 per call she answers. What is the least average daily cost per operator (male & female combined), assuming he can employ a maximum of 20 female and male employees each? a. ₹827.5 b. ₹807.5 c. ₹847.5 d. ₹817.5 e. None of these ---

The answer to the question regarding the least average daily cost per operator (male & female combined) is ₹807.50. Explanation: 1. Cost Calculation for Female Operators: - A female operator can handle 110 calls per day. - The fixed wage for a female operator is ₹650. - Bonus: Each female operator earns an additional ₹2.25 per call. - Therefore, for 110 calls, the bonus is: \( \text{Bonus} = 110 \times 2.25 = ₹247.50 \). - Hence, total cost for a female operator: \( \text{Total cost (female)} = 650 + 247.50 = ₹897.50 \). 2. Cost Calculation for Male Operators: - A male operator can handle 75 calls per day. - The fixed wage for a male operator is ₹500. - Bonus: Each male operator earns an additional ₹3.50 per call. - So for 75 calls, the bonus is: \( \text{Bonus} = 75 \times 3.50 = ₹262.50 \). - Thus, total cost for a male operator: \( \text{Total cost (male)} = 500 + 262.50 = ₹762.50 \). 3. Operational Needs: - The BPO needs to handle 2600 calls per day. - First, utilize 20 male operators. Hence, calls answered by males: \( 20 \times 75 = 1500 \) calls. - Remaining calls to be handled by female operators: \( 2600 - 1500 = 1100 \) calls. Hence, necessary female operators: \( \frac{1100}{110} = 10 \) female operators. 4. Total Cost Calculation: - Total cost for 20 male operators: \( 20 \times 762.50 = ₹15,250 \). - Total cost for 10 female operators: \( 10 \times 897.50 = ₹8,975 \). - Overall total cost: \( 15,250 + 8,975 = ₹24,225 \). 5. Average Daily Cost: - Total number of operators: 30 (20 male + 10 female). - The average daily cost = \( \frac{₹24,225}{30} = ₹807.50 \). Thus, the least average daily cost per operator is ₹807.50【4:5†source】.

com Q.27) A BPO service provider engages male and female operators for answering 2600 calls per day. A male operator can handle 75 calls per day whereas a female operator can handle 110 calls per day. The male and the female operators get a fixed wage of *500 and 650 per day respectively. In addition, a male operator gets 3.50 per call he answers and a female operator gets *2.25 per call she answers. What is the least average daily cost per operator (male & female combined. assuming he can employ a maximum of 20 female and male employees each? a. 827.5 b. 807.5 c. *847.5 d. 817.5 e. None of these riddhijain8

The least average daily cost per operator (male & female combined) for the BPO service provider, based on the calculations from the document, is ₹807.50. Explanation: 1. Male Operators: - Handle 75 calls per day. - Fixed wage: ₹500. - Bonus: ₹3.50 per call, totaling ₹3.50 * 75 = ₹262.50. - Total daily cost for a male operator: ₹500 + ₹262.50 = ₹762.50. 2. Female Operators: - Handle 110 calls per day. - Fixed wage: ₹650. - Bonus: ₹2.25 per call, totaling ₹2.25 * 110 = ₹247.50. - Total daily cost for a female operator: ₹650 + ₹247.50 = ₹897.50. 3. Total Call Handling: - Total calls to be answered: 2600. - If using 20 male operators, they can handle 20 * 75 = 1500 calls. - Remaining calls: 2600 - 1500 = 1100 that need to be handled by female operators. - Number of female operators required: 1100 / 110 = 10. 4. Total Cost Calculation: - Cost for 20 male operators: 20 * ₹762.50 = ₹15,250. - Cost for 10 female operators: 10 * ₹897.50 = ₹8,975. - Total cost for all operators = ₹15,250 + ₹8,975 = ₹24,225. 5. Average Cost Per Operator: - Total number of operators = 20 male + 10 female = 30. - Average cost = Total cost / Number of operators = ₹24,225 / 30 = ₹807.50【4:1†source】.

com Q.27) A BPO service provider engages male and female operators for answering 2600 calls per day. A male operator can handle 75 calls per day whereas a female operator can handle 110 calls per day. The male and the female operators get a fixed wage of *500 and 650 per day respectively. In addition, a male operator gets 3.50 per call he answers and a female operator gets *2.25 per call she answers. What is the least average daily cost per operator (male & female combined. assuming he can employ a maximum of 20 female and male employees each? a. 827.5 b. 807.5 c. *847.5 d. 817.5 e. None of these riddhijain8

Sure! Here’s the extracted text: --- Q.80) ₹16/- in the form of ₹1/- coins are to be distributed among A, B, C, and D such that each boy receives at least one rupee and each of them receives a different amount. D receives 4 more rupees than B. B should get more rupees than C but less rupees than A. What is the difference between the maximum and minimum rupees that A can have? a) 4 b) 3 c) 2 d) 1 e) None of these --- Let me know if you need any further assistance!

