Quantitative Aptitude: Data Interpretation Questions Set 99

Directions(1-5): The profit percentage of a company over different months have been given below. Refer the following graph & answer the following questions.

  1. If the expenditures in February and May are equal, then the approximate ratio of the income in February and May respectively ?
    31:35
    27:31
    17:19
    31:34
    11:20
    Option D
    For February 55 =( I1-x)/x*100
    I1 = 155x/100
    For May 70 =( I2-x)/x*100
    I2 = 170x/100
    Required ratio = I1/I2 = 31:34

     

  2. What is the average profit earned for the given months?
    55(4/5)
    57(2/5)
    41(5/6)
    50(3/4)
    55(5/6)
    Option E
    Average percent profit earned for the given month
    = 1/6*(40+55+45+65+70+60)
    = 55(5/6)

     

  3. If the profit in May was Rs.4 crores, what was the profit in June?
    11.12%
    33.33%
    50%
    25%
    Cannot be determined.
    Option E
    From the obtain information about the percentage profit only.
    To find the profit in June we must have the data for the income or expenditure in June.
    Therefore, the profit for June cannot be determined.

     

  4. If the income in April was Rs.264 crores, what was the expenditure in April?
    168 crores
    152 crores
    135 crores
    171 crores
    160 crores
    Option E
    For April 65 = (264-x)/x*100
    x = 160 crores

     

  5. In which month is the expenditure minimum?
    April
    March
    January
    June
    Cannot be determined.
    Option E
    The comparison of percent profit for different months the comparison of the expenditures is not possible without more data. Therefore, the year with minimum expenditure cannot be determined.

     


  6. Directions(6-10):
    Study the following graph carefully to answer the given questions
    Time taken by the pipes to fill a tank/cistern (hours/minutes).

  7. Two pipes, P3 and P4 are opened simultaneously and it is found that due to the leakage in the bottom, 17/7 minutes are taken extra to fill the tank. If the tank is full, in what approximate time would the leak empty it?
    29 min.
    35 min.
    22 min.
    40 min.
    39 min.
    Option E
    Total time taken by both pipes before the leak was developed = 60/7 minutes
    Now, leaks is developed which will take T time to empty the tank.
    So, (1/15 +1/20 – 1/T) = 60/7 + 17/7
    => (1/15 +1/20 – 1/T) = 1/11
    solve for T 660/17 minutes == 39 minutes

     

  8. Two pipes P7 and P8 can fill a cistern. If they are opened on alternate minutes and if pipe P7 is opened first, in how many minutes will the tank be full?
    7 min.
    5 min.
    6 min.
    4 min.
    3 min.
    Option C
    Pipe P7 can fill = 1/12
    Pipe P8 can fill = ¼
    For every two minutes, 1/12 + 1/4 = 1/3 Part filled
    Total = 6 minutes

     

  9. A large cistern can be filled by two pipes P1 and P2. How many minutes will it take to fill the Cistern from an empty state if P2 is used for half the time and P1 and P2 fill it together for the other half?
    4.8 min.
    6.0 min.
    7.5 min.
    6.2 min.
    5.5 min.
    Option C
    Part filled by P1 and P2 = 1/15 + 1/10 = 1/6 Part filled by P2 = 1/10
    Required time = x/2(1/6 + 1/10) = 2/15
    = 15/2 = 7.5 minutes

     

  10. Two pipes P5 and P6 can fill a tank. If both the pipes are opened simultaneously, after how much time should P6 be closed so that the tank is full in 8 minutes?
    18 min.
    12 min.
    20 min.
    15 min.
    32 min.
    Option A
    Required time = y (1-(t/x)) = 27(1-(8/24))
    = 18 minutes

     

  11. Three pipes P9, P10, and P18 can fill a tank. If Pipe P18 alone can fill a tank in 24 minutes then the pipe P18 is closed 12 minutes before the tank is filled. In what time the tank is full?
    6 (3/10)
    3 (4/15)
    5 (2/11)
    8 (4/13)
    7 (2/13)
    Option D
    Let T is the time taken by the pipes to fill the tank
    => (1/12 + 1/18 + 1/24)*(T – 12) + (1/12 + 1/18)*12 = 1
    => Time = 108/13 = 8(4/13)

     


Related posts

Leave a Comment