Question
Download Solution PDFप्रत्येक 9 और 13 से कौन - सी सबसे छोटी धनात्मक संख्या घटाई जानी चाहिए। ताकि उनका तृतीयानुपाती 18 हो?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFदिया गया है:
18 तृतीयानुपाती है।
प्रयुक्त सूत्र:
1. यदि A : B :: C : D है, तो (B × C) = (A × D)
2. (a - b)2 = a2 - 2ab + b2
गणना:
माना सबसे छोटी धनात्मक संख्या Q है।
अनुपात (9 - Q) : (13 - Q) : : (13 - Q) : 18 है।
प्रश्नानुसार,
(13 - Q)2 = (9 - Q) × 18
⇒ 169 + Q2 - 26Q = 162 - 18Q
⇒ Q2 - 8Q + 7 = 0
⇒ (Q - 7)(Q - 1) = 0
इसलिए, सबसे छोटी संख्या = Q = 1
∴ सबसे छोटी संख्या 1 है।
Last updated on Jul 7, 2025
-> The SSC CGL Notification 2025 for the Combined Graduate Level Examination has been officially released on the SSC's new portal – www.ssc.gov.in.
-> This year, the Staff Selection Commission (SSC) has announced approximately 14,582 vacancies for various Group B and C posts across government departments.
-> The SSC CGL Tier 1 exam is scheduled to take place from 13th to 30th August 2025.
-> Aspirants should visit ssc.gov.in 2025 regularly for updates and ensure timely submission of the CGL exam form.
-> Candidates can refer to the CGL syllabus for a better understanding of the exam structure and pattern.
-> The CGL Eligibility is a bachelor’s degree in any discipline.
-> Candidates selected through the SSC CGL exam will receive an attractive salary. Learn more about the SSC CGL Salary Structure.
-> Attempt SSC CGL Free English Mock Test and SSC CGL Current Affairs Mock Test.
-> Candidates should also use the SSC CGL previous year papers for a good revision.