The problem states that ₹16/- in coins are to be distributed among four individuals A, B, C, and D such that each receives a different amount, with specific conditions: 1. Each person receives at least one rupee. 2. D receives 4 more rupees than B. 3. B receives more than C but less than A. To solve this, let's represent: - A's amount as \(a\), - B's amount as \(b\), - C's amount as \(c\), - D's amount as \(d\). Given: - \(a + b + c + d = 16\) - \(d = b + 4\) - \(b > c\) - \(b < a\) Substituting \(d\) in the total equation gives: \[ a + b + c + (b + 4) = 16 \] This simplifies to: \[ a + 2b + c = 12 \quad (1) \] Considering the constraints Since each must receive a different positive integer: 1. Start with \(b\) being the smallest integer greater than \(c\). 2. Set minimum values accordingly to find maximum and minimum for \(a\). Let's test a few values for \(b\): - Assume \(b = 2\): - Then \(d = 6\), i.e., \(d = b + 4\) - From equation (1): \[ a + 2 + c + 6 = 16 \implies a + c = 8 \] This results in possible pairs: - \(c = 1\), then \(a = 7\) (valid). - Values are \(c = 1, b = 2, d = 6, a = 7\). For maximum value of \(a\), we should keep \(b\) to lowest: - Checking values like: - \(b = 3\) results in valid amounts but decreases \(a\). Final Evaluation After checking variations, we find the only viable pairings resulting lead to: 1. Minimum \(a = 7\) 2. Maximum \(a = 5\) if \(b = 3\). Thus: The maximum value of \(a\) = 7 and minimum = 5, giving: - The difference = \(4\). Hence the solution is 4. The difference between the maximum and minimum rupees that A can have is 4【4:0†source】.

Here’s the extracted text from the image: --- Q.83) Aman started a job which he could finish, working alone, in 36 days. After 4 days, when he was asked to finish the job within 27 days, he asked his friend Bala to help him. Bala joined him from the very next day and they worked for 2 days. Then after Aman left the job due to family emergency and asked Bala to complete the job. Bala’s cousin, Chetan also accompanied him after 6 more days. In those, Bala took 2 holidays due to bad health. If Chetan alone can finish the job at most in 60 days, which of the following can’t be a number of days in which Bala alone can do the job? (The job was done on at the end of 27 days.) a) 40 b) 32 c) 36 d) 48 e) None of these --- If you need further assistance, feel free to ask!

The answer to the question is that Bala cannot finish the job in 36 days. Here's how the calculations were made: 1. Work Done by Aman: - Aman can finish the job alone in 36 days. - In 4 days of work, Aman completes \( \frac{4}{36} = \frac{1}{9} \) of the job. 2. Remaining Work After 4 Days: - Remaining work after Aman has worked for 4 days is \( 1 - \frac{1}{9} = \frac{8}{9} \). 3. Bala Joins: - Bala and Aman work together for 2 days. - Combined work rate of Aman and Bala needs to be determined by introducing the variable \( x \) for the number of days Bala needs to finish the work alone. 4. Work Done in 2 Days: - Let \( x \) be the number of days Bala would take to finish it alone. - Their work rate for 2 days become \( \frac{1}{36} + \frac{1}{x} \) for 2 days. - Therefore, combined work in 2 days is \( 2 \left(\frac{1}{36} + \frac{1}{x}\right) \). 5. Total Work Rate: - After 6 days of total work (4 days by Aman and 2 days with Aman and Bala), Aman leaves, and Bala continues for 21 days. - Chetan joins after 6 days and works for 15 days in total. 6. Chetan's Work Contribution: - Given Chetan can finish the job in 60 days, his contribution can be calculated. 7. Setting up the Equation: - The equation formed from all work would be: \[ \frac{6}{36} + 2 \left( \frac{1}{36} + \frac{1}{x} \right) + \frac{15}{60} = 1 \] Simplifying this leads to the conclusion that \( x \) must be greater than 36 for a solution to be consistent. Hence, any value for \( x \) less than 36 is not possible, narrowing down the answer to option (b) 36 days【4:3†source】.

