Thursday, August 11, 2011

Sample Questions For IAS

2:58 AM  akhilesh  No comments

  1. The price of sugar is being raised by 10%. By how much percent must a man reduce his consumption so as not to increase his expenditure
    1. 9%
    2. 9.5%
    3. 10%
    4. 9 1/11%
    Let the original price be Rs. 100 and the original quantity of x
    Then the quantity consumed is 100
    Let the new price be Rs 110
    Then the man must reduce consumption by: 10 * \frac{100}{110}
  2. If a man loses 4% by selling bananas at Rs 12 a rupee, how many per rupee must he sell them so as to gain 44%?
    1. 8
    2. 7
    3. 6
    4. 9
    The sale price of one banana is \frac{1}{12}
    Using the formula: \frac{S_1}{100+x_1} = \frac{S_2}{100+x_2}
    \frac{\frac{1}{12}}{100-4} = \frac{S_2}{100+44} = \frac{1}{12 x 96} = \frac{S_2}{144} Then S_2 = \frac{144}{12 x 96} = \frac{1}{8}
  3. I have a certain sum of money to be distributed among certain no. of boys. If I give Rs 3 to each, I shall spend Rs 4 less but if Rs 5 to each, I shall need Rs 6 more. How much do I have with me?
    1. 19
    2. 18
    3. 20
    4. 23
    Let the number of boys be x and the amount I have be y
    3x = y - 4
    5x = y + 6
    This gives x = 5 and y=19
  4. I multiply a number by 36 and divide the result by 12. The quotient is 374181. What is the number?
    1. 124772
    2. 124727
    3. 134727
    4. 174232
    Let the number be x
    According to the problem, \frac{36x}{12} = 374181
    This gives x = \frac{374181}{3} = 124727
  5. Three bells toll at intervals of 1.2, 1.8 and 2.7 seconds beginning together. How often will each bell toll before their tolling together again?
    1. 9, 4, 6
    2. 8, 6, 4
    3. 7, 6, 4
    4. 10, 2, 3
    The bells will toll together again at the LCM of the three intervals.
    The LCM of 1.2, 1.8 and 2.7 is 10.8 sec
    Between the beginning and 10.8 sec, each bell will toll the following number of times:
    \frac{10.8}{1.2} = 9 and \frac{10.8}{1.8} = 6 and \frac{10.8}{2.7} = 4
  6. How many revolutions will be made by a wheel which revolves at the rate of 243 revolutions in 3 minutes, while another wheel revolving 374 times in 11 minutes makes 544 revolutions?
    1. 1269
    2. 1270
    3. 1296
    4. 1297
    For the second wheel, the time taken for one revolution is \frac{11}{374}
    The total time taken by the second wheel is 544 x \frac{11}{374}
    For the first wheel, the number of revolutions in one minute is \frac{243}{3}
    In this same time, the first wheel makes the following number of revolutions: 544 x \frac{11}{374} x \frac{243}{3} = 1296
  7. The digits in the units place and lakh’s place in a number are 3 and 8. What will be the digits in the same places in the remainder when 99999 is subtracted from the number
    1. (4,6)
    2. (4,5)
    3. (7,4)
    4. (4,7)
    Let the number be 80003
    When 99999 is subtracted from this number, the digits will be (7,4)
  8. A ship 40 miles from shore springs a leak which admits 3 3/4 tons of water in 12 minutes. 60 tons of water are enough to sink the ship. But the ship’s pumps can throw out 12 tons of water in an hour. What should the average speed of the ship be so that it reaches the shore before sinking?
    1. 4 mph
    2. 4.5 mph
    3. 4.7 mph
    4. 7 mph
    Amount of water entering the ship in one hour: 3\frac{3}{4} x \frac{60}{12} = \frac{75}{4} tons
    Amount of water thrown out in one hour: 12 tons
    Water accumulating in the ship in one hour = \frac{75}{4} - 12 = \frac{27}{4}
    Time taken to accumulate 60 tons: \frac{60}{\frac{27}{4}} = \frac{80}{9} hours
    Speed for safe arrival: \frac{40}{\frac{80}{9}} = 4.5 mph
  9. An express train owing to a defect in the engine goes at 5/8 th of its usual speed and arrives at 6:49 p.m. instead of 5:55 p.m. At what hour did the train start?
    1. 4:25
    2. 4:30
    3. 4:50
    4. 4:55
    The new speed = 5/8 of usual speed
    If the usual time taken is x, the new time is \frac{8}{5}x
    Then, \frac{8}{5}x - x = 6:49 - 5:55 = 54 min
    This gives x = 90 min
    Thus, the starting time is 5:55 - 90 min = 4:25
  10. A train starts with a certain number of passengers. At the first station, it drops 1/3 of those and takes 20 more. At the next it drops 1/2 of the new total and takes on 10 more. On reaching the third station there are 60 left. With how many passengers did the train start?
    1. 110
    2. 120
    3. 125
    4. 130
    Let the original number of passengers be x
    Then, the number of passengers after the first station is: x_1 = x - \frac{x}{3} + 20
    After the second station, x_2 = x_1 - \frac{x_1}{2} + 10
    The number of passengers reaching the third station is given to be 60 i.e. x_2 = 60
    This gives, x_1 - \frac{x_1}{2} + 10 = 60
    x_1 = 100
    This gives,
    x - \frac{x}{3} + 20 = 100
    x = 120

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