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



  • Solution: Get the diagram correct first, where G1, G2 and G3 are the positions of the 3 gutters.

    0_1514262602527_827b36fa-8207-4ae9-94a7-3ed699e65597-image.png

    From the diagram, it is clear that:
    AG1 = BG3 = x (given in question)
    2 * G1G2 = G2G3 = 2y (given in question).
    At 30kmph, AG1 takes 5 minutes.
    So, AG1 = 30 * 5/60 = 2.5kms.
    So, 2.5 + y + 2y + 2.5 = 20 (As total distance is 20km)
    Hence y = 5.
    So, G1G2 = 5 and G2G3 = 10.
    Now, let us calculate the time taken:
    AG1 = 5 mins (given in the question)
    From G1, velocity is 60kmph, and to reach G3, he has to travel 15 kms.
    It takes 15 minutes for that.
    Now, from G3 to A, it is a 17.5 km journey at 60kmph. It takes 17.5 minutes.
    So, total time taken is 5 + 15 + 17.5 = 37.5
    Also, 1 minute for taking patient in and out of ambulance.
    So, total time taken is 38.5
    He should be saved in 40 minutes.
    So, the doctor gets 40 – 38.5 = 1.5 minutes.



  • Q17. (CAT 2002)
    It takes 6 technicians a total of 10 hours to build a new server from Direct Computer, with each working at the same rate. If six technicians start to build the server at 11 am and one technician per hour is added beginning at 5 pm, at what time will the server be complete?
    (1) 6:40 pm
    (2) 7:00 pm
    (3) 7:20 pm
    (4) 8:00 pm



  • Solution: 6 technicians take 10 hours to build the server. So, total man-hours required are 60.
    So, let us take here that the work that needs to be completed is “60”.
    In 10hrs, 6 technicians complete 10*6 = “60” amount of work, which is the whole work.
    So, in 10hrs, 1 technician completes “10” amount of work.
    Hence, each technician completes in 1hr, “1” amount of work.
    Now, from 11am – 5pm that is for 6hrs there are 6 technicians.
    1 technician 1 hr = “1” amount of work.
    Hence 6 technicians in 6hrs complete “36” amount of work.
    Remaining is 60 – 36 = “24” amount of work.
    Next hour, there are 7 employees.
    Hence the work done is “7”, and total work till now is 36 + 7 = 43.
    Next hour, there are 8 employees.
    Hence the work done is “7”, and total work till now is 43 + 8 = 51.
    Next hour, there are 9 employees.
    Hence the work done is “9”, and total work till now is 51 + 9 = 60.
    Hence, the work is completed 3hrs after 5pm.



  • Q18. (CAT 2002)
    Three small pumps and one large pump are filling a tank. Each of the three small pumps works at 2/3rd the rate of the large pump. If all 4 pumps work at the same time, then they should fill the tank in what fraction of time that it would have taken the large pump alone?
    (1) 4/7
    (2) 1/3
    (3) 2/3
    (4) 3/4



  • Solution : Let the large pump fill at the rate of x litres/hour.
    So, each of the small tank fills at the rate of 2x/3 litres/hour.
    The problem can be solved in many ways, I will do it in the least time consuming way.
    We have :
    L = x litres/hr (Large tank)
    S1 = S2 = S3 = 2x/3 litres/hr ( 3 small tank).
    Let x = 3.
    Then S1 = S2 = S3 = 2.
    Let the capacity of the tank be 9l ( 3 + 2 + 2 + 2)
    Large tank alone will take 3hrs (as its rate is 3 litres/hr)
    Large + 3 small tank will take 1hr (As rate will be 3 +2 + 2 + 2 = 9 litres/hr )
    Hence, ratio is 1:3



