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MBA Aspirant

@Vaibhav

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Created new thread Here's the extracted text: Q.29 Rahul has ‘x’ kg of Rice with him. Ritika buys 1/3rd of this amount plus (1/2) kg of rice. Rima buys 1/4th of the remaining amount plus (1/5) kg of rice. Thereafter, no rice is left in the shop. Find out the range of the possible value of ‘x’. a. 0 < x < 0.5 b. 1.5 < x < 2 c. 0.5 < x < 1.5 d. Cannot be determined e. None of these

1 month ago

Here's the extracted text: Q.29 Rahul has ‘x’ kg of Rice with him. Ritika buys 1/3rd of this amount plus (1/2) kg of rice. Rima buys 1/4th of the remaining amount plus (1/5) kg of rice. Thereafter, no rice is left in the shop. Find out the range of the possible value of ‘x’. a. 0 < x < 0.5 b. 1.5 < x < 2 c. 0.5 < x < 1.5 d. Cannot be determined e. None of these
Created new thread Here's the extracted text: Q.29 Rahul has ‘x’ kg of Rice with him. Ritika buys 1/3rd of this amount plus (1/2) kg of rice. Rima buys 1/4th of the remaining amount plus (1/5) kg of rice. Thereafter, no rice is left in the shop. Find out the range of the possible value of ‘x’. a. 0 < x < 0.5 b. 1.5 < x < 2 c. 0.5 < x < 1.5 d. Cannot be determined e. None of these

1 month ago

Here's the extracted text: Q.29 Rahul has ‘x’ kg of Rice with him. Ritika buys 1/3rd of this amount plus (1/2) kg of rice. Rima buys 1/4th of the remaining amount plus (1/5) kg of rice. Thereafter, no rice is left in the shop. Find out the range of the possible value of ‘x’. a. 0 < x < 0.5 b. 1.5 < x < 2 c. 0.5 < x < 1.5 d. Cannot be determined e. None of these

The correct answer to the question is c. 0.5 < x < 1.5. Explanation: 1. Ritika's Purchase: Ritika buys \( \frac{x}{3} + \frac{1}{2} \) kg of rice. - Remaining rice after Ritika's purchase: \[ x - \left(\frac{x}{3} + \frac{1}{2}\right) = \frac{2x}{3} - \frac{1}{2} \] 2. Rima's Purchase: Rima buys \( \frac{1}{4} \) of the remaining amount plus \( \frac{1}{5} \) kg: - Rima's purchase is: \[ \frac{1}{4} \left(\frac{2x}{3} - \frac{1}{2}\right) + \frac{1}{5} \] - Remaining amount after Rima's purchase: \[ \left(\frac{2x}{3} - \frac{1}{2}\right) - \left( \frac{1}{4} \left(\frac{2x}{3} - \frac{1}{2}\right) + \frac{1}{5} \right) = 0 \] 3. Equating and simplifying provides a range for \( x \): - After this computation, it results in: \[ 20x - 23 = 0 \Rightarrow x = 1.15 \] Thus, the value of \( x \) can lie within the range 0.5 < x < 1.5【4:7†source】.
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IIM Mumbai Admissions Queries 2025-27

Welcome to the Official Admissions Thread for Indian Institute of Management Mumbai - 2025-27 This thread is dedicated to candidates who have been shortlist...
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