Calm Interview Questions, Process, and Tips

let the total work to be done =1
so a+b+c=1 (say)
then a and b are supposed to do 7/11 th of work
so c= 1- (7/11)
c=4/11
therefore c gets (4/11)*550
=200 is the answer

let (xy) be original number.
then, 10y + x +45 = 10x + y; eq 1
and x + y = 13; eq 2
solve ab equation,
then xy = 94;

126
cos if one of the factor is given for hcf and lcm and multiply hcf and lcm and them divide it by the given factor
so 18*3780/540=18*7=126

these are question tht are seem to be like it consumes
time. so dont see jus the question and run to the next .

ans. a) 12

if 9 balls are added then the ratio to the combination
becomes 2:4:3.

9 balls make the ratio 3 for grey balls in tht mixture.
so the factor is 9/3 = 3 (in tht mixture)

so the same factor has to be maintained through out the
ratio. so black balls is 4 * 3 = 12 and white balls is 2 *
3 = 6.

let t = total no of students.. then
students who passed one or both subjects,
n(e U h) = n(e) + n(h) – n(e intersection h)
=> t = 0.8t + 0.7t – 144
=> t = 1.5t – 144
students who failed both subjects is 10% i.e. 0.1t
=>t-n(e U h) = 0.1t,
=>t -(1.5t – 144) = 0.1t
=>t- 1.5t- 0.1t = -144
=> -0.6t = -144
=>t = 240

16

1+1=2
2+2=4
3+4-7
4+7=11
5+11=16

3 hours ago.

Thin candle melts 3/4 in 3 hours leaving 1/4
Where as in the same time thick candle melts 3/6 leaving 3/6 which is 1/2. Now thick candle is exactly twice than the thin candle.

Or via modeling:

We need to find time at which the length of the thin candle is half the thick candle. Let x be the time. Thin candle melts at 1/4 an hour and thick candle melts at 1/6 an hour. In x hours they melt at x/4 and x/6 respectively. What’s left will be 1 – x/4 and 1 – x/6. We need to find x at which :

2 * (1 – (x/4)) = 1 – (x/6)

This equation results in x = 3