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Quantitative Aptitude Test - 13

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Quantitative Aptitude Test - 13
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  • Question 1
    4 / -1

    A management institute has six senior professors and four junior professors. Three professors are selected at random for a Government project. The probability that at least one of the junior professors would get selected is:

    Solution

    At least one junior professor can be selected means the number of selected junior professors will be either 1 or 2 or 3.

    Number of ways this can happen = 4C1 x 6C2 + 4C2 x 6C1 + 4C3 = 100

    Probability = 100/10C3 = 100/120 = 5/6

  • Question 2
    4 / -1

    Five small pumps and two large pumps are filling a tank. Each small pump works at 3/5ᵗʰ the rate of the large pump. If all seven work at the same time, they should fill the tank in what fraction of the time it would have taken by one large pump alone?

    Solution

    Given, small pump = 3/5ᵗʰ the rate of large pump

    => 5 small pumps = 3 large pumps

    ∴ 5 small pumps + 2 large pumps = 5 large pumps.

    ∴ Required fraction = 1/5

  • Question 3
    4 / -1

    Find the average percentage change of production of rice from 2013 to 2016 if the percentage increase in production in 2014, 2015 and 2016 are 20%, 10% and 15% respectively when calculated with respect to previous year.

    Solution

    If the value in 2013 is 100 then the value in 2016 will be,

    100 x 1.2 x 1.1 x 1.15 = 151.8

    So the increase = 51.8 and the average increase = 51.8/3 = 17.26%

  • Question 4
    4 / -1

    The number of people voting for two candidates A & B in an election is a prime number less than 250. The ratio of people who vote for A to those who vote for B, could be

    Solution

    In this case the total of two numbers should always give us a prime number. Using the options:

    82 + 83 = 165 (not a prime number)

    75 + 78 = 153 (not a prime number)

    76 + 81 = 157 (prime number)

    91 + 78 = 169 (not a prime number)

    36 + 169 = 205 (not a prime number)

  • Question 5
    4 / -1

    Montu is preparing for law entrance exam through an online test portal, which calculates his average score after every mock paper. In his last two papers P and Q, he got 83 and 75 marks respectively. As a result, his average score in mocks increase by 2 and 1 marks respectively. Now, he took one more test R and his average further increases by 3 marks. How much he scored in mock R?

    Solution

    Let the total marks scored and number of mocks taken by Montu be 't' and 'a' respectively, before taking mocks P, Q and R.

    Average, A = t/a

    After taking mock P, average increases by 2.

    Thus,

    A + 2 = (t + 83)/(a+1)

    (t/a) + 2 = (t + 83)/(a+1)

    Or, t = 81a - 2a2

    After taking mock Q, average further increases by 1

    Thus,

    (t/a) + 3 = (t + 83 + 75)/(a+2)

    Substituting t from above, we have

    (81 - 2a) + 3 = (81a - 2a2 + 158)/(a+2)

    Solving, a = 10

    So, t = 810 - 200 = 610

    Now, let the marks obtained in mock R be m.

    After taking mock R, average increases by 3,

    Thus,

    768 + m = 13x67

    m = 103

  • Question 6
    4 / -1

    There is an army camp of 60 soldiers. 1000 kg of wheat is supplied to the camp which will last for 60 days. However after a few days 20 more soldiers joined the camp due to which wheat lasted for 30 more days from the day they joined. After how many days did the additional 20 soldiers join the camp?

    Solution

    Let the quantity of wheat consumed by a soldier in one day be n kg.

    Since 1000 kg of the wheat was sufficient for 60 soldiers for 60 days.

    Hence, 1000 = 60 X 60 X n

    ⇒ n = (5/18) kg

    Let's assume that after p days 20 more soldiers joined the camp.

    Hence, consumption of wheat by 60 soldiers for p days + consumption of wheat by (60 + 20) 80 soldiers for 30 days = 1000

    ⇒ 60 X p X n + 80 X n X 30 = 1000

    ⇒ (60p + 2400) X n = 1000

    ⇒ (60p + 2400) X (5/18) = 1000

    ⇒ (60p + 2400) = 3600

    ⇒ p = 20

    Hence the 20 additional soldiers joined the camp after 20 days.

  • Question 7
    4 / -1

    Car 'A' travels at the speed of 52 km/hr and reaches its destination in 9 hours. Car 'B' travels at the speed of 42 km/hr and reaches its destination in 6 hours. What is the ratio of the distance covered by car A and car B respectively?

    Solution

    Distance travelled by car A = 52x9 = 468 km

    Distance travelled by car B = 42x6 = 252 km

    Required ratio = 468:252 = 13:7

  • Question 8
    4 / -1

    What should come in place of the question mark (?) in the following number series?

    13, 49, 113, ?, 357, 553

    Solution

    The pattern is 13 + 62 = 49, 49 + 82 = 113, 113 + 102 = 213, 213 + 122 = 357, 357 + 142 = 553

  • Question 9
    4 / -1

    What should come in place of the question mark (?) in the following number series?

    41, 46, 56, 67, 80, 88, ?

    Solution

    The next term is obtained by adding the sum of digits to the number. Hence the next number is 88 + 8 + 8 = 104.

  • Question 10
    4 / -1

    What should come in place of the question mark (?) in the following number series?

    9, 4, 11, 2, 13, ?

    Solution

    9 - 5 = 4, 4 + 7 = 11, 11 - 9 = 2 , 2 + 11 = 13, 13 - 13 = 0

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