SSC IMD SA Mock 03 – Question 41

Question:

‘L’, ‘M’, ‘N’ and ‘O’ have different amounts. ‘M’ has 3 times the amount ‘L’ has. ‘O’ has 50% less than ‘M’. The amount ‘N’ has equals the average of all four persons. If the sum of amounts ‘L’ and ‘N’ is Rs. 5,600, find the amount ‘N’ has (approximate value).

‘L’, ‘M’, ‘N’ और ‘O’ के पास अलग-अलग राशियाँ हैं। ‘M’ के पास ‘L’ से 3 गुना राशि है। ‘O’ के पास ‘M’ से 50% कम है। ‘N’ के पास की राशि चारों व्यक्तियों के औसत के बराबर है। यदि ‘L’ और ‘N’ की राशियों का योग ₹5,600 है, तो ‘N’ के पास कितनी राशि है (लगभग मान)?

Options:

  • Rs. 2,460
  • Rs. 2,805
  • Rs. 3,200
  • Rs. 3,623
  • ₹2,460
  • ₹2,805
  • ₹3,200
  • ₹3,623

Correct Answer: d)

Explanation:

Let \(L = x\). Then \(M = 3x\) and \(O = 1.5x\).
\(N = \dfrac{x + 3x + 1.5x + N}{4} \implies 4N = 5.5x + N \implies 3N = 5.5x \implies N = \dfrac{11}{6}x\).
Given \(L + N = x + \dfrac{11}{6}x = \dfrac{17}{6}x = 5600 \implies x = \dfrac{5600 \times 6}{17}\).
\(N = \dfrac{11}{6} \times \dfrac{5600 \times 6}{17} = \dfrac{61600}{17} \approx \text{Rs. } 3,623.53 \approx \text{Rs. } 3,623\).

माना \(L = x\)। तब \(M = 3x\) तथा \(O = 1.5x\)।
\(N = \dfrac{5.5x + N}{4} \implies 3N = 5.5x \implies N = \dfrac{11}{6}x\)।
\(L + N = \dfrac{17}{6}x = 5600 \implies N = \dfrac{61600}{17} \approx ₹3,623\)।

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