Larry H Miller Group Interview Questions, Process, and Tips

The Answer is c

Let’s assume the length of each train is ‘L’ and the speeds of the two trains are ‘V₁’ and ‘V₂’ respectively.

When the trains are moving in the opposite direction, their relative speed is the sum of their individual speeds. The total distance they need to cover is the sum of their lengths. Since they cross each other completely in 5 seconds, we can set up the following equation:

(V₁ + V₂) × 5 = 2L

When the trains are moving in the same direction, their relative speed is the difference between their individual speeds. The total distance they need to cover is the difference between their lengths. Since they cross each other completely in 15 seconds, we can set up the following equation:

(V₁ – V₂) × 15 = 2L

Now, let’s solve these equations to find the ratio of their speeds.

From the first equation, we have:
(V₁ + V₂) × 5 = 2L
V₁ + V₂ = (2L) / 5

From the second equation, we have:
(V₁ – V₂) × 15 = 2L
V₁ – V₂ = (2L) / 15

Let’s add these two equations together:
V₁ + V₂ + V₁ – V₂ = (2L) / 5 + (2L) / 15
2V₁ = (6L + 2L) / 15
2V₁ = (8L) / 15
V₁ = (4L) / 15

So, the speed of the first train is (4L) / 15.

Now, let’s substitute this value back into the first equation to find V₂:
(4L) / 15 + V₂ = (2L) / 5
V₂ = (2L) / 5 – (4L) / 15
V₂ = (6L – 4L) / 15
V₂ = (2L) / 15

Therefore, the speed of the second train is (2L) / 15.

The ratio of their speeds is given by:
(V₁ / V₂) = ((4L) / 15) / ((2L) / 15)
(V₁ / V₂) = 4L / 2L
(V₁ / V₂) = 2

So, the ratio of their speeds is 2:1.

Burn Rope One first, it will burn in 30 mins.Mark 30 mins and at the end of 30 mins now burn second rope from both sides (15 mins it will take ),and simultaneously burn third and last rope until it is half left.30+15= 45 , Remaining time 7.5 mins.By this time last rope is burnt half ,and now burn it from both sides as well it will exhaust in 7.5 mins. 30+15+7.5=52.5 measured.

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

-1

x+8y=20, x=-3y
(-3y)+8y=20
5y=20
y=4
The answer for y is 4

1:2

answer = 16.5Rs

A=3*9=27Rs
B=7*15=105Rs
132 R for 10 kg
therefore,
1kg amount = 132/10=13.2 so,
5kg mixture price = 13.2*5=66Rs
mixture will sell 25%profit then the price =66*(25/10)=33/2=16.5Rs

26 rupees and 53 paise

Initially (2*33 + 1*24)/(2+1) = 30p a gm
So, cost for 100g = 100*30 = 3000p

Later, (1*33 + 2*24)/(1+2) = 71/3 p a gm.
So, cost for 100g = 100*(71/3) = 2366.67p

He saves 3000 – 2366.67 = 633.33p

Concept of weighted average.

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