Mia Interview Questions, Process, and Tips

Let length of tunnel is x meter
Distance = 800+x meter
Time = 1 minute = 60 seconds

Speed = 78 km/hr = 78*5/18 m/s = 65/3 m/s

Distance = Speed*Time
=>800+x=653∗60=>800+x=20∗65=1300=>x=1300−800=500

5c2*4c1/9c3 +5c3/9c3=25/42

Let\: the\: total\: price \: be\: Rs. X\: then,

B= 2x/7 & A= \left ( x-2 \right )/7

So, A:B = 5x/7 : 2x/7 = 5 : 2

Let \: B’s \: capital\; be\: Rs.Y\: then\: \left ( 16000\ast 8 \right )/4y= 5/2

=> (16000\ast 8\ast 2)/(4\ast 5) = y

=> y = Rs. 12800

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

3.1

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.

Daughter-in-law

8days

find out what is the problem and taking the bad out always works.

₹1 today increases to (1+1/8) = 9/8 after 1year.

After 2 years it increases to 9/8 of 9/8 = (9/8)^2 =81/64 times.

Hence ₹64000 increase to 64000×81/64 =₹81000

(N * 1.1) * 0.9 = 7920
N=8000

The batsman on 98 is on strike. He hits the ball and they run 3. UNFORTUNATELY one of the batsmen doesn`t turn correctly for one of the runs and the umpire calls ONE SHORT and awards only two runs. Therefore the first batsman has his century. There is now 1 ball remaining and one run is required to win. The batsman on strike, however is now the one on 97 runs. He now either hits a 4 or a 6. They win the game and both batsmen scored centuries.

Read more: 3 runs required in 3 balls to win with only a wicket left. The batsmen is on 98 and the runner is on 97. How will both the batsmen score centuries … – 3 runs required in 3 balls to win with only a wicket left. The batsmen is on 98 and the runner is on 97. How will both the batsmen score centuries as well win the match ?

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