Solved CAT Questions (Arithmetic) - Set 4

  • Skadoooosh!!!

    Q1. (CAT 1999)
    Forty per cent of the employees of a certain company are men, and 75% of the men earn more than Rs. 25,000 per year. If 45% of the company’s employees earn more than Rs. 25,000 per year, what fraction of the women employed by the company earn less than or equal to Rs. 25,000 per year?
    a. 2/11
    b. 1/4
    c. 1/3
    d. 3/4

    Let say total number of employees = 100
    So men employees = 40
    Women employees = 100 - 40 = 60
    75% of men earn more than 25,000 => 40 x 75/100 = 30 men.
    45% of total employees earn more than 25,000 => 45 Employees
    So 45 - 30 = 15 women earn more than 25000 which is 1/4th of the total women employee
    => 1 - 1/4 = 3/4th of the women employee earn less than 25000.

    Even though we wrote many steps for making the solution clear, this can be solved quickly in our mind with in 10 seconds.

    Note: In most of the questions involving it would be easier to assume total as 100 and then solve.

    Q2. (CAT 2000)
    The table below shows the age-wise distribution of the population of Reposia.

    Age GroupPercentage
    Below 15 years30
    15 - 2417.75
    25 - 3417
    35 - 4414.5
    45 - 5412.5
    55 - 647.1
    65 and above1.15

    The number of people aged below 35 years is 400 million. If the ratio of females to males in the ‘below 15 years’ age group is 0.96, then what is the number of females (in millions) in that age group?
    a) 82.8
    b) 90.8
    c) 80.0
    d) 90.0

    If you look at the table, people aged below 35 years is 30 + 17.75 + 17 = 64.75% of the total population (which is given as 400 million)
    So total population = 400 x 100 / 64.75 = 617.76 million
    Female/Male (in below 15 years age group) = 0.96 : 1
    Total population below 15 years age group = 617.76 x 30 / 100 = 185.33 million
    Number of females = (0.96/1.96) x 185.33 million = 90.77 million (~ Option b)

    Q3. (CAT 1999)
    Navjivan Express from Ahmedabad to Chennai leaves Ahmedabad at 6.30 a.m. and travels at 50 kmph towards Baroda situated 100 km away. At 7.00 a.m. Howrah-Ahmedabad Express leaves Baroda towards Ahmedabad and travels at 40 kmph. At 7.30 a.m. Mr Shah, the traffic controller at Baroda realizes that both the trains are running on the same track. How much time does he have to avert a head-on collision between the two trains?
    a. 15 min
    b. 20 min
    c. 25 min
    d. 30 min

    I know this inspector and before solving the question will tell you the something that happened in the interview of this traffic controller.
    Our guy meets the inspector for interview.
    Inspector puts this question to him: "What would you do if you realized that two trains were heading for each other on the same track?"
    He says, "I would switch the points for one of the trains."
    "What if the lever broke?" asked the inspector.
    "Then I'd dash down out of the signal box," he said, "and I'd use the manual lever over there."
    "What if that had been struck by lightning?" - Inspector didn't let him go that easily!
    "Then," he continues, "I'd run back into the signal box and phone the next signal box."
    "What if the phone was engaged?" - asked the inspector.
    "Well in that case, I'd rush down out of the box and use the public emergency phone at the level crossing up there."
    "What if that was vandalized?" - Inspector strikes again.
    "Oh well, then I'd ask my uncle to come to station as soon as possible."
    This puzzles the inspector, so he asks, "Why would you do that?"
    "Because he's never seen a train crash." ;)

    Now back to the question. At 7:30, train leaves from A to B would have covered 50 km while train from B to A would have covered 20 km since it travels for only half an hour.
    Distance remaining between the trains = 100 - (50+20) i.e. 30km and the relative speed = 90kmph.
    Therefore, time = 30/90 hours = 20 mins.

