Aptitude miscellaneous


  1. 2 km 5 m is equal to









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    As we know that ,
    ∵ 1000 metre = 1 km

    ∴ 5 metre =
    5
    = 0.005 km
    1000

    Correct Option: C

    As we know that ,
    ∵ 1000 metre = 1 km

    ∴ 5 metre =
    5
    = 0.005 km
    1000

    ∴ 2 km 2 metre = (2 + 0.005) km = 2.005 km


  1. For Rs. 25,500, a furniture shop sells 3 computer tables and 5 chairs OR 2 computer tables and 9 chairs. If one wants to buy a set of only 1 computer table and 1 chair, how much does he need to pay?









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    Let C.P. of 1 computer table = Rs. p
    C.P. of 1 chair = Rs. q
    According to the question,
    3p + 5q = 25500 ..... (i)
    2p + 9q = 25500 ..... (ii)
    Clearly, cost of 1 computer table = cost of 4 chairs From equation (i),
    3 × 4q + 5q = 25500 ⇒ 17q = 25500

    ⇒ q =
    25500
    = 1500
    17

    Correct Option: D

    Let C.P. of 1 computer table = Rs. p
    C.P. of 1 chair = Rs. q
    According to the question,
    3p + 5q = 25500 ..... (i)
    2p + 9q = 25500 ..... (ii)
    Clearly, cost of 1 computer table = cost of 4 chairs From equation (i),
    3 × 4q + 5q = 25500 ⇒ 17q = 25500

    ⇒ q =
    25500
    = 1500
    17

    ∴ Cost of 1 computer table = 4 × 1500 = Rs. 6000
    ∴ Required answer = 6000 + 1500 = Rs. 7500



  1. Every Sunday, Gin jogs 3 miles. For he rest of the week, each day he jogs 1 mile more than the previous day. How many miles Gin jogs in 2 weeks?









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    As per the given in question ,
    Total distance covered in two weeks = 2 (3 + 4 + 5 + 6 + 7 + 8 + 9) miles
    Total distance covered in two weeks = 2 × 42

    Correct Option: C

    As per the given in question ,
    Total distance covered in two weeks = 2 (3 + 4 + 5 + 6 + 7 + 8 + 9) miles
    Total distance covered in two weeks = 2 × 42 = 84 miles
    Therefore , Gin jogs 84 miles in 2 weeks .


  1. Three runners A, B and C run a race, with runner A finishing 12 metres ahead of runner B and 18 metres ahead of runner C, while runner B finishes 8 metres ahead of runner C. Each runner travels the entire distance at a constant speed. The length of the race is









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    Let the length of the race be y km.
    When A finishes the race,
    ∴ A : B : C = y : (y – 12) : (y – 18)
    When B covers 12 km. , C covers 10 km.

    y - 12
    =
    12
    =
    6
    y - 18105

    Correct Option: B

    Let the length of the race be y km.
    When A finishes the race,
    ∴ A : B : C = y : (y – 12) : (y – 18)
    When B covers 12 km. , C covers 10 km.

    y - 12
    =
    12
    =
    6
    y - 18105

    ⇒ 6y – 108 = 5y – 60
    ⇒ 6y – 5y = 108 – 60
    ⇒ y = 48 km.
    Hence , required length of the race be 48 km.



  1. An hour-long test has 60 problems. If a student completes 30 problems in 25 minutes, then the required seconds he has taken on average for computing each of the remaining problems is









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    From the question,
    The student is required to answer 30 questions in 35 minutes.

    ∴ Required average time =
    35
    × 60seconds
    30

    Correct Option: A

    From the question,
    The student is required to answer 30 questions in 35 minutes.

    ∴ Required average time =
    35
    × 60seconds = 70 seconds
    30