## SAP Aptitude Interview Questions & Answers

1. Question 1. The Average Score Of A Cricketer For 13 Matches Is 42 Runs. If His Average Score For The First 5 Matches Is 54, Then What Was His Average Score (in Runs) For Last 8 Matches?

Total Score = Average * Number of matches

Total score of 13 matches = 13 × 42 = 546

Total score of first 5 matches = 5 × 54 = 270

Therefore, total score of last 8 matches = 546 – 270 = 276

Average = 276/8 = 34.5

2. Question 2. If The Graphs Of The Equations X + Y = 0 And 5y + 7x = 24 Intersect At (m, N), Then The Value Of M +n Is?

By solving the given two equations, we get the intersection point (12, – 12).

So, m = 12, n = -12

Hence, m + n = 12 – 12 = 0.

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4. Question 3. A Function F(x) Is Defined As F(x) = F(x – 2) – X(x + 2) For All The Integer Values Of X And F(1) + F(4) = 0. What Is The Value Of F(1) + F(2) + F(3) + F(4) + F(5) + F(6)?

Let S = f(1) + f(2) + f(3) + f(4) + f(5) + f(6)

As f(1) + f(4) = 0, therefore S = f(2) + f(3) + f(5) + f(6) —— (1)

f(2) = f(0) – 8

f(3) = f(1) – 15

f(4) = f(2) – 24 = f(0) – 32

f(5) = f(3) – 35 = f(1) – 50

f(6) = f(4) – 48 = f(0) – 80

Put the above values in equation (1), we get

S = f(0) – 8 + f(1) – 15 + f(1) – 50 + f(0) – 80

S = 2(f(0) + f(1)) – 153 —— (2)

As we already know f(1) + f(4) = 0 ⇒f(1) + f(0) – 32 = 0 ⇒f(1) + f(0) = 32

Putting this value in equation 2, we get S = 2(32) – 153 = -89

5. Question 4. What Annual Payment Will Discharge A Debt Of Rs. 6,450 Due In 4 Years At 5% Per Annum Simple Interest?

Let the annual installment be rs. x

therefore (x + x*3*5/100)+(x + x*2*5/100)

+(x + x*1*5/100)+x =6450

=>115x/100+110x/100+105x/100+x=6450

=>115x+110x+105x+100x=6450*100

=>430x=6450*100

x=6450*100/430=rs.1500

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7. Question 5. The Average Of The First 100 Positive Integers Is?

n(n+1)/2

1+2+3+….+n

therefore average of these numbers=n+1/2

therefore required average

100+1/2=50.5

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9. Question 6. In A Family, The Average Age Of A Father And A Mother Is 35 Years. The Average Age Of The Father, Mother And Their Only Son Is 27 Years. What Is The Age Of The Son?

Father+Mother=2*35=70 years

Father+Mother+Son=27*3=81 years

therefore Son’s age=81-70=11 years

10. Question 7. The Length And Breadth Of A Rectangle Are Increased By 20% And 40% Respectively. What Is The Percentage Increase In Its Area?

Apply the percentage formula.

The percentage increase in the area will be P + Q + PQ / 100.

So we get the answer as 20 +40 + 20 × 40/100 = 68%.

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SAP FICO Interview Questions

12. Question 8. The Diameter Of A Circle Is 21 Metres. It Will Take How Many Revolutions To Cover A Distance Of 6.6 Km?

No. of revolutions = Distance/circumference.

Distance = 6.6 × 1000 = 6600 metres.

Circumference = 2 × 22/7 × 21/2 = 66 metres.

No. of revolutions = 6600/66 = 100 revolutions.

13. Question 9. If Shalu Buys 6 More Apples, His Carton Will Weigh 14.5 Kilograms.if The Weight Of One Apple Is 250 Grams.how Many Apples Did He Initially Had In His Carton?

Weight of 1 apple = 250 g

Weight of 6 apples = 250 * 6= 1.5 kg

Initial weight = 14.5 – 1.5= 13 kg

Number of apples in his carton initially :

= 13000/250

=52 apples

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15. Question 10. A Tailor Has 37.5 Metres Of Cloth And He Has To Make 8 Piecesout Of A Metre Of Cloth. How Many Pieces Can He Make Out Of This Cloth?

The tailor has to make 8 piece from 1 metre.

