BMT Defence & Security UK Interview Questions, Process, and Tips

i Guess 8 mtrs.

d

2 Leap years
+2 to tuedsay
Thursday is the correct answer

68 is the least number when divided by 8, 12, 18 and 24 leaves the remainders 4, 8, 14 and 20

Take 5 pills from jar 1, 4 pills from jar2,…. and 1 pill
from jar 5 and put altogether in the scale. the ideal
weight should be (1+2+3+4+5)x10= 150gms. for eg if the jar2
is contaminated then the weight will be 146 gms. so
depending on the amount of weight loss we can identify the
contaminated jar.

C.1175

A. Rs. 410

57 and 3
12 x 57 = 684
-8 x 3 = -24
684 – 24 = 660
660/ 60 = 11

I think experience because if experience brings pay. If someone having experience in any work then he/she can work easily. Although pay is important but not more then experience

100/8=12-4
12/8=1-4
ans:144

In 3 hours thinker candle becomes half
In the same time thinner candle becomes one fourth. Now the
required condition is satisfied. So 3 Hours

Chambal

20- buy 15 books
15- buy 20 books

Original price is 20.

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

Answer: 3121 gold coins
Let total no of coins be M
Let the disbursement D to each son:

D1 = 1 + (M – 1)/5 = (M + 4)/5
D2 = 1 + ( M – D1 -1)/5 = (D1) * 4/5
D3= (D2) * 4/5
D4= (D3) * 4/5
D5= (D4) * 4/5
Total disbursements to sons=
= ∑D= (M+4)*1/5[ 1+4/5+(4/5)(4/5)+ (4/5)(4/5)(4/5)+(4/5)(4/5)(4/5)(4/5) ]
= (2101/3125)*(M+4)
Thus balance left for daughters =M-{(2101/3125)*(M+4)}
=(1024M-8404)/3125
This balance should be a positive integer ( assuming M and all disbursements are full coins )
Thus 1024M-8404 should be a multiple of 3125….so….
1024M – 8404 = N*3125 where N is an integer
Using Python code:
n=int(input(“Enter num n: “))
X=int()
a=int()
a=0
X=’ ‘
for a in range(0,n+1):
a=a+1
X= (3125*a + 8404)/1024
if (3125*a + 8404)% 1024== 0:
print(X,a)
Enter num n: 10000
3121.0 1020
6246.0 2044
9371.0 3068
12496.0 4092
15621.0 5116
18746.0 6140
21871.0 7164
24996.0 8188
28121.0 9212
We get minimum value of N = 1021 and M = 3121 gold coins