Time and Work


  1. A contractor undertook to finish a work in 92 days and employed 110 men. After 48 days, he found that he had already done 3/5 part of the work, the number of men he can withdraw so that the work may still be finished in time is :









  1. View Hint View Answer Discuss in Forum

    Using Rule 1,

    M1D1
    =
    M2D2
    W1W2

    110 × 48
    =
    M2 × 44
    3
    2
    55

    ⇒ M2 × 44 × 3 = 110 × 48 × 2
    ⇒ M2 =
    110 × 48 × 2
    = 80
    44 × 3

    ∴ Number of men can be withdrawn
    = 110 – 80 = 30

    Correct Option: D

    Using Rule 1,

    M1D1
    =
    M2D2
    W1W2

    110 × 48
    =
    M2 × 44
    3
    2
    55

    ⇒ M2 × 44 × 3 = 110 × 48 × 2
    ⇒ M2 =
    110 × 48 × 2
    = 80
    44 × 3

    ∴ Number of men can be withdrawn
    = 110 – 80 = 30


  1. 3 men and 7 women can do a job in 5 days, while 4 men and 6 women can do it in 4 days. The number of days required for a group of 10 women working together, at the same rate as before, to finish the same job is :









  1. View Hint View Answer Discuss in Forum

    3 × 5 men + 7 × 5 women
    = 4 × 4 men + 6 × 4 women
    ⇒ 16 men – 15 men = 35 women – 24 women
    ∴ 1 man = 11 women
    ⇒ 3 men + 7 women = 40 women
    ∴ M1D1 = M2D2
    ⇒ 40 × 5 = 10 × D2
    ⇒ D2 = 20 days
    Aliter : Using Rule 11,
    Here, A1 = 3, B1 = 7, D1 = 5
    A2 = 4, B2 = 6, D2 = 4
    A3 = 0, B3 = 10
    Required days

    =
    D1D2(A1B2 - A2B1)
    days
    D1(A1B3 - A3B1) - D2(A2B3 - A3B2)

    =
    5 × 4(3 × 6 - 4 × 7)
    5(3 × 10 - 0) - 4(4 × 10 - 0)

    =
    20 × (-10)
    = 20 days
    150 - 160

    Correct Option: D

    3 × 5 men + 7 × 5 women
    = 4 × 4 men + 6 × 4 women
    ⇒ 16 men – 15 men = 35 women – 24 women
    ∴ 1 man = 11 women
    ⇒ 3 men + 7 women = 40 women
    ∴ M1D1 = M2D2
    ⇒ 40 × 5 = 10 × D2
    ⇒ D2 = 20 days
    Aliter : Using Rule 11,
    Here, A1 = 3, B1 = 7, D1 = 5
    A2 = 4, B2 = 6, D2 = 4
    A3 = 0, B3 = 10
    Required days

    =
    D1D2(A1B2 - A2B1)
    days
    D1(A1B3 - A3B1) - D2(A2B3 - A3B2)

    =
    5 × 4(3 × 6 - 4 × 7)
    5(3 × 10 - 0) - 4(4 × 10 - 0)

    =
    20 × (-10)
    = 20 days
    150 - 160



  1. Either 8 men or 17 women can paint a house in 33 days. The number of days required to paint three such houses by 12 men and 24 women working at the same rate is :









  1. View Hint View Answer Discuss in Forum

    8 men ≡ 17 women

    ⇒ 12 men ≡
    17
    × 12
    8

    =
    51
    women
    2

    ∴ 12 men + 24 women
    =
    51
    + 24 =
    99
    women
    22

    By
    M1D1
    =
    M2D2
    W1W2

    =
    17 × 33
    =
    99 × D2
    12 × 3

    ⇒ D2 =
    17 × 33 × 6
    = 34 days
    99

    Aliter : Using Rule 12,
    Here, A = 8, B = 17, a = 33
    A1 = 12, B1 = 24
    Number of days=
    a(A × B)
    A1B + B1A

    =
    33(8 × 17)
    12 × 17 + 24 × 8

    =
    4488
    396

    No. of days to paint 3 houses
    =
    4488
    × 3 = 34 days
    396

    Correct Option: C

    8 men ≡ 17 women

    ⇒ 12 men ≡
    17
    × 12
    8

    =
    51
    women
    2

    ∴ 12 men + 24 women
    =
    51
    + 24 =
    99
    women
    22

    By
    M1D1
    =
    M2D2
    W1W2

    =
    17 × 33
    =
    99 × D2
    12 × 3

    ⇒ D2 =
    17 × 33 × 6
    = 34 days
    99

    Aliter : Using Rule 12,
    Here, A = 8, B = 17, a = 33
    A1 = 12, B1 = 24
    Number of days=
    a(A × B)
    A1B + B1A

    =
    33(8 × 17)
    12 × 17 + 24 × 8

    =
    4488
    396

    No. of days to paint 3 houses
    =
    4488
    × 3 = 34 days
    396


  1. 20 men or 24 women can complete a piece of work in 20 days. If 30 men and 12 women undertake to complete the work, the work will be completed in









  1. View Hint View Answer Discuss in Forum

    20 men ≡ 24 women
    ⇒ 5 men ≡ 6 women
    ∴ 30 men + 12 women = 40 men
    ∴ M1D1 = M2D2
    ⇒ 20 × 20 = 40 × D2

    ⇒ D2 =
    20 × 20
    = 10 days
    40

    Aliter : Using Rule 12,
    Here, A = 20, B = 24, a = 20
    A1 = 30, B1 = 12
    Required time =
    a(A × B)
    A1B + B1A

    =
    20(20 × 24)
    30 × 24 + 12 × 20

    =
    9600
    720 + 240

    =
    9600
    = 10 days
    960

    Correct Option: A

    20 men ≡ 24 women
    ⇒ 5 men ≡ 6 women
    ∴ 30 men + 12 women = 40 men
    ∴ M1D1 = M2D2
    ⇒ 20 × 20 = 40 × D2

    ⇒ D2 =
    20 × 20
    = 10 days
    40

    Aliter : Using Rule 12,
    Here, A = 20, B = 24, a = 20
    A1 = 30, B1 = 12
    Required time =
    a(A × B)
    A1B + B1A

    =
    20(20 × 24)
    30 × 24 + 12 × 20

    =
    9600
    720 + 240

    =
    9600
    = 10 days
    960



  1. 30 men can repair a road in 18 days. They are joined by 6 more workers. Now the road can be repaired in









  1. View Hint View Answer Discuss in Forum

    Using Rule 1,
    M1D1 = M2D2
    ⇒ 30 × 18 = 36 × D2

    ⇒ D2 =
    30 × 18
    = 15 days
    36

    Correct Option: B

    Using Rule 1,
    M1D1 = M2D2
    ⇒ 30 × 18 = 36 × D2

    ⇒ D2 =
    30 × 18
    = 15 days
    36