So for 1 piece ,the measurement will be 1/8=0.125 m

from 37.5 m tailor can get 37.5/0.125 = 300 pieces

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17. Question 11. How Many Digits Will Be There To The Right Of The Decimal Point In The Product Of 95.75 And .02554 ?

Sum of decimal places = 7.

Since the last digit to the extreme right will be zero (since 5 x 4 = 20),

so there will be 6 significant digits to the right of the decimal point.

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19. Question 12. If A Person Walks At 14 Km/hr Instead Of 10 Km/hr, He Would Have Walked 20 Km More. The Actual Distance Travelled By Him Is?

Let the actual distance travelled be x km.

Then, x/10=x + 20/14

14x = 10x + 200

4x = 200

x= 50 km.

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21. Question 13. In A Flight Of 600 Km, An Aircraft Was Slowed Down Due To Bad Weather. Its Average Speed For The Trip Was Reduced By 200 Km/hr And The Time Of Flight Increased By 30 Minutes. The Duration Of The Flight Is?

Let the duration of the flight be x hours.

Then,600/x-600/x + (1/2)= 200

600/x-1200/2x + 1= 200

x(2x + 1) = 3

2×2 + x – 3 = 0

(2x + 3)(x – 1) = 0

x= 1 hr.

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23. Question 14. 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?

Let Abhay’s speed be x km/hr.

Then,30/x-30/2x= 3

6x = 30

x = 5 km/hr.

24. Question 15. Robert Is Travelling On His Cycle And Has Calculated To Reach Point A At 2 P.m. If He Travels At 10 Kmph, He Will Reach There At 12 Noon If He Travels At 15 Kmph. At What Speed Must He Travel To Reach A At 1 P.m.?

Let the distance travelled by x km.

Then,x/10-x/15= 2

3x – 2x = 60

x = 60 km.

Time taken to travel 60 km at 10 km/hr =[60/10]hrs= 6 hrs.

So, Robert started 6 hours before 2 P.M. i.e., at 8 A.M.

Required speed =[60/5]kmph.= 12 kmph.

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26. Question 16. It Takes Eight Hours For A 600 Km Journey, If 120 Km Is Done By Train And The Rest By Car. It Takes 20 Minutes More, If 200 Km Is Done By Train And The Rest By Car. The Ratio Of The Speed Of The Train To That Of The Cars Is?

Let the speed of the train be x km/hr and that of the car be y km/hr.

Then,120/x+480/y= 8  -> 1/x+4/y =1/15…(1)

And,200/x+400/y=25/3-> 1/x+2/y =1/24….(2)

solving(1)and(2)we get: x = 60 and y = 80.

Ratio of speeds=60:80=3:4.

27. Question 17. A Farmer Travelled A Distance Of 61 Km In 9 Hours. He Travelled Partly On Foot @ 4 Km/hr And Partly On Bicycle @ 9 Km/hr. The Distance Travelled On Foot Is?

Let the distance travelled on foot be x km.

Then, distance travelled on bicycle=(61 -x)km.

So,x/4+(61 -x)/9= 9

9x + 4(61 -x) = 9 x 36

5x = 80

x = 16 km.

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29. Question 18. A Man Covered A Certain Distance At Some Speed. Had He Moved 3 Kmph Faster, He Would Have Taken 40 Minutes Less. If He Had Moved 2 Kmph Slower, He Would Have Taken 40 Minutes More. The Distance (in Km) Is?

Let distance = x km and usual rate = y kmph.

Then,x/y-x/y+3=40/60 -> 2y(y + 3) = 9x ….(i)

And,x/y -2-x/y=40/60 -> y(y – 2) = 3x ….(ii)

On dividing (i) by (ii), we get: x = 40.

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31. Question 19. A Bike Runs At The Speed Of 50 Km/h When Not Serviced And Runs At 60 Km/h When Serviced. After Servicing The Bike Covers A Certain Distance In 6 H. How Much Time Will The Car Take To Cover The Same Distance When Not Serviced?

From the given data, after servicing speed of the bike = 60 km/h

Distance covered in 6h. = (60 * 6) km = 360km

When it’s not service,the time taken to cover 360 km = (360 / 50) = 7.2h

32. Question 20. Present Ages Of Sameer And Anand Are In The Ratio Of 5 : 4 Respectively. Three Years Hence, The Ratio Of Their Ages Will Become 11 : 9 Respectively. What Is Anand’s Present Age In Years?

