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Let Suvarna, Tara, Uma and Vibha be S,T,U,V respectively
initially in the beginning each persons share be
V = x U = y T = z
S = w = (x+y+z+32) Reason: She has to double others share, so she should have each and everyone’s share and still should be left out with 32
after 1st Round of game
S loses and is out with 32 and doubles the others share
V = 2x U = 2y T = 2z
After 2nd Round of game
T loses and is out with 32 and doubles the others share
V = 4x U = 4y
This means T had 2z = 2x + 2y + 32
After 3rd round of game
U looses and is out with 32 and doubles others share
V = 8x
This means U initially has 4y = 4x + 32
In the end V = 8x = 32
Solving this we get x = 4, y = 12, z = 32 and w = 80
There fore Suvarna had highest share in the beginning
1600 years contain 0 odd day.
300 years contain 1 odd day.
94 years = (23 leap years + 71 ordinary years)
= (46 + 71) odd days
= 117 odd days, i.e., 5 odd days
Days from 1st January 1995 to 28th February 1995
= (31 + 28) days = 59 days
= (8 weeks + 3 days) = 3 odd days
∴ Total number of odd days
= (0 + 1 + 5 + 3) = 9 odd days i.e., 2 odd days.
So, the required day is Tuesday.
Traffic Jams
Public Speaking
What to Wear to a Party
Noises on an Airplane
Bees
Circular Track sis of 11 km
Speed of Mens are
4 , 5.5 , 8 km/hr
Time at Which they are at starting Points again
11/4 , 11/5 , 11/8
11/4 , 2.2 , 11/8
We need to find LCM of these
to find at what time they Meet again at starting point
LCM 11 * 2 = 22
After 22 Hrs they will meet at starting point
2hrs
Solution # 15-18
Sitting solution be- D T M B H R
Ans 15 :Mr. D
ans 16 : 1 2 3
Ans 17 : 3
Ans 18 : Mr B Suffering from anemia.
2*3*5*7*11*13*17*19=9699690
8:20
I believe it will be 30 km/hr
40
40
30
60
– 10 cows
+ 20 days
if 40 cows can eat for forty days
then 20 cows can eat for 80 days
Let:
1. speed of boat in still water = x
2. speed of stream = y;
Then net speed of boat upstream = x – y = 16km / 4 h = 4kmph
Net speed of boat downstream = x + y = 16 / 2 = 8kmph
Now solve for x to get x = 6kmph in still water.
let x be lenghth of candles.
thicker one—- in 1 hr 1/6th will go.
thinner one— in 1 hr 1/4th will go.
let he lit for n hrs.
so in n hrs thicker one goes n/6.
and thinner one goes n/4.
so remainini length are x-n/6 and x-n/4.so x-n/6=2(x-n/4)
implie n=3x. if x=1 n=3hrs.
Let ‘N’ is the smallest number which divided by 13 and 16 leaves respective remainders of 2 and 5.
Required number = (LCM of 13 and 16) – (common difference of divisors and remainders)
= (208) – (11) = 197.
24000
Answer:
11 days.
Step-by-step explanation:
In the question,
Time taken by Ramesh to finish a piece of work = 20 days
Time taken by Sushil to finish a work = 25 days
Time for which they worked together = 5 days
Sushil left after = 5 days
So,
One day work of Ramesh is,
One day work of Sushil is,
So,
Work done in 5 days is given by,
Therefore, Remaining work is given by,
Now, as the Sushil left the remaining work was done by Ramesh,
Time taken by Ramesh for the remaining work is,
Therefore, the remaining work will be completed in 11 days by Ramesh.