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20
451 times.
Explanation: There are 60 minutes in an hour.
In ¾ of an hour there are (60 * ¾) minutes = 45 minutes.
In ¾ of an hour there are (60 * 45) seconds = 2700 seconds.
Light flashed for every 6 seconds.
In 2700 seconds 2700/6 = 450 times.
The count start after the first flash, the light will
flashes 451 times in ¾ of an hour.
Mixture is 1.2 L of A + 2.4L of B which is 3.6L total. 1.2L / 3.6L = 1/3 (0.33)
8 days
13. One group of each = 2+3+5 = 10 and 1 remainder for each = 1+1+1 = 3. 10+3 = 13
The answer is A)
y1 = 62 Rs/kg
y2 = 72 Rs/kg
y = 64.5 Rs/kg
y2 – y1 = 10 Rs/kg
The distance between the y and y1 is
y – y1 = 64.5 – 62 = 2.5
x1 = (y – y1)/(y2 – y1) = 2.5/10 = 0.25
x2 = 1 – x1 = 1 – 0.25 = 0.75
The target price is calculated by the lever method.
x1 * y2 + x2 * y1 = 0.25 * 72 + 0.75 * 62.5 = 64.5
The ratio is of y1 to y2 is
0.75 : 0.25
Divide by both by 0.25
3 : 1
5x/4
if am standing on a ballconi so i see the people what they
are doing and how they are handling with the people and
what they are taking to each and everybody.
the ratio of present age of x:y is 5:6
And,we have to find the present age of x.
so,the present age of X is definitely will be multiple of 5.
X=35
so,35 is the answer.
Call from HR for interview in an unexpected time!
mausi
U is the answer as
T is sitting opposite to R and W is sitting opposite to U
A rude person who doesn’t respect me
Let the revolutins made by bigger wheel is: x
then revolution by smaller wheel will be: x+10
Now, the distance covered by both wheels will be same.
So,
7(x+10)=9x
7x+70=9x
70=2x
35=x
so distance travelled=9*35=315
or 7*45=315
answer is 12
bcoz
there are 36 people and atmost he can shake for 3 people so
36/3=
12
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