जर एका अपूर्णांकाचा अंश 20% ने वाढविला आणि छेद 30% ने कमी केला, तर मिळणारा अपूर्णांक \(\frac{{39}}{{25}}\) आहे. तर मूळ अपूर्णांक किती?

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SSC Matric Level Previous Paper (Held on: 9 Nov 2020 Shift 1)
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  1. \(\frac{{67}}{{75}}\)
  2. \(\frac{{100}}{{91}}\)
  3. \(\frac{{75}}{{68}}\)
  4. \(\frac{{91}}{{100}}\)

Answer (Detailed Solution Below)

Option 4 : \(\frac{{91}}{{100}}\)
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दिलेल्याप्रमाणे:

जर एका अपूर्णांकाचा अंश 20% ने वाढविला आणि छेद 30% ने कमी केला, तर मिळणारा अपूर्णांक 39/25 आहे.

वापरलेली संकल्पना:

टक्केवारी

गणना:

अपूर्णांक x/y आहे असे मानू

प्रश्नानुसार,

⇒ \(\frac{{x + \frac{{20x}}{{100}}}}{{y - \frac{{30y}}{{100}}}} = \frac{{39}}{{25}}\)

⇒ \(\frac{{\frac{{6x}}{5}}}{{\frac{{7y}}{{10}}}} = \frac{{39}}{{25}}\)

⇒ \(\frac{{6x}}{{7y}} = \frac{{39}}{{25}} \times \frac{5}{{10}}\)

⇒ \(\frac{x}{y} = \frac{{39}}{{50}} \times \frac{7}{6}\)

⇒ \(\frac{x}{y} = \frac{{273}}{{300}}\)

∴ \(\frac{x}{y} = \frac{{91}}{{100}}\)

∴ मूळ अपूर्णांक 91/100 आहे.
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