Engineering Maths - Time and Work

Refer to the data below and answer the questions that follow.
The variation in the speed of a car on a particular day at the respective times is shown in the table below:

 s(km/hr)04050851010
 t(hr)11.00 am11.30 am1.00 pm1.30 pm3.30 pm4.30 pm

21. If car maintains the speed it has at 4.30 p.m. then at what time will the car cover the same distance as it had covered from 11.00 a.m. to 1.00 p.m.?

  • Option : D
  • Explanation : Distance covered are the same.
    Speed in the latter is maintained at 10 kmph.
    ∴ 77.5 = 10 × x
    => x= 7.75 hours.
    Hence, clock time will be 4.30 + 7 hours 45 mins
    = 12.15 a.m.
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Refer to the data below and answer the questions that follow.
The variation in the speed of a car on a particular day at the respective times is shown in the table below:

 s(km/hr)04050851010
 t(hr)11.00 am11.30 am1.00 pm1.30 pm3.30 pm4.30 pm

22. Average speed of the car from 11.00 a.m. to 4.30 p.m. is:

  • Option : C
  • Explanation :

    Distance travelled from 1.00 pm to 4.30 p.m.
    = A(ΔBQRY) + A(ΔBYC) + A(ΔCZD) + A(ΔZRTE)

    = 130 + 8.75
    = 138.75 km.
    Total distance travelled = 77.5 + 138.75
    = 216.25 km.
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23. A cheetah a deer which is 100m ahead. The time in which the deer takes 10 leaps the cheetah takes only 6 leaps. In one leap, the deer covers 1 m while the cheetah covers 2 m. In how many leaps would the cheetah catch up the deer?

  • Option : A
  • Explanation :
    While deer makes 10 leaps, cheetah makes 6 leaps.
    ∴ Distance covered in 10 leaps by deer
    = 10 x 1 = 10m
    Distance covered in 6 h~aps by cheetah
    = 6 x 2 = 12 m
    Hence in six leaps, cheetah gains
    12 - 10 = 2 m over the deer.
    Cheetah has to gain 100 m over the deer.
    So, total leaps required by cheetah
    = 6/2 x 100 = 300
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Refer to the graph below and answer the questions that follow.
The engine of the new car Palio brought by Fiat in the market follows certain mileage characteristics as shown in the diagram below:

Vijay bought the new car Palio and filled exactly 10 litres in his car. He goes at a speed of 40 km/hr for first 20 km and then at a speed of 60 km/hr for next 60 km and driving at a constant speed reaches his destination which was 115 km away from his starting point.

25. What is the time taken by Vijay if he reaches his destination in minimum possible time and finishes all the fuel?

  • Option : C
  • Explanation :
    Fuel required for first 20 km = 20/8 = 2.5 litres
    Fuel required for next 60 km = 60/15 = 4 litres.
    ∴ Petrol left = 10 - 6.5 = 3.5 litres.
    Now, he has to travel = 115 - (60 + 20)
    = 35 km in 3.5 litres.
    So he can go at a speed of either 45 km/hr or 80 km/hr as mileage is 10 km/litre in both cases. But since he reaches in minimum time, he travels at 80 km/hr

    = 1 hours 56 minutes and 25 seconds.
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