Teknovance solutions pvt ltd Interview Questions, Process, and Tips

9

Suppose the lengthier arm of weighing pan is of x cm and other arm is y cm .Also let weight if each melon be m kg.
so applying equilibrium of torque principles ,we get
case 1:-
1x-8my=0
case 2:-
2mx-1y=0
using case 1 equation , we substitute value of x into case 2 equation;
16mmy-1y=0
(16mm-1)y=0
y is length of weighing arm and cannot be 0,
therefore ,
16mm-1 =0
16mm=1
mm=1/16
m= square root (1/16)
m=+-1/4
m is weight of melon and cant be negative.
Hence m, weight of one melon is 1/4 kg

None

P1/P2 = C1*T1/ C2*T2
where P1 = Partner 1’s Profit.
C1 = Partner 1’s Capital.
T1 = Time period for which Partner 1 contributed his capital.
P2 = Partner 2’s Profit.
C2 = Partner 2’s Capital.
T2 = Time period for which Partner 2 contributed his capital
6000/3000= 20,000*6/x*12
x=5000

192 Zeroes

Answer: z and u

Explanation:

Y is to the right of U and exactly in front of V. Therefore,

U Y
V

Z is behind W and W and X are at extreme ends. So, W has to be to the right of Y. The final arrangement is as follows.
U Y W
X V Z Therefore, Z and U are at extreme ends is true.

12 month=>900Rs+Uniform
1month:75 => 9month:675
675-650=25 =>price of Uniform

to reach land
first time it travels 8 mtrs
next time (half=4) 4+4 mtrs (up&down)
next time (half=4) 2+2 mtrs (up&down)
next time (half=4) 1+1 mtrs (up&down)
and so on.
Altogether,close to 24 mtrs.

How will you know the odd is in lighter one or heavier one from only one weighing. It will require 2 weighing to find the odd set and one weighing for odd coin in that set i.e total 3 weighings.

(c)UW

0237

A’s speed=6 mph ,B’s speed=8 mph
Let, after x hrs, they will meet.
so, the distance traveled by A in x hrs should be the same as the distance traveled by B in (x-1/2)hrs [as B started the journey after 30 min of A]
Thus, 6x=8(x-1/2)[as distance=speed*time]
=>8x-6x=4
=>2x=4
=>x=2
after 2 hrs they will meet so time=(9+2)=11.00 a.m

4) QEOKTWG

1:2:3 or 1:2:4

coz steam has more heat content than hot water