# Top 31 Sap Aptitude Interview Questions You Must Prepare 29.Nov.2023

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

Number of white balls = 8.

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

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

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.

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

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

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.

n(n+1)/2

1+2+3+....+n

therefore average of these numbers=n+1/2

therefore required average

100+1/2=50.5

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.

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.

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

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..

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

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

2x2 + x - 3 = 0

(2x + 3)(x - 1) = 0

x= 1 hr.

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.

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.

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.

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.

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.

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.

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.

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

So, m = 12, n = -12

Hence, m + n = 12 – 12 = 0.

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

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

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.

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

=44950/5

= 8990.

Let the actual distance travelled be x km.

Then, x/10=x + 20/14

14x = 10x + 200

4x = 200

x= 50 km.

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.

Let the number of hens be x and the number of cows be y.

Then, x + y = 48 .... (i)

and 2x + 4y = 140      x + 2y = 70 .... (ii)

Solving (i) and (ii) we get: x = 26, y = 22.

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.

Let Abhay's speed be x km/hr.

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

6x = 30

x = 5 km/hr.

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%.

Father+Mother=2*35=70 years

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

therefore Son's age=81-70=11 years