IBM Aptitude Interview Questions & Answers

  1. Question 1. If 20 Men Can Construct A Divider Of 112 Meters In Length In 6 Days, What Length Of A Comparable Divider Can Be Worked By 25 Men In 3 Days ?

    Answer :

    20 men in 6 days can build 112 meters

    25 men in 30 days can build=112*(25/20)*(3/6)

    = 70 meters

  2. Question 2. 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 ?

    Answer :

    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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  4. Question 3. In A Race Of 600 Meters, A Can Beat B By 60 Meters And In A Race Of 500 Meters; B Can Beat C By 50 Meters. By What Number Of Meters Will A Beat C In A Race Of 400 Meters ?

    Answer :

    Let’s assume A finishes the 600 m race in 60 sec, then

    600/60 = 10 m/sec is his speed

    B traveled (600-60 = 540 m in 60 sec, therefore

    540/60 = 9 m/sec is B’s speed

    “in a race of 500 metres, B can beat C by 50 metres.”

    500/9 = 55.56 sec is B’s time to finish a 500 m race

    C traveled 500-50 = 450 m in 55.56 sec, therefore

    450/55.56 = 8.1 m/sec is C’s speed

    By how many will A beat C in a race of 400 metres?

    400/10 = 40 sec for A to run a 400 m race

    C will travel 8.1*40 = 324 m in 40 sec therefore

    C will be 400-324 = 76 m behind when A crosses the finish line

  5. Question 4. On The Off Chance That The Accumulated Dividends On A Specific Total Of Cash For A Long Time At 10% For Each Annum Be Rs. 993, What Might Be The Basic Intrigue?

    Answer :

    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/ – 

    Presently SI @ 10% on Rs 3000/ – for 3 yrs = (3000 x 10 x 3)/100 

    = Rs 900/ –

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  7. Question 5. What Annual Installment Will Discharge A Debt Of Rs. 4600 Due In 4 Years At 10% Simple Interest ?

    Answer :

    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.

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  9. Question 6. A Number Whose Fifth Part Expanded By 5 Is Equivalent To Its Fourth Part Lessened By 5, Is

    Answer :

    X/5 + 5 = x/4 – 5 

    ⇒ x/5 – x/4 = 10 

    X/20 = 10 

    ⇒ x = 200

  10. Question 7. A Man Pushes Downstream 30 Km And Upstream 18 Km, Taking 5 Hours Each Time. What Is The Speed Of The Stream (current) ?

    Answer :

    Let x=speed of boat and y=speed of current

    =30/ (x+y)=18/(x-y)=5 by solving y=1.2 km/hr

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  12. Question 8. A Train 125 Meter Long Is Running At 50 Km/hr. In What Time Will It Pass A Man Running At 5 Km/hr In A Similar Bearing In Which The Train Is Going ?

    Answer :

    Distance=125 meter speed=50-5=45km/hr=>45*5/18=12.5 m/s

    Time=125/12.5=10sec

  13. Question 9. A Is Twice As Fast As B Is Thrice As Fast As C. The Journey Covered By C In 42 Minutes, What Will Be Covered By A Is

    Answer :

    B is thrice as fast as C

    C covered in 42 minutes

    B covered in 42/3=14 min

    A is twice as fast as B

    A covers in 14*(1/2) = 7 min

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  15. Question 10. A Can Complete A Work In 40 Days And B In 28 Days. In The Event That A And B Together Take Every Necessary Step, At That Point Roughly In How Long Will A Similar Function Be Finished ?

    Answer :

    A’s 1day’s work = 1/40 

    B’s 1day’s work = 1/28 

    They can cooperate in = 1/40 + 1/28 = 16 days (estimate)

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  17. Question 11. Teena Is More Youthful Than Rani By 6 Years. On The Off Chance That The Proportion Of Their Ages Is 6:8, Discover The Time Of Teena:

    Answer :

    On the off chance that Rani Age is x, at that point Teena age is x-6, 

    So (x-6)/x = 6/8 

    => 8x-48 = 6x 

    => 2x = 48 

    => x = 24 

    So, Teena age is 24-6 = 18 years

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  19. Question 12. A Man Purchases A Book For Rs.29.50 And Offers It For Rs 31.10. Discover His Gain Percent.

    Answer :

    So we have C.P. = 29.50 

    S.P. = 31.10 

    Gain = 31.10 – 29.50 = Rs. 1.6 

    Gain %=( Gain/Cost*100)% 

    = (1.6/29.50*100)%=5.4%

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  21. Question 13. Look At The Series: A4, __, C16, D32, E64. What Number Should Fill The Blank ?

    Answer :

    The letters Increase by 1; the numbers are duplicated by 2.

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  23. Question 14. A Quick Typist Can Type Some Issue In 2 Hours And A Moderate Typist Can Type The Same In 3 Hours. In The Event That Both Kinds Consolidate, In What Amount Of Time Will They Wrap Up ?

