Thursday, August 11, 2011

Sample Questions For IAS

  1. How many litres of a solution that is 15% salt must be added to another solution that is 8% salt so that the resulting solution is 10% salt?
    1. 2 litres
    2. 3 litres
    3. 4 litres
    4. 1 litre
    Let n represent the number of litres of the 15% solution. Then the amount of salt in the 15% solution = 0.15n
    Also the amount of the salt in the 8% solution = 0.08 x 5
    Also, the amount of salt in the resulting 10% mixture = 0.10 (n+5)
    Now, the total amount of salt in both the solutions combined must equal the amount of salt in the mixture
    0.15n + (0.08 x 5) = 0.10 (n+5)
    This gives  n = 2
  2. If positive integers x and y are not both odd, which of the following must be even?
    1. xy
    2. x + y
    3. x – y
    4. x + y – 1
    Since it is given that x and y are not both odd, either both must be even or one even and one odd. Using these two alternatives, we can test the outcome of each answer choice to determine when both must be even.
    1. (even)(even) = even; (even)(odd) = even. The answer must be
      even for both alternatives
    2. even + even = even; even + odd = odd. The answer need not be even for both alternatives
    3. even – even = even; even – odd = odd. Need not be even
    4. even + even – 1 = odd; even + odd – 1 = even. Need not be even
  3. Equal amounts of water are poured into two empty jars of different capacities, which made one jar ¼ full and the other jar 1/3 full. If the water in the jar with the lesser capacity is poured into the jar with greater capacity, what fraction of the larger jar will be filled with water?
    1. 1/7
    2. 2/7
    3. 1/12
    4. 7/12
    It is given that the amounts of water in the two jars are equal. Thus, when the smaller jar is poured into the larger jar, the water in the larger jar will be doubled i.e. the larger jar will become 2 x ¼ = ½ full
  4. If u > t, r > q, s > t and t > r, which of the following must be true?
    1. u > s
    2. s > q
    3. u > s and s > q
    4. s > q and u > r
going from the given information, u > t > r > q. While it is given that s > t it is not known whether s is between t and u or greater than u. Now, looking at each choice given
  1. It is known that u > t and s > t. Select t=2, u=3, s=4. It follows that u < s. Hence the given choice is Not true
  2. Since s > t, t > r, r > q, it follows that s > q. Must be true
  3. Just shown in A that u > s is not true
  4. Since u > t, t > r, it follows that u > r. must be true
  1. A certain clock marks every hour by striking a number of times equal to the hour, and the time required for a stroke is exactly equal to the time interval between strokes. At 6:00 the time lapse between the beginning of the first stroke and the end of the last stroke is exactly 22 seconds. At 12:00, how many seconds lapse between the beginning of the first stroke and the end of the last stroke?
    1. 72 seconds
    2. 50
    3. 48
    4. 46
    At 6:00, there are 6 strokes and 5 intervals between strokes. Thus, there are 11 equal time intervals in the 22 seconds between the beginning and end.
    Since the total time lapse is given as 22 seconds, each interval must last 22/11 = 2 seconds
    At 12:00, there are 12 strokes and 11 intervals between strokes. Thus, there are 23 equal intervals, each of duration 2 seconds.
    Thus, the total time lapsed between beginning and end must be 23 x 2 = 46 seconds
  2. A company that ships boxes to 12 distribution centres uses colour coding to indentify each centre. If either one colour or a pair of colours is used to represent each centre, and each centre is uniquely identified by that colour code, what is the minimum number of colours needed to uniquely identify all centres? Assume that the order of the colours in a pair does not matter.
    1. 4
    2. 5
    3. 6
    4. 12
    Since the question asks for the minimum number of colours needed, let us start with the lowest answer choice available. We can calculate each successive option until finding the minimum number of colours that can identify the 12 distribution centres.
    For any given number of colours, the number of centres it can identify is given by the number of combinations possible.
