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Time and Work Test 21
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Time and Work Test 21
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  • Question 1/10
    1 / -0

    A and B alone can do a piece of wok in 8 and 18 days respectively. In how many days the work will be completed if they both work on alternate days starting with B?

    Solutions

    Explanation: 

    A = 8 days, B = 18 days

    Total work = LCM(8,18) = 72

    So efficiency of A = 72/8 = 9, efficiency of B = 72/18 = 4

    2 days work of (A+B) = 9+4 = 13

    2*5(10) days work of (A+B) = 9+4 = 13*5 = 65

    So remaining work = 72-65 = 7

    Now A’s turn on 6th day, he will do remaining work(7) in 7/9 day

    So total 10 7/9 days

     

  • Question 2/10
    1 / -0

    A, B and C can all together do piece of work in 10 days, in which B takes three times as long as A and C together do the work and C takes twice as long as A and B together take to do the work. In how many days B can alone do the work? 

    Solutions

    Explanation: 

    (A+C) in x days so B completes in 3x days

    then (1/x) + (1/3x) = 1/10

    solve, x = 40/3

    so B in 3x = 3*(40/3)= 40 days

    OR

    Given A+B+C = 10 and that B takes 3 times as A+C, so A+C is three times stronger than B

    So this means that 4 times stronger can do work in 10 days

    So 1 time stronger(B) in 4*10 = 40 days

     

  • Question 3/10
    1 / -0

    20 men can complete a piece of work in 14 days. 7 men started the work and after 20 days, 7 more men joined the work. In how many days the remaining work will be completed?

    Solutions

    Explanation: 

    Let (7+7) complete remaining work in x days. So

    20*14 = 7*20 + 14*x

    x = 10 days

     

  • Question 4/10
    1 / -0

    20 men can complete a work in 14 days and 20 women can complete the same work in 18 days. 7 men and 9 women started the work. After working for some days, they were replaced by 10 men and 10 women who complete the remaining work in 9 days. How much work was completed by initially employed men and women?

    Solutions

    Explanation: 

    20 m in 14 days so 10 men in (20*14)/10 = 28 days

    20 w in 18 days so 10 women in (20*18)/10 = 36 days

    So (1/28 + 1/36)*9 = 4/7

    So 1 – 4/7 = 3/7 work was done by 7 men and 9 women

     

  • Question 5/10
    1 / -0

    A, B and C can alone complete a work in 10, 12 and 15 days respectively. A and C started the work and after working for 4 days, A left and B joined. In how many days the total work was completed? 

    Solutions

    Explanation: 

    (A+C) = (1/10 + 1/15) = 1/6. They worked for 4 days so did (1/6)*4 = 2/3rd of work

    Remaining work = 1 – 2/3 = 1/3

    Now A left , B and C working

    (B+C) = (1/12 + 1/15) = 9/60 = 3/20. They worked for x days and completed 1/3rd of work so (3/20)*x = 1/3, so x = 20/9 days

    Total = 4 + 20/9

     

  • Question 6/10
    1 / -0

    A, B and C can alone complete a work in 10, 12 and 15 days respectively. All started the work but B left the work 3 days before completion. How much work was then done by A and B together in the total work? 

    Solutions

    Explanation: 

    Let work completed in x days, so A and C worked for all x days, and B for (x-3) days. So

    (1/10 + 1/15)*x + (1/12)*(x-3) = 1

    Solve, x = 5 days

    In 5 days, A did 5/10 = 1/2 of work

    In (5-3) = 2 days, B did 2/12 = 1/6 of work

    So total by A and B = (1/2 + 1/6) = 2/3

     

  • Question 7/10
    1 / -0

    2 men and 3 women can together complete a piece of work in 4 days and 3 men and 2 women together can complete work in 3 days. In how many days 10 women will complete this work? 

    Solutions

    Explanation: 

    2m + 3w = 4, 3m + 2w = 3

    So 4(2m + 3w) = 3(3m + 2w)

    8m + 12w = 9m + 6w

    6w = 1m

    Given 2m + 3w = 4, so 2*(6w) + 3w = 4, so 15 women in 4 days, so 10 women in (15*4)/10 = 6 days

     

  • Question 8/10
    1 / -0

    A alone can complete a work in 5 days more than A+B together and B alone can complete a work in 45 days more than A+B together. Then in how many days A and B together can complete the work? 

    Solutions

    Explanation: 

    Shortcut = √5×45 = 15

    OR

    Let (A+B) can do in x days, so

    1/(x+5) + 1/(x+45) = 1/x

    Solve, x2 = 225, x = 15

     

  • Question 9/10
    1 / -0

    20 men can complete a work in 14 days and 20 women can complete the same work in 18 days. 8 men start the work and after working for 21 days, they are replaced by x women. If the remaining work is to be completed by x women in 9 days, then how many women should be employed? 

    Solutions

    Explanation: 

    20 m in 14 days so 8 men in (20*14)/8 = 35 days

    In 21 days 8 men complete (1/35)*21 = 3/5 work

    Remaining work = 1 – 3/5 = 2/5

    20 women do 1 work in 18 days so x women will do 2/5 work in 9 days

    10*(2/5)*18 = x*1*9

     

  • Question 10/10
    1 / -0

    A alone can complete a work in 21 days. If B is 40% more efficient than A, then in how many days A and B together can complete the work? 

    Solutions

    Explanation: 

    Let efficiency of A is x, so of B = (140/100)*x = 7x/5

    So ratio of efficiencies = x : 7x/5 = 5 : 7

    So ratio of days = 7 : 5

    A can do in 21 days, so 7y = 21, y = 3

    So B can do in 5*3 = 15 days

    A+B in (21*15)/(21+15) = 8 ¾ days

     

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