यदि (J + K), 25 है और (J × K), 150 है, तब (J - K) का मान ज्ञात कीजिये|

  1. ±3
  2. ±6
  3. ±5
  4. ±8

Answer (Detailed Solution Below)

Option 3 : ±5
Free
RRB Group D Full Test 1
3.3 Lakh Users
100 Questions 100 Marks 90 Mins

Detailed Solution

Download Solution PDF

दिया है,

(J + K) = 25      ----(1)

और J × K = 150

उपरोक्त समीकरण का वर्ग करने पर हमें प्राप्त होता है

⇒ (J + K)2 = 252

⇒ J2 + K2 + 2JK = 625

⇒ J+ K2 = 625 - 300

⇒ J2 + K= 325      ----(2)

(J - K) प्राप्त करने के लिए, हमें समीकरण (2) से 2JK घटाना है

⇒ J2 + K2 - 2JK = 325 - 2JK

⇒ (J - K)2 = 325 - 300 = 25

∴ (J - K) = ±5

Latest RRB Group D Updates

Last updated on Jul 18, 2025

-> A total of 1,08,22,423 applications have been received for the RRB Group D Exam 2025. 

-> The RRB Group D Exam Date will be announced on the official website. It is expected that the Group D Exam will be conducted in August-September 2025. 

-> The RRB Group D Admit Card 2025 will be released 4 days before the exam date.

-> The RRB Group D Recruitment 2025 Notification was released for 32438 vacancies of various level 1 posts like Assistant Pointsman, Track Maintainer (Grade-IV), Assistant, S&T, etc.

-> The minimum educational qualification for RRB Group D Recruitment (Level-1 posts) has been updated to have at least a 10th pass, ITI, or an equivalent qualification, or a NAC granted by the NCVT.

-> Check the latest RRB Group D Syllabus 2025, along with Exam Pattern.

-> The selection of the candidates is based on the CBT, Physical Test, and Document Verification.

-> Prepare for the exam with RRB Group D Previous Year Papers.

More Function Questions

More Algebra Questions

Get Free Access Now
Hot Links: teen patti casino apk teen patti cash game teen patti neta