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Divide 150 into two parts such that the sum of their reciprocals is 3/112. Calculate both the parts.
- A50, 90
- B70, 80
- C60, 90
- D50, 100
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CCommunity
Let the two parts be ‘x’ and ‘(150 – x)’
Sum of their reciprocals = 3/112
⇒ 1/x + 1/(150 – x) = 3/112
⇒ (150 – x + x)/x(150 – x) = 3/112
⇒ 150/x(150 – x) = 3/112
⇒ 150 × 112/3 = x(150 – x)
⇒ 5600 = 150x – x2
⇒ x2 – 150x + 5600 = 0
⇒ x2 – 70x – 80x + 5600 = 0
⇒ x(x – 70) – 80(x – 70) = 0
⇒ (x – 70)(x – 80) = 0
⇒ x = 70, 80
∴ The two parts are 70 and 80
Sum of their reciprocals = 3/112
⇒ 1/x + 1/(150 – x) = 3/112
⇒ (150 – x + x)/x(150 – x) = 3/112
⇒ 150/x(150 – x) = 3/112
⇒ 150 × 112/3 = x(150 – x)
⇒ 5600 = 150x – x2
⇒ x2 – 150x + 5600 = 0
⇒ x2 – 70x – 80x + 5600 = 0
⇒ x(x – 70) – 80(x – 70) = 0
⇒ (x – 70)(x – 80) = 0
⇒ x = 70, 80
∴ The two parts are 70 and 80