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UGC-NET-Paper1

Maths

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A man travels three-fifth of a distance AB at a speed of 3a and the remaining at the speed of 2b. If he goes from B to A and returns at a speed of 5 in the same time then, :

1. 1/a + 1/b = 1/c
2. a + b = c
3. 1/a + 1/b = 2/c
4. 1/a + 1/b = 1/2c

एक व्यक्ति AB दूरी का तीन-पांचवां भाग 3a की गति से तथा शेष भाग 2b की गति से तय करता है। यदि वह B से A तक जाता है तथा उसी समय में 5 की गति से वापस आता है, तो:

1. 1/a + 1/b = 1/c
2. a + b = c
3. 1/a + 1/b = 2/c
4. 1/a + 1/b = 1/2c
This Question came in
UGC-NET-EnvSci-21-January-2025-Shift-1-Q33
UGC-NET Paper 1 Full Course

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Detailed Explanation & Answer
Step 1: Define Variables
Let total distance be **D**
First part: (3/5)D at speed **3a**
Time taken = (3/5)D / 3a = (1/5)D / a

Second part: (2/5)D at speed **2b**
Time taken = (2/5)D / 2b = (1/5)D / b

Total time for AB = (1/5)D (1/a + 1/b)

Step 2: Compute Return Time
Speed for BA and back = **5**
Total distance = **2D**
Total time = 2D / 5 = (2/5)D

Step 3: Equating Times
(1/5)D (1/a + 1/b) = (2/5)D / c

Cancel **D/5** on both sides:
1/a + 1/b = 2/c

Final Answer:
**(3) 1/a + 1/b = 2/c**
चरण 1: चर परिभाषित करें
कुल दूरी **D** मान लें
पहला भाग: (3/5)D गति **3a** पर
लगा समय = (3/5)D / 3a = (1/5)D / a

दूसरा भाग: (2/5)D गति **2b** पर
लगा समय = (2/5)D / 2b = (1/5)D / b

AB के लिए कुल समय = (1/5)D (1/a + 1/b)

चरण 2: वापसी समय की गणना करें
BA और वापस जाने की गति = **5**
कुल दूरी = **2D**
कुल समय = 2D / 5 = (2/5)D

चरण 3: समय की बराबरी करना
(1/5)D (1/a + 1/b) = (2/5)D / c

**D/5** को रद्द करें दोनों पक्ष:
1/a + 1/b = 2/c

अंतिम उत्तर:
**(3) 1/a + 1/b = 2/c**
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