SSC GD Constable practice question
A cistern already has one-fourth of its capacity filled with water. Pipe A can fill the empty cistern in 12 hours, and a leak can empty a full cistern in 18 hours. If pipe A is opened and the leak continues to act, in how many hours will the cistern become full?
From a SSC GD Constable Quantitative Aptitude practice set · Quantitative Aptitude
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Correct answer: (D) 27 hours
Net rate = 1/12 - 1/18 = (3-2)/36 = 1/36 per hour. Remaining part = 1 - 1/4 = 3/4. Time = (3/4) ÷ (1/36) = (3/4) × 36 = 27 hours.
हिंदी में प्रश्न
एक टंकी में पहले से ही उसकी क्षमता का एक-चौथाई पानी भरा है। पाइप A खाली टंकी को 12 घंटे में भर सकता है, और एक रिसाव पूरी भरी टंकी को 18 घंटे में खाली कर सकता है। यदि पाइप A खोला जाए और रिसाव लगातार सक्रिय रहे, तो टंकी कितने घंटे में पूरी भर जाएगी?
उत्तर जाँचने के लिए किसी विकल्प पर टैप करें।
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सही उत्तर: (D) 27 घंटे
शुद्ध दर = 1/12 - 1/18 = (3-2)/36 = 1/36 प्रति घंटा। शेष भाग = 1 - 1/4 = 3/4। समय = (3/4) ÷ (1/36) = (3/4) × 36 = 27 घंटे।
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