Texas oncology Interview Questions, Process, and Tips

$20

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.

Answer is 400.(14^2=196, 16^2=256, 18^2=324, 20^2=400.)

C

If n = no. of revolutions for back wheel
9n = 7n + 10*7
n=315

Answer: 5400

Explanation:
100x + 400 = 12(x – 5)

x = 50

100 * 50 + 400 = 5400

Let the number of one rupee coins in the bag be x.
Number of 50 paise coins in the bag is 93 – x.
Total value of coins
[100x + 50(93 – x)]paise = 5600 paise
=> x = 74

ANS = 74

2*3(15+4)=114
114*1/2=57 rs

x = 5/3y
x – 4 = y + 4
x – 8 = y
x = 5/3 * (x-8)
5/3x – 40/3 = x
2/3x = 40/3
2x = 40
x = 20
y = 20 – 8
y = 12
x + 4 = 24
y – 4 = 8
Ratio is 24:8 , so 3:1

0.255555…..

mother

( C ) 6

27 root 3