  • Q19. (CAT 2002)
    There is a tunnel connecting city A & B. There is a cat which is standing at 3/8 thelength of the tunnel from A. It listens a whistle of the train and starts running towards the entrance where, the train and the cat meet. In another case, the cat started running towards the exit and the train again met the cat at the exit. What is the ratio of their speeds?
    (1) 4:1
    (2) 1:2
    (3) 8:1
    (4) None of these



  • Solution: Let ‘c’ and ‘t’ be the velocity of train and cat.
    Get the diagram right (if diagram is required)

    0_1514262673312_341a9d51-d7ce-428c-bbb4-b44d542b225f-image.png

    1st case :
    Train and cat meet at point A in the diagram.
    Let train be at a distance of “y” from the entrance of the tunnel.
    Cat is at a distance of 3x/8 from the entrance of the tunnel (Cat is between A&B. i.e. 3x/8 from A and 5x/8 from B )
    So, by the time train runs a distance of y, cat runs a distance of 3x/8 (As they reach entrance at the same time)
    Hence, equating the time taken, we have, y/t = 3x/8c

    2nd case:
    Train and cat meet at point B in the diagram.
    Train travels a distance of y + x (Till the end of the tunnel)
    Cat travels a distance of 5x/8 (Till the end of the tunnel)
    Both reach at the same time, hence we can equate the time. We get, y + x/t = 5x/8c.

    Subtracting the 2 equations, we have, (2nd case – 1st case) :
    [(y +x)/t] – [y/t] = (5x/8c) – (3x/8c)
    x/t = 2x/8c
    c/t = 1/4.
    t/c = 4/1.



  • Q20. (CAT 2001)
    A student took five papers in an examination, where the full marks were the same for each paper. His marks in these papers were in the proportion of 6 : 7 : 8 : 9 : 10. In all papers together, the candidate obtained 60% of the total marks. Then the number of papers in which he got more than 50% marks is:
    a. 2
    b. 3
    c. 4
    d. 5



  • Solution: Let the maximum mark in each subject be 100. Hence, total marks (maximum) is 500, as there are 5 subjects.
    Candidate obtained 60% of total marks, hence 60% of 500 = 300.
    His individual marks were 6x, 7x, 8x, 9x and 10x.
    So, total marks will be 6x + 7x + 8x + 9x + 10x = 40x
    40x = 300.
    x = 300/40 = 7.5
    His marks are 6x, 7x, 8x, 9x and 10x.
    So his marks were 45, 52.5 and all others are 50+ (as 7*7.5 itself is 50+)
    So, 4 subjects he has 50+ marks out of 100, which means more than 50%.



  • Q21. (CAT 2001)
    A can complete a piece of work in 4 days. B takes double the time taken by A, C takes double that of B, and D takes double that of C to complete the same task. They are paired in groups of two each. One pair takes two-thirds the time needed by the second pair to complete the work. Which is the first pair?
    a. A, B
    b. A, C
    c. B, C
    d. A, D



  • Solution: In a day, A would do 1/4 th of work ; B would d 1/8th of work, C would do 1/16th of work, D would do 1/32th of work.
    One pair would take > > 2/3 rd of time take by second pair.
    Out of all combinations ; we get A+D together in a day can do 9/32 th of work.
    B + C can do 3/16th of work.
    3/16 = 2/3 * 9/32
    B + C = 2/3 (A + D )
    Hence the first pair is B,C and the second pair is A,D.

    Note: A, D or B, C should be the answer because for other combinations this ratio will be less than 1/2.now we can safely assume it to be A,D combination and start checking with it - saves time!



  • Q22 (CAT 1997)
    A certain race is made up of three stretches: A, B and C, each 2 km long, and to be covered by a certain mode of transport. The following table gives these modes of transport for the stretches, and the minimum and maximum possible speeds (in km/hr) over these stretches. The speed over a particular stretch is assumed to be constant. The previous record for the race is 10 min.

