- Virtualmind - software house General Aptitude Interview Questions
- Virtualmind - software house Trainee Interview Questions
- Virtualmind - software house Personal Questions round Interview Questions
- Virtualmind - software house HR Interview Questions
- Virtualmind - software house Lead Interview Questions
24*4= x(x+4)
solve this
answer is x=8
X+4= 12
20
xy+2y-x=6
xy+2y-x-2=6-2
y(x+2)-1(x+2)=4
(y-1)(x+2)=4
so c is 4
94
copy cat
The 3-digit number can be written as the sequence [n, 2n, 3n]
n = 1 → [1, 2, 3] → valid
n = 2 → [2, 4, 6] → valid
n = 3 → [3, 6, 9] → valid
n = 4 → [4, 8, 12] is not valid because this would lead to a 4-digit number
any value of n > 4 would also produce invalid answer
Answer: Three numbers: {123, 246, 369}
5x/4
45
9 years
Bcz 3×3=9
5×3=15
7×3=21
Total = 45
Avg of this 45/3 = 15
343 : 729 :: 125 : ?
7^3 : 9^3 :: 5^3:7^3
32 Km/hr
12
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
Answer: 8:10, 7:10
Explanation:
The bus b1, which started at P, reached S at 10:40, passing through the intermediary cities Q and R.
The time taken to travel from P to S
= 3 * 40 + 2 * 15 = 150minute
(journey)+(stoppage) = 2 hrs 30 minutes.
Hence, b1 started at 10:40 – 2:30 = 8:10 at P.
b2 reached Q, starting at U, through the city T, S and R.
The time taken by it to reach S = 4 * 40 + 3 * 15 = 205 minutes = 3 hr 25 minutes.
Hence, b2 started at, 10:35 – 3:25 = 7:10, at U.
73.
one year back father = 72 and son = 36.