Solved CAT Questions (Arithmetic) - Set 10

  • Skadoooosh!!!

    Q1. (CAT 1993)
    ABC forms an equilateral triangle in which B is 2 km from A. A person starts walking from B in a direction parallel to AC and stops when he reaches a point D directly east of C. He, then, reverses direction and walks till he reaches a point E directly south of C. Then D is
    a) 3 km east and 1 km north of A
    b) 3 km east and √3 km north of A
    c) √3 km east and 1 km south of A
    d) √3 km west and 3 km north of A


    From the figure, D is 3 km east and √3 km north of A
    Note - DBG forms a right triangle with hypotenuse, DB = 2 and BG = 1. Apply Pythagoras, DG = √3

    Q2. (CAT 1993)
    In the above question, total distance walked by the person is
    a) 3 km
    b) 4 km
    c) 2√3 km
    d) 6 km

    From the figure, person travels 2 + 2 + 2 = 6 km

    Q3. (CAT 1993)
    A group of workers was put on a job. From the second day onwards, one worker was withdrawn each day., The job was finished when the last worker was withdrawn. Had no worker been withdrawn at any stage, the group would have finished the job in two-thirds the time. How many workers were there in the group?
    (a) 2
    (b) 3
    (c) 5
    (d) 11

    say there were n workers in total. So it took n days to finish the job
    If 1 unit of work is completed by each worker then
    day 1 = n units of work is done
    day 2 = n - 1 units of work is done
    day 3 = n - 2 units of work is done
    day n - 2 = 2 units of work is done
    day n - 1 = 1 unit of work is done
    day n - work completed

    So total = 1 + 2 + 3 + ... + n = n(n + 1)/2 units of work

    If all worked together, then work would be completed in 2n/3 days.
    => unit of work = ( 1 + 1 + ... n times) 2n/3 = 2n^2/3

    2n^2/3 = n(n + 1)/2
    4n = 3n + 3
    n = 3

    Q4. (CAT 1993)
    A ship leave on a long voyage. When it is 18 miles from the shore, a seaplane, whose speed is ten times that of the ship, is sent to deliver mail. How far from the shore does the seaplane catch up with the ship?
    (a) 24 miles
    (b) 25 miles
    (c) 22 miles
    (d) 20 miles

    Speed of ship = S
    Speed of seaplane = 10S
    Let say we consider the travel starts 18 miles from the shore (both ship and plane are travelling now)
    When they meet Ship travels x (say) miles and plane travels 18 + x miles.
    Time is same for both
    So, x/S = (18 + x)/10S
    10x = 18 + x
    x = 2
    So they will meet 18 + 2 = 20 miles from the shore.

    Q5. (CAT 1991)
    Three machines, A, B and C can be used to produce a product. Machine A will take 60 hours to produce a million units. Machine B is twice as fast as Machine A. Machine C will take the same amount of time to produce a million units as A and B running together. How much time will be required to produce a million units if all the three machines are used simultaneously?
    (a) 12 hours
    (b) 10 hours
    (c) 8 hours
    (d) 6 hour

    Let say A does n unit of work per day
    So B does 2n units of work per day
    C does 2n + n = 3n units of work per day
    So combined A, B and C can do 6n units of work per day
    As A (n units of work per day) takes 60 hours to produce a million units, we can say it will take 60/6 = 10 hours to produce a million units by all three machines working together (6n units of work per day)

    Q6. (CAT 1991)
    A player rolls a die and receives the same number of rupees as the number of dots on the face that turns up. What should the player pay for each roll if he wants to make a profit of one rupee per throw of the die in the long run?
    (a) Rs. 2.50
    (b) Rs. 2
    (c) Rs.3.50
    (d) Rs. 4

    For an unbiased dice, we can get 1, 2, 3, 4, 5 and 6
    so for long run, the average earning could be (1 + 2 + 3 + 4 + 5 + 6)/6 = 21/6 = 3.5
    To make a profit of 1 rupee per throw, he should pay 3.5 - 1 = 2.50 for each throw.

    Q7. (CAT 1991)
    Every day Neera’s husband meets her at the city railway station at 6.00 p.m. and drives her to their residence. One day she left early from the office and reached the railway station at 5.00 p.m. She started walking towards her home, met her husband coming from their residence on the way and they reached home 10 minutes earlier than the usual time. For how long did she walk?
    (a) 1 hour
    (b) 50 minutes
    (c) 1/2 hour
    (d) 55 minutes

    They arrived home 10 minutes earlier. So they saved 5 minutes in each direction.
    Means they saved 5 minutes in the Station - Home journey
    As the husband met her 5 minutes early, Neeraja would have walked for 1 hour - 5 minutes = 55 minutes

    Q8. (CAT 1991)
    A sum of money compounded annually becomes Rs.625 in two years and Rs.675 in three years. The rate of interest per annum is
    (a) 7%
    (b) 8%
    (c) 6%
    (d) 5%

    625 * 1 * r/100 = 675 - 625 = 50
    r = 8%

    Q9. (CAT 1990)
    In a stockpile of products produced by three machines M1, M2 and M3, 40% and 30% were manufactured by M1 and M2 respectively. 3% of the products of M1 are defective, 1% of products of M2 defective, while 95% of the products of M3 are not defective. What is the percentage of defective in the stockpile?
    (a) 3%
    (b) 5%
    (c) 2.5%
    (d) 4%

    Let say we have 100 products
    40 are from M1 and 30 are from M2 => 30 are from M3
    1.2 product from M1 are defective
    0.3 product from M2 are defective
    1.5 product from M3 are defective
    So total 1.2 + 0.3 + 1.5 = 3 out of 100 are defective
    3% is the answer.

    Q10. (CAT 1990)
    A car after traveling 18 km from a point A developed some problem in the engine and speed became 4/5 of its original speed As a result, the car reached point B 45 minutes late. If the engine had developed the same problem after traveling 30 km from A, then it would have reached B only 36 minutes late. The original speed of the car (in km per hour) and the distance between the points A and B (in km.) is
    (a) 25, 130
    (b) 30,150
    (c) 20, 90
    (d) None of these

    Let's consider the points between 18 km and 30 km from A (the magic happens here!)
    Speed of Car = S
    In the first case, due to the problem, car covered this distance in 4S/5 speed. (1/5 decrease in speed)
    This should amount to 1/4 increase in time.
    We know this difference in time is 45 - 36 = 9 minutes
    so t/4 = 9 => t = 36 (t being the time required to cover the 30 - 18 = 12 km stretch)
    12 km in 36 minutes => speed = 20.

    Let d be the distance between A and B
    In case 1, first 18 km was covered in 20 kmph
    and the rest, (d - 18) km was covered in 4/5th of speed = 16 kmph.
    Had the total distance is covered in 20 kmph, we would have reached 45 minutes early.
    18/20 + (d - 18)/16 - d/20 = 45 minutes = 3/4 hours
    d = 78 km.

    Answer is none of these.

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