    0_1514262850005_bd94dc22-9d19-41b4-a5ff-a10c29e4fb53-image.png

    Mr Tortoise completes the race at an average speed of 20 km/hr. His average speed for the first two stretches is four times that for the last stretch. Find the speed over stretch C.
    a. 15 km/hr
    b. 12 km/hr
    c. 10 km/hr
    d. This is not possible

    Mr Hare completes the first stretch at the minimum speed and takes the same time for stretch B. He takes 50% more time than the previous record to complete the race. What is Mr Hare's speed for the stretch C?
    a. 10.9 km/hr
    b. 13.3 km/hr
    c. 17.1 km/hr
    d. None of these

    Anshuman travels at minimum speed by car over A and completes stretch B at the fastest speed. At what speed should he cover stretch C in order to break the previous record?
    a. Maximum speed for C
    b. Minimum speed for C
    c. This is not possible
    d. None of these



  • Part 1 : Solution: Total distance = 2 + 2 + 2 = 6KM
    Average speed for the whole journey = 20 KM/hr
    Time taken = 6/20 = 0.3 hr ----> (1)
    Speed for C = S
    Then avg speed for A + B = 4S
    Time taken for A + B = (2+2)/4S = 1/S
    Time Taken for C = 2/S
    Total Time = 1/S + 2/S = 0.3 ( from (1) )
    => S = 10 Km/hr

    Part 2 : Solution : Hare is taking 2/40 hours = 1/20 hours = 3 minutes each for first 2 sets.
    Total time taken = 10 + 10/2 = 15 min.
    Time taken for third set = 9 min.
    2 km in 9 min for C - speed = 13.3kmph

    Part 3 : Solution : Previous record is 10 min.
    Stretch A is covered at minimum speed - 2 km at 40 kmph = 3 minutes
    Stretch B is covered at fastest speed - 2 km at 50 kmph = 2.4 minutes
    Minimum time required to cover C = 2 km at 20 kmph = 6 minutes
    So we cannot break the previous record!



  • Q23 (CAT 1997)
    The Weirdo Holiday Resort follows a particular system of holidays for its employees. People are given holidays on the days where the first letter of the day of the week is the same as the first letter of their names. All employees work at the same rate.

    Raja starts working on February 25, 1996, and finishes the job on March 2, 1996. How much time would T and J take to finish the same job if both start on the same day as Raja?
    a. 4 days
    b. 5 days
    c. Either (a) or (b)
    d. Cannot be determined

    Starting on February 25, 1996, if Raja had finished his job on April 2, 1996, when would T and S together likely to have completed the job, had they started on the same day as Raja?
    a. March 15, 1996
    b. March 14, 1996
    c. March 22, 1996
    d. Data insufficient



  • Part 1 : Solution: 1996 is a leap year.
    Raja takes 7 days to complete the job (So our work package is 7 person days)
    Now we'll apply Zeller's rule (Rule is explained as part of Q7) to determine that Feb 25th,1996 is a Sunday.
    If T and J starts on Sunday, by Wednesday they can complete the work. (Sunday - 2 Person days, Monday - 2 Person days, Tuesday - 1 Person day (T is on leave) and Wednesday - 2 Person days)
    So 4 days is sufficient.

    Part 2 : Solution: Raja doesn't have his name's first letter common to any of the day, so he won't rest.
    He started his work on 25th Feb and finished on April 2 => He took 38 days to complete the job ( 1996 is a leap year) and he finished 1/38th of the work each day.

    Now S will take break on Saturday & Sunday and T will take a break on Tuesday & Thursday.
    So every week each of them work for 5 days and together they work for 10 days.
    so 10/38 of the work will be completed in one week. In three weeks they complete 30/38 of the work.

    Feb 25th, 1996 is a Sunday. (Refer Zeller's rule in Q7)
    8 person days is required to finish the job.
    Monday - Both work together - 2 person days
    Tuesday - Only S - 1 person day
    Wednesday - Both work together - 2 person days
    Thursday - Only S - 1 person day
    Friday - 2 person days
    So by Friday they will complete the work.
    Total 21 + 6 = 27 days is required. March 22, 1996 would be the answer.



