TS Police SI Mains 2023 Paper 3 · Question 49
Question 49 of the TSLPRB SI Final Written Examination 2023, Paper III (Arithmetic & Reasoning) (8 April 2023). Try it first, then open the answer and explanation.
[3(2/5 of 5/7) + (5 ÷ 2/9 × 6/5 + 1)7^(−1)] / √289 =
- 3/17
- 2/7
- 13/17
- 12/7
[3(2/5 of 5/7) + (5 ÷ 2/9 × 6/5 + 1)7^(−1)] / √289 =
- 3/17
- 2/7
- 13/17
- 12/7
Show answer
Answer: (2) 2/7
సమాధానం: (2) 2/7
2/5 of 5/7 = 2/7, so the first term is 6/7; 5 ÷ 2/9 × 6/5 = 45/2 × 6/5 = 27, so the bracket is 28 and 28 × 7^(−1) = 4. The numerator is 6/7 + 4 = 34/7, and dividing by √289 = 17 gives 2/7.
2/5 of 5/7 = 2/7, కాబట్టి మొదటి పదం 6/7; 5 ÷ 2/9 × 6/5 = 45/2 × 6/5 = 27, కాబట్టి కుండలీకరణంలోని విలువ 28 మరియు 28 × 7^(−1) = 4. లవం 6/7 + 4 = 34/7, దీనిని √289 = 17 చే భాగిస్తే 2/7 వస్తుంది.