    Answer :

    The quick typist’s work done in 1 hr = 1/2

    The moderate typist’s work done in 1 hr = 1/3

    If they work to join, work is done in 1 hr = 1/2+1/3 = 5/6 So,

    the work will be finished in 6/5 hours. i.e., 1+1/5 hours = 1hr 12 min

  24. Question 15. What Is The Aggregate Of All Numbers Somewhere In The Range Of 100 And 1000 Which Are Distinct By 14 ?

    Answer :

    The number nearest to 100 which is more noteworthy than 100 and divisible by 14 is 112, which is the principal term of the arrangement which must be summed. The number nearest to 1000 which is under 1000 and distinct by 14 is 994, which is the last term of the arrangement. 112 + 126 + …. + 994 = 14(8+9+ … + 71) = 35392

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  26. Question 16. Gavaskar’s Average In His Initial 50 Innings Was 50. After The 51st Innings, His Average Was 51. What Number Of Runs Did He Score In His 51st Inning? (assuming That He Lost His Wicket In His 51st Innings)

    Answer :

    Add up to score after 50 innings = 50*50 = 2500

    Total score after 51 innings = 51*51 = 2601.

    So, runs made in the 51st innings = 2601-2500 = 101

    If he had not lost his wicket in his 51st innings, he would have scored an unbeaten 50 in his 51st innings.

  27. Question 17. I Drove 60 Km At 30 Kmph And After That An Extra 60 Km At 50 Km Ph. Register My Normal Speed Over My 120 Km.

    Answer :

    37 ½ km ph Solution: Time required for the initial 60 km = 120 min.

    The Time required for the second 60 km = 72 min.

    Add up to time required = 192 min

    Average speed = (60*120)/192 = 37 1/2

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  29. Question 18. A Boat Can Go At A Speed Of 13 Km/hr In Still Water. On The Off Chance That The Speed Of The Stream Is 4 Km/hr, Discover The Time Taken By The Vessel To Go 68 Km Downstream.

    Answer :

    Speed downstream = (13 + 4) km/hr = 17 km/hr.

    Time taken to movement 68 km downstream = (68/17) hrs = 4 hrs

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  31. Question 19. A, B, C Are The Partner In A Business. During A Specific Year. A Got 33% Of The Benefit. B Got One-fourth Of The Benefit And C Got The Rest Of The Rs. 5000. What Amount Of Measure Of Cash Did A Get?

    Answer :

    Lets expect Total benefit x 

    x * (1-1/3-1/4) = 5000 

    => x*(12-4-3)/12 = 5000 

    x = 5000*12/5 = Rs. 12000 

    so An’s offer = Rs. (1/3*12000) = Rs. 4000

  32. Question 20. A Man Possesses 2/3 Of The Statistical Surveying Bureau Business And Offers 3/4 Of His Offers For Rs. 75000. What Is The Value Of Business?

    Answer :

    3/4 of his offer = 75000 

    So his offer = 100000. 

    2/3 of business esteem = 100000 

    So add up to esteem = 150000

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  34. Question 21. Nirmal And Kapil Began A Business Contributing Rs. 9000 And Rs. 12000 Separately. Following A Half Year, Kapil Pulled Back Portion Of His Speculation. In The Event That Following A Year, The Aggregate Benefit Was Rs. 4600, What Was Kapil’s Share Initially ?

    Answer :

    Nirmal: Kapil = 9000*12:(12000*6+6000*6) = 1:1 

    Kapil share = Rs. [4600 *(1/2)) = Rs. 2300

  35. Question 22. Anirudh, Harish, And Sahil Put A Sum Of Rs.1, 35,000 In The Proportion 5:6:4 Anirudh Contributed Has The Capital For 8 Months. Harish Contributed For A Half Year And Sahil Contributed For 4 Months. On The Off Chance That They Acquire A Benefit Of Rs.75, 900, Then What Is The Offer Of Sahil In The Profit ?

    Answer :

    Anirudh contribute for 8 months, Harish contributed for 6 and 

    sahil for 4 months in the proportion of 5:6:4 

    so proportion = 5*8 : 6*6 : 4*4 

    => 40:36:16 

    => 10:9:4 

    So sahil’s profit= (4/23)*75900 = 13200

  36. Question 23. A Begins Riding His Bicycle At 10 Am With A Speed Of 20kmph And B Likewise Begins At 10 Am With A Speed Of 40kmph From A Similar Point In A Similar Way. Returns South At 12 O’clock And B Turns North At 11 Am. What Will Be The Distance Between A And B At 2 Pm ?

    Answer :

    At 12 O’clock, A cover 40km and on the opposite side B at 11 o clock cover 40km, again they went towards each other (which is really the separation between them), that is A needs to make a trip 2hr (From 12 to 2 at 20km/hr.) i.e. 2*20=40km and opposite side B needs to Travelled out 3hr (From 11 to 2 at 40km/hr.) i.e. 3*40=120Km.

    At that point Then the total distance traveled by them is the Actual distance between them i.e. 40+120= 160Km 

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  38. Question 24. 60 Liters Of Diesel Is Required To Movement 600 Km Utilizing An 800 Cc Motor. In The Event That The Volume Of Diesel Required To Cover A Separation Changes Specifically As The Limit Of The Motor, At That Point What Number Of Liters Of Diesel Is Required To Movement 800 Km Utilizing 1200 Cc Motor ?

    Answer :

    Let keep 800 cc steady and compute the measure of diesel for 800 km

    800*60/600=80 liters. 

    Presently, ascertain diesel required for new separation i.e. 800 km, 

    80*1200/800=120 liters.

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  40. Question 25. If A Person Walks At 14 Km/hr Instead Of 10 Km/hr, He Would Have Walked 20 Km More. The Actual Distance Traveled By Him Is:

    Answer :

    Let the real separation voyaged be x km. 

    At that point, x/10 = (x + 20)/14. 

    => 14x = 10x + 200 

    => 4x = 200. 

    => x = 50 km.