CGPSC 2025 Paper 2 (CSAT) Questions with Solutions PART I
Q1. Sadharan byaj (Simple Interest) ki ek nishchit dar se koi dhan 10 varsh me doguna ho jata hai, toh kitne varsh me yeh tiguna ho jayega?
(A) 15 varsh
(B) 20 varsh
(C) 25 varsh
(D) 30 varsh
- Answer: (B) 20 varsh
- Solution: Mano Mooladhan (Principal) = P. 10 varsh me doguna (2P) hota hai, yani Byaj (Interest) = 2P - P = P. Formula: \text{Byaj} = \frac{P \times R \times T}{100} \implies P = \frac{P \times R \times 10}{100} \implies R = 10\%. Ab dhan ko tiguna (3P) hona hai, toh naya Byaj = 3P - P = 2P. 2P = \frac{P \times 10 \times T}{100} \implies T = 20\text{ varsh}.
Q2. Yadi Shaloo, Rakhi se lambi hai lekin Prerna se chhoti hai aur Rakhi, Meghna jitni lambi hai lekin Komal se lambi hai, toh Meghna hai:
(A) Shaloo se badi
(B) Komal se chhoti
(C) Shaloo jitni lambi
(D) Shaloo se chhoti
- Answer: (D) Shaloo se chhoti
- Solution: Kathan ke anusar lambai ka kram (Order of height): \text{Prerna} > \text{Shaloo} > \text{Rakhi} = \text{Meghna} > \text{Komal}. Is kram se saaf hai ki Meghna, Shaloo se chhoti hai.
Q3. (8 - 4\sqrt{3}) ka vargmool (Square Root) kya hai?
(A) (2 - \sqrt{3})
(B) (2\sqrt{3} - 1)
(C) (\sqrt{6} - \sqrt{2})
(D) (\sqrt{6} - \sqrt{3})
- Answer: (C) (\sqrt{6} - \sqrt{2})
- Solution: Hume \sqrt{8 - 4\sqrt{3}} nikalna hai. Ise hum (a-b)^2 = a^2 + b^2 - 2ab ke roop me likh sakte hain: 8 - 4\sqrt{3} = 6 + 2 - 2(\sqrt{6})(\sqrt{2}) = (\sqrt{6})^2 + (\sqrt{2})^2 - 2\sqrt{6}\sqrt{2} = (\sqrt{6} - \sqrt{2})^2. Isliye vargmool = (\sqrt{6} - \sqrt{2}) hoga.
Q4. Yadi A, B, C ek vritt (circle) par teen bindu hain jiska kendra O hai tatha \angle AOB = 90^\circ evam \angle BOC = 120^\circ hai, tab \angle ABC barabar hai:
(A) 60^\circ
(B) 75^\circ
(C) 90^\circ
(D) 135^\circ
- Answer: (B) 75^\circ
- Solution: Kendra par bana kul kon \angle AOC = 360^\circ - (90^\circ + 120^\circ) = 150^\circ. Vritt ke niyam ke anusar, paridhi (circumference) par bana kon kendra par bane kon ka aadha hota hai. Isliye, \angle ABC = \frac{\angle AOC}{2} =\frac{150^\circ}{2} = 75^\circ.
Q6. Decision Making: Aap ek shodhlaboratory (research lab) me karyarat hain. Aapke adhinasthon (subordinates) me se ek ne apne dam par kuch achha karya kiya hai aur yeh karya ek pratishthit patrika me prakashan ke yogya hai. Aap kya karenge?
(A) Sabhi aankde lekar apne naam se prakashit karenge
(B) Adhinasth se lekhako me apna naam bhi jodne ke liye kahenge
(C) Apne adhinasth ko batayenge ki jab tak aapko sah-lekhak nahi banaya jata, aap use paper prakashit karne ki anumati nahi denge
(D) Adhinasth ko apne dam par paper prakashit karne ke liye protsahit karenge
- Answer: (D) Adhinasth ko apne dam par paper prakashit karne ke liye protsahit karenge
- Solution: Ek aadarsh prabandhak aur naitik (ethical) leader ke roop me aapko apne subordinates ke hard work ko credit dena chahiye aur unke bhabishya ke liye unhe motivate karna chahiye.
Q7. Ek aadmi naav ko dhara ke anukul (downstream) 20 km, 2 ghante me tatha dhara ke pratikul (upstream) 4 km, 2 ghante me chala sakta hai. Sthir jal me naav chalane ki uski gati hogi:
(A) 4 km/h
(B) 6 km/h
(C) 8 km/h
(D) 10 km/h
- Answer: (B) 6 km/h
- Solution: Dhara ke anukul gati (v + u) = \frac{20}{2} = 10\text{ km/h}. Dhara ke pratikul gati (v - u) = \frac{4}{2} = 2\text{ km/h}. Sthir jal me naav ki gati (v) = \frac{(10 + 2)}{2} = 6\text{ km/h}.
Q8. Ek छात्र apne ghar se 4 km/h ki gati se chalta hai aur apne school 5 minute deri se pahunchta hai. Yadi uski gati 5 km/h ho, toh vah 10 minute pehle pahunch jata hai. Uske ghar se school ki doori hai:
(A) 3 km
(B) 5 km
(C) 6 km
(D) 4 km
- Answer: (B) 5 km
- Solution: Formula: \text{Doori (Distance)} = \frac{S_1 \times S_2}{S_2 - S_1} \times (\text{Samay ka antar}). Yahan, S_1 = 4\text{ km/h}, S_2 = 5\text{ km/h}. Samay ka antar = 5 min deri + 10 min pehle = 15 minute = \frac{15}{60}\text{ ghante} = \frac{1}{4}\text{ ghante}. \text{Doori} = \frac{4 \times 5}{5 - 4} \times \frac{1}{4} = 20 \times \frac{1}{4} = 5\text{ km}.
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