Let the present ages of Sameer and Anand be 5x years and 4x years respectively.

Then, 5x + 3/4x + 3=11/9

9(5x + 3) = 11(4x + 3)

45x + 27 = 44x + 33

45x – 44x = 33 – 27

x = 6.

Anand’s present age = 4x = 24 years.

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34. Question 21. A Man Is 24 Years Older Than His Son. In Two Years, His Age Will Be Twice The Age Of His Son. The Present Age Of His Son Is?

Let the son’s present age be x years. Then, man’s present age = (x + 24) years.

(x + 24) + 2 = 2(x + 2)

x + 26 = 2x + 4

x = 22.

35. Question 22. 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 Age At Present?

Let the ages of Kunal and Sagar 6 years ago be 6x and 5x years respectively.

Then, (6x + 6) + 4/(5x + 6) + 4  =11/10

10(6x + 10) = 11(5x + 10)

5x = 10

x = 2.

Sagar’s present age = (5x + 6) = 16 years.

36. Question 23. 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?

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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38. Question 24. Two, Trains, One From Howrah To Patna And The Other From Patna To Howrah, Start Simultaneously. After They Meet, The Trains Reach Their Destinations After 9 Hours And 16 Hours Respectively. The Ratio Of Their Speeds Is?

Let us name the trains as A and B. Then,

(A’s speed) : (B’s speed) = b : a = 16 : 9 = 4 : 3.

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40. Question 25. Tickets Numbered 1 To 20 Are Mixed Up And Then A Ticket Is Drawn At Random. What Is The Probability That The Ticket Drawn Has A Number Which Is A Multiple Of 3 Or 5?

Here, S = {1, 2, 3, 4, …., 19, 20}.

Let E = event of getting a multiple of 3 or 5 = {3, 6 , 9, 12, 15, 18, 5, 10, 20}.

P(E) =   n(E)/ n(S)=9/20.

41. Question 26. What Is The Probability Of Getting A Sum 9 From Two Throws Of A Dice?

In two throws of a dice, n(S) = (6 x 6) = 36.

Let E = event of getting a sum ={(3, 6), (4, 5), (5, 4), (6, 3)}.

P(E) =   n(E)/ n(S)=4/36 =1/9.

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43. Question 27. A Bag Contains 6 Black And 8 White Balls. One Ball Is Drawn At Random. What Is The Probability That The Ball Drawn Is White?

Let number of balls = (6 + 8) = 14.

Number of white balls = 8.

P (drawing a white ball) =8/14=4/7.

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45. Question 28. What Is The Average Candidates Who Appeared From State Q During The Given Years?

Required average=8100 + 9500 + 8700 + 9700 + 8950/5

=44950/5

= 8990.

46. Question 29. What Are The Average Marks Obtained By All The Seven Students In Physics?

Average marks obtained in Physics by all the seven students

=1/7x [ (90% of 120) + (80% of 120) + (70% of 120) + (80% of 120) + (85% of 120) + (65% of 120) + (50% of 120) ]

=1/7x [ (90 + 80 + 70 + 80 + 85 + 65 + 50)% of 120 ]

=1/7x [ 520% of 120 ]

= 89.14.

47. Question 30. What Was The Aggregate Of Marks Obtained By Sajal In All The Six Subjects?

Aggregate marks obtained by Sajal

= [ (90% of 150) + (60% of 130) + (70% of 120) + (70% of 100) + (90% of 60) + (70% of 40) ]

= [ 135 + 78 + 84 + 70 + 54 + 28 ]

= 449.

48. Question 31. There Are Two Examinations Rooms A And B. If 10 Students Are Sent From A To B, Then The Number Of Students In Each Room Is The Same. If 20 Candidates Are Sent From B To A, Then The Number Of Students In A Is Double The Number Of Students In B. The Number Of Students In Room A Is?

Let the number of students in rooms A and B be x and y respectively.

Then, x – 10 = y + 10      x – y = 20 …. (i)

and x + 20 = 2(y – 20)      x – 2y = -60 …. (ii)

Solving (i) and (ii) we get: x = 100 , y = 80.

The required answer A = 100.

49. Question 32. A Man Has Some Hens And Cows. If The Number Of Heads Be 48 And The Number Of Feet Equals 140, Then The Number Of Hens Will Be?