Home›MPPSC Prelims›Paper 2: General Aptitude Test (CSAT)›General Aptitude›Basic Numeracy and Data Interpretation›Basic Numeracy: Numbers and their Relations, Order of Magnitude (Class X level)›The average of 4 consecutive odd numbers is 18. Then the product of

The average of 4 consecutive odd numbers is 18. Then the product of the largest and smallest numbers among them is :

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A.(A) 187
B.(B) 247
C.(C) 315Correct
D.(D) 391
Explanation & Solution

The correct answer is (C) 315.

Step-by-step algebra:

  • Let the 4 consecutive odd numbers be (x−3),(x−1),(x+1),(x+3)(x - 3), (x - 1), (x + 1), (x + 3)(x−3),(x−1),(x+1),(x+3).
  • Their average is x=18x = 18x=18.
  • The four numbers are: 18−3=15,  18−1=17,  18+1=19,  18+3=2118 - 3 = 15, \; 18 - 1 = 17, \; 18 + 1 = 19, \; 18 + 3 = 2118−3=15,18−1=17,18+1=19,18+3=21.
  • Smallest number = 15, Largest number = 21.
  • Product of the largest and smallest numbers =15×21=<strong>315</strong>= 15 ×\times× 21 = <strong>315</strong>=15×21=<strong>315</strong>.

हिंदी में प्रश्न एवं आदर्श उत्तर

4 क्रमागत विषम संख्याओं का औसत 18 है। इन संख्याओं में से सबसे बड़ी तथा सबसे छोटी संख्याओं का गुणनफल है :

A.(A) 187
B.(B) 247
C.(C) 315सही
D.(D) 391

सही उत्तर (C) 315 है।

चरणबद्ध बीजगणितीय हल:

  • माना 4 क्रमागत विषम संख्याएं (x−3),(x−1),(x+1),(x+3)(x - 3), (x - 1), (x + 1), (x + 3)(x−3),(x−1),(x+1),(x+3) हैं।
  • इनका औसत x=18x = 18x=18 है।
  • अतः चारों संख्याएं क्रमशः: 15,17,19,2115, 17, 19, 2115,17,19,21 हैं।
  • सबसे छोटी संख्या = 15 तथा सबसे बड़ी संख्या = 21।
  • गुणनफल =15×21=<strong>315</strong>= 15 ×\times× 21 = <strong>315</strong>=15×21=<strong>315</strong>।
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