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A fraction becomes \(\frac{1}{3}\) , if 2 is added to both of its numerator and denominator. The same fraction becomes \(\frac{2}{5}\), when 3 is added to both its numerator and denominator. Let the original fraction be \(\frac{x}{y}\).(a) \(\frac{x+2}{y+2} = \frac{1}{3}\, implies:\) (i) 3x + 6y = 2 (ii) 3x – 6y = –4 (iii) 3x – y = 4 (iv) None. (b) \(\frac{x+3}{y+3}\)= \(\frac{2}{5}\) implies: (i) 5x + 15y = 6 (ii) 5x – 2y = –9 (iii) 5x – 2y = 9 (iv) None. (c) The value of x is: (i) 1 (ii) 2(iii) 3 (iv) None. (d) The value of y is: (i) 5 (ii) 6 (iii) 7(iv) None. (e) Required (original) fraction is: (i) \(\frac{1}{5}\)(ii) \(\frac{2}{7}\)(iii) \(\frac{3}{5}\)(iv) \(\frac{1}{7}\). |
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Answer» (a) \(\frac{x+2}{y+2} = \frac{1}{3}\) ⇒ 3(x +2) = y + 2 ⇒ 3x + 6 = y + 2 ⇒ 3x – y + 4 = 0 ⇒ 3x – y = – 4. ... (1) Hence, option (ii) is correct. (b) \(\frac{x+3}{y+3} = \frac{2}{5}\) ⇒ 5(x+3) = 2(y+3) ⇒ 5x + 15 = 2y + 6 ⇒ 5x – 2y + 9 = 0 ⇒ 5x – 2y = – 9. ... (2) Hence, option (ii) is correct. (c) Now, multiplying equation (1) by 2, we get 6x – 2y = – 8. ... (3) Now, subtracting equation (2) from equation (3), we get (6x – 2y) – (5x – 2y) = – 8 – (– 9) ⇒ 6x – 5x – 2y + 2y = – 8 + 9 ⇒ x = 1. Hence, the value of x is 1. Hence, option (i) is correct. (d) By putting x = 1 in equation (1), we get 3 × 1 – y = – 4 ⇒ 3 – y = – 4 ⇒ y = 3 + 4 = 7. Hence, the value of y is 7. Hence, option (iii) is correct. (e) Required (original) fraction is \(\frac{x}{y} = \frac{1}{7}.\) Hence, option (iv) is correct. |
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