Question Bank - 100 Arithmetic Questions From Previous CAT Papers (Solved)


  • Skadoooosh!!!


    Q99. (CAT 2001)
    The owner of an art shop conducts his business in the following manner: Every once in a while he raises his prices by X %, then a while later he reduces all the new prices by X %. After one such up-down cycle, the price of a painting decreased by Rs. 441. After a second up-down cycle the painting was sold for Rs. 1,944.81. What was the original price of the painting?
    (1) Rs 2,756.25
    (2) Rs 2,256.25
    (3) Rs 2,500
    (4) Rs 2,000


  • Skadoooosh!!!


    Q100. (CAT 2001)
    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. What was the length of the race?
    (1) 36 meters
    (2) 48 meters
    (3) 60 meters
    (4) 72 meters


  • Skadoooosh!!!


    Let speed of the slowest runner be x and fastest runner be 2x
    Let d be the distance covered by the slowest runner in 5 min
    Then the distance covered by the fastest runner in 5 min is (d + 1000) metre
    d/x = (d + 1000)/2x
    => d = 1000 metre
    fastest runner covers (d + 1000) = 2000 metre in 5 minute
    therefore he needs 10 minute to cover 4000 metre


  • Skadoooosh!!!


    If we start forming equations here, we are trapped!

    If you increase and decrease by the same percentage (say x) then the final result will be always lesser than than the initial value (net change is -x^2/100%). So to get the same result the increase percentage should be greater than the decrease percentage. Answer is p > q

    Note - Number of years doesn't matter here. So it would be easy to just take one year and observe the trend. For example, let say he had 100 goats at the beginning and increased by 25%, to make the final count as 125. Now to bring it back to 100, we have to decrease by 20%. 25% > 20%


  • Skadoooosh!!!


    Let d = distance of race.
    Between A's finish and B's finish, B runs 12 meters while C runs 10 meters.
    B's rate to C's rate = 6/5.
    When B finishes, he is 8 meters ahead of C.
    6/5 = d/(d-8)
    d = 48 meters


  • Skadoooosh!!!


    First cycle :
    P(1 - x/100)(1 + x/100) = P - 441
    so (x^2/100) = 441/P -- (1)

    Second cycle :
    (P - 441) (1 - x^2/100) = 1944.81 -- (2)

    From (1) & (2)
    (P - 441) (1 - 441/P) = 1944.81
    (P - 441)(P - 441) = 1944.81P
    P^2 - 2826.81P + 194481 = 0
    P = 2756.25



  • @master_shifu ...IS their any other way to solve this question?


  • Being MBAtious!


    @deepalis727

    Work with ratios

    Ratio of speed of N & S = 4 : 1
    => Ratio of time taken by N & S = 1 : 4
    Total time taken = 1 hour
    Time taken by N to cover the distance AB = 1/5 hour = 12 minutes
    Time taken by S to cover the distance AB = 4/5 hour = 48 minutes
    When the train was late by 20 minutes, total time available = 40 minutes
    N doubled the speed, so time taken to cover distance AB = 12/2 = 6 minutes
    So, S should cover the distance in 40 - 6 = 34 minutes
    ratio = 6 : 34 = 1 : 6 (approx)



  • @master_shifu
    Can we not calculate the average speed by adding the total distance, i.e 9000 kms and then adding the total time taken , i.e. 25 hours + 24 hours + 1.55(halt in India) and then dividing those figures. Total distance/total time?



  • @master_shifu In 19 th question CAT 2002 in arithmetic, the answer comes is 43.5


 

Looks like your connection to MBAtious was lost, please wait while we try to reconnect.