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Time And Work Test - 6

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Time And Work Test - 6
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  • Question 1
    2 / -0.83

    In a stream, Q lies in between P and R such that it is equidistant from both P and R. A boat can go from P to Q and back in 6 hours 30 minutes while it goes from P to R in 9 hours. How long would it take to go from R to P?

    Solution

    Since P to R is double the distance of P to Q,
    Therefore, it is evident that the time  taken from P to R and back would bedouble the time taken from P to Q and back (i.e. double of 6.5 hours = 13 hours).

    Since going from P to R takes 9 hours, coming back from R to P would take 4 hoursi.e. 13-  9 = 4

    So OptionA is correct

  • Question 2
    2 / -0.83

    If A and B together can complete a piece of work in 15 days and B alone in 20 days, in how many days can A alone complete the work?

    Solution

    A and B complete a work in = 15 days
    ⇒One day 's work of (A + B) = 1/ 15

    B complete the work in = 20 days
    ⇒ One day 's work of B = 1/20

    ⇒A 's one day 's work = 1/15 −1/20 = (4 −3)/6 = 1/60

    Thus, A can complete the work in = 60 days .

    So OptionA is correct

  • Question 3
    2 / -0.83

    A and B together can do a piece of work in 50 days. If A is 40% less efficient than B, in how many days can A  working alone complete 60% of the work?

    Solution

    Given:

    A and B together can do a piece of work in 50 days.

    A is 40% less efficient than B

    Concept used:

    Total work = Efficiency of the workers  ×time taken by them

    Calculation:

    Let the efficiency of B be 5a

    So, efficiency of A = 5a  ×60%

    ⇒3a

    So, total efficiency of them = 8a

    Total work = 8a  ×50

    ⇒400a

    Now,

    60% of the work = 400a  ×60%

    ⇒240a

    Now,

    Required time = 240a/3a

    ⇒80 days

    ∴ A  can complete 60% of the work  working alone in 80 days.

  • Question 4
    2 / -0.83

    Sheldon had to cover a distance of 60 km. However, he started 6 minutes later than his scheduled time and raced at a speed 1 km/h higher than his originally planned speed and reached the finish at the time he would reach it if he began to race strictly at the appointed time and raced with the assumed speed. Find the speed at which he travelled during the journey described.

    Solution

    Solve this question through options.
    ⇒  For instance, if he travelled at 25 km/h, his original speed would have been 24 km/h.
    ⇒The time difference can be seen to be 6 minutes in this case = 60 / 24 –60 / 25 = 0.1 hrs = 6 mins

    Thus,25 km/h is the correct answer. 

    So OptionA is correct

  • Question 5
    2 / -0.83

    Two sprinters run the same race of100  m One runs at a speed of  10  m/s and the other runs at  8  m/s. By what time will the first sprinter beat the other sprinter?

    Solution

    Correct option is C
    Time taken by first sprinter  
    = 100/10 = 10sec
    Time taken by second sprinter  
    = 100/80 = 12.5sec
    Difference = 12.5 - 10 = 2.5 sec

  • Question 6
    2 / -0.83

    X can do a piece of work in 20 days. He worked at it for 5 days and then Y finished it in 15 days. In how many days can X and Y together finish the work?

    Solution

    • X ’s five day work = 5/20 = 1/4. Remaining work = 1 –1/4 = 3/4.
    • This work was done by Y in 15 days. Y does 3/4th of the work in 15 days, he will finish the work in 15 ×4/3 = 20 days.  
    • X &Y together would take 1/20 + 1/20 = 2/20 = 1/10 i.e. 10 days to complete the work.

    So Option C   is correct

  • Question 7
    2 / -0.83

    A dog sees a cat. It estimates that the cat is 25 leaps away. The cat sees the dog and starts running with the dog in hot pursuit. If in every minute, the dog makes 5 leaps and the cat makes 6 leaps and one leap of the dog is equal to 2 leaps of the cat. Find the time in which the cat is caught by the dog (assume an open field with no trees).

    Solution

    Initial distance = 25 dog leaps
    Per-minute dog makes 5 dog leaps and  cat makes 6 cat leaps = 3 dog leaps
    ⇒  Relative speed = 2 dog leaps / minutes
    ⇒  An initial distance of 25 dog leaps would get covered in 12.5 minutes .

    So Option D   is correct

  • Question 8
    2 / -0.83

    Two cars started simultaneously towards each other and met each other 3 h 20 min later. How much time will it take the slower car to cover the whole distance if the first arrived at the place of departure of the second 5 hours later than the second arrived at the point of departure of the first?

    Solution

    1. total time to meet : The two cars meet after 3 hours and 20 minutes, which is 10/3 hours.

    2. Relative speed of the two cars : Since both cars are moving towards each other, the combined speed of both cars determines how quickly they cover the distance between them. If we let   S1  be the speed of the first car, and   S2  be the speed of the second car, their combined speed would be   S1 + S2 . We can use the time it takes to meet (10/3 hours) and the total distance between the two cars to form the equation: (S1 + S2 ) ×(10/3) = D (where D is the total distance).

    3. Time differences : We are also told that the first car arrives at the second car 's starting point 5 hours after the second car arrives at the first car 's starting point. This gives us another relationship between the times it takes each car to complete the journey: T1 = T2 + 5 (where T1 is the time for the first car and T2 is the time for the second car to cover the full distance).

    4. Speeds and times : The speed of each car is the total distance (D) divided by the time it takes for the car to complete the journey. So, we can express the speeds of the cars as: S1 = D / T1 S2 = D / T2

    5. Combining the equations : Now, we substitute these expressions for S1 and S2 into the first equation (from step 2): (D / T1 ) + (D / T2 ) = 3D / 10. This simplifies to: (1 / T1 ) + (1 / T2 ) = 3 / 10.

    6. Substitute the time difference : We already know that T1 = T2 + 5, so we substitute this into the equation: (1 / (T2 + 5)) + (1 / T2 ) = 3 / 10.

    7. Solving for T2 : We now solve this equation to find the time taken by the slower car (T2 ). Solving the equation gives us T2 = 5 hours.

  • Question 9
    2 / -0.83

    Charlie and Alan run a race between points A and B, 5 km apart. Charlie starts at 9 a.m. from A at a speed of 5 km/hr, reaches B, and returns to A at the same speed. Alan starts at 9:45 a.m. from A at a speed of 10 km/hr, reaches B and comes back to A at the same speed. At what time do Charlie and Alan first meet each other?

    Solution

    Time take for reaching B for ram is T = 5/5 = 1hr
    Time taken for reaching B for Shyam is T = 5/10 = 1/2 hr

    Its 10 am when ram reaches B and 10: 15 when Shyam reaches B , so they must meet each after 10 and before 10:15 for sure as ram starts back after reaching B

    if we see for options 10: 10 am is the only answer

  • Question 10
    2 / -0.83

    Two pipes, when working one at a time, can fill a cistern in 3 hours and 4 hours, respectively while a third pipe can drain the cistern empty in 8 hours. All the three pipes were opened together when the cistern was 1/12 full. How long did it take for the cistern to be completely full?

    Solution

    Let the total amount of work in filling a cistern be 24 units. (LCM of 3, 4 and 8)

    Work done by pipe 1 in 1 hour = 24/3 = 8 units.

    Work done by pipe 2 in 1 hour = 24/4 = 6 units.

    Work done by pipe 3 in 1 hour = 24/ (-8) = -3 units

    Total work done in 1 hour = 8 + 6 –3 = 11 units

    The time required to complete 11/12th  of the work = 11/12 ×24/ 11 = 2 hours

    ∴The correct answer is 2 hours.

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