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Mathematics

A and B can do a piece of work in 40 days, B and C in 30 days, and C and A in 24 days.

(i) How long will it take them to do the work, working together?

(ii) In what time can each finish it working alone?

Direct & Inverse Variations

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Answer

(A + B)'s 1 day work = 140\dfrac{1}{40}

(B + C)'s 1 day work = 130\dfrac{1}{30}

(C + A)'s 1 day work = 124\dfrac{1}{24}

2(A + B + C)'s 1 day work = 140+130+124\dfrac{1}{40} + \dfrac{1}{30} + \dfrac{1}{24}

= (3+4+5)120\dfrac{(3 + 4 + 5)}{120}

= 12120\dfrac{12}{120}

= 110\dfrac{1}{10}

(A + B + C)'s 1 day work = 12×10\dfrac{1}{2 \times 10}

= 120\dfrac{1}{20}

No. of days required to complete the work when A, B and C are working together = 20 days

Hence, A, B and C working together can complete the work in 20 days.

(ii) A's 1 day work = (A + B + C)'s 1 day work - (B + C)'s 1 day work

= 120−130\dfrac{1}{20} - \dfrac{1}{30}

= (3−2)60\dfrac{(3 - 2)}{60}

= 160\dfrac{1}{60}

∴ A alone will complete the work in 60 days.

B's 1 day work = (A + B + C)'s 1 day work - (C + A)'s 1 day work

= 120−124\dfrac{1}{20} - \dfrac{1}{24}

= (6−5)120\dfrac{(6 - 5)}{120}

= 1120\dfrac{1}{120}

∴ B alone will complete the work in 120 days.

C's 1 day work = (A + B + C)'s 1 day work - (A + B)'s 1 day work

= 120−140\dfrac{1}{20} - \dfrac{1}{40}

= (2−1)40\dfrac{(2 - 1)}{40}

= 140\dfrac{1}{40}

∴ C alone will complete the work in 40 days.

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