send mail to support@abhimanu.com mentioning your email id and mobileno registered with us! if details not recieved
Resend Opt after 60 Sec.
By Loging in you agree to Terms of Services and Privacy Policy
Claim your free MCQ
Please specify
Sorry for the inconvenience but we’re performing some maintenance at the moment. Website can be slow during this phase..
Please verify your mobile number
Login not allowed, Please logout from existing browser
Please update your name
Subscribe to Notifications
Stay updated with the latest Current affairs and other important updates regarding video Lectures, Test Schedules, live sessions etc..
Your Free user account at abhipedia has been created.
Remember, success is a journey, not a destination. Stay motivated and keep moving forward!
Refer & Earn
Enquire Now
My Abhipedia Earning
Kindly Login to view your earning
Support
Type your modal answer and submitt for approval
A sum of Rsx is divided among A, B and C such that the ratio of the shares of A and B is 7:12 and that of B and C is 8:5. If the
difference in the shares of A and C is 219, then the value of x is:
17,231
15,321
11,607
21,901
Let’s break down what’s happening with the ratios and the shares:
- The ratio of A:B is 7:12.
- The ratio of B:C is 8:5.
- The difference between A and C’s shares = 219.
- Find the total sum (x).
Here’s how you connect those ratios:
- Combine A:B and B:C using B as the link.
- Make B’s term common:
- A:B = 7:12
- B:C = 8:5
- If you adjust so B matches, you get:
- B’s LCM = 24 (12×2 & 8×3)
- So, rewrite:
- A:B = 7×2:12×2 = 14:24
- B:C = 8×3:5×3 = 24:15
- So now, A:B:C = 14:24:15
Let’s name their shares 14k, 24k, and 15k. The difference between A and C is 14k - 15k = -1k (take absolute value if needed).
- So, |14k – 15k| = 219, so k = 219.
Total sum: 14k + 24k + 15k = 53k
That’s 53 × 219 = 11,607
Option 3 is absolutely correct.
- Option 1: 17,231 – Not correct.
- Option 2: 15,321 – Not correct.
- Option 3: 11,607 – THIS IS THE ONE.
- Option 4: 21,901 – Nope.
By: santosh ProfileResourcesReport error
Access to prime resources
New Courses