Sample Aptitude Questions of IBM

Aptitude questions are generally asked in placement tests of top companies. In case of the written test conducted by IBM, aptitude is tested in form of cognitive ability games. This section presents questions as visual problems to be solved based on the mathematical knowledge and numerical abilities of the test-taker. For your practice, we have given some sample questions as per the concepts tested in IBM placement tests.
IBM Aptitude Questions
  1. In a race of 600 metres, A can beat B by 60 metres and in a race of 500 metres, B can beat C by 50 metres. By how many metres will A beat C in a race of 400 metres?
    1. 76 metres
    2. 80 metres
    3. 70 metres
    4. 84 metres
    Answer: Option A
    A runs B runs C runs
    600 metres race 600m 540 m
    500 metres race 500 m 450m
    Combing ratio A runs B runs C runs
    300metres - 2700meters - 2430metres
    Unitary A runs B runs C runs
    Method 400mtres - 360 metres - 324 metres
    ∴ A beats C by 400-324 = 76 metres.
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  3. A man travelled a distance of 80 km in 7 hours partly on foot at the rate of 8 km per hour and partly on bicycle at 16 km per hour. Find the distance travelled on foot.
    1. 26 km
    2. 32 km
    3. 30 km
    4. 28 km
    Answer: Option B
    Total Time = 7hrs
    Let the distance travelled by foot @ 8kmph be x kms
    ∴ distance travlled by bicycle @ 16kmph be (80-x)kms
    ATQ. 7hr = x/8 + (80-x)/16
    ∴ 7 = (2x + 80 - x)/16
    ∴ x = 32kms
  4. If 20 men can build a wall 112 metres long in 6 days, what length of a similar wall can be built by 25 men in 3 days?
    1. 65mtr.
    2. 52mtr.
    3. 70mtr.
    4. 78mtr.
    Answer: Option C
    20 men is 6 days can build 112 metres
    25 men in 3 days can build = 112*(25/20)x(3/6)
    = 70 meters
  5. If the compound interest on a certain sum of money for 3 years at 10% per annum be Rs. 993, what would be the simple interest?
    1. Rs. 880
    2. Rs. 890
    3. Rs. 895
    4. Rs. 900
    Answer: Option D
    Let P = Principal
    A - Amount
    We have a = P(1 + R/100)3 and CI = A - P
    ATQ 993 = P(1 + R/100)3 - P
    ∴ P = Rs 3000/-
    Now SI @ 10% on Rs 3000/- for 3 yrs = (3000 x 10 x 3)/100
    = Rs 900/-
  6. What annual installment will discharge a debt of Rs. 4600 due in 4 years at 10% simple interest?
    1. 1000
    2. 1030
    3. 1100
    4. None of these
    Answer: Option A
    Let the annual instalment be Rs. 100. The first instalment will be paid one year from now i.e. 3 years before it is actually due. The second instalment will be paid two years from now i.e. 2 years before it is actually due.
    The third instalment will be paid 1 year before it is actually due.
    The fourth instalment will be paid on the day the amount is actually due.
    On the first instalment the interest will be paid for 3 years, on the second for 2 years, on the third for 1 year, on the fourth for 0 year. In total an interest for 6 years will be paid (3 + 2 + 1 + 0) on Rs. 100 @ 10%. Interest = (100 × 6 × 10)/100 = Rs. 60 and the principal is Rs 100 × 4 = Rs 400. The total loan that can be discharged is Rs. 400 + 60 = Rs. 460. Here the technique of Chain Rule will be applied. I.e. for Rs. 460 the instalment required is Rs. 100, for Rs. 4600 the instalment required is 4600 × 100/460 = Rs. 1000.
  7. A number whose fifth part increased by 5 is equal to its fourth part diminished by 5, is
    1. 160
    2. 180
    3. 200
    4. 220
    Answer: Option C
    x/5 + 5 = x/4 - 5
    ⇒ x/5 - x/4 = 10
    x/20 = 10
    ⇒ x = 200
  8. Two numbers are such that the ratio between them is 3 : 5, but if each is increased by 10, the ratio between them becomes 5:7. The numbers are
    1. 3, 5
    2. 7, 9
    3. 13, 22
    4. 15, 25
    Answer: Option D
    No's are in the ratio 3:5
    Le the No's be 3x and 5x
    ATQ. (3x+10) : (5x+10) = 5:7
    ∴ x = 5
    No's are (15,25)
  9. A man rows downstream 30 km and upstream 18 km, taking 5 hours each time. What is the velocity of the stream (current)?
    1. 1.2 km/hr
    2. 1.5 km/hr
    3. 2.5 km/hr
    4. 1.8 km/hr
    Answer: Option A
    LET X =SPEED OF BOAT AND Y = SPEED OF CURRENT.
    ⇒ 30/(X+Y)=18/(X-Y)=5 BY SOLVING Y = 1.2
  10. A train 125 metre long is running at 50 km/hr. In what time will it pass a man running at 5 km/hr in the same direction in which the train is going?
    1. 25 sec
    2. 10 sec
    3. 20 sec
    4. 15 sec
    Answer: Option B
    Distance = 125 metres Speed = 50-5 = 45kmph => 45 * (5/18) = 12.5 m/s
    Time = 125 / 12.5 = 10 sec.
  11. A is twice as fast as B and B is thrice as fast as C is. The journey covered by C in 42 minutes, will be covered by A in
    1. 11 min
    2. 14 min
    3. 7 min
    4. 17 min
    Answer: Option C
    B is thrice as fast as C
    C covers in 42min
    ∴ B covers in 42/3 = 14 min
    A is twice as fast as B
    ∴ A covers in 14 * (1/2) = 7 min
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