Nagarro Aptitude Interview Questions & Answers

  1. Question 1. A Sphere And A Cube Have The Same Surface Area. Find The Ratio Of Their Volumes?

    Answer :

    SA of a sphere: 4πr² 

    SA of a cube: 6x² 

    4πr² = 6x² 

    r²/x² = 6 / 4π 

    (r/x)² = 3 / 2π 

    r/x = (3 / 2π)^0.5 

    Volume of a sphere: 4/3 πr³ 

    Volume of a cube: x³ 

    Find the ratio meaning (4/3 πr³)/x³ 

    = (4π/3)(r³/x³) 

    = (4π/3)(r/x)³ 

    = (4π/3)(3 / 2π)^1.5 

    = (2²π/3)[3^1.5 / (2^1.5)(π^1.5)] 

    = √2√3 / √π 

    = √(6/π).

  2. Question 2. 4 Horses Are Tethered At 4 Corners Of A Square Plot Of Side 63 Meters So That They Just Cannot Reach One Another. The Area Left Ungrazed Is?

    Answer :

    area of square = 63*63=3969 m^2

    area inside the square that is grazed =4*area of quadrants of 4 circles 

    = 4*(1/4)*(22/7)*(63/2)*(63/2) =3118.5

    area left ungrazed = 3969-3118.5=850.5.

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  4. Question 3. How Many No Between 1000 To 9900 In Which Four Digits Of Number Are Different?

    Answer :

    Number of 4 digit numbers having all different digits = 9*9*8*7 = 4536 

    smallest number : 1234, largest number : 9876, all are between 1000 and 9900.

  5. Question 4. A Train 360 M Long Is Running At A Speed Of 45 Km/hr. In What Time Will It Pass A Bridge 140 M Long?

    Answer :

    Formula for converting from km/hr to m/s: X km/hr = X *( 5/18) m/s.

    so Speed = 45 * 5/18 m/sec = 25/2 m/sec.

    Total distance to be covered = (360 + 140) m = 500 m.

    Required time = 500 * 2/25 sec = 40 sec.

  6. Question 5. The G.c.d. Of 1.08, 0.36 And 0.9 Is?

    Answer :

    Given numbers are 1.08 , 0.36 and 0.90

    H.C.F of 108, 36 and 90 is 18 ( G.C.D is nothing but H.C.F)

    So H.C.F of given numbers = 0.18.

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  8. Question 6. The Slant Height Of A Right Circular Cone Is 10 M And Its Height Is 8 M. Find The Area Of Its Curved Surface?

    Answer :

    l = 10m

    h = 8m

    so r = √(l^2 – h^2) = √(10^2 – 8^2) = 6m 

    So Curved surface area = (π * r * l) = ( π * 6 * 10) m2 = 60π m2.

  9. Question 7. How Many Times In A Day, Are The Hands Of A Clock In Straight Line But Opposite In Direction?

    Answer :

    The hands of a clock point in opposite directions (in the same straight line) 11 times in every 12 hours. (Because between 5 and 7 they point in opposite directions at 6 o’clock only).

    So, in a day, the hands point in the opposite directions 22 times.

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  11. Question 8. In Covering A Distance Of 30 Km, Abhay Takes 2 Hours More Than Sameer. If Abhay Doubles His Speed, Then He Would Take 1 Hour Less Than Sameer. Abhay’s Speed Is?

    Answer :

    Let Abhay’s speed be x km/hr.

    So 30/x – 30/2x = 3

    => 6x = 30

    x = 5 km/hr.

  12. Question 9. The Sum Of The Present Ages Of A Father And His Son Is 60 Years. Six Years Ago, Father’s Age Was Five Times The Age Of The Son. After 6 Years, Son’s Age Will Be?

    Answer :

    Let the present ages of son and father be x and (60 -x) years respectively.

    Then, (60 – x) – 6 = 5(x – 6)

    => 54 – x = 5x – 30

    => 6x = 84

    => x = 14.

    Son’s age after 6 years = (x+ 6) = 20 years.

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  14. Question 10. Sum Of Three Even Consecutive Numbers Is 48, And Then Least Number Is?

    Answer :

    Let the numbers be 2n, 2n+2 and 2n+4

    2n + (2n+2) + (2n+4) = 48

    6n = 48-6 = 42, n = 7

    Hence the numbers are — > 14, 16 and 18

    The least number is 14.

  15. Question 11. Walking At 3/4 Of His Usual Speed ,a Man Is Late By 1/2 Hr. The Usual Time Is?

    Answer :

    Usual speed = S

    Usual time = T

    Distance = D

    New Speed is ¾ S

    New time is 4/3 T

    4/3 T – T = 5/2

    T=15/2 = 7 ½.