  • Q24. (CAT 2001)
    If 09/12/2001 happens to be Sunday, then 09/12/1971 would have been a
    a. Wednesday
    b. Tuesday
    c. Saturday
    d. Thursday



  • Solution: Apply Zeller's rule directly and we'll get the answer as Thursday. Wait, if you don't know Zeller's rule then it is a very handy trick for Date related problems. Rule is explained below [credits - MathForum]

    Zeller's Rule

    Here's the formula:
    f = k + [(13 * m-1)/5] + D + [D/4] + [C/4] - 2 * C.

    Eg: January 29, 2064 is a ?

    -- k is the day of the month. Let's use January 29, 2064 as an example. For this date, k = 29.
    -- m is the month number. Months have to be counted specially for Zeller's Rule: March is 1, April is 2, and so on to February, which is 12. (This makes the formula simpler, because on leap years February 29 is counted as the last day of the year.) Because of this rule, January and February are always counted as the 11th and 12th months of the previous year. In our example, m = 11.
    -- D is the last two digits of the year. Because in our example we are using January (see previous bullet) D = 63 even though we are using a date from 2064.
    -- C stands for century: it's the first two digits of the year. In our case, C = 20.
    Now let's substitute our example numbers into the formula.
    f = k + [(13 * m-1)/5] + D + [D/4] + [C/4] - 2 * C
    = 29 + [(13 * 11-1)/5] + 63 + [63/4] + [20/4] - 2 * 20
    = 29 + [28.4] + 63 + [15.75] + [5] - 40
    = 29 + 28 + 63 + 15 + 5 - 40
    = 100.

    Once we have found f, we divide it by 7 and take the remainder. Note that if the result for f is negative, care must be taken in calculating the proper remainder. Suppose f = -17. When we divide by 7, we have to follow the same rules as for the greatest integer function; namely we find the greatest multiple of 7 less than -17, so the remainder will be positive (or zero). -21 is the greatest multiple of 7 less than -17, so the remainder is 4 since -21 + 4 = -17. Alternatively, we can say that -7 goes into -17 twice, making -14 and leaving a remainder of -3, then add 7 since the remainder is negative, so -3 + 7 is again a remainder of 4.

    A remainder of 0 corresponds to Sunday, 1 means Monday, etc. For our example, 100/7 = 14, remainder 2, so January 29, 2064 will be a Tuesday.



  • Q25. (CAT 2001)
    At his usual rowing rate, Rahul can travel 12 miles downstream in a certain river in six hours less than it takes him to travel the same distance upstream. But if he could double his usual rowing rate for this 24 mile round trip, the downstream 12 miles would then take only one hour less than the upstream 12 miles. What is the speed of the current in miles per hour?
    a. 7/3
    b. 4/3
    c. 5/3
    d. 8/3



  • Solution: Let speed in stagnant water is x and speed of current=y.
    12/(x - y) - 12/(x + y) = 6..
    1/(x - y) - 1/(x + y) = 1/2
    2y/(x^2-y^2) = 1/2 --- (1)

    Similarly from second condition 2y/(4x^2 - y^2) = 1/12 -- (2)

    (1) / (2) = (4x^2 - y^2)/(x^2-y^2) = 6

    2x^2 = 5y^2
    Substituting this value in any of the condition we'll get y=8/3.
    so option D



  • Q26. (CAT 2000)
    A truck travelling at 70 kilometres per hour uses30% more diesel to travel a certain distance than it does when it travels at the speed of 50 kilometres per hour. If the truck can travel 19.5 kilometres on a litre of diesel at 50 kilometres per hour, how far can the truck travel on 10 litres of diesel at a speed of 70 kilometres per hour?
    (1) 130
    (2) 140
    (3) 150
    (4) 175


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