Q.82) Apollo 2020 was sent into space to confirm the presence of aliens on a planet Naeem. The astronaut was wearing a watch which was running perfectly fine till he was on earth. After arriving on the planet, Naeem he noticed that the minute and hour hands are meeting every hour. Approximately, how much time did the watch gained/lost in an earth day terms? a) Lost 120 minutes in a day b) Gained 120 minutes in a day c) Gained 100 minutes in a day d) Lost 90 minutes in a day e) None of these

The answer to the question is that the watch gained 120 minutes in a day. Explanation: The watch's hands meet every hour, which means that instead of the usual 12 meetings in 12 hours for a normal watch, it meets every hour on the planet Naeem. On Earth, in 12 hours, there are 11 such meetings (at 0, 1, 2, ... 10, 11). Thus, on Earth, the minute hand traverses 360° – so every hour is equal to 30° gained over one hour of the clock face. On planet Naeem, however, to meet every hour means the minute hand must traverse a full 360° in every hour. Therefore, for the hands to meet, the minute hand effectively gains more relative angular speed compared to the Earth. Calculating this: 1. The minute hand on Earth moves at 6° per minute and takes 60 minutes to traverse 360°. 2. On Naeem, the watch gains 30° extra every hour (due to moving from 360° back to 0°), which translates into: \[ \text{Gains in 60 minutes} = 30° \] \[ \text{Total gains in one day (24 hours)} = 24 \times 30 = 720° \] This corresponds to: \[ \text{Total minutes gained} = \frac{720°}{6° \text{ per minute}} = 120 \text{ minutes} \] Thus, the watch gained 120 minutes in an Earth day, leading to option b) being the correct answer .

Q.80) ₹16/- in the form of ₹1/- coins are to be distributed among A, B, C, and D such that each boy receives at least one rupee and each of them receives a different amount. D receives 4 more rupees than B. B should get more rupees than C but less rupees than A. What is the difference between the maximum and minimum rupees that A can have? a) 4 b) 3 c) 2 d) 1 e) None of these

To solve the problem, we can summarize the information given and set up equations based on the requirements: 1. Let the amounts received by A, B, C, and D be represented as \(a\), \(b\), \(c\), and \(d\) respectively. 2. According to the problem: - Each boy receives at least one rupee and receives a different amount. - D receives 4 more rupees than B: \(d = b + 4\) - B receives more rupees than C but less than A: \(c < b < a\) 3. The total amount distributed is ₹16: \(a + b + c + d = 16\). From these conditions, substituting \(d\) in the total amount: \[ a + b + c + (b + 4) = 16 \] This can be simplified to: \[ a + 2b + c + 4 = 16 \quad \Rightarrow \quad a + 2b + c = 12 \] Now, we know: - Each amount is different, and \(a\), \(b\), \(c\), and \(d\) must be at least 1 rupee each. To explore the possible amounts: 1. The smallest values for \(c\) and \(b\) could be: - For \(c = 1\): Then \(b\) would need to be at least 2, - Thus \(a\) needs to satisfy: \(a + 2(2) + 1 = 12 \rightarrow a = 8\) (which is valid, \(d = 6\)). So one configuration is: - \(a = 8\), \(b = 2\), \(c = 1\), \(d = 6\). From this setup, we can explore the maximum and minimum amounts that A can get. - Exploring higher values for \(b\): - \(b = 3\) -> \(c = 2\) -> \(d = 7\) -> \(a + 2(3) + 2 = 12 \rightarrow a = 4\). - . . . - Adjusting lower values will find more configurations while keeping distinct values. Finally: - Minimum possible \(a = 4\) - Maximum possible \(a = 8\) Calculating the difference: \[ \text{Difference} = \text{Maximum} - \text{Minimum} = 8 - 4 = 4 \] Hence, the answer is 4. The answer choice is (a) 4, and the difference between the maximum and minimum rupees that A can receive is indeed 4【4:0†source】.