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  17. Question 12. If 7 Spiders Make 7 Webs In 7 Days, Then 1 Spider Will Make 1 Web In How Many Days?

    Answer :

    Let the required number days be x.

    Less spiders, More days (Indirect Proportion)

    Less webs, Less days (Direct Proportion)

    Spiders 1 : 7 } 

    > :: 7 : x

    Webs 7 : 1 }

    1 x 7 x x = 7 x 1 x 7

    => x = 7.

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  19. Question 13. A Student Multiplied A Number By 3/5 Instead Of 5/3. What Is The Percentage Error In The Calculation?

    Answer :

    Let the number be x.

    Then, error = x*5/3 – x*3/5 = x * 16/15

    So Error% = ( x*16/15 * 3/5x * 100)% = 64%.

  20. Question 14. If A’s Height Is 40% Less Than That Of B, How Much Percent B’s Height Is More Than That Of A?

    Answer :

    Excess of B’s height over A’s = [(40/(100 – 40)] x 100%

    = 66.66%

  21. Question 15. From A Group Of 7 Men And 6 Women, Five Persons Are To Be Selected To Form A Committee So That At Least 3 Men Are There On The Committee. In How Many Ways Can It Be Done?

    Answer :

    We may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only).

    Required number of ways = (7C3 * 6C2)+(7C4 * 6C1)+(7C5)

    =>[(7*6*5)/(3*2*1) * (6*5)/(2*1)]+(7C3 * 6C1) + (7C2)

    => 525 + [(7*6*5)/(3*2*1)* 6]+[(7*6)/(2*1)] 

    = (525 + 210 + 21)

    = 756.

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  23. Question 16. In How Many Different Ways Can The Letters Of The Word ‘mathematics’ Be Arranged So That The Vowels Always Come Together?

    Answer :

    In the word ‘MATHEMATICS’, we treat the vowels AEAI as one letter.

    Thus, we have MTHMTCS (AEAI).

    Now, we have to arrange 8 letters, out of which M occurs twice, T occurs twice and the rest are different.

    So,Number of ways of arranging these letters = (8!/2!*2!)=10080.

    Now, AEAI has 4 letters in which A occurs 2 times and the rest are different.

    Number of ways of arranging these letters = (4!/2)=12.

    => Required number of words = (10080 * 12) = 120960.

  24. Question 17. What Will Be The Day Of The Week 15th August, 2010?

    Answer :

    15th August, 2010 = (2009 years + Period 1.1.2010 to 15.8.2010)

    Odd days in 1600 years = 0

    Odd days in 400 years = 0

    9 years = (2 leap years + 7 ordinary years) = (2 * 2 + 7 * 1) = 11 odd days 4 odd days.

    Jan. Feb. March April May June July Aug. 

    (31 + 28 + 31 + 30 + 31 + 30 + 31 + 15) = 227 days

    227 days = (32 weeks + 3 days) 3 odd days.

    Total number of odd days = (0 + 0 + 4 + 3) = 7 0 odd days.

    Given day is Sunday.

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  26. Question 18. A Thief Is Noticed By A Policeman From A Distance Of 200 M. The Thief Starts Running And The Policeman Chases Him. The Thief And The Policeman Run At The Rate Of 10 Km And 11 Km Per Hour Respectively. What Is The Distance Between Them After 6 Minutes ?

    Answer :

    find the relative speed of the thief and policeman = (11-10)km/hr = 1 km/hr

    Distance covered in 6 minutes = (1/60)*6=1/10km => 100meters

    So distance between them after 6 minutes = (200-100)=100 meters.

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  28. Question 19. An Industrial Loom Weaves 0.128 Metres Of Cloth Every Second. Approximately, How Many Seconds Will It Take For The Loom To Weave 25 Metre Of Cloth ?

    Answer :

    Lets assume the time required to weave 25 meters = x sec.

    Rule: More cloth means More time (Direct Proportion)

    => 0.128:1::25:x

    =>x=[(25*1)/0.128]

    =>x=195.31

    So time needed ~195 seconds.

  29. Question 20. There Are Three Numbers, These Are Co-prime To Each Other Are Such That The Product Of The First Two Is 551 And That Of The Last Two Is 1073. What Will Be The Sum Of Three Numbers?

    Answer :

    Given that numbers are co primes,

    and two products have the middle number in common.

    => Middle number = H.C.F. of 551 and 1073 = 29

    so first number is = 551/29 = 19

    => Third number = 1073/29 = 37

    Therefore, sum of these numbers is = (19 + 29 + 37) = 85.

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  31. Question 21. When A Student Weighing 45 Kgs Left A Class, The Average Weight Of The Remaining 59 Students Increased By 200g. What Is The Average Weight Of The Remaining 59 Students?

