Ogden school district Interview Questions, Process, and Tips

A

12

Option c

The last person covered 120.71 meters.
It is given that the platoon and the last person moved with
uniform speed. Also, they both moved for the identical
amount of time. Hence, the ratio of the distance they
covered – while person moving forward and backword – are
equal.
Let’s assume that when the last person reached the first
person, the platoon moved X meters forward.
Thus, while moving forward the last person moved (50+X)
meters whereas the platoon moved X meters.
Similarly, while moving back the last person moved [50-(50-
X)] X meters whereas the platoon moved (50-X) meters.
Now, as the ratios are equal,
(50+X)/X = X/(50-X)
(50+X)*(50-X) = X*X
Solving, X=35.355 meters
Thus, total distance covered by the last person
= (50+X) + X
= 2*X + 50
= 2*(35.355) + 50
= 120.71 meters
Note that at first glance, one might think that the total
distance covered by the last person is 100 meters, as he
ran the total lenght of the platoon (50 meters) twice.
TRUE, but that’s the relative distance covered by the last
person i.e. assuming that the platoon is stationary.

C. 96

2hrs

total sum = 205 Rs.

answer is 306.04

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;

consider the tank capacity as 90 litres.
to fill the tank in 3 hrs first pipe must flow at a rate of 30 litres/hr
2nd pipe has to flow at a rate of 45 litres/hr
if two pipes are opened at a time the flow rate will become 75 litres/hr
90/75= 6/5 = 1.2= (i.e) 1hr and 12 minutes 1 1/5 hr

64

42

Diagonal of square=1.414* a, where a is side of square.
ans=1.414*20
ans=28.28