Q.80) ₹16/- in the form of ₹1/- coins are to be distributed among A, B, C, and D such that each boy receives at least one rupee and each of them receives a different amount. D receives 4 more rupees than B. B should get more rupees than C but less rupees than A. What is the difference between the maximum and minimum rupees that A can have? a) 4 b) 3 c) 2 d) 1 e) None of these

The answer to the question regarding the distribution of ₹16 among A, B, C, and D is 3. Explanation: To distribute the coins, we need to assign different amounts to each boy under the following constraints: 1. Each boy receives at least 1 rupee. 2. D receives 4 more rupees than B. 3. B receives more than C but less than A. Let’s denote: - C receives \( c \) rupees. - B receives \( b \) rupees (where \( b > c \)). - A receives \( a \) rupees (where \( a > b \)). - D receives \( d \) rupees (where \( d = b + 4 \)). The equation governing the total distribution can be set up as: \[ a + b + c + d = 16 \] Substituting \( d \) with \( b + 4 \): \[ a + b + c + (b + 4) = 16 \] \[ a + 2b + c + 4 = 16 \] \[ a + 2b + c = 12 \] To explore different distributions that satisfy all conditions: 1. The minimum value for \( b \) can be 2 (since B should get more than C). If \( b \) is 2, \( d \) becomes 6. Hence, we can explore further: - If \( b = 2 \), then: - \( d = 6 \) - Substitute back to find \( c \) and \( a \): \[ a + 2 + c = 12 \] \[ a + c = 10 \] The distribution could be: - If \( c = 1 \), then \( a = 9 \) (values: A=9, B=2, C=1, D=6). - If \( c = 2 \), then \( a = 8 \) (values: A=8, B=2, C=2, D=6 - not valid since C and B must be different). - If \( c = 3 \), then \( a = 7 \) (values: A=7, B=3, C=3, D=7 - not valid). Continuing this way, valid distributions yield \( A \) as either 7 or 5 while keeping \( b \) as high as 3 or 2. Evaluating the difference for \( A \): - Maximum value of \( A \): 7 - Minimum value of \( A \): 4 Thus, the difference between the maximum and minimum rupees that A can receive is: \[ 7 - 4 = 3 \] This concludes the reasoning, confirming that the difference between the maximum and minimum amounts that A can have is 3【4:12†source】.

Sure! Here is the extracted text from the image: "Q.80) ₹16/- in the form of ₹1/- coins are to be distributed among A, B, C, and D such that each boy receives at least one rupee and each of them receives a different amount. D receives 4 more rupees than B. B should get more rupees than C but less rupees than A. What is the difference between the maximum and minimum rupees that A can have? a) 4 b) 3 c) 2 d) 1 e) None of these"

To solve the problem regarding distributing ₹16 among A, B, C, and D under the given conditions, let's summarize the requirements: 1. Each person receives at least one rupee and different amounts. 2. D receives 4 more rupees than B. 3. B receives more than C but less than A. Let's denote the amounts received by A, B, C, and D as a, b, c, and d respectively. From the conditions, we can express them as follows: - \( d = b + 4 \) (D receives 4 more than B) - \( b > c \) (B receives more than C) - \( b < a \) (B receives less than A) The total amount distributed among A, B, C, and D is: \[ a + b + c + d = 16 \] Substituting \( d \) gives us: \[ a + b + c + (b + 4) = 16 \] This simplifies to: \[ a + 2b + c = 12 \] Considering the constraints discussed, let's find the maximum and minimum values for A while complying with the conditions. Calculation Steps 1. Identify the possible values: - Since \( b \) (the amount for B) must be greater than C, let's start by assigning the minimum values for each: - If C receives 1 rupee, B must receive more than this (at least 2). - Assume values: - Let \( c = 1 \) - Then \( b \) can be 2 (the smallest possible). - Hence, D then becomes \( d = 2 + 4 = 6 \). 2. Calculate A: - Plugging in the values: \[ a + 2 + 1 + 6 = 16 \implies a + 9 = 16 \implies a = 7 \] 3. Check feasibility with other values for B and adjustments: - If \( c = 1 \), and B = 3 (next smallest): - Then D = 3 + 4 = 7. - This results in: \[ a + 3 + 1 + 7 = 16 \implies a + 11 = 16 \implies a = 5 \] 4. Minimum value for A: - Continuing this pattern, one could conclude that for \( b = 3\), we can test: - \( b = 4\) making D = 8, yielding impossibility along constraints. Accordingly, as B increases, A's maximum could go down or base rises close to its values, leading to a minimum possible value. After testing through these calculations while incrementing both B and D, we find A has the maximum valid count down from 9 vp and can fluctuate at 5 or 6 max, with C at its base. Answer The difference between the maximum and minimum rupees A can have leads to: - Maximum: 7 - Minimum: 5 - Thus, the difference is \( 7 - 5 = 2 \). Therefore, the option is \( 2 \). The correct answer is \( 2 \)【4:0†source】.