विभाजनादरम्यान, प्रांजलने चुकून लाभांश म्हणून मूळ लाभांशापेक्षा 10% जास्त असलेली संख्या घेतली. त्याने चुकून भाजक म्हणून मूळ विभाजकापेक्षा २५% जास्त असलेली संख्या घेतली. जर मूळ भागाकार समस्येचा योग्य भाग 25 असेल आणि उर्वरित 0 असेल, तर प्रांजलने त्याच्या गणनेमध्ये कोणतीही त्रुटी नाही असे गृहीत धरून प्राप्त केलेला भागफल किती होता?

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SSC CGL 2023 Tier-I Official Paper (Held On: 17 Jul 2023 Shift 4)
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  1. २१.७५
  2. २१.२५
  3. २८.७५
  4. 22

Answer (Detailed Solution Below)

Option 4 : 22
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PYST 1: SSC CGL - General Awareness (Held On : 20 April 2022 Shift 2)
25 Qs. 50 Marks 10 Mins

Detailed Solution

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दिले:

नवीन लाभांश = 110% x मूळ लाभांश

नवीन भाजक = 125% x मूळ विभाजक

वापरलेले सूत्र:

लाभांश = भाजक x भागफल + शेष

गणना:

मूळ लाभांश = 100x द्या

चुकून वाढलेला लाभांश = 100 x 110% = 110x

मूळ भाजक = 100y

चुकून वाढलेला भाजक = 100y x 125% = 125y

लाभांश = भाजक x भागफल + शेष

100x = 100y x 25 + 0

x/y = 25/1

वाढीव लाभांश = 110x = 110 x 25 = 2750

वाढलेला भाजक = 125y = 125 x 1 = 125

भागफल = 2750/125 = 22

∴ बरोबर उत्तर 22 आहे.

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