    Q4 - Q6 (CAT 1999)
    A road network (shown in the figure below) connects cities A, B, C and D. All road segments are straight lines. D is the mid-point on the road connecting A and C. Roads AB and BC are at right angles to each other with BC shorter than AB. The segment AB is 100 km long. Ms X and Mr Y leave A at 8.00 a.m., take different routes to city C and reach at the same time. X takes the highway from A to B to C and travels at an average speed of 61.875 kmph. Y takes the direct route AC and travels at 45 kmph on segment AD. Y’s speed on segment DC is 55 kmph


    Q4. What is the average speed of Y?
    a. 47.5 kmph
    b. 49.5 kmph
    c. 50 kmph
    d. 52 kmph

    Route of X = A - B - C ; Average speed = 61.785 kmph
    Route of Y = A - C ; Travels 45 kmph on segment AD and at 55 kmph on DC
    Average speed (distance in both cases are same) = 2 x 45 x 55 / (45 + 55) = 49.50

    Note: When one travels at speed a for half the distance and speed b for other half of the distance, Average speed = 2ab/(a + b)

    Q5. The total distance travelled by Y during the journey is approximately
    a. 105 km
    b. 150 km
    c. 130 km
    d. Cannot be determined

    Total distance travelled by Y = AC
    Apply Pythagoras theorem : 100^2 + BC^2 = AC^2 -- (1)
    Now another condition given is that both X and Y reach at the same time ( T = D/S)
    So (100 + BC)/61.785 = AC/49.50 ( 49.50 is the avg speed for Y we calculated before)
    (100 + BC)/AC = 61.785 / 49.50 ~ 1.25
    BC = 1.25AC - 100 -- (2)
    Solve (1) and (2), AC = 105 and BC = 31
    Y travelled 105 KM.

    Q6. What is the length of the road segment BD?
    a. 50 km
    b. 52.5 km
    c. 55 km
    d. Cannot be determined

    The midpoint of the hypotenuse (D) is equidistant from all the three vertices.
    So BD = AD = CD = AC/2 = 105/2 = 52.5 KM

    Q7 - Q8 (CAT 1999)
    Rajiv reaches city B from city A in 4 hours, driving at speed of 35 kmph for the first two hour and at 45 kmph for the next two hours. Aditi follows the same route, but drives at three different speeds: 30, 40 and 50 kmph, covering an equal distance in each speed segment. The two cars are similar with petrol consumption characteristics (km per litre) shown in the figure below.
    Q7. Zoheb would like to drive Aditi’s car over the same route from A to B and minimize the petrol consumption for the trip. What is the quantity of petrol required by him?
    a. 6.67 L
    b. 7 L
    c. 6.33 L
    d. 6.0 L

    To minimize the petrol consumption, Zoheb should maximize the mileage. He needs to drive at 40 kmph which gives him 24 km per litre.
    Total distance covered will be 35 x 2 + 45 x 2 = 160 km.
    Petrol consumption = 160/24 = 6.67 Litres.

    Q8. The quantity of petrol consumed by Aditi for the journey is
    a. 8.3 L
    b. 8.6 L
    c. 8.9 L
    d. 9.2 L

    Aditi follows the same route so the distance is same as calculated before - 160 KM.
    Petrol consumption = (160/3)/16 + (160/3)/24 + (160/3)/50 = 8.9 litres

    Q9. (CAT 1998)
    A company has a job to prepare certain number cans and there are three machines A, B and C for this job. A can complete the job in 3 days, B can complete the job in 4 days, and C can complete the job in 6 days. How many days will the company take to complete the job if all the machines are used simultaneously?
    a. 4 days
    b. 4/3 days
    c. 3 days
    d. 12 days

    It is easy to assume the work package consist of 12 cans ( LCM of 3, 4 and 6)
    A will make 4 cans, B will make 3 cans and C will make 2 cans per day.
    A, B and C together will make 9 cans a day
    So for 12 cans, 12/9 = 4/3 days.

    Q10. (CAT 1998)
    Distance between A and B is 72 km. Two men started walking from A and B at the same time towards each other. The person who started from A travelled uniformly with average speed of 4 km/hr. While the other man travelled with varying speed as follows: in the first hour his speed was 2 km/hr, in the second hour it was 2.5 km/hr, in the third hour it was 3 km/hr, and so on. When will they meet each other?
    a. 7 hr
    b. 10 hr
    c. 35 km from A
    d. Mid-way between A and B

    M1 - Uniform speed of 4 km/hr
    So distance covered in nth hour = 4n

    M2 - distance covered in 1st, 2nd, 3rd hour.. is 2, 2.5, 3, 3.5 etc.. respectively
    This is an AP with a = 2 and d = 0.5
    Total distance covered in nth hour = Sum of n terms of the above AP = n/2 [ 4 + (n - 1)/2]
    n/2 [ 4 + (n - 1)/2] = 4n
    [ 4 + n/2 - 1/2] = 8
    n/2 = 4 + 1/2 = 9/2
    n = 9 (So they will meet in the 9th hour)
    Distance covered by M1 (or M2) when they meet = 4 x 9 = 36 (half way between A and B )

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