Socan Interview Questions, Process, and Tips

189

E

Now as per question, Kamal invested for 12 months and Sameer invested for 7 months.
So, Kamal:Sameer=(9000×12​):(8000×7​)
=108.56=27:14
Sameer ratio in profit will be =6970×4114​=Rs.2380

Answer: 66.67 km approx.

Solution:

Let the first train A move at u km/h.
Let the second train B move at v km/h.
Let the distance between two trains be d km
Let the speed of bee be b km/h
Therefore, the time taken by trains to collide = d/(u+v)
Now putting all the known values into the above equation, we get,

u = 50 km/hr
v = 70 km/hr
d = 100 km
b = 80 km/hr

Therfore, the total distance travelled by bee
= b*d/(u+v)
= 80 * 100/(50+70)
= 66.67 km (approx)

3630

1+0^3=1
1+1^3=2
2+2^3=10
10+3^3=37
37+4^3=101
101+5^3=226
1,2,10,37,101,226……..

A

Answer is 274 .

Let the required term be x.

Pattern :- 1) Find the difference between consecutive terms as :

2 – 1 = 1

4 – 2 = 2

13 – 4 = 9

31 – 13 = 18

112 – 31 = 81

2) Now write the above numbers obtained by differences in series as :

1, 2, 9, 18, 81, y

(Suppose y be next term after 81.)

3) Look the difference series carefully and it follows the below pattern as :

(9 * 2)/1 = 18

(18 * 9)/2 = 81

(81 * 18)/9 = 162 = y (the next term after 81)

Thus the required answer is :

x = 112 + y

x = 112 +162 =274

x+y+z
—– = 6800
3
x>=6400
6400 + y + z = 20400
y + z = 14000
to get the greatest of y and z, lets assume y = 6400
so, z = 7600
so ANS is 7600

a)0
b)206
c)0
d)250
e)39
f)92

7 months

Let A, B and C be the three 6-faced dice.
Then, according to the question,
Since two dices has to be equal, that value can be any of the 6 faces, i.e., 6C1​ cases.
Now for each case, 2 equal dices can be selected from 3 dices in 3C2​ i.e., 3 ways.
And for each of the above, the third dice can have any of the 5 remaining faces
The possible outcomes are P(A)=61​,P(B)=61​,P(C)=65​,P(A)=61​,P(B)=65​,P(C)=61​ and P(A)=65​,P(B)=61​,P(C)=61​
Hence the required probability = 61​×61​×65​×6×3=21690​=125​

2^a = 2 x 2+a
a^3= 2 x a + 3
=> 4+a= 2a + 3
=> a = 1

ans is C.
let H be present age of husaband.W be present age of wife.
H+W=91.
now let diff between their ages be x.that is H-W=x.
now when husband is W yaers old, his wife must be W-x years
old.and it is given that H=2(W-x). so 2W-H=2x. and H-
W=x.eliminating x we get 4W=3H. but H+W=91, so solving thse
two H=52 W=39.

TWICE