    Answer :

    Lets assume average weight of the 59 students = x kgs.

    => Total weight of the 59 = 59*x

    Given when the weight of this student who left is added, the total weight of the class = (59x + 45)

    The average weight decreases by 0.2 kgs, when these students included.

    => (59x+45)/60=(x-0.2)

    => (59x+45)= 60(x-0.2)

    => 59x + 45 = (60A – 12)

    => 45 + 12 = 60x – 59x

    => x = 57.

  32. Question 22. Nirmal And Kapil Started A Business Investing Rs. 9000 And Rs. 12000 Respectively. After 6 Months, Kapil Withdrew Half Of His Investment. If After A Year, The Total Profit Was Rs. 4600, What Was Kapil’s Share Initially ?

    Answer :

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

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

  33. Question 23. A Is Able To Do A Piece Of Work In 15 Days And B Can Do The Same Work In 20 Days. If They Can Work Together For 4 Days, What Is The Fraction Of Work Left?

    Answer :

    Total work done by A + B in 1 day = 1/15 + 1/(20 ) = 7/60

    Work done in 4 days = 7/60 × 4 = 7/15

    Therefore, fraction of work left = 1 – 7/15 = 8/15.

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  35. Question 24. Population Of A Village Increased By 5% From 2007 To 2008 And By 25% From 2005 To 2009. If The Population Of The Village Was 480 In 2007, What Was Its Population In 2009?

    Answer :

    Population in 2007 = 480

    In 2008 = 1.05 × 480 = 504

    In 2009 = 1.25 × 504 = 630.

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  37. Question 25. 24% Of 150 × 3/4= ?

    Answer :

    24% of 150 × 3/4= x = 36 × 4/3 = 48.

    3/17 of 20% of 510 + 7 = x2

    3/17 of 20% of 510 + 7 = x2

    3/17 × 20/100 × 510 + 7 = 18 + 7 = 25 => x = 5.

  38. Question 26. In The Month Of March, Hiten Spent 45% Of His Monthly Salary On Paying Bill And Rent. Out Of The Remaining Salary, He Invested 60% In Ppf And The Remaining He Deposited In Bank. He Deposited Rs. 15,400 In Bank. If In April, He Got An Increment Of 10%, What Was His Salary In April?

    Answer :

    Let the initial salary of Hiten be ‘S’. Then:

    (0.55 × 0.4) S = 15400 => S = 70000

    After increment, Hiten’s salary = 1.1 × 70000 = 77000.

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  40. Question 27. A Person Covers A Certain Distance By Travelling At A Uniform Speed Of 120 Km/h For 90 Minutes. At What Speed Will He Have To Travel In Order To Cover The Same Distance In 1 Hour 20 Minutes? (in Km/h)

    Answer :

    Distance = 120 × 90/60 = 180 km

    Speed required to cover 180 km in 1 hr 20 mins

    ( 4/3 hrs) = 180 × ¾ = 135 km/hr.

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  42. Question 28. In Jar A, 120 Litres Milk Was Mixed With 24 Litre Water. 12 Litre Of This Mixture Was Taken Out And 3 Litre Water Was Added. If 27 Litre Of Newly Formed Mixture Is Taken Out, What Will Be The Resultant Quantity Of Water In The Jar? (in Litre)

    Answer :

    Ratio of milk : water in Jar A = 120 : 24 = 5 : 1

    12 lts of this mixture is taken out => milk = 5/6 × 12 = 10 lts and water = 2 lts taken out

    3 lts of water added = 24 – 2 + 3 = 25 lts => new ratio of milk : water

    = 110 : 25 = 22 : 5

    Now 27 lts of this mixture is again taken out =>

    water taken out = 5/27 × 27 = 5 lts

    water left = 25 – 5 = 20 lts.

  43. Question 29. A Boat, Whose Speed In 15 Km/hr In Still Water Goes 30 Km Downstream And Comes Back In A Total Of 4 Hours 30 Minutes. What Is The Speed Of The Stream? (in Km/hr)?

    Answer :

    Speed of boat = 15 k/h

    Let the speed of stream be ‘S’

    Given : 30/(15+S) + 30/(15-S) = 9/2

    Solving we get, S = 5 km/h.

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  45. Question 30. Six Years Ago, The Ratio Of The Ages Of Kunal And Sagar Was 6 : 5. Four Years Hence, The Ratio Of Their Ages Will Be 11:10. What Is Sagar’s Present Age?

    Answer :

    Let the ages of Kunal and Sagar be K and S respectively.

    Given : (K-6)/(S-6) = 6/5 => 5K – 6S = -6 ……(i)

    And: (K+4)/(S+4) = 11/10 => 10K – 11 S = 4 …..(ii)

    Solving (i) & (ii) we get: S = 16 years.