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)›Number 198 is divided into three parts such that 1/2 of the first

Number 198 is divided into three parts such that 1/2 of the first part, 1/3 of the second part and 1/4 of the third part shall all be equal. Then the value of the largest part will be :

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A.(A) 68
B.(B) 78
C.(C) 88Correct
D.(D) 98
Explanation & Solution

The correct answer is (C) 88.

Step-by-step mathematical working:

  • Let the three parts of 198 be x,y,zx, y, zx,y,z.
  • Given: x2=y3=z4=kx2\frac{x}{2}2x​ = y3\frac{y}{3}3y​ = z4\frac{z}{4}4z​ = k2x​=3y​=4z​=k.
  • Therefore: x=2k,  y=3k,  z=4kx = 2k, \; y = 3k, \; z = 4kx=2k,y=3k,z=4k.
  • Sum of parts =2k+3k+4k=9k=198  ⟹  k=1989=22= 2k + 3k + 4k = 9k = 198   ⟹  \implies⟹ k = 1989\frac{198}{9}9198​ = 22=2k+3k+4k=9k=198⟹k=9198​=22.
  • The parts are: x=2(22)=44,  y=3(22)=66,  z=4(22)=88x = 2(22) = 44, \; y = 3(22) = 66, \; z = 4(22) = 88x=2(22)=44,y=3(22)=66,z=4(22)=88.
  • The value of the largest part is 88.

हिंदी में प्रश्न एवं आदर्श उत्तर

संख्या 198 को तीन भागों में इस प्रकार विभाजित किया जाता है कि पहले भाग का 1/2, दूसरे भाग का 1/3 एवं तीसरे भाग का 1/4, सभी बराबर हैं। तब सबसे बड़े भाग का मान होगा :

A.(A) 68
B.(B) 78
C.(C) 88सही
D.(D) 98

सही उत्तर (C) 88 है।

चरणबद्ध गणितीय हल:

  • माना 198 के तीन भाग x,y,zx, y, zx,y,z हैं।
  • प्रश्नानुसार: x2=y3=z4=kx2\frac{x}{2}2x​ = y3\frac{y}{3}3y​ = z4\frac{z}{4}4z​ = k2x​=3y​=4z​=k।
  • अतः: x=2k,  y=3k,  z=4kx = 2k, \; y = 3k, \; z = 4kx=2k,y=3k,z=4k।
  • तीनों भागों का योग =2k+3k+4k=9k=198  ⟹  k=22= 2k + 3k + 4k = 9k = 198   ⟹  \implies⟹ k = 22=2k+3k+4k=9k=198⟹k=22।
  • तीनों भाग: x=44,  y=66,  z=88x = 44, \; y = 66, \; z = 88x=44,y=66,z=88 हैं।
  • अतः सबसे बड़े भाग का मान 88 है।
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