Question 33 (A) - CBSE Class 10 Sample Paper for 2026 Boards - Maths Basic - Solutions of Sample Papers for Class 10 Boards

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Question 33 (A) The numerator of a fraction is 3 less than its denominator. If 2 is added to both of its numerator and denominator then the sum of the new fraction and original fraction is 29/20. Find the original fraction.Let Numerator be x & Denominator be y So, fraction is 𝒙/𝒚 Given that The numerator of a fraction is 3 less than its denominator. x = y – 3 Also, If 2 is added to both of its numerator and denominator then the sum of the new fraction and original fraction is 29/20. New Fraction = (𝑵𝒖𝒎𝒆𝒓𝒂𝒕𝒐𝒓 + 𝟐)/(𝑫𝒆𝒏𝒐𝒎𝒊𝒏𝒂𝒕𝒐𝒓 + 𝟐) = (𝑥 + 2)/(𝑦 + 2) Now, Old fraction + New Fraction =𝟐𝟗/𝟐𝟎 𝑥/𝑦+(𝑥 + 2)/(𝑦 + 2)=29/20 Putting x = y – 3 from (1) ((𝑦 − 3))/𝑦+((𝑦 − 3) + 2)/(𝑦 + 2)=29/20 (𝒚 − 𝟑)/𝒚+(𝒚 − 𝟏)/(𝒚 + 𝟐)=𝟐𝟗/𝟐𝟎 ((𝑦 − 3) (𝑦 + 2) + 𝑦(𝑦 − 1))/(𝑦(𝑦 + 2))=29/20 (𝑦(𝑦 + 2) − 3(𝑦 + 2) + 𝑦(𝑦 − 1))/(𝑦(𝑦 + 2))=29/20 (𝑦^2 + 2𝑦 − 3𝑦 − 6 + 𝑦^2 − 𝑦)/(𝑦^2 + 2𝑦)=29/20 (𝑦^2+ 𝑦^2 + 2𝑦 − 3𝑦 − 𝑦 − 6)/(𝑦^2 + 2𝑦)=29/20 (𝟐𝒚^𝟐 − 𝟐𝒚 − 𝟔)/(𝒚^𝟐 + 𝟐𝒚)=𝟐𝟗/𝟐𝟎 20(2𝑦^2 − 2𝑦 − 6)=29(𝑦^2 + 2𝑦) 40𝑦^2−40𝑦 −120=29𝑦^2+58𝑦 40𝑦^2−29𝑦^2−40𝑦−58𝑦−120=0 𝟏𝟏𝒚^𝟐−𝟗𝟖𝒚−𝟏𝟐𝟎=𝟎 We find roots using splitting the middle term method Splitting the middle term method We need to find two numbers where Sum = –98 Product = 11 × –120 = –1320 𝟏𝟏𝒚^𝟐−𝟏𝟏𝟎𝒚+𝟏𝟐𝒚−𝟏𝟐𝟎=𝟎 11𝑦(𝑦−10)+12(𝑦−10)=0 (𝟏𝟏𝒚+𝟏𝟐)(𝒚−𝟏𝟎) =𝟎 So, 𝑦=−12/11 𝑦=10 Since y is denominator, it cannot be in fractions ∴ 𝒚=𝟏𝟎 is only possible Now, x = y – 3 x = 10 – 3 x = 7 ∴ Our required fraction = 𝒙/𝒚=𝟕/𝟏𝟎