    If n is the number of colours available, and r is the number used for a given code, the number of possible combinations is _nC_r = \frac{n!}{r!(n-r)!}
No. of colours
Number represented by one colour
Number represented by two colours
Total represented
Decision
4
4
_4C_2=\frac{4!}{2!(4-2)!} = 6
4 + 6 = 10
Not sufficient
5
5
_5C_2 = \frac{5!}{2!(5-2)!} = 10
5 + 10 = 15
Sufficient
  1. A box contains 100 balls, numbered 1 to 100. If three balls are selected at random and with replacement from the box, what is the probability that the sum of the three numbers will be odd?
    1. ¼
    2. 3/8
    3. ½
    4. 5/8
    Since there are 50 odd and 50 even balls, the probability of selecting an odd or an even ball is ½.
    For the sum of the three numbers to be odd, the numbers must all be odd or one number must be odd and the other two even.
    Probability of selecting three odd balls = \frac{1}{2}\frac{1}{2}\frac{1}{2} = \frac{1}{8}
    Probability of selecting one odd and two even balls = P{odd, even, even} + P{even, odd, even} + P{even, even, odd}
    P{odd,even,even} = P{even, odd, even} = P{even, even, odd} = \frac{1}{2}\frac{1}{2}\frac{1}{2} = \frac{1}{8}
    Thus, the probability of one odd and two even balls = \frac{1}{8} + \frac{1}{8} + \frac{1}{8} = \frac{3}{8}
    Hence, the probability that the sum of the three balls will be odd is \frac{1}{8} + \frac{3}{8} = \frac{1}{2}
  2. A box contains 14 apples and 23 oranges. How many oranges must be removed from the box so that 70% of the contents of the box will be apples?
    1. 3
    2. 6
    3. 14
    4. 17
    From the given information it can be determined that there is a total of 37 fruit pieces in the box.
    If x is the number of oranges that need to be removed, \frac{14}{37-x} = 0.70
    This gives x = 17
  3. An airline passenger is planning a trip that involves three connecting flights that leave from airports A, B and C. The first flight leaves airport A every hour beginning at 08:00 and arrives at airport B 2.5 hours later. The second flight leaves airport B every 20 minutes beginning at 08:00 and arrives at C 1 1/6 hours later. The third flight leaves airport C every hour beginning at 08:45. What is the least total amount of time the passenger must spend between flights?
    1. 25 min
    2. 1 hr 5 min
    3. 1 hr 15 min
    4. 2 hr 20 min
Assume that the passenger takes the first flight from airport A at 08:00. This would take him to airport B at 10:30
The earliest flight he can take from airport B will leave at 10:40. There is a 10 min wait here. This flight would take him to airport C 1 1/6 hours later i.e. 1 hr 10 min later i.e. at 11:50
The earliest flight he can take from airport C will leave at 12:45. There is a 55 min wait here.
Thus the total wait time is 10 + 55 = 1 hr 5 min
  1. Each of 25 people is enrolled in history, mathematics or both. If 20 are enrolled in history and 18 in mathematics, how many are enrolled in both?
    1. 12
    2. 10
    3. 13
    4. 15
    The 25 people can be divided into three sets: those who history only, those in mathematics only, and those in both
    If n is the number of people enrolled in both courses, 20-n is the number in enrolled in history only and 18-n is the number in mathematics only.
    Since the total number of students is 25, n + (20-n) + (18-n) = 25
    This gives n = 13
  2. In a certain production lot, 40% of toys are red and the remaining is green. Half the toys are small and half is large. If 10% of the toys are red and small, and 40% of toys are green and large, how many of the toys are red and large?
    1. 60
    2. 40
    3. 50
    4. 70
    Let us organize the information in a table:

    Red
    Green
    Total
    Small
    10%

    50%
    Large


    50%
    Total
    40%
    60%
    100%
    Based on what is given, we can compute the missing information:

    Red
    Green
    Total
    Small
    10%
    40%
    50%
    Large
    30%
    20%
    50%
    Total
    40%
    60%
    100%
    Let us assume that n is the total number of toys
    It is also given that 40 toys are green and large. Then 0.20n = 40
    This gives n = 200
    Thus, the number of toys that are red and large is 0.30n = 60

0 comments:

Post a Comment