Fachhochschule wiener neustadt Interview Questions, Process, and Tips

Let’s say,

I have x coins of 50 paise and (80-x) coins of 100 paise,so the equation is like this ,
50x + (80-x)*100 = 64*100
x = 32
So ,I have 32 coins of 50 paise

Here is the solution to the given version of the puzzle (9 balls, one is heavier, need to identify oddball), where we label the balls A, B, …, I:

1. Weigh ABC versus DEF.

Scenario a: If these (1) balance, then we know the oddball is one of G, H, I.

2. Weigh G versus H.

Scenario a.i: If these (2) balance, the oddball is I.

Scenario a.ii: If these (2) do not balance, the heavier one is the oddball.

Scenario b: If these (1) do not balance, then the oddball is on the heavier side. For simplicity, assume the ABC side is heavier, so the oddball is one of A, B, C.

2. Weigh A versus B.

Scenario b.i: If these (2) balance, the oddball is C.

Scenario b.ii: If these (2) do not balance, the heavier one is the oddball.

answer is maximum of 2.

4x:5x:6x
100:50:25
100×4x +50×5x+25×6x
then 400x + 250x + 150x = 16000
800x=16000
x = 20

25 paisa=6x = 20×6=120

so the ans is 120 coins

mausi

20km

6561 → 6560/2 [-1]
3280
1640
820
410
205 [+1] =206
103 → [-1] = 102
51 [+1] = 52
26
13 [-1] = 12
6
3 [+1] 4
2
1

5 (3+2)
if he wins then it becomes 4 double than 2
and if looses then opponent becomes 3 which is equal.

In the example first we have to find the area of the field

we have given the following values ,

cost of fencing per meter = Rs 1.50

Diameter of circular field = 28

We have to find the area , a = ?

We know that,

Area = 2 π r

= 2 x 22/7 x 14

= 2 x 22 x 2

= 44 x 2

= 88 sq. m

So, area of circle is 88 sq.m

Now just multiply it with 1.50

cost of fencing = 88 x 1.50

= Rs. 132

So, the cost of fencing the circular field is Rs. 132

When none of the digits are repeated:

The hundred’s place can be filled by any of the digits: 2, 3, 5, 6, 7 or 8 except the one which has already been used at the thousand’s place, so it can be filled in 5 ways.

Similarly tens’ place can be filled in 4 ways: only those 4 numbers which have not been use either at hundred’s or thousand’s place.
Unit’s place can be filled in only 3 ways. So, total number of nos. Possible =4×5×4